class notes

McGILL UNIVERSITYFACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATH 141 2010 01CALCULUS 2

Information for Students(Winter Term, 2009/2010)

Pages 1 - 20 of these notes may be considered the Course Outline for this course. The page

numbers shown in the table of contents and in the upper right hand corners of pages are not the

same as the numbers of pages in the PDF document. If you wish to print out specific pages, you

should first view the relevant pages at your screen, and determine what are the numbers of the

corresponding PDF pages.

W. G. Brown

April 17, 2010

Information for Students in MATH 141 2010 01

Contents1 General Information 1

1.1 Force Majeure . . . . . . . . . . 11.2 Instructors and Times . . . . . . 21.3 Calendar Description . . . . . . 2

1.3.1 Calendar Description . . 21.3.2 Late transfer from MATH

151/MATH 152 . . . . . 31.4 Tutorials . . . . . . . . . . . . . 3

1.4.1 Tutorial Times, Locations,and Personnel (subject tochange) . . . . . . . . . 3

1.4.2 Teaching Assistants (TA’s) 31.4.3 Friday, April 02nd, 2010

and Monday, April 05th,2010 . . . . . . . . . . 4

1.5 Evaluation of Your Progress . . 51.5.1 Your final grade (See Ta-

ble 3, p. 11) In the eventof extraordinary circum-stances beyond the Uni-versity’s control, the con-tent and/or evaluation schemein this course is subjectto change. . . . . . . . . 5

1.5.2 WeBWorK . . . . . . . 61.5.3 Written Submissions. . . 71.5.4 Quizzes at the Tutorials. 71.5.5 Final Examination . . . 81.5.6 Supplemental Assessments 81.5.7 Machine Scoring: “Will

the final examination bemachine scored?” . . . . 9

1.5.8 Plagiarism. . . . . . . . 91.5.9 Corrections to grades . . 10

1.6 Published Materials . . . . . . . 101.6.1 Required Text-Book . . 101.6.2 Optional Reference Books 101.6.3 Recommended Video Ma-

terials . . . . . . . . . . 121.6.4 Other Calculus Textbooks 13

1.6.5 Website . . . . . . . . . 131.7 Syllabus . . . . . . . . . . . . . 141.8 Preparation and Workload . . . 15

1.8.1 Prerequisites. . . . . . . 151.8.2 Calculators . . . . . . . 151.8.3 Self-Supervision . . . . 161.8.4 Escape Routes . . . . . 171.8.5 Terminology . . . . . . 18

1.9 Communication with Instructorsand TA’s . . . . . . . . . . . . . 18

1.10 Commercial tutorial and exampreparation services . . . . . . . 19

1.11 Special Office Hours and Tutorials 20

2 Draft Solutions to Quiz Q1 212.1 Instructions to Students . . . . . 212.2 Monday Versions . . . . . . . . 212.3 Tuesday Versions . . . . . . . . 232.4 Wednesday Versions . . . . . . 242.5 Thursday Versions . . . . . . . 252.6 Friday Versions . . . . . . . . . 26

3 Draft Solutions to Quiz Q2 283.1 Instructions to Students . . . . . 283.2 Monday Versions . . . . . . . . 293.3 Most Tuesday Versions . . . . . 313.4 Most Wednesday Versions . . . 333.5 Thursday Versions . . . . . . . 353.6 Friday Versions . . . . . . . . . 38

4 References 2014.1 Stewart Calculus Series . . . . . 2014.2 Other Calculus Textbooks . . . . 202

4.2.1 R. A. Adams . . . . . . 2024.2.2 Larson, Hostetler, et al. . 2034.2.3 Edwards and Penney . . 2034.2.4 Others, not “Early Tran-

scendentals” . . . . . . 2044.3 Other References . . . . . . . . 204

A Timetable for Lecture Section 001 ofMATH 141 2010 01 1001


B Timetable for Lecture Section 002 ofMATH 141 2009 01 2001

C Supplementary Notes for Students inSection 001 of MATH 141 2010 01 3001C.1 Lecture style in Lecture Section

001 . . . . . . . . . . . . . . . 3001C.2 Supplementary Notes for the Lec-

ture of January 04th, 2010 . . . 3002C.2.1 §5.1 Areas and Distances. 3002

C.3 Supplementary Notes for the Lec-ture of January 06th, 2010 . . . 3005C.3.1 §5.1 Areas and Distances

(conclusion). . . . . . . 3005C.3.2 §5.2 The Definite Integral 3007

C.4 Supplementary Notes for the Lec-ture of January 08th, 2010 . . . 3013C.4.1 Summary of the last lec-

tures . . . . . . . . . . . 3013C.4.2 §5.2 The Definite Inte-

gral (conclusion) . . . . 3014C.4.3 §5.3 The Fundamental The-

orem of Calculus . . . . 3019C.5 Supplementary Notes for the Lec-

ture of January 11th, 2010 . . . 3021C.5.1 §5.3 The Fundamental The-

orem of Calculus (con-clusion) . . . . . . . . . 3021

C.5.2 §5.4 Indefinite Integralsand the “Net Change” The-orem . . . . . . . . . . 3023

C.6 Supplementary Notes for the Lec-ture of January 13th, 2010 . . . 3028C.6.1 §5.4 Indefinite Integrals

and the “Net Change” The-orem (conclusion) . . . 3028

C.6.2 §5.5 The Substitution Rule 3033C.7 Supplementary Notes for the Lec-

ture of January 15th, 2010 . . . 3036C.7.1 §5.5 The Substitution Rule

(conclusion) . . . . . . . 3036C.7.2 5 Review . . . . . . . . 3049

C.8 Supplementary Notes for the Lec-ture of January 18th, 2010 . . . 3052C.8.1 §6.1 Areas between Curves 3052

C.9 Supplementary Notes for the Lec-ture of January 20th, 2010 . . . 3061C.9.1 §6.2 Volumes . . . . . . 3061

C.10 Supplementary Notes for the Lec-ture of January 22nd, 2010 . . . 3069C.10.1 §6.3 Volumes by Cylin-

drical Shells . . . . . . . 3069C.10.2 §6.4 Work . . . . . . . . 3075

C.11 Supplementary Notes for the Lec-ture of January 25th, 2010 . . . 3076C.11.1 §6.5 Average value of a

function . . . . . . . . . 3077C.12 Supplementary Notes for the Lec-

ture of January 27th, 2010 . . . 3082C.12.1 §7.1 Integration by Parts 3082

C.13 Supplementary Notes for the Lec-ture of January 29th, 2010 . . . 3089C.13.1 §7.1 Integration by Parts

(conclusion) . . . . . . . 3089C.13.2 §7.2 Trigonometric Inte-

grals . . . . . . . . . . . 3092C.14 Supplementary Notes for the Lec-

ture of February 01st, 2010 . . . 3095C.14.1 §7.2 Trigonometric Inte-

grals (conclusion) . . . . 3095C.15 Supplementary Notes for the Lec-

ture of February 03rd, 2010 . . . 3101C.15.1 §7.3 Trigonometric Sub-

stitution . . . . . . . . . 3101C.16 Supplementary Notes for the Lec-

ture of February 05th, 2010 . . . 3106C.16.1 §7.3 Trigonometric Sub-

stitution (conclusion) . . 3106C.17 Supplementary Notes for the Lec-

ture of February 08th, 2010 . . . 3111C.17.1 §7.4 Integration of Ra-

tional Functions by Par-tial Fractions . . . . . . 3111


C.18 Supplementary Notes for the Lec-ture of February 10th, 2010 . . . 3117C.18.1 §7.4 Integration of Ra-

tional Functions by Par-tial Fractions (conclusion) 3117

C.19 Supplementary Notes for the Lec-ture of February 12th, 2010 . . . 3124C.19.1 §7.5 Strategy for Integra-

tion . . . . . . . . . . . 3124C.19.2 §7.6 Integration Using Ta-

bles and Computer Al-gebra Systems (OMIT) . 3129

C.19.3 §7.7 Approximate Inte-gration (OMIT) . . . . . 3129

C.20 Supplementary Notes for the Lec-ture of February 15th, 2010 . . . 3130C.20.1 §7.8 Improper Integrals . 3130

C.21 Supplementary Notes for the Lec-ture of February 17th, 2010 . . . 3140C.21.1 §8.1 Arc Length . . . . 3140C.21.2 §8.2 Area of a Surface

of Revolution . . . . . . 3146C.22 Supplementary Notes for the Lec-

ture of February 19th, 2010 . . . 3147C.22.1 §8.2 Area of a Surface

of Revolution (conclusion) 3147C.22.2 §8.3 Applications to Physics

and Engineering (OMIT) 3150C.22.3 §8.4 Applications to Eco-

nomics and Biology (OMIT) 3151C.22.4 §8.5 Probability (OMIT) 3151

C.23 Supplementary Notes for the Lec-ture of March 01st, 2010 . . . . 3152C.23.1 §10.1 Curves Defined by

Parametric Equations . . 3152C.23.2 §10.2 Calculus with Para-

metric Curves . . . . . . 3156C.24 Supplementary Notes for the Lec-

ture of March 03rd, 2010 . . . . 3158C.24.1 §10.2 Calculus with Para-

metric Curves (continued) 3158

C.25 Supplementary Notes for the Lec-ture of March 05th, 2010 . . . . 3164C.25.1 §10.2 Calculus with Para-

metric Curves (conclusion) 3164C.25.2 §10.3 Polar Coordinates 3165

C.26 Supplementary Notes for the Lec-ture of March 08th, 2010 . . . . 3172C.26.1 §10.3 Polar Coordinates

(continued) . . . . . . . 3172C.26.2 §10.4 Areas and Lengths

in Polar Coordinates . . 3179C.27 Supplementary Notes for the Lec-

ture of March 10th, 2010 . . . . 3181C.27.1 §10.4 Areas and Lengths

in Polar Coordinates (con-tinued) . . . . . . . . . 3181

C.28 Supplementary Notes for the Lec-ture of March 12th, 2010 . . . . 3195C.28.1 §10.4 Areas and Lengths

in Polar Coordinates (con-clusion) . . . . . . . . . 3195

C.28.2 §10.5 Conic Sections . . 3196C.28.3 §11.1 Sequences . . . . 3197C.28.4 Sketch of Solutions to Prob-

lems on the Final Exam-ination in MATH 141 200501 . . . . . . . . . . . . 3198

C.29 Supplementary Notes for the Lec-ture of March 15th, 2010 . . . . 3207C.29.1 §11.1 Sequences (conclu-

sion) . . . . . . . . . . 3207C.29.2 §11.2 Series . . . . . . . 3210

C.30 Supplementary Notes for the Lec-ture of March 17th, 2010 . . . . 3213C.30.1 §11.2 Series (conclusion) 3213C.30.2 §11.3 The Integral Test

and Estimates of Sums . 3217C.31 Supplementary Notes for the Lec-

ture of March 19th, 2010 . . . . 3219C.31.1 §11.3 The Integral Test

and Estimates of Sums(conclusion) . . . . . . . 3219


C.32 Supplementary Notes for the Lec-ture of March 22nd, 2010 . . . . 3225C.32.1 §11.4 The Comparison

Tests . . . . . . . . . . 3225C.32.2 Sketch of Solutions to Prob-

lems on the Final Exam-ination in MATH 141 200601 . . . . . . . . . . . . 3230

C.33 Supplementary Notes for the Lec-ture of March 24th, 2010 . . . . 3245C.33.1 §11.5 Alternating Series 3245C.33.2 Solutions to Problems on

the Final Examination inMATH 141 2007 01 . . 3249

C.34 Supplementary Notes for the Lec-ture of March 26th, 2010 . . . . 3265C.34.1 §11.6 Absolute Conver-

gence and the Ratio andRoot Tests . . . . . . . . 3265

C.35 Supplementary Notes for the Lec-ture of March 29th, 2010 . . . . 3268C.35.1 §11.6 Absolute Conver-

gence and the Ratio andRoot Tests (conclusion) . 3268

C.36 Supplementary Notes for the Lec-ture of Wednesday, March 31st,2010 . . . . . . . . . . . . . . . 3272C.36.1 §11.7 Strategy for Test-

ing Series . . . . . . . . 3272C.37 Supplementary Notes for the Lec-

ture of Wednesday, April 7th, 2010 3278C.37.1 Final Examination in MATH

141 2008 01 (one version) 3278C.37.2 Draft Solutions to the Fi-

nal Examination in MATH141 2009 01 (Version 4) 3293

C.38 Supplementary Notes for the Lec-ture of Friday, April 09th, 2010 . 3297C.38.1 Final Examination in MATH

141 2009 01 (Version 4,continued) . . . . . . . 3297

C.39 Supplementary Notes for the Lec-ture of Monday, April 12th, 2010 3301C.39.1 Final Examination in MATH

141 2009 01 (Version 4,continued) . . . . . . . 3301

C.40 Supplementary Notes for the Lec-ture of Wednesday, April 14th,2010 . . . . . . . . . . . . . . . 3308C.40.1 Final Examination in MATH

141 2009 01 (Version 4,concclusion) . . . . . . 3308

D Problem Assignments from Previous Years 5001D.1 1998/1999 . . . . . . . . . . . . 5001

D.1.1 Assignment 1 . . . . . . 5001D.1.2 Assignment 2 . . . . . . 5001D.1.3 Assignment 3 . . . . . . 5002D.1.4 Assignment 4 . . . . . . 5002D.1.5 Assignment 5 . . . . . . 5002

D.2 1999/2000 . . . . . . . . . . . . 5003D.2.1 Assignment 1 . . . . . . 5003D.2.2 Assignment 2 . . . . . . 5004D.2.3 Assignment 3 . . . . . . 5006D.2.4 Assignment 4 . . . . . . 5007D.2.5 Assignment 5 . . . . . . 5009D.2.6 Assignment 6 . . . . . . 5010

D.3 2000/2001 . . . . . . . . . . . . 5012D.4 2001/2002 . . . . . . . . . . . . 5012D.5 MATH 141 2003 01 . . . . . . . 5012D.6 MATH 141 2004 01 . . . . . . . 5012D.7 MATH 141 2005 01 . . . . . . . 5013

D.7.1 Written Assignment W1 5013D.7.2 Written Assignment W2 5014D.7.3 Written Assignment W3 5016D.7.4 Written Assignment W4 5017D.7.5 Written Assignment W5 5019

D.8 MATH 141 2006 01 . . . . . . . 5021D.8.1 Solution to Written As-

signment W1 . . . . . . 5021D.8.2 Solution to Written As-

signment W2 . . . . . . 5024


D.8.3 Solutions to Written As-signment W3 . . . . . . 5025



D.9 MATH 141 2007 01 . . . . . . . 5032

E Quizzes from Previous Years 5033E.1 MATH 141 2007 01 . . . . . . . 5033

E.1.1 Draft Solutions to Quiz Q1 5033E.1.2 Draft Solutions to Quiz Q2 5043E.1.3 Draft Solutions to Quiz Q3 5055E.1.4 Draft Solutions to Quiz Q4 5070

E.2 MATH 141 2008 01 . . . . . . . 5086E.2.1 Draft Solutions to Quiz Q1 5086E.2.2 Draft Solutions to Quiz Q2 5099E.2.3 Draft Solutions to Quiz Q3 5110E.2.4 Draft Solutions to Quiz Q4 5121

E.3 MATH 141 2009 01 . . . . . . . 5134E.3.1 Draft Solutions to Quiz Q1 5134E.3.2 Draft Solutions to Quiz Q2 5140E.3.3 Draft Solutions to Quiz Q3 5146

F Final Examinations from Previous Years 5152F.1 Final Examination in Mathemat-

ics 189-121B (1996/1997) . . . 5152F.2 Final Examination in Mathemat-

ics 189-141B (1997/1998) . . . 5153F.3 Supplemental/Deferred Examina-

tion in Mathematics 189-141B(1997/1998) . . . . . . . . . . . 5155

F.4 Final Examination in Mathemat-ics 189-141B (1998/1999) . . . 5156

F.5 Supplemental/Deferred Examina-tion in Mathematics 189-141B(1998/1999) . . . . . . . . . . . 5158







F.12 Final Examination in MATH 1412003 01 . . . . . . . . . . . . . 5169

F.13 Supplemental/Deferred Examina-tion in MATH 141 2003 01 . . . 5171





F.18 Final Examination in MATH 1412006 01 (One version) . . . . . 5192


F.20 Final Examination in MATH 1412007 01 (One version) . . . . . 5199

F.21 Supplemental/Deferred Examina-tion in MATH 141 2007 01 (Oneversion) . . . . . . . . . . . . . 5203

F.22 Final Examination in MATH 1412008 01 (one version) . . . . . . 5207

F.23 Supplemental/Deferred Examina-tion in MATH 141 2008 01 (oneversion) . . . . . . . . . . . . . 5217

F.24 Final Examination in MATH 1412009 01 (one version) . . . . . . 5221

G WeBWorK 6001G.1 Frequently Asked Questions (FAQ) 6001

G.1.1 Where is WeBWorK? . 6001


G.1.2 Do I need a password touse WeBWorK? . . . . 6001

G.1.3 Do I have to pay an ad-ditional fee to use WeB-WorK? . . . . . . . . . 6001

G.1.4 When will assignmentsbe available on WeBWorK? 6002

G.1.5 Do WeBWorK assign-ments cover the full rangeof problems that I shouldbe able to solve in thiscourse? . . . . . . . . . 6002

G.1.6 May I assume that thedistribution of topics onquizzes and final exam-inations will parallel thedistribution of topics inthe WeBWorK assign-ments? . . . . . . . . . 6002

G.1.7 WeBWorK provides fordifferent kinds of “Dis-play Mode”. Which shouldI use? . . . . . . . . . . 6002

G.1.8 WeBWorK provides forprinting assignments in“Portable Document Format”(.pdf), “PostScript” (.ps)and “TEXSource” forms.Which should I use? . . 6003

G.1.9 What is the relation be-tween WeBWorK and We-bCT? . . . . . . . . . . 6003

G.1.10 What do I have to do onWeBWorK? . . . . . . 6003

G.1.11 How can I learn how touse WeBWorK? . . . . 6004

G.1.12 Where should I go if Ihave difficulties with WeB-WorK ? . . . . . . . . . 6004

G.1.13 Can the WeBWorK sys-tem ever break down ordegrade? . . . . . . . . 6004

G.1.14 How many attempts mayI make to solve a partic-ular problem on WeB-WorK? . . . . . . . . . 6005

G.1.15 Will all WeBWorK as-signments have the samelength? the same value? 6005

G.1.16 Is WeBWorK a good in-dicator of examination per-formance? . . . . . . . . 6005

H Contents of the DVD disks forLarson/Hostetler/Edwards 6101

List of Tables1 Schedule and Locations of Tu-

torials, as of April 17, 2010. . . 42 Tutors’ Coordinates, as of April

17, 2010 . . . . . . . . . . . . . 53 Summary of Course Requirements,

as of April 17, 2010; (all datesare subject to change) . . . . . . 11

4 Some Antiderivatives . . . . . . 30255 Very Short Table of Indefinite In-

tegrals . . . . . . . . . . . . . . 3026

List of Figures1 The region(s) bounded by y =

x2 and y = x4 . . . . . . . . . . 30532 The region(s) bounded by y =

sin x, y = sin 2x between x = 0and x = π

2 . . . . . . . . . . . . 30553 The region(s) bounded by y =

8 − x2, y = x2 between x = ±3 . 30564 The region(s) bounded by y =√

x + 2, y = x between x = 0and x = 4 . . . . . . . . . . . . 3058

5 Regions for Example C.31 . . . 3061


6 The region(s) bounded by x +

y = 3 and x = 4 − (y − 1)2 . . . 30707 The curve x = cos θ+sin 2θ, y =

sin θ + cos 2θ . . . . . . . . . . 31588 The cardioid with equation r =

2(1 − sin θ) . . . . . . . . . . . 31709 The limacon r = 1 − 3 cos θ, . . 317210 The spiral with equation r = θ,

(θ ≥ 0) . . . . . . . . . . . . . . 317411 The spiral with equation r = θ,

(θ ≤ 0) . . . . . . . . . . . . . . 317512 The full spiral with equation r =

θ, −∞ < θ < +∞ . . . . . . . . 317613 The “4-leafed rose” with equa-

tion r = sin 2θ, . . . . . . . . . . 317714 The “5-leafed rose” with equa-

tion r = sin 5θ, . . . . . . . . . . 317815 The lemniscates r2 = sin 2θ, r2 =

cos 2θ . . . . . . . . . . . . . . 317916 Intersecting polar curves r = 1+

sin θ, r2 = 4 sin θ . . . . . . . . 318217 Curves r = sin θ, r = cos θ . . . 318418 The strophoid r = 2 cos θ − sec θ 318619 Curves r = 2 + sin θ, r = 3 sin θ . 318820 Intersections of the limacon r =

1 − 2 cos θ with the circle r = 1 . 319121 Intersections of the curve r =

sec θ with the circle r = 1 . . . . 319322 The curves with equations r =

1 − cos θ, (θ ≤ 0), and r = 1 +

sin θ, and the point(

12 ,

π2

). . . . 3243

23 The cardioids with equations r =

2 + 2 sin θ, r = 6 − 6 sin θ . . . . 326224 The region bounded by cardioids

r = 2+2 sin θ, r = 6−6 sin θ andcontaining the point (r, θ) = (1, 0) 3263

25 The curves with equations r =

4 + 2 cos θ, r = 4 cos θ + 5 . . . 329126 The curves with equations r =

3 + 3 cos θ, (0 ≤ θ ≤ 2π), andr = 9 cos θ, (0 ≤ θ ≤ π) . . . . . 3304

27 The limacon r = 1 + 2 sin θ . . . 5071

Information for Students in MATH 141 2010 01 1

1 General InformationDistribution Date: January 04th, 2010(all information is subject to change)

Pages 1 - 20 of these notes may be considered the Course Outline for this course.

These notes may undergo minor corrections or updates during the term: the defini-tive version will be the version accessible at

http://www.math.mcgill.ca/brown/math141b.html

or on myCourses, at

http://www.mcgill.ca/mycourses/

Students are advised not to make assumptions based on past years’ operations,as some of the details concerning this course could be different from past years.Publications other than this document may contain unreliable information aboutthis course.All details of the course could be subject to discretionary change in case of forcemajeure.

1

1.1 Force MajeureIn the event of extraordinary circumstances beyond the University’s control, all details of thiscourse, including the content and/or evaluation scheme are subject to change.

1Please note that the statements about MATH 141 in an SUS publication called Absolute Zero were not givento instructors of this course to check, and some of them may not be currently correct.


1.2 Instructors and TimesINSTRUCTOR: Prof. W. G. Brown Dr. S. Shahabi Dr. A. Hundemer

(Course Coordinator)LECTURE SECTION: 1 2 3

CRN: 576 577 578OFFICE: BURN 1224 BURN 1243 BURN 1128

OFFICE HOURS: W 15:45→16:45 F 09:30→11:30 MW 15:30→16:25(subject to change) F 10:00→11:00 (tentative)

or by appointmentTELEPHONE: (514)-398-3836 (514)-398-3803 (514)-398-5318

E-MAIL:2 BROWN@ SHAHABI@ [email protected] MATH.MCGILL.CA MATH.MCGILL.CA

CLASSROOM: ADAMS AUD LEA 219 ADAMS AUDCLASS HOURS: MWF 11:35–12:25 h. MWF 11:35–12:25 h. MW 16:35–17:55 h.

1.3 Calendar Description1.3.1 Calendar Description

MATH 1414 CALCULUS 2. (4 credits; 3 hours lecture; 2 hours tutorial. Prerequisites:MATH 139 or MATH 140 or MATH 150. Restriction: Not open to students who have takenMATH 121 or CEGEP objective 00UP or equivalent; not open to students who have taken orare taking MATH 122 or MATH 130 or MATH 131, except by permission of the Department ofMathematics and Statistics. Each Tutorial section is enrolment limited.) The definite integral.Techniques of integration. Applications. Introduction to sequences and series.

Students Lacking the Prerequisite will, when discovered, be removed from the course.Students without the prerequisite (or standing in a course recognized by the Admissions Officeas being equivalent to MATH 140) should not assume that, in possibly permitting MINERVAto accept their registration for MATH 141, the University was tacitly approving their regis-tration without the prerequisite. In particular, students who obtained a grade of F in MATH139/140/150 are expressly excluded from registration in MATH 141, even if they registered inthe course before the failed or missed examination.

2Please do not send e-mail messages to your instructors through the WebCT or WeBWorK3 systems; rather,use the addresses given in §1.2 on page 2.

3E-mail messages generated by the Feedback command in WeBWorK should be used sparingly, and con-fined to specific inquiries about WeBWorK assignments.

4The previous designation for this course was 189-141, and the version given in the winter was labelled189-141B; an earlier number for a similar course was 189-121.


1.3.2 Late transfer from MATH 151/MATH 152

Some students from MATH 151 or MATH 152 may be permitted to transfer into MATH 141after the end of the Change of Course Period. If your instructor in MATH 151 or 152 advisesyou that you are in this category, please send an e-mail message to Professor Brown as soonas your transfer has been approved.5

1.4 Tutorials1.4.1 Tutorial Times, Locations, and Personnel (subject to change)

Every student must be registered in one lecture section and one tutorial section for this course.Tutorials begin in the week of January 11th, 2010. The last tutorials in all Tuesday, Wednesday,and Thursday tutorial sections will be in the week beginning Monday, April 05th, 2010; thelast tutorial in Monday tutorial sections will be on Monday, April 12th, 2010; the date of thelast tutorial in Friday tutorial sections will be announced later in the term. Table 1 gives times,locations, and the tutor’s name for each of the tutorials; Table 2 gives the tutors’ coordinates.The information in these tables is subject to change. We try to publicize changes butsometimes we are not informed in advance.6

You are expected to write quizzes only in the tutorial section in which you are reg-istered.7 You do not have a licence to move from one tutorial section to another at will,even if you find the time, location, or personnel of your tutorials either temporarily or perma-nently inconvenient; in the latter case the onus is on you to transfer formally to another tutorialsection, to change your other classes, or to drop MATH 141 2010 01. Please remember thattransfers must be completed by the Course Change (drop/add) deadline (January 19th, 2010),and are subject to the maximum capacities established for each tutorial section8

1.4.2 Teaching Assistants (TA’s)

The tutors in MATH 141 2010 01 are graduate students in Mathematics and Statistics. Likeyou, they are students, albeit at the graduate level; they have deadlines and commitments andpersonal lives, and the time they have available for MATH 141 is limited and controlled by acollective agreement (union contract). Please respect the important functions that our tutorsprovide, and do not ask them for services they are not expected to perform:

5This is to ensure that your WeBWorK account is opened, and that your date of entry to the course is recorded.6The current room for your tutorial should always be available by clicking on “Class Schedule” on MINERVA

FOR STUDENTS, http://www.mcgill.ca/minerva-students/.7In some time slots there may be several tutorial sections, meeting in different rooms.8Your instructors do not have the ability to change the maximum capacities of tutorials.


# CRN Day Begins Ends Room TutorT004 579 Fri 13:35 15:25 BURN 1B39 J. FeysT005 580 Tue 14:05 15:55 ARTS 260 H. BigdelyT006 581 Tue 16:05 17:55 BURN 1B23 L. CandeloriT007 582 Tue 16:05 17:55 BURN 1B24 F. CastellaT008 583 Thurs 14:05 15:55 BURN 1B39 Y. CanzaniT009 584 Thurs 16:05 17:55 BURN 1B39 X. ZhangT010 585 Thurs 16:05 17:55 BURN 1B23 A.-P. GrecianuT011 586 Mon 13:35 15:25 ARTS W-20 J. MacdonaldT012 587 Mon 14:35 16:25 BURN 1B36 M. PrevostT013 588 Mon 14:35 16:25 BURN 1B24 J. Tousignant-BarnesT014 589 Wed 13:35 15:25 ARTS W-20 J. RestrepoT015 590 Wed 14:35 16:25 LEA 14 B. TajiT016 591 Wed 14:35 16:25 BURN 1B36 A. TchengT017 2071 Mon 13:35 15:25 ENGMD 276 A. TombergT018 2072 Wed 13:35 15:25 BURN 1B24 P. RempelT019 8194 Tue 08:05 09:55 BURN 1B23 Y. RabhiT020 8840 Wed 15:35 17:25 ENGMD 279 A. FarooquiT021 8841 Fri 15:35 17:25 ENGMD 256 Y. Zhao

Some of these room assignments could change before or early in the beginning of the term, as wehave a pending request to upgrade some of the rooms. In any case, all assignments are subject to

change.

Table 1: Schedule and Locations of Tutorials, as of April 17, 2010.

• Outside of the normal quiz times in their tutorials, tutors are neither expected nor au-thorized to administer a special quiz or a quiz that has already been administered toothers.

• Tutors in MATH 141 2010 01 are not permitted to offer paid, private tuition to studentsin any tutorial section of this course.

1.4.3 Friday, April 02nd, 2010 and Monday, April 05th, 2010

These two lecture/tutorial days are lost because of the Easter holidays. While the lectures willresume on Wednesday, April 07th, 2010, and the total number of lecture hours is similar topast years, there will be some disruption to Monday and Friday tutorials: the Monday tutorialswill meet on Monday, April 12th, 2010, but there is no scheduled Friday available to completethe Friday tutorials; alternative arrangements for students in Friday tutorials will be announcedlater in the term.


Tutor E-mail address Office Office HoursBURN Day(s) Begins Ends Day Begins Ends

Bigdely, H. [email protected] 1030 M 12:00 15:00Candelori, L. [email protected] 1032 T 11:30 13:00 T 18:30 20:00Canzani, Y. [email protected] 1133 T 12:30 14:30 Th 13:00 14:00Castella, F. [email protected] 1008 M 08:30 10:30 M 14:30 15:30Farooqui, A. [email protected] 1023 TTh 16:00 17:30

Feys, J. [email protected] 1020 T 17:00 18:30 W 09:00 10:30Grecianu, A.-P. [email protected] 1007 Th 14:30 16:00 F 13:00 14:30Macdonald, J. [email protected] 1030 M 09:30 11:30 M 15:30 16:30

Prevost, M. [email protected] 1117 TTh 13:00 14:30Rabhi, Y. [email protected] 1021 M 13:00 14:30 T 10:00 11:30

Rempel, P. [email protected] 1140 Th 10:00 13:00Restrepo, J. [email protected] 1117 T 10:30 12:00 T 14:30 16:00

Taji, B. [email protected] 1031 F 15:00 18:00Tcheng, A. [email protected] 1029 TTh 14:30 16:00Tomberg, A. [email protected] 1031 W 12:30 15:30

Tousignant-Barnes, J. [email protected] 1032 TTh 09:00 10:30Zhang, X. [email protected] 1020 M 08:30 11:30Zhao, Y. [email protected] 1034 F 12:30 15:30

During her/his office hours, a tutor is available to all students in the course,not only to the students of her/his tutorial section.

For last minute changes, see myCourses (WebCT).

Table 2: Tutors’ Coordinates, as of April 17, 2010

1.5 Evaluation of Your Progress1.5.1 Your final grade (See Table 3, p. 11) In the event of extraordinary circumstances

beyond the University’s control, the content and/or evaluation scheme in thiscourse is subject to change.

Your grade in this course will be a letter grade, based on a percentage grade computed fromthe following components:

1. Assignments submitted over the Web: Six9 (6) WeBWorK homework assignments —counting together for 10%.

2. Quizzes given at the tutorials: Four (4), counting together for 20%.10

9Numbers of assignments, quizzes, etc., are as planned as of the date of this version of these notes. Studentsmust be prepared for the possibility that it could be necessary to adjust these numbers during the term. If thereare any changes, these will be announced on myCourses, by broadcast e-mail messages, or by announcements atthe lectures.

10But be warned: students who fail to write quizzes are often at risk in this course. The quizzes are mainlya learning, rather than a testing experience. You need the information that comes from writing quizzes in a


3. The final examination — counting for 70%.

Where a student’s performance on the final examination is superior to her performance on thetutorial quizzes, the final examination grade will replace the quiz grades in the calculations;in that case the grade on the final examination will count for 90% of the final grade. It is notplanned to permit the examination grade to replace the grades on WeBWorK assignments.

1.5.2 WeBWorK

1. The WeBWorK system, developed at the University of Rochester — is designed toexpose you to a large number of drill problems, and where plagiarism is discouraged.WeBWorK is accessible only over the Internet. Details on how to sign on to WeBWorKare contained in Appendix G to these notes, page 6001.

Only answers submitted by the due date and time will count. The WeBWorK assign-ments which count in your term mark will be labelled A1, . . ., A6.

2. Due dates and times for WeBWorK assignments. Most due dates for WeBWorKassignments will be on specified Sundays, about 23:30h; last minute changes in the duedates may be announced either on WeBWorK, on myCourses, or by an e-mail message11

As mentioned in the WeBWorK FAQ (cf. Appendix G), if you leave your WeBWorKassignment until the hours close to the due time on the due date, you should not besurprised if the system is slow to respond. This is not a malfunction, but is simplya reflection of the fact that other students have also been procrastinating! To benefitfrom the speed that the system can deliver under normal conditions, do not delay yourWeBWorK until the last possible day! If a systems failure interferes with the due date ofan assignment, arrangements may be made to change that date, and an e-mail messagemay be broadcast to all users (to the e-mail addresses on record), or a note posted in thecourse announcements on myCourses; but slowness in the system just before the duetime will not normally be considered a systems failure.12

3. Numbers of permitted attempts at WeBWorK questions. While the number of timesyou may attempt each problem on WeBWorK An will be limited, there will be a com-panion “Practice” Assignment Pn (n = 1, 2, . . . , 6) with an unlimited number of attemptsat similar problems, but in which the specific data may be different. Thus you have

group, and observing whether your performance was at an appropriate level. Students who deny themselves thisexperience often undergo a rude awakening at the final examination.

11Be sure that your e-mail addresses are correctly recorded. See 4, p. 19 of these notes.12Should you find that the system is responding slowly, do not submit your solutions more than once; you may

deplete the number of attempts that have been allowed to you for a problem: this will not be considered a systemsfailure.


the opportunity to prepare yourself on the Practice assignment before attempting the ac-tual assignment. The practice assignments DO NOT COUNT in your term mark, eventhough a grade is recorded.. Practice assignment Pn is normally due one week beforeassignment An. Another assignment which will not count will be Practice AssignmentP0, which is directed to students who are not familiar with the WeBWorK system.

1.5.3 Written Submissions.

In accord with McGill University’s Charter of Students’ Rights, students in this course have theright to submit in English or in French any written work that is to be graded; course materialsare normally provided only in English.

Written Assignments. There will be no Written Assignments in MATH 141 2010 01.13

1.5.4 Quizzes at the Tutorials.

1. There will be 4 quizzes, numbered Q1, Q2, Q3, Q4, administered at the tutorials. Thesequizzes will be graded, and returned. The primary purpose of a quiz is to diagnosepossible gaps in your understanding. In the grading formula the quiz component ofthe final grade will be replaced by the final examination grade, if that is to a student’sadvantage.

2. Students may write a quiz only in the tutorial in which they are registered.

3. Medical absences. If you have missed or expect to miss a quiz for a valid reason (med-ical or otherwise), please communicate directly with Professor Brown, providing a copyof the medical or other supporting documents; do not contact your TA. Authorized med-ical absences can be accommodated only through averaging, as students in MATH 141are never permitted to write a quiz in any tutorial section other than the one in whichthey are registered. We cannot offer “makeup” sessions for quizzes.

4. To prepare for a quiz you should be working exercises in the textbook based on the ma-terial currently under discussion at the lectures, and you should have attempted any openWeBWorK assignments. But, unlike the WeBWorK assignments — where the empha-sis is on correct answers alone — students may be expected to provide full solutions to

13While it is not required for grading purposes, students are urged to keep a systematic record of writtensolutions to problems in the textbook. This could be in the form of a workbook, or a file folder, but should beorderly enough that you can look back at a later time to see your solutions. You are invited to bring such a file toTA’s or instructors at their office hours, to receive advice about the quality and correctness of your solutions.


some or all problems on quizzes.14

The quizzes may examine on only a sampling of topics. Students should not assume thattopics not examined are in any subsidiary parts of the syllabus.

5. Your tutors will normally bring graded quizzes to the tutorial to be returned to you.University regulations do not permit us to leave unclaimed materials bearing names andstudent numbers in unsupervised locations; you may be able to recover an unclaimedquiz from the tutor who graded it, during her/his regular office hours. Be sure to attendthe tutorial following a quiz15, as claims of incorrect recording of a quiz or assignmentgrade will need to be substantiated by a graded paper.

6. Your quiz grades on assignments and quizzes will be posted on myCourses within about2 weeks after they become available. Your WeBWorK grades may not be transferred tomyCourses until the end of the term, but will be visible on the WeBWorK site.

1.5.5 Final Examination

A 3-hour-long final examination will be scheduled during the regular examination period forthe winter term (April 15th, 2010 through April 30th, 2010). You are advised not to make anytravel arrangements that would prevent you from being present on campus at any time duringthis period.16

1.5.6 Supplemental Assessments

1. Supplemental Examination. There will be a supplemental examination in this course.(For information about Supplemental Examinations, see

http://www.mcgill.ca/artscisao/departmental/examination/supplemental/.)

2. There is No Additional Work Option. “Will students with marks of D, F, or J have theoption of doing additional work to upgrade their mark?” No. (“Additional Work” refersto an option available in certain Arts and Science courses, but not available in MATH141 2010 01.)

14In Math 141 the general rule for quizzes is that full solutions are expected to all problems, unless you receiveexplicit instructions to the contrary: ALWAYS SHOW YOUR WORK! The solutions in the Student SolutionsManual [9] to the textbook can serve as a guide to what should be included in a “full” solution.

15The return of Quiz Q1 may be delayed to the 2nd week after the quiz was written.16Your instructors learn the date of your examination at the same time as you do — when the Provisional

examination timetable is published.


1.5.7 Machine Scoring: “Will the final examination be machine scored?”

Multiple-Choice Problems It is possible that the final examination, or part of it, could bemachine scored. Multiple choice problems, possibly machine scored, could also appear onsome quizzes. (Machine grading, if implemented in whole or in part, would be a change fromthe practice of past years in this course. Such a change was introduced into the most recentexamination in MATH 140, which included a substantial number of machine-graded multiplechoice questions.)

Answer-Only Problems. Some of the problems on quizzes and/or the final examination —possibly a substantial number of them — may request that the answer only be given, and maynot carry part marks which could be based on the work leading up to the answer.

1.5.8 Plagiarism.

While students are not discouraged from discussing methods for solving WeBWorK assign-ment problems with their colleagues, all work that you submit must be your own. The Senateof the University requires the following message in all course outlines:

“McGill University values academic integrity. Therefore all students must understand themeaning and consequences of cheating, plagiarism and other academic offences under theCode of Student Conduct and Disciplinary Procedures. (See http://www.mcgill.ca/integrityfor more information).

“L’universite McGill attache une haute importance a l’honnetete academique. Il incombepar consequent a tous les etudiants de comprendre ce que l’on entend par tricherie, plagiatet autres infractions academiques, ainsi que les consequences que peuvent avoir de tellesactions, selon le Code de conduite de l’etudiant et des procedures disciplinaires. (Pour deplus amples renseignements, veuillez consulter le site http://www.mcgill.ca/integrity).”

It is a violation of University regulations to permit others to solve your WeB-WorK problems, or to extend such assistance to others; you could be askedto sign a statement attesting to the originality of your work. The Handbookon Student Rights and Responsibilities17 states in ¶A.I.15(a) that

“No student shall, with intent to deceive, represent the work of another personas his or her own in any academic writing, essay, thesis, research report, projector assignment submitted in a course or program of study or represent as his orher own an entire essay or work of another, whether the material so representedconstitutes a part or the entirety of the work submitted.”

17http://upload.mcgill.ca/secretariat/greenbookenglish.pdf


You are also referred to the following URL:

http://www.mcgill.ca/integrity/studentguide/

Other Fraud. It is a serious offence to alter a graded quiz paper and return it to the tutorunder the pretense that the work was not graded properly.

1.5.9 Corrections to grades

Grades will eventually be posted on myCourses. If you believe a grade has been recordedincorrectly, you must advise your tutor not later than 4 weeks after the grade has been posted,and not later than the day before of the final examination whichever of these dates is earlier. Itis hoped that grades will be posted within 2 weeks of the due date. You will have to present thegraded quiz to support your claim, which must be submitted to the tutor that graded the quiz. Ifhe/she believes there has been an error, the tutor will advise Professor Brown. New correctionsto the myCourses posting will appear the next time grades are uploaded to myCourses.

1.6 Published Materials1.6.1 Required Text-Book

The textbook for the course is J. Stewart, SINGLE VARIABLE CALCULUS: Early Tran-scendentals, Sixth Edition, Brooks/Cole (2008), ISBN 0-495-01169-X, [1]. This book is thefirst half of J. Stewart, CALCULUS: Early Transcendentals, Sixth Edition, Brooks/Cole(2008), ISBN 0-495-01166-5, [2]; this edition covers the material for Calculus 3 (MATH 222)as well, but is not the text-book for that course at the present time. The textbook will be soldin the McGill Bookstore bundled with its Student Solutions Manual (see below). The ISBNnumber for the entire bundle is 0-495-42966-X.

1.6.2 Optional Reference Books

Students are urged to make use of the Student Solution Manual:

• D. Anderson, J. A. Cole, D. Drucker, STUDENT SOLUTIONS MANUAL FOR STEW-ART’S SINGLE VARIABLE CALCULUS: Early Transcendentals, Sixth Edition, Brooks/Cole(2008), ISBN 0-495-01240-8, [3]. This book is also sold “bundled” with the text book;we expect the Bookstore to stock the bundle numbered ISBN 0-495-42966-X [4].

The publishers of the textbook and Student Solutions Manual also produce

UPDATED TO April 17, 2010


Item # Due Date DetailsP0 DOES NOT COUNT: introduces WeBWorK

WeBWorK P1 17 Jan 10 DOES NOT COUNT; practice for A1Assignments A1 24 Jan 10(cf. §1.5.2) P2 31 Jan 10 DOES NOT COUNT; practice for A2

10% A2 07 Feb 10P3 14 Feb 10 DOES NOT COUNT; practice for A3A3 21 Feb 10P4 28 Feb 10 DOES NOT COUNT; practice for A4A4 07 Mar 10P5 14 Mar 10 DOES NOT COUNT; practice for A5A5 28 Mar 10P6 28 Mar 10 DOES NOT COUNT; practice for A6A6 12 Apr 10 A1–A6 count equally, but may have different num-

bers of problems.Quizzes Q1 18–22 Jan 10 Quizzes Q1 — Q4 count equally, but(cf. §1.5.4) Q2 08–12 Feb 10 the quizzes may be of different lengths.20% or 0% Q3 08–12 Mar 10

Q4 22–26 Mar 10Final Exam 15–30 Apr 10 Date of exam to be announced by Faculty70% or 90%

SupplementalExam(cf. §1.5.6.1)

18–19 Aug 10 Only for students who do not obtain standing at thefinal. Supplemental exams count in your averagelike taking the course again; exam counts for 100%.

Table 3: Summary of Course Requirements, as of April 17, 2010; (all dates are subject tochange)

• a “Study Guide”, designed to provide additional help for students who believe theyrequire it: R. St. Andre, STUDY GUIDE FOR STEWART’S SINGLE VARIABLECALCULUS: Early Transcendentals, Sixth Edition, Brooks/Cole (2008), ISBN 0-495-01239-4, [5]. (The “Study Guide” resembles the Student Solution Manual in ap-pearance: be sure you know what you are buying.)

• a “Companion” which integrates a review of pre-calculus concepts with the contentsof Math 140, including exercises with solutions: D. Ebersole, D. Schattschneider, A.Sevilla, K. Somers, A COMPANION TO CALCULUS. Brooks/Cole (1995), ISBN 0-534-26592-8 [39].


1.6.3 Recommended Video Materials

Use of the following materials is recommended, but is not mandatory18.

Text-specific DVDs for Stewarts Calculus, early transcendentals, 6th edition [videorecord-ing]. The publisher of Stewart’s Calculus has produced a series of videodisks, [?]. These willinitially be available for reserve loan at the Schulich Library. There may not be DVD viewingequipment freely available in the library; the intention is that interested students borrow disksfor viewing on their own equipment at home. Disk 1 covers Chapters 1-6 of the textbook.

Videotapes for Stewart’s Calculus The publisher of Stewart’s Calculus had earlier pro-duced a series of videotapes, [14] Video Outline for Stewart’s Calculus (Early Transcenden-tals), Fifth Edition. These will be available for reserve loan at the Schulich Library. There maynot be VCR viewing equipment in the library; the intention is that interested students borrowa tape for viewing on their own equipment at home.

Larson/Hostetler/Edwards DVD Disks A set of video DVD disks produced for anothercalculus book, [28] Calculus Instructional DVD Program, for use with (inter alia) Larson /

Hostetler / Edwards, Calculus of a Single Variable: Early Transcendental Functions, ThirdEdition [29] is produced by the Houghton Mifflin Company. A copy has been requested to beplaced on reserve in the Schulich Library. In Appendix H of these notes there are charts thatindicate the contents of these disks that pertain to MATH 141.

Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcendentals, 6thEdition, (similar to [15]) 19This CD-ROM is included with certain new copies of the text-book. It contains, after an enlightening “pep-talk” by the author, a discussion of some of theworked examples in the text-book, followed by a quiz for each section in the book. Some stu-dents may find the animations of the examples helpful, although the examples are all workedin the book. You might wish to try some of the quiz questions using paper and pencil, andthen check your answers with those given on the CD. It is not recommended that you attemptto enter your answers digitally, as this is a time-consuming process, and uses a different inputmethod from your WeBWorK assignments, which serve the same purpose.

18No one will check whether you have used any of these aids; a student can obtain a perfect grade in the coursewithout ever consulting any of them. No audio-visual or calculator aid can replace the systematic use of paperand pencil as you work your way through problems. But the intelligent use of some of these aids can deepen yourunderstanding of the subject. However, the most important aid is the Student Solutions Manual to the textbook!

19The version of this CD-ROM for the 6th edition is being catalogued by the Library; it may not be availableat the beginning of the term.


1.6.4 Other Calculus Textbooks

While students may wish to consult other textbooks, instructors and teaching assistants inMath 141 will normally refer only to the prescribed edition of the prescribed textbook for thecourse. Other books can be very useful, but the onus is on you to ensure that your book coversthe syllabus to at least the required depth; where there are differences of terminology, you areexpected to be familiar with the terminology of the textbook.20

In your previous calculus course(s) you may have learned methods of solving problems thatappear to differ from those you find in the current textbook. Your instructors will be pleasedto discuss any such methods with you personally, to ascertain whether they are appropriate tothe present course. In particular, any methods that depend upon the use of a calculator, or theplotting of multiple points, or the tabulation of function values, or the inference of a trend froma graph should be regarded with scepticism.

1.6.5 Website

These notes, and other materials distributed to students in this course, will be accessiblethrough a link on the myCourses page for the course, as well as at the following URL:

http://www.math.mcgill.ca/brown/math141b.html

The notes will be in “pdf” (.pdf) form, and can be read using the Adobe Acrobat reader, whichmany users have on their computers. This free software may be downloaded from the followingURL:

http://www.adobe.com/prodindex/acrobat/readstep.html 21

The questions on some old examinations will also be available as an appendix to these noteson the Web.22

Where revisions are made to distributed printed materials — for example these informationsheets — we expect that the last version will be posted on the Web.

The notes and WeBWorK will also be available via a link from the myCourses (WebCT)URL:

http://mycourses.mcgill.ca

20There should be multiple copies of the textbook on reserve in the Schulich library.21At the time of this writing the current version appears to be 8.n.22There is no reason to expect the distribution of problems on quizzes or in assignments and examinations from

previous years be related to the frequencies of any types of problems on the examination that you will be writingat the end of the term.


1.7 SyllabusSection numbers in the following list refer to the text-book [1]. The syllabus will include allof Chapters 5, 6, 7, 8, 10, 11 with omissions, as listed below.23

Chapter 5: Integrals. §§5.1 – 5.5. The Midpoint Rule, defined in §5.2, and appearing fromtime to time subsequently, is not examination material.

Chapter 6: Applications of Integration. §§6.1 – 6.3; §6.5. (§6.4 is not examination mate-rial, but Science students are urged to read it.)

Chapter 7: Techniques of Integration. §§7.1 – 7.3; §7.4, excluding the Weierstrass substitu-tion [1, Exercises 57-61]; §7.5 §7.8. (§7.6, intended for use in conjunction with integraltables and/or computer algebra systems, is not examination material, but students areadvised to try to solve the problems manually; §7.7 requires the use of a calculator or acomputer, and consequently is not examination material.)

Chapter 8: Further Applications of Integration. §8.1, §8.2 only. (§§8.3, 8.4 are not exam-ination material, but students are urged to read the applications relevant to their courseof study; §8.5 is not examination material.)

Chapter 9: Differential Equations. (No part of this chapter is examination material; how-ever, students are urged to read §9.4 Exponential Growth and Decay).

Chapter 10: Parametric Equations and Polar Coordinates. §§10.1 – 10.4.(§§10.5, 10.6 are not examination material.

Chapter 11. Infinite Sequences and Series. §§11.1 – 11.7. (§§11.8–11.12 are not examina-tion material; however, students are urged to peruse these sections.)

Appendices Appendix G contains material shifted from [22, §5.6]. Students are expected toknow the properties themselves, as they were discussed in MATH 139 and MATH 140.After the class has studied Chapter 5, the definition of the natural logarithm ln x will

thenceforth be taken to be that given in the Appendix, as

x∫

0

dtt

.

23If a textbook section is listed below, you should assume that all material in that section is examinationmaterial even if the instructor has not discussed every topic in his lectures; however, the instructors may give youinformation during the term concerning topics that may be considered subsidiary.

Do not assume that a topic is omitted from the syllabus if it has not been tested in a WeBWorK assign-ment or a quiz, or if it has not appeared on any of the old examinations in the course! Some topics to not lendthemselves to this type of testing; others may have been omitted simply because of lack of space, or oversight.By the same token, you need not expect every topic in the course to be examined on the final examination.


Please do not ask the tutors to provide information as to which textbook sections should beemphasized. Unless you are informed otherwise by the instructors in the lecture sections orpublished notes — printed, or mounted on the Web — you should assume that all materialslisted are included in the syllabus. You are not expected to be able to reproduce proofs of thetheorems in the textbook. However, you could be expected to solve problems in which theremight be unspecific real variables, rather than specific numbers, and which problems mightlook like textbook theorems.24

1.8 Preparation and Workload1.8.1 Prerequisites.

It is your responsibility as a student to verify that you have the necessary prerequisite. It wouldbe foolish25 to attempt to take the course without it.

Students who obtained only a grade of C in MATH 139 or MATH 140 would be advisedto make a special effort to reinforce their foundations in differential calculus; if weaknessin MATH 139 or MATH 140 was a consequence of poor preparation for that course, it isnot too late to strengthen those foundations as well.26 The fact that MINERVA may permityou to register does not relieve you from the responsibility to observe university regulationsconcerning prerequisites, and exposes you to the risk of failure in a course for which youare nor properly prepared; students with an F in MATH 139 or MATH 140 could have theirregistration in MATH 141 annulled. The regulations are in place to protect you!

1.8.2 Calculators

The use of calculators is not permitted in either quizzes or the examination in this course.Students whose previous mathematics courses have been calculator-oriented would be advisedto make particular efforts to avoid the use of a calculator in solving problems in this course,in order to develop a minimal facility in manual calculation. This means that you are urged todo all arithmetic by hand. Students who use calculators when they answer their WeBWorKproblems are undermining the usefulness of the programme to themselves: learn to use thebuilt-in calculation capabilities that are present in WeBWorK.

24The intention is that you should be learning how to solve problems, but should not have to memorize wholeproofs from the textbook.

25and contrary to McGill regulations26The reality of inflated grading at McGill or at your previous institution must not be overlooked: it could

happen that students who obtained a grade higher than C in the prerequisite course do not have adequateskills to succeed in MATH 141! The onus is on you to seek help and to take remedial actions where necessary.


1.8.3 Self-Supervision

This is not a high-school course, and McGill is not a high school. The monitoring of yourprogress before the final examination is largely your own responsibility. Students must notassume that they will be exposed in lectures and tutorials to detailed model solutions for everytype of Calculus 2 problem. It is essential that you supplement these classes with serious workon your own, carefully reading the textbook and solving problems therein.

While the tutors and instructors are available to help you, they cannot do so unless and untilyou identify the need for help. WeBWorK and quizzes are designed to assist you in doing this.If you encounter difficulties, take them to the tutors during one of their many office hours: youmay attend the office hours of any tutor in the course, and are not restricted to those of the tutorof the tutorial in which you are registered.

Time Demands of your Other Courses. Be sure to budget enough time to attend lecturesand tutorials, for private study, and for the solution of many problems. Don’t be tempted todivert calculus study time to courses which offer instant gratification. While the significance ofthe tutorial quizzes in the computation of your grade is minimal, these are important learningexperiences, and can assist you in gauging your progress in the course. This is not a coursethat can be crammed for: you must work steadily through the term if you wish to develop thefacilities needed for a strong performance on the final examination.

Lecture Times, and Preparation for the Lectures The lecture sections in MATH 141 201001 meet at the times that have been made available to us: early in the morning, or late inthe afternoon. While these times may not appeal to you, you should not underestimate thedamage you do to your expectations in the course by missing lectures, either occasionally —when you find it convenient to divert calculus time to other purposes — or systematically. Toextract maximum benefit from the lectures, you should peruse the scheduled material beforecoming to class, trying some of the textbook problems; your instructors invite you to drawtheir attention to specific difficulties that you encounter before class in the textbook — it maybe possible to respond to these difficulties during the lecture.

Working Problems on Your Own. An effective way to master the calculus is through work-ing large numbers of problems from the textbook. Your textbook was selected partly becauseof the availability of an excellent Student Solutions Manual [9]; this manual has brief but com-plete solutions to most of the odd-numbered exercises in the textbook. The skills you acquirein solving textbook problems could have much more influence on your final grade than eitherWeBWorK or the quizzes.


When to do the WeBWorK assignment. I recommend that you defer working WeBWorKproblems until you have tried some of the easier odd-numbered problems in the textbook. Forthese you (should) have the Student Solutions Manual to help you check your work. Onceyou know that you have the basic concepts mastered, then is a good time to start workingWeBWorK problems. But these should be done first from a printed copy of your assignment— not worked during real time online.

The real uses of WeBWorK and the quizzes. Students often misunderstand the true signif-icance of WeBWorK assignments and the quizzes. While both contribute to your grade, theycan help you estimate the quality of your progress in the course. Quizzes are administered un-der examination conditions, so poor performance or non-performance on quizzes can providean indicator of your expectations at the final examination; take proper remedial action if youare obtaining low grades on quizzes27. Since WeBWorK is not completed under examinationconditions, the grades you obtain may not be a good indicator of your expectations on the ex-amination; if you require many attempts before being able to solve a problem on WeBWorK,you should use that information to direct you to areas requiring extra study: the WeBWorKgrades themselves have little predictive use, unless they are unusually low. However, whileboth WeBWorK and the quizzes have a role to play in learning the calculus, neither is as im-portant as reading your textbook, working problems yourself, and attending and listening atlectures and tutorials.

What to strive for on WeBWorK assignments. Since the practice assignments give youample opportunity to experiment, your success rate on the assignments “that count” should beclose to 100%. If you are needing more than 2 attempts to solve a WeBWorK problem, thenyou are probably not ready to work the assignment. In order to be able to solve a WeBWorKproblem successfully on the first attempt you will need to check your work, and this is a skillthat you will need on the final examination, and in the advanced studies or the real world whereyou may eventually be applying the calculus.

1.8.4 Escape Routes

At any time, even after the last date for dropping the course, students who are experiencingmedical or personal difficulties should not hesitate to consult their advisors or the StudentAffairs office of their faculty. Don’t allow yourself to be overwhelmed by such problems; theUniversity has resource persons who may be able to help you.

27The worst action is to miss the quizzes, and thereby block out an unwelcome message.


1.8.5 Terminology

Do not be surprised if your instructors and tutors use different terminology from what youhave heard in your previous calculus course, particularly if that course was at a high school.Sometimes the differences are purely due to different traditions in the professions.

“Negative x” Your instructors and tutors will often read a formula −x as minus x, not asnegative x. To a mathematician the term negative refers to real numbers which are not squares,i.e. which are less than 0, and −x can be positive if x itself is negative.

However, mathematicians will sometimes refer to the operation of changing a sign as thereplacement of x by “its negative”; this is not entirely consistent with the usual practice, but isan “abuse of language” that has crept into the professional jargon.

Inverse trigonometric functions A formula like sin−1 x will be read as the inverse sine ofx — never as “sine to the minus 1” or “sine to the negative 1”. However, if we write sinn x,where n is a positive integer, it will always mean (sin x)n. These conventions apply to any ofthe functions sin, cos, tan, cot, sec, csc; they also apply to the hyperbolic functions, which wehave met on general functions, so a formula like f 2(x) does not have an obvious meaning, andwe will avoid writing it when f is other than a trigonometric or hyperbolic function.

Logarithms Mathematicians these days rarely use logarithms to base 10. If you were taughtto interpret log x as being the logarithm to base 10, you should now forget that — although itcould be the labelling convention of your calculator. Most often, if your instructor speaks of alogarithm, and writes log x, he will be referring to the base e, i.e. to loge; that is, he is referringto the function that calculus books call ln. When a logarithm to some other base is intended,it will either be denoted by an explicit subscript, as log2, or some comment will be made atthe beginning of the discussion, as “all logarithms in this discussion are to the base 2”. Yourinstructors try to think like mathematicians even when lecturing to their classes, and so we usethe language and terminology we use when talking to each other.

1.9 Communication with Instructors and TA’s1. E-mail messages to your instructor or your TA should be sent to the addresses shown in

Table 1.2 and Table 2. Please show your full name and/or student number, so that wecan clearly identify you.

2. The only messages sent through WeBWorK should be those generated by the Feedbackfacility: this means a message that refers to a specific problem on a specific WeBWorK


assignment, generated by clicking on the Feedback button while you are working thatproblem, and after you have entered your proposed answer(s) into the answer box(es).28

3. Please do not send instructors messages using the Mail facility of myCourses. Thisfacility is difficult for instructors to use, since it is not integrated with the other mailservices. We normally disable myCourses mail for that reason. If you had a need to senda message while you are connected to myCourses, just open another window and send amessage with your regular e-mail client.

4. Keep your e-mail address up to date Both myCourses and WeBWorK contain an e-mail address where we may assume you can be reached. If you prefer to use anothere-mail address, the most convenient way is to forward your mail from your studentmailbox, leaving the recorded addresses in these two systems unchanged.

1.10 Commercial tutorial and exam preparation servicesWe do not endorse any commercial tutorial service, nor any service that claims to prepare stu-dents to write examinations. We have no way of evaluating the quality of any such operations,nor whether they are conforming to the University’s general practices. Caveat emptor!

28This facility should be used sparingly; you should not expect instant response, so questions sent close to thedue time on the due date will not likely receive a reply before the assignment becomes due.


1.11 Special Office Hours and TutorialsThe following chart will show any special activities that are scheduled during the term. Thistable was last updated on April 09, 2010.

Review Tutorial TA/Instructor location Date TimePolar Coordinates Dr. Y. Zhao ARTS 145 13 April 15:05–16:55

Sequences + Series J. Feys BURN 1B45 16 April 18:05–19:55



2 Draft Solutions to Quiz Q1

For each of the days of the week a sample quiz is given below. Our policy is that, to earn partmarks on any separately numbered part of any question a student must have, in the opinion ofthe grader, earned at least half of the marks available.

2.1 Instructions to Students1. Show all your work. Marks may not be given for answers not supported by a full solu-

tion. For future reference, the form of your solutions should be similar to those shownin the textbook or Student Solutions Manual for similar problems.

2. In your folded answer sheet you must enclose this question sheet: it will be returnedwith your graded paper. (WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH0.) All submissions should carry your name and student number.

3. Time = 20 minutes.

4. No calculators are permitted.

2.2 Monday Versions1. [10 MARKS] Use a right Riemann sum (i.e. a Riemann sum where the component

rectangles hang by their upper right-hand corner from the graph) to compute

3∫

0

(10x −

2) dx . No other method will be accepted!

Solution: If we divide the interval [0, 3] into n subintervals of equal lengths ∆x = 3n , the

right end-points of these intervals are ∆x, 2∆x, . . . , n∆x, so the definite integral is equalto the following limit of a (right) Riemann sum:

limn→∞

n∑

i=1

(10i∆x − 2) ∆x = limn→∞

10(∆x)2n∑

i=1

i − 2∆xn∑

i=1

1

= limn→∞

(10 · 32

n2 ·n(n + 1)

2− 2 · 3

n· n

)

= limn→∞

(45 · n(n + 1)

n2 − 6 · nn

)

= limn→∞

(45 · 1 ·

(1 +

1n

)− 6

)= 39 .


It would not have been acceptable to evaluate this integral using the Fundamental Theo-rem, but that theorem could be used by a student to verify her work:

3∫

0

(10x − 2) dx =[5x2 − 2x

]3

0= 5 · 32 − 2 · 3 = 45 − 6 = 39.

2. [10 MARKS] Compute

(a)

π3∫

0

(sec x)(7 sec x + 3 tan x) dx .

(b)

1∫

0

√x ·

(5x2 − 5x − 4

)dx .

Solution: For these problems the use of the Fundamental Theorem was not excluded.

(a)

π3∫

0

(sec x)(7 sec x + 3 tan x) dx =

π3∫

0

(7 sec2 x + 3 sec x · tan x

)dx

= [7 tan x + 3 sec x]π30

=

(7 tan

π

3+ 3 sec

π

3

)− (7 tan 0 + 3 sec 0)

= 7√

3 + 3 · 2 − 7 · 0 − 3 · 1 = 7√

3 + 3 .

(b)

1∫

0

√x ·

(5x2 − 5x − 4

)dx =

∫ 1

0

(5x

52 − 5x

32 − 4x

12)

=

[5 · 2

7· x 7

2 − 5 · 25· x 5

2 − 4 · 23· x 3

2

]1

0

=107− 2 − 8

3= −68

21.


2.3 Tuesday Versions1. [10 MARKS] Use a left Riemann sum (i.e. a Riemann sum where the component rect-

angles hang by their upper left-hand corner from the graph) to compute

2∫

0

(3x + 2) dx .

No other method will be accepted!


left end-points of these intervals are 0,∆x, 2∆x, . . . , (n − 1)∆x, so the definite integral isequal to the following limit of a (left) Riemann sum:

limn→∞

n∑

i=1

(3(i − 1)∆x + 2) · ∆x = limn→∞

3(∆x)2n∑

i=1

(i − 1) + 2∆xn∑

i=1

1

= limn→∞

(12n2 ·

(n − 1)n2

+4n

)

limn→∞

(6(1 − 1

n

)+ 4

)= 6 + 4 = 10 .


2∫

0

(3x + 2) dx =

[3 · 1

2· x2 + 2x

]3

0=

32· 4 + 2 · 2 = 6 + 4 = 10.

2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find the

derivative of the function f (x) =

√x∫

2

cosh tt3 dt. Fully simplify your answer!

Solution: Define u =√

x. Then

ddx

∫ √x

2

cosh tt3 dt =

ddu

∫ u

2

cosh tt3 dt · du

dtby the Chain Rule

=cosh u

u3 · dudt

by the Fundamental Theorem

=cosh

√x

x32

dt · 12

x−12

=cosh

√x

2x2 .


2.4 Wednesday Versions1. [10 MARKS] Use a right Riemann sum (i.e. a Riemann sum where the component

rectangles hang by their upper right-hand corner from the graph) to compute

0∫

−2

2x2 dx .


Solution: If we divide the interval [−2, 0] into n subintervals of equal lengths ∆x =0−(−2)

n = 2n , the right end-points of these intervals are −2 + ∆x,−2 + 2∆x, . . . ,−2 + n∆x,

so the definite integral is equal to the following limit of a (right) Riemann sum:

limn→∞

n∑

i=1

2 (−2 + i∆x)2 ∆x = 2 limn→∞

(4∆x − 4i(∆x)2 + i2(∆x)3

)

= 2 limn→∞

4∆xn∑

i=1

1 − 4(∆x)2n∑

i=1

i + (∆x)3n∑

i=1

i2

= 2 limn→∞

4 · 2n· n − 4 ·

(2n

)2

· n(n + 1)2

+

(2n

)3

· n(n + 1)(2n + 1)6

= 2 limn→∞

(8 − 8

(1 +

1n

)+

43·(1 +

1n

) (2 +

1n

))

= 16 − 16 +163

=163.


0∫

−2

2x2 dx = 2[13

x3]0

−2=

23

(03 − (−2)3

)=

163.

2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find the

derivative of the function

e4x∫

e2x

√(ln t)2 + 7 dt. Fully simplify your answer!

Solution: Define v = e4x and u = e2x. We begin by dividing the interval of integrationat some convenient point, which in this proof may not even lie within the interval, andthen reverse the limits of integration in the first summand:

e4x∫

e2x

√(ln t)2 + 7 dt =

1∫

e2x

√(ln t)2 + 7 dt +

e4x∫

1

√(ln t)2 + 7 dt


= −∫ e2x

1

√(ln t)2 + 7 dt +

∫ e4x

1

√(ln t)2 + 7 dt

Now we apply the Chain Rule in differentiating each of the summand integrals:

ddx

∫ e4x

e2x

√(ln t)2 + 7 dt = − d

dx

∫ e2x

1

√(ln t)2 + 7 dt +

ddx

∫ e4x

1

√(ln t)2 + 7 dt

= − ddu

∫ e2x

1

√(ln t)2 + 7 dt

·dudx

+ddv

∫ e4x

1

√(ln t)2 + 7 dt

·dvdx

= −√

(ln e2x)2 + 7 · 2e2x +√

(ln e4x)2 + 7 · 4e4x

= −√

(2x)2 + 7 · 2e2x +√

(4x)2 + 7 · 4e4x

= −√

4x2 + 7 · 2e2x +√

16x2 + 7 · 4e4x .

2.5 Thursday Versions1. [10 MARKS] Use a left Riemann sum (i.e. a Riemann sum where the component rect-

angles hang by their upper left-hand corner from the graph) to compute

2∫

−1

(x2 + 1

)dx.


Solution: If we divide the interval [−1, 2] into n subintervals of equal lengths ∆x =2−(−1)

n = 3n , the left end-points of these intervals are −1,−1 + ∆x,−1 + 2∆x, . . . ,−1 + (n−

1)∆x, so the definite integral is equal to the following limit of a (left) Riemann sum:

limn→∞

n∑

i=1

((−1 + (i − 1)∆x)2 + 1

)∆x

= limn→∞

n∑

i=1

(2 − 2(i − 1)∆x + (i − 1)2(∆x)2

)· ∆x

= limn→∞

2∆xn∑

i=1

1 − 2(∆x)2n∑

i=1

(i − 1) + (∆x)3n∑

i=1

(i − 1)2

= limn→∞

2 · 3n· n − 2 ·

(3n

)2

· (n − 1)n2

+

(3n

)3

· (n − 1)n(2(n − 1) + 1)6

= limn→∞

(6 − 9

(1 − 1

n

)+

92

(1 − 1

n

) (2 − 1

n

))= 6 − 9 + 9 = 6 .

It would not have been acceptable to evaluate this integral using the Fundamental Theo-


rem, but that theorem could be used by a student to verify her work:

2∫

−1

(x2 + 1

)dx = 2

[13

x3 + x]2

−1=

(83

+ 2)−

(−1

3− 1

)= 6 .

2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find all

critical numbers of the function

9∫

ex2+7x+12

ln t dt.

Solution: We first determine the derivative, using the Fundamental Theorem and the

Chain Rule. Define u = ex2+7x+12. Thendudx

= (2x + 7)ex2+7x+12.

ddx

9∫

ex2+7x+12

ln t dt =ddx

−

ex2+7x+12∫

9

ln t dt

= − ddu

ex2+7x+12∫

9

ln t dt

· du

dx

= −(ln ex2+7x+12

)· (2x + 7)ex2+7x+12

= −(x2 + 7x + 12

)(2x + 7)ex2+7x+12

= −(x + 3)(x + 4)(2x + 7)ex2+7x+12 .

This derivative has an exponential factor which cannot equal 0. The function is differen-tiable for all x, and the derivative is 0 when x = −4,−7

2 ,−3, so the latter 3 points are thecritical numbers of the function.

2.6 Friday Versions1. [10 MARKS] Use a right Riemann sum (i.e. a Riemann sum where the component rect-

angles hang by their upper right-hand corner from the graph) to compute

2∫

0

(x2 − 3

)dx.



right end-points of these intervals are ∆x, 2∆x, . . . , n∆x, so the definite integral is equal


to the following limit of a (right) Riemann sum:

limn→∞

n∑

i=1

((i∆x)3 − 3

)∆x = lim

n→∞

(∆x)4n∑

i=1

i3 − 3∆xn∑

i=1

1

= limn→∞

(2n

)4

· n2(n + 1)2

4− 3 · 2

n· n

= limn→∞

4(1 +

1n

)2

− 2 = 4 − 6 = −2 .


2∫

0

(x3 − 3

)dx =

[14· x4 − 3x

]2

0= 4 − 6 = −2.

2. [10 MARKS] Use the Fundamental Theorem of Calculus and the Chain Rule to find all

critical numbers of the function f (x) =

4∫

x2−4x−5

sinh t dt.

Solution: Define u = x2 − 4x − 5, sodudx

= 2x − 4. Then

ddx

∫ 4

x2−4x−5sinh t dt =

ddx

−∫ x2−4x−5

4sinh t dt

= − ddu

∫ x2−4x−5

4sinh t dt

·dudx

= −(sinh

(x2 − 4x − 5

))· (2x − 4) .

The sinh function is 0 if and only if its argument is 0, here if and only if (x − 5)(x + 1) =

x2 − 4x − 5 = 0, i.e., if and only if x = 5 or x = −1. Thus the derivative — which isdefined for all x — is 0 when x = −1, 2, 5; these are the critical numbers.


3 Draft Solutions to Quiz Q2

For each of the days of the week a sample quiz is given below. Our policy is that, to earn partmarks on any separately numbered part of any question a student must have, in the opinion ofthe grader, earned at least half of the marks available.

3.1 Instructions to Students1. Show all your work! To be awarded partial marks on a part of a question a student’s

answer for that part must be deemed to be more than 50% correct. For future reference,the form of your solutions should be similar to those shown in the textbook or StudentSolutions Manual for similar problems.





3.2 Monday Versions1. [10 MARKS] Use appropriate substitutions to compute

(a)∫

x3 + 3x4 + 12x − 2

dx

(b)∫ et

et − 8e−t dt (Hint: Start with the substitution u = et.)

Solution:

(a) In general it’s a long procedure to integrate rational function (=ratios of polynomi-als). However, in this case, one can observe that the numerator is precisely 1

4 of thederivative of the denominator. So, if we use a substitution u = x4 + 12x − 2, wehave du =

(4x3 + 12

)dx = 4

(x3 + 3x

)dx. Thus

∫x3 + 3

x4 + 12x − 2dx =

∫1u

du4

=14

ln u + C =14

ln(x4 + 12x − 2

)+ C .

(b) The integrand is a rational function, but the variable is an exponential. Whilewe could simplify the integrand before proceeding with a substitution, it appearspromising to choose a substitution like u = et. Differentiation yields du = et dt. Sowe have

∫et

et − 8e−t dt =

∫du

u − 8u

=

∫u

u2 − 8du .

Now we have a rational function in u, which can be simplified by a second substi-tution like v = u2 − 8, where dv = 2u du. We obtain

∫u

u2 − 8du =

∫dv2v

=ln v2

+ C =ln

(u2 − 8

)

2+ C = ln

√e2t − 8 + C .

2. [10 MARKS] Let R be the region bounded by the lines x = 1, y = ln 5, and and the curvey = ln x. Compute the volume of the solid of revolution obtained by revolving R aboutthe x-axis. You may either use the method of cross-sections (or washer method) or themethod of cylindrical shells.


Solution:

1. (Using Washers=Cross-sections):

Volume = π

∫ 5

1

((ln 5)2 − (ln x)2

)dx .

We can integrate by parts, taking u = (ln 5)2 − (ln x)2, dv = dx, so du = −2 ln xx

dx,v = x. We obtain

π

∫ 5

1

((ln 5)2 − (ln x)2

)dx = π

[x((ln 5)2 − (ln x)2

)]5

1−

∫ 5

1x ·

(−2 ln x

x

)dx

= π[x((ln 5)2 − (ln x)2

)+ 2(x ln x − x)

]5

1

= π((0 + 10 ln 5 − 10) −

((ln 5)2 + 0 − 2

))

= π(10 ln 5 − (ln 5)2 − 8

).

Where the integral of ln x was required, the student could either have quoted it frommemory, or could have integrated it by parts as is done in the textbook.

2. (Using Cylindrical Shells): For this purpose we will need the equation of the curvey = ln x in the form x = ey, in order to calculate the horizontal length of the shell of

radius y to be ex − 1. Thus the volume is 2π

ln 5∫

0

y (ey − 1) dy. To evaluate this integral

we may again use integration by parts, here taking U = y, dV = ey − 1, so dU = dy,V = ey − y.

2π∫ ln 5

0y (ey − 1) dy = 2π

([y (ey − y)

]ln 50 −

∫ ln 5

0(ey − y) dy

)

= 2π[y (ey − y) −

(ey − y2

2

)]ln 5

0

= 2π(5 ln 5 − (ln 5)2

2− 4

).


3.3 Most Tuesday Versions1. [10 MARKS] Use appropriate substitutions to compute

(a)∫

e5t + 5e5t + 25t

dt

(b)∫

x3√

x2 − 1 dx (Hint: Start with the substitution u = x2 − 1.)

Solution:

(a) Try the substitution u = e5t + 25t. Then du =(5e5t + 25

)dt = 5

(e5t + 5

)dt. Hence

∫e5t + 5

e5t + 25tdt =

15

∫duu

=15

ln u + C

=15

ln(e5t + 25t

)+ C .

(b) The hint suggests using the substitution u = x2 − 1⇒ du = 2x dx. Then∫

x3√

x2 − 1 dx =

∫(u + 1)

√u · du

2

=15· u 5

2 +13

u32 + C

=15

(x2 − 1)52 +

13

(x2 − 1)32 + C.

(There are variations of this substitution that could have been used. For example,we could try the substitution u =

√x2 − 1, so u2 = x2 − 1, x2 = u2 + 1, u du = x dx.

Then∫

x3√

x2 − 1 dx =

∫(u2 + 1)u · u du

=

∫ (u4 + u2

)du

=u5

5+

u3

3+ C

=15

(x2 − 1

) 52

+13

(x2 − 1

) 32

+ C .

2. [10 MARKS] Let R be the region in the first quadrant bounded by the lines y = 0, y = 5x,

and the curve y =6x− 1. Compute the volume of the solid of revolution obtained by


revolving R about the x-axis. You may either use the method of cross-sections (or diskmethod) or the method of cylindrical shells.

Solution: To determine the point of intersection of the lines y = 5x y =6x− 1 we solve

the equations simultaneously: eliminating y between the equations yields 5x2 +x−6 = 0,i.e., (5x + 6)(x − 1) = 0, so the curves intersect in points with x = −6

5 or x = 1. Butx = − 6

5 corresponds to an intersection point in the 3rd quadrant: the only intersectionpoint in the first quadrant is (x, y) = (1, 5).

(a) (Using Washers=Cross-sections): While we can integrate using either of the meth-ods we know, using “washers” will require two separate integrals, since the de-scription of the outer radius of the washer changes at x = 1.

Volume =

∫ 1

0π(5x)2 dx +

∫ 6

1π

(6x− 1

)2

dx

= π

[253

x3]1

0+ π

[−36

x− 12 ln x + x

]6

1

= π

(130

3− 12 ln 6

).

(b) (Using Cylindrical Shells): The right boundary of the region is the curve whose

equation can be reformulated as x =6

y + 1. The length of the cylinder with radius y

is, therefore,6

y + 1− y

5. The volume is, therefore, 2π

5∫

0

y(

6y + 1

− y5

)dy . Because

we haven’t yet studied integration of rational functions in general, the easiest wayto integrate this now is to use a substitution like u = y + 1 to make the denominatorinto powers of the variable:

2π

5∫

0

y(

6y + 1

− y5

)dy = 2π

∫ 6

1(u − 1)

(6u− u − 1

5

)du

=2π5

∫ 6

1

(−u2 + 2u + 29 − 30

u

)du

=2π5

[−u3

5+ u2 + 29u − 30 ln |u|

]6

1

= π

(130

3− 12 ln 6

).


3.4 Most Wednesday Versions1. [10 MARKS]

(a) Compute the average value of f (x) = cos3( x16

)· sin

( x16

)on the interval

(0,

83· π

).

(b) Use an appropriate substitution to compute the integral∫

1e5x + e−5x dx . (Hint:

Start with the substitution u = e5x.)

Solution:

(a) The average is

∫ 8π3

0cos3

( x16

)· sin

( x16

)dx

∫ 8π3

01 dx

. The integral in the numerator can be

evaluated by a substitution u = cos(

x16

), which implies that

du = − sin( x16

)· 1

16dx .

Hence

∫ 8π3

0cos3

( x16

)· sin

( x16

)dx =

∫ √3

2

1

(−16u3

)du = − 4u4

] √32

1=

74.

Hence the average is74

83 · π − 0

=21

32π.

(b) Try the substitution u = e5x; du = u · 5 · dx⇒ dx =du5u

. Then

∫1

e5x + e−5x dx =

∫1

u +1u

· du5u

=15

∫du

u2 + 1

=15

arctan u + C =15

arctan e5x + C .


2. [10 MARKS] Use integration by parts to compute the integral∫

arctan(6x

)dx.

Solution: We know that the integration of arctan x can be accomplished by integrationby parts, so we can try the same tactic here, with u = arctan

(6x

), and dv = dx. Then

v = x, and

du =1

1 +(

6x

)2 ·−6x2 dx =

−636 + x2 · dx .

Hence ∫arctan

(6x

)dx = arctan

(6x

)· x + 6

∫x

36 + x2 dx .

The new integrand can be integrated by using a substitution like w = 36 + x2, where

dw = 2x dx , so x dx =dw2

:

∫arctan

(6x

)dx = arctan

(6x

)· x + 6

∫1

2wdw

= arctan(6x

)· x + 3 · ln |w| + C

= arctan(6x

)· x + 3 · ln

(36 + x2

)+ C .


3.5 Thursday Versions

1. [10 MARKS] Compute the integral∫

cos(5t) · cosh t dt.

Solution: We can integrate this product by two successive integrations by parts, judi-ciously chosen; care is needed in the second application, since a poor decision couldreverse the effect of the first integration. We can begin with u = cos 5t, dv = cosh t dt,which imply that du = −5 sin 5t, v = sinh t. Then

∫cos(5t) · cosh t dt = (cos 5t) · sinh t −

∫(−5 sin 5t) sinh t dt

= (cos 5t) · sinh t + 5∫

(sin 5t) sinh t dt .

In a second application we can take U = sin 5t, dV = sinh t dt, which imply that dU =

5 cos 5t dt, V = cosh t. Then∫

cos(5t) · cosh t dt

= (cos 5t) · sinh t + 5∫

(sin 5t) sinh t dt

= (cos 5t) · sinh t + 5((sin 5t) cosh t −

∫5(cos 5t) cosh t

)

= (cos 5t) · sinh t + 5(sin 5t) cosh t) − 25∫

(cos 5t) cosh t dt

While we haven’t completed the solution yet, we have expressed the original integral interms of itself; if we shift the integral from the right side of the equation to the left, andcombine the 25 copies with the one, we obtain

(1 + 25)∫

cos(5t) · cosh t dt = (cos 5t) · sinh t + 5(sin 5t) cosh t + C ,

which we can solve by division by 26, to obtain∫

cos(5t) · cosh t dt =1

26(cos 5t) · sinh t +

526

(sin 5t) cosh t + C .

(The name of the constant was changed, since this constant is different from the preced-

ing: of course, we could also have writtenC26

.)

2. [10 MARKS] Use a suitable substitution and then integration by parts to compute the

integral∫

x ln(x+5) dx. (Hint: Start with the substitution u = x+5, then use integration

by parts.)


Solution: One can start by defining a substitution u = x + 5, but the substitution couldbe deferred, or even entirely avoided. For example, just take u = ln(x + 5), dv = x dx, so

du =dx

x + 5, v =

x2

2.

Then ∫x ln(x + 5) dx = (ln(x + 5)) · x2

2−

∫x2

2· dx

x + 5.

The last integral could be integrated by simply using long division, where x2 = (x +

5)(x − 5) + 25, so

12

∫x2

x + 5=

∫x − 5

2dx +

252

∫1

x + 5dx =

x2

4− 5x

2+

252

ln |x + 5| + C . (1)

Hence∫

x ln(x + 5) dx =x2

2· ln(x + 5)) − x2

4+

5x2− 25

2ln |x + 5| + C1 .

The long division step is one of the standard operations we will see when we study theintegration of general rational functions. But one could have used, in the second integralin equation (1), a substitution u = x + 5, to obtain∫

x2

2(x + 5)dx =

∫(u − 5)2

2udu

=

∫ (u2− 5 +

252u

)du

=u2

4− 5u +

252

ln |u| + C =(x + 5)2

4− 5(x + 5) +

252

ln |x + 5| + C2 .

The substitution could also have been effected at the beginning, and it was this sequencethat was contemplated in the problem:

∫x ln(x + 5) dx =

∫(u − 5)(ln u) du .

Now we can integrate by parts, taking U = ln u, dV = (u − 5) du, so that dU =duu

,

V =u2

2− 5u,

∫x ln(x + 5) dx =

∫(u − 5)(ln u) du


= (ln u)(u2

2− 5u

)−

∫ (u2

2− 5u

)· du

u

= (ln u)(u2

2− 5u

)−

∫ (u2− 5

)du

= (ln u)(u2

2− 5u

)− u2

4+ 5u + C

= (ln(x + 5))((x + 5)2

2− 5(x + 5)

)− (x + 5)2

4+ 5(x + 5) + C3 .


3.6 Friday Versions1. [10 MARKS] Let R be the region bounded by the lines y = 0, y = ln 7, x = 0, and

the curve y = ln x. Compute the volume of the solid obtained by revolving R about thex-axis. [Hint: Sketch the region; then choose the appropriate method.]

Solution:

(a) (Using Cylindrical Shells): We will need to find the height of the cylindrical shellof radius y, and this will require transforming the equation y = ln x into x = ey.

The distance from x = 0 to x = ey is then ey − 0, so the volume is

ln 7∫

0

2πy · ey dy ,

which we will integrate by parts, taking u = y, dv = ey dy, so that du = dy, v = ey:

ln 7∫

0

2πy · ey dy = 2π[yey]ln 7

0 − 2π∫ ln 7

0ey dy

= 2π[yey]ln 7

0 − 2π [ey]ln 70

= 2π((ln 7) · 7 − 0) − 2π(7 − 1) = 2π(7 ln 7 − 6) .

(b) (Using Washers): One drawback of this method is that there will be two kinds ofwashers: For 0 ≤ x ≤ 1 the washers have no hole in the middle, and the radius is aconstant, ln 7; but for 1 ≤ x ≤ 7 there will be a hole whose radius is ln x. The areais, therefore, a sum

∫ 1

0π((ln 7)2 − 02

)dx +

∫ 7

1π((ln 7)2 − (ln x)2

)dx

=

∫ 7

0π((ln 7)2 − 02

)dx − π

∫ 7

1(ln x)2 dx

= π(ln 7)2∫ 7

0dx − π

∫ 7

1(ln x)2 dx

= 7(ln 7)2 − π∫ 7

1(ln x)2 dx .

To integrate (ln x)2 we could apply integration by parts twice. Start by taking u =

(ln x)2, v′ = 1, so that u′ = 2ln x

x, v = x, and

∫ 7

1(ln x)2 dx =

[(ln x)2 · x

]7

1− 2

∫ 7

1

ln xx· x dx


=[(ln x)2 · x

]7

1− 2

∫ 7

1ln x dx

=[(ln x)2 · x

]7

1− 2[x · ln x − x]7

1

= [7(ln 7)2 − 0] + 2[(7 ln 7 − 7) − (0 − 1)]

which leads to the same result as found earlier.

2. [10 MARKS] Use an appropriate substitution to compute the integral

9∫

0

16 +√

xdx .

(Hint: Start with the substitution u = 6 +√

x.)

Solution: One substitution that would simplify this integral is u = 6 +√

x ⇒ √x =

u − 6⇒ x = (u − 6)2 ⇒ dx = 2(u − 6) du. Thus

9∫

0

16 +√

xdx =

∫ 6+√

9

6

2(u − 6)u

du

= 2∫ 9

6

(1 − 6

u

)du

= 2[u − 6 ln u]96 = 2(9 − 6 ln 9 − 6 + 6 ln 6) = 6 + 12 ln

23.


4 References

4.1 Stewart Calculus Series[1] J. Stewart, Single Variable Calculus (Early Transcendentals), Sixth Edition. Thomson *

Brooks/Cole (2008). ISBN 0-495-01169-X.

[2] J. Stewart, Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole(2008). ISBN 0-495-01166-5.

[3] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s Sin-gle Variable Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole(2008). ISBN 0-495-01240-8.

[4] J. Stewart, Single Variable Calculus (Early Transcendentals), Sixth Edition. Thomson *Brooks/Cole (2008); bundled with Student Solutions Manual for Stewart’s Single Vari-able Calculus (Early Transcendentals), Sixth Edition. Thomson * Brooks/Cole (2008).ISBN 0-495-42966-X.

[5] R. St. Andre, Study Guide for Stewart’s Single Variable Calculus (Early Transcenden-tals), Sixth Edition. Thomson * Brooks/Cole (2008). ISBN 0-495-01239-4.

[6] J. Stewart, Multivariable Calculus (Early Transcendentals), Sixth Edition. Thomson *Brooks/Cole (2008). ISBN 0-495-?????-?.

[7] J. Stewart, Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson *Brooks/Cole (2003). ISBN 0-534-39330-6.

[8] J. Stewart, Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole(2003). ISBN 0-534-39321-7.

[9] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s Sin-gle Variable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole(2003). ISBN 0-534-39333-0.

[10] J. Stewart, Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson *Brooks/Cole (2003); bundled with Student Solutions Manual for Stewart’s Single Vari-able Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003).ISBN 0-17-6425411.

[11] J. Stewart, Single Variable Essential Calculus (Early Transcendentals). Thomson *Brooks/Cole (2006). Thomson * Brooks/Cole (2003). ISBN 0-495-10957-6.


[12] J. Stewart, Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole(2003); bundled with Student Solutions Manual for Stewart’s Single Variable Calculus(Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-10307-3.

[13] R. St. Andre, Study Guide for Stewart’s Single Variable Calculus (Early Transcenden-tals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39331-4.

[14] Video Outline for Stewart’s Calculus (Early Transcendentals), Fifth Edition. Thomson* Brooks/Cole (2003). ISBN 0-534-39325-X. 17 VCR tapes.

[15] Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcendentals, 5thEdition. Thomson * Brooks/Cole (2003). ISBN 0-534-39326-8.

[16] H. Keynes, J. Stewart, D. Clegg, Tools for Enriching Calculus, CD to accompany [7]and [8]. Thomson * Brooks/Cole (2003). ISBN 0-534-39731-X.

[17] J. Stewart, Single Variable Calculus (Early Transcendentals), Fourth Edition.Brooks/Cole (1999). ISBN 0-534-35563-3.

[18] J. Stewart, Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole (1999). ISBN0-534-36298-2.

[19] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s SingleVariable Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole (1999). ISBN0-534-36301-6.

[20] J. Stewart, L. Redlin, S. Watson, Precalculus: Mathematics for Calculus, EnhancedReview Edition. Thomson * Brooks/Cole. (2006). ISBN: 0-495-39276-6.

[21] J. Stewart, Trigonometry for Calculus. Thomson * Brooks/Cole. ISBN: 0-17-641227-1.

4.2 Other Calculus Textbooks

4.2.1 R. A. Adams

[22] R. A. Adams, Calculus, Single Variable, Fifth Edition. Addison, Wesley, Longman,Toronto (2003). ISBN 0-201-79805-0.

[23] R. A. Adams, Calculus of Several Variables, Fifth Edition. Addison, Wesley, Longman,Toronto (2003). ISBN 0-201-79802-6.


[24] R. A. Adams, Calculus: A Complete Course, Fifth Edition. Addison, Wesley, Longman,Toronto (2003). ISBN 0-201-79131-5.

[25] R. A. Adams, Student Solution Manual for Adams’, Calculus: A Complete Course, FifthEdition. Addison, Wesley, Longman, Toronto (2003). ISBN 0-201-79803-4.

[26] R. A. Adams, Calculus: A Complete Course, with Solution Manual, Fifth Edition. Ad-dison, Wesley, Longman, Toronto (2003). ISBN 0-131-30565-4.

[27] R. A. Adams, Calculus: A Complete Course Manual, Sixth Edition. Addison, Wesley,Longman, Toronto (2006). ISBN 0-321-27000-2.

4.2.2 Larson, Hostetler, et al.

[28] Calculus Instructional DVD Program, for use with (inter alia) Lar-son/Hostetler/Edwards, Calculus of a Single Variable: Early Transcendental Functions,Third Edition [29]. Houghton Mifflin (2003). ISBN 0-618-25097-2.

[29] R. Larson, R. P. Hostetler, B. H. Edwards, D. E. Heyd, Calculus, Early Transcenden-tal Functions, Third Edition. Houghton Mifflin Company, Boston (2003). ISBN 0-618-22307-X.

4.2.3 Edwards and Penney

[30] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus, Early Transcendentals,Sixth Edition. Prentice Hall, Englewood Cliffs, NJ (2002). ISBN 0-13-041407-7.

[31] C. H. Edwards, Jr., and D. E. Penney, Calculus with Analytic Geometry, Early Tran-scendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997). ISBN0-13-793076-3.

[32] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Calculus with Ana-lytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, EnglewoodCliffs, NJ (1997). ISBN 0-13-079875-4.

[33] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus with Analytic Geome-try, Early Transcendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ(1997). ISBN 0-13-793092-5.

[34] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Single VariableCalculus with Analytic Geometry, Early Transcendentals Version, Fifth Edition. Pren-tice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-095247-1.


4.2.4 Others, not “Early Transcendentals”

[35] G. H. Hardy, A Course of Pure Mathematics, 10th edition. Cambridge University Press(1967).

[36] H. S. Hall, S. R. Knight, Elementary Trigonometry, Fourth Edition. Macmillan andCompany, London (1905).

[37] S. L. Salas, E. Hille, G. J. Etgen, Calculus, One and Several Variables, 10th Edition.John Wiley & Sons, Inc. (2007). ISBN 0471-69804-0.

4.3 Other References

[38] G. N. Berman, A Problem Book in Mathematical Analysis. Mir Publishers, Moscow,(1975) 1977

[39] D. Ebersole, D. Schattschneider, A. Sevilla, K. Somers, A Companion to Calculus.Brooks/Cole (1995). ISBN 0-534-26592-8.

[40] McGill Undergraduate Programs Calendar 2009/2010. Also accessible athttp://coursecalendar.mcgill.ca/ug200910/wwhelp/wwhimpl/js/html/wwhelp.htm

Information for Students in Lecture Section 1 of MATH 141 2010 01 1001

A Timetable for Lecture Section 001 of MATH 141 2010 01Distribution Date: Monday, January 04th, 2010

(Last updated on March 29th, 2010. Subject to further correction and change.)Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAYJANUARY

04 §5.1, §5.2 06 §5.1, §5.2 08 §5.3Tutorials begin week of January 11th, 2010

11 §5.3, §5.4 13 §5.4, §5.5 15 §5.5Course changes must be completed on MINERVA by Tuesday, Jan. 19, 2010

18 §6.1 Q1 20 §6.2 Q1 22 §6.3 Q1

WeBWorK Assignment A1 due Jan. 24, 2010Deadline for withdrawal with fee refund = Jan. 24, 2010

Verification Period: January 25 – 29, 201025 §6.5 A1 27 §7.1 29 §7.1, §7.2

FEBRUARY01 §7.2 03 §7.3 05 §7.3

WeBWorK Assignment A2 due Feb. 07, 201008 §7.4 Q2 A2 10 §7.4 Q2 12 §7.5, §7.8 Q2

Deadline for web withdrawal (with W) from course via MINERVA = Feb. 14, 201015 §7.8 17 §8.1, §8.2 19 §8.2

Study Break: February 21 – 27, 2010No lectures, no regular office hours, no regular tutorials!

WeBWorK Assignment A3 due Feb. 21, 201022 NO LECTURE, NO TU-

TORIALS A3

24 NO LECTURE, NO TU-TORIALS


Notation: An = Regular WeBWorK Assignment An due about 23:30 hourson the Sunday preceding this Monday

Qn = Quiz Qn will be administered at the tutorials this week.X = reserved for eXpansion or review


Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAYMARCH

01 §10.1,§10.2 03 §10.2 05 §§10.2–§10.4WeBWorK Assignment A4 due Mar. 07, 2010

08 §10.4 A4 Q3 10 §10.4 Q3 12 §10.4, §11.1 Q3

15 §11.2, §11.3 17 §11.3 19 §11.3WeBWorK Assignment A5 due Monday, Mar. 21, 2010

22 §11.4 A5 Q4 24 §11.5 Q4 26 §11.6 Q4

29 §11.6, §11.7 31 §11.7, XAPRIL

02 NO LECTURE Q4

This week’s tutorials are the last (except for Monday tutorials).WeBWorK Assignment A6 due Apr. 04, 2010

05 NO LECTURE 07 X 09 X12 X 14 X





B Timetable for Lecture Section 002 of MATH 141 2009 01Distribution Date: Monday, January 04th, 2010

(Subject to further correction and change.)Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAYJANUARY

04 §5.1, §5.2 06 §5.3 08 §5.4Tutorials begin week of January 11th, 2010

11 §5.4, §5.5 13 §5.5 15 §6.1Course changes must be completed on MINERVA by Tuesday, Jan. 19, 2010

18 §6.2 Q1 20 §6.2, §6.3 Q1 22 §6.3 Q1

WeBWorK Assignment A1 due Jan. 24, 2010Deadline for withdrawal with fee refund = Jan. 24, 2010

Verification Period: January 25 – 29, 201025 §6.5 A1 27 §6.5, §7.1 29 §7.1, §7.2

FEBRUARY01 §7.2 03 §7.2, §7.3 05 §7.3

WeBWorK Assignment A2 due Feb. 07, 201008 §7.3, §7.4 Q2 A2 10 §7.4 Q2 12 §7.5, §7.8 Q2

Deadline for web withdrawal (with W) from course via MINERVA = Feb. 14, 201015 §7.8 17 §8.1, §8.2 19 §8.2

Study Break: February 21 – 27, 2010No lectures, no regular office hours, no regular tutorials!

WeBWorK Assignment A3 due Feb. 21, 201022 NO LECTURE, NO TU-

TORIALS A3







Section numbers refer to the text-book.

MONDAY WEDNESDAY FRIDAYMARCH

01 §10.1,§10.2 03 §10.3 05 §10.3, §10.4WeBWorK Assignment A4 due Mar. 07, 2010

08 §10.4 A4 Q3 10 §11.1, §11.2 Q3 12 §11.2 Q3

15 §11.3 17 §11.4 19 §11.4, §11.5WeBWorK Assignment A5 due Sunday, Mar. 28, 2010

22 §11.5 A5 Q4 24 §11.6 Q4 26 §11.6 Q4

29 §11.6 31 §11.7, XAPRIL

02 NO LECTURE Q4

This week’s tutorials are the last (except for Monday tutorials).WeBWorK Assignment A6 due Apr. 04, 2010

05 NO LECTURE 07 X 09 X12 X 14 X




C Supplementary Notes for Students in Section 001 of MATH141 2010 01

C.1 Lecture style in Lecture Section 001Lecture content. The timetable on pages 1001, 1002 will show you approximately what Iplan to discuss at each lecture. I suggest that you look through the material in advance. Ifyou have time to try some of the exercises, and find some that cause you difficulty, you arewelcome to bring them to my attention; perhaps I may be able to work some of these examplesinto the lecture.

What goes on the chalkboard? — Should I take notes? I believe strongly that studentsshould not sit in the lecture feverishly copying notes for fear of missing some essential topic;in this course most of what you need to know is contained in the textbook. You should takenotes, but you should be trying to think at the same time. The chalkboard will be used for

• statement/illustration of specific definitions and theorems

• sketching solutions to problems, or classes of problems

• a scratchpad

Some of this material will be useful to you in learning the material in the course. Even whenthe material on the board is equivalent to something in your textbook, the act of writing mayhelp you remember it. But much of the material will be restatements of your textbook, so youshould normally not panic if you miss something.

Graphs My emphasis is on qualitative properties of the graphs of functions, but not on theproduction of extremely precise graphs. You can expect to see me draw on the chalkboardsketches that are extremely crude approximations of functions, sometimes even caricaturesof the true graph. Mathematicians do not base proofs on sketches of graphs — the role of asketch is usually only to assist the reader to visualize the verbal or symbolic reasoning whichaccompanies it. Sometimes a graph is used help one discover a phenomenon, but the resultwould not be acceptable to a mathematician unless it could be proved in a non-graphical way.29

These supplementary notes Section and paragraph headings follow the order of topics inthe textbook. While some of the comments or explanations may be helpful in understanding thebook, the notes are not required reading for examination purposes. Sometimes it may happenthat the discussion of a topic or an exercise evolves during the lecture into one which requires

29This is why I usually avoid problems in the textbook that appear to be drawing inferences from graphs.


more detail than is practical to write on the chalkboard. In such cases you may be referredat the following lecture to supplementary material that will be contained in the notes placedon the Web. Such evolutions are spontaneous and not planned, and cannot be announced inadvance.

Timing and corrections The notes will usually not be posted until after the lecture. WhileI do try to check the notes before posting them, there will inevitably be errors: if you seesomething that doesn’t look right, please ask. The notes will be progressively corrected asmisprints and other errors are discovered.

C.2 Supplementary Notes for the Lecture of January 04th, 2010Release Date: January 04th, 2010,

subject to revision

Textbook Chapter 5. INTEGRALS.(There will be examples, etc., in these notes that were not discussed specifically at the lecture,because of time constraints.)

C.2.1 §5.1 Areas and Distances.

When, in [1, §2.1], the textbook was motivating the differential calculus, it presented two appli-cations: “The Tangent Problem”, which was geometric; and “The Velocity Problem”, whichwas physical. Now, in motivating the integral calculus, the author presents two analogousproblems: “The Area Problem”, which is geometric; and “The Distance Problem”, which isphysical.

The Area Problem. The textbook discusses approximation of the area under the graph ofy = f (x) between x = a and x = b — more precisely, the area between the graph, the x-axis,and the vertical lines x = a and x = b, as a limit of a sum of areas of narrow vertical rectangles.The approximation is first motivated with simple functions where the area can be boundedabove and below by easily computable sums, which together converge to the same value astheir number approaches ∞ and their width approaches 0. You should become comfortableusing the “sigma notation”, where, for integers n1 and n2 with n1 ≤ n2, we write

n2∑

i=n1

f (i)

to mean the sum f (n1) + f (n1 + 1) + f (n1 + 2) + . . . + f (n2) . The textbook then proposes thefollowing definition of the area between the graph of a function, two vertical lines, and thex-axis:


Definition C.1 (Preliminary) Let f be a function which is continuous on the interval [a, b].The area of the region bounded by the graph of f , the x-axis, and the vertical lines x = aand x = b (where a ≤ b) is the limit of a sum of the areas of rectangles “hanging” from thegraph, as follows: subdivide the interval [a, b] into n smaller intervals by points x1, x2, . . . , xn−1,where x1 < x2 < . . . < xn−1, and, for convenience, we define x0 = a, xn = b; select points x∗i

(i = 1, . . . , n) such that xi ≤ x∗i+1 ≤ xi+1, and consider the sum Rn =

n∑

i=1

f(x∗i

)(xi − xi−1).

(This definition is “preliminary” in that we haven’t yet argued that such a limit need exist.This is not the type of limit — of a function of one variable — studied in MATH 140. Wealso need to clarify what restrictions hold for the points x1, x2, . . . xn−1, and how we select thepoints x∗1, x∗2, . . . x∗n.)

Example C.1 (cf. [1, Example 5.1.2, p. 357]) Let a and b be non-negative real numbers, andconsider the area under the parabola y = x2 between the vertical lines x = a, x = b, and abovethe line y = 0.

1. Let’s first consider the special case where the left side of the region is along the y=axis,i.e., where a = 0. From this special case we will be able to derive the general solution.

2. Divide the interval [0, b] into n intervals of equal widthb − 0

n; thus the points x1, x2, . . .

are chosen to be xi =b − 0

n· i (i = 1, 2, . . . , n). For the sample points x∗i , let’s consider

the points at the right end of each of the subintervals: so x∗i = xi (i = 1, 2, . . . , n). Then

Rn =

n∑

i=1

(b − 0

n· i)2

·((

b − 0n· i)−

(b − 0

n· (i − 1)

))(2)

where the second factor is simply the common length of all the subintervals, i.e.,bn

.

The first factor is squared because we are evaluating the function f (x) = x2 at the pointb − 0

n· i. In order to evaluate this sum we need to know the sum of the squares of the

first n positive integers (=natural numbers). If you didn’t learn this in high school, hereit is:

n∑

i=1

i2 =n(n + 1)(2n + 1)

6. (3)

We won’t be able to prove this formula here: a proof requires use of a tool like Mathe-matical Induction, which we are omitting from the syllabus. Applying (3) to (2) yields

Rn = b3 ·n(n + 1)(2n + 1)

6n3 = b3 ·

(2 +

3n

+1n2

)

6.


As n is permitted to become arbitrarily large positively, Rn → b3

3.

3. If the region starts at the origin and extends to the line x = a, then the area will bea3

3.

So, if the region we wish to study starts at the line x = a, and ends at the line x = b,

where b ≥ a, we need only subtract one of these areas from the other, obtainingb3 − a3

3.


C.3 Supplementary Notes for the Lecture of January 06th, 2010Release Date: January 06th, 2010,

subject to revision

C.3.1 §5.1 Areas and Distances (conclusion).

The Distance Problem. Here the textbook considers what appears to be a different typeof problem, and shows that the solution is the same type of sum met in the Area Problemabove. In this case the problem is to determine the distance travelled by a particle moving sothat its velocity at time t is a given function f (t). It will be seen that the distance can again beinterpreted as a limit of a sum — the same sum that would be seen if we attempted to determinethe area in the xt-plane under the graph x = f (t).

If a particle is known to be moving along the x-axis at a velocity of v(t) = x′(t) =dxdt

(t)cm/s, how much distance is traversed between time t = a and time t = b? If, by distance, wemean displacement or “signed distance”, where movement to the right counts positively, andto the left negatively30, then the distance is x(b)− x(a); we shall see that this can be interpreted

as the area under the graph y =ddt

x(t) between x = a and x = b, which we will be denoting

by

b∫

a

ddt

x(t) dt . When it is intended to consider all motion as non-negative — the way the

odometer of an automobile measures distance, then we would want to find the area under thegraph not of the velocity, v(t) = x′(t), but of the speed, |v(t)| = |x′(t)|, and the distance travelledwould be ∫ b

a|v′(t)| dt =

∫ b

a

∣∣∣∣∣ddt

x(t)∣∣∣∣∣ dt .

But in practice the word distance is often used with either meaning, so care is required.I have shown that the area under the graph of the velocity represents the directed dis-

tance travelled by the moving particle. But we expressed the velocity as the derivative of thedisplacement of the particle relative to some fixed origin; and the distance travelled can beexpressed as the difference between two values of the displacement. This is a special case ofthe Fundamental Theorem of the (Integral) Calculus, which will be introduced below.

5.1 Exercises30Note that, with this definition, a particle that moves around and then returns to the same point will have

travelled a distance of 0.


[1, Exercise 20, p. 365] Determine a region whose area is equal to the limit

limn→∞

n∑

i=1

2n

(5 +

2in

)10

.

Do not evaluate the limit at this time.

Solution: Take the widths of the approximating rectangles to be constant,

4x =2n

;

then n of these constant widths would span a distance of length 2. The limit can be

seen to be limn→∞4x ·

n∑i=1

(5 + i4x)10 . The rectangles could be interpreted, for example,

as “hanging by their upper right hand corners” from the curve y = (5 + x)10 abovethe interval 0 ≤ x ≤ 2. (Later in the course we shall see that the area is (5+2)11−511

11 =1,928,498,618

11 .)

In the lecture I mentioned that the area of this region could be represented in other ways,for example, by hanging the component rectangular elements by their left upper corner

from the graph. In that case the sum could have been written as limn→∞

n∑

i=1

2n

(5 +

2(i − 1)n

)10

;

or, alternatively, as limn→∞

n−1∑

i=0

2n

(5 +

2in

)10

.

The original limit sum, or the limits of either of the latter sums, could also be interpretedas being the area under the graph of f (x) = x10 between x = 5 and x = 7. The transitionfrom this interpretation to the earlier one is represented by moving the location of they-axis — without changing the area of the region.

[1, Exercise 21, p. 365] Determine a region whose area is equal to the limit

limn→∞

n∑

i=1

π

4ntan

iπ4n

.

Do not evaluate the limit at this time.

Solution: You should read the discussion of this problem in your Student Solution Man-ual. What is different from the preceding example to this one, is that we don’t have anystraightforward algebraic way of evaluating the limit of the sum of rectangles obtainedto approximate this area. Thus there is going to be something new in our theory if wewill be able to determine this area exactly. And, in fact, we shall be able to evaluate this


area exactly! The upper boundary of the region in this case could be the graph of thefunction y = tan x, and the other boundaries of the region could be, in addition to thex-axis, the vertical lines x = 0 and x =

π

4.

C.3.2 §5.2 The Definite Integral

The formal definition of the integral involves a number of technical difficulties which I shallnot consider in detail in this course. You should read the definition the textbook gives of theintegral [1, p. 366], but you are not going to be asked to work with it in full generality; in factthe definition given in the textbook is simpler than the definition that is normally used for theRiemann Integral. We would need to appeal to this definition if we wished to formally proveall the properties that the author is going to ascribe to the integral; but we shall not attempt toprovide such proofs.

The usual definition of the integral would permit the widths of the subintervals, here de-noted by ∆x, to be different: ∆x1 for the first subinterval, ∆x2 for the second, etc., and wouldthen require that the largest of them should approach zero. This technicality is needed forgeneral functions, but will not be discussed further in this course.

Read the book and be sure you know the definitions of each of the following terms:

• sample points

• definite integral of f from a to b

• integral sign

• integrand

• limits of integration

• lower limit, upper limit

• Riemann sum

Where we take, as the sample points in the subintervals, global maximum points for the func-tion on the subintervals, we have what is called the upper Riemann sum; analogously, we mayspeak of the lower Riemann sum. To prove the existence of the definite integral we would wantto show that the difference between these sums approaches 0 in the limit. This can be shownto be the case in particular when the function is continuous everywhere, and it is even true incertain more general situations. In this course we will normally be taking the functions to beeither continuous or, more generally, “piecewise continuous”; that is, we will permit functionswhich can be obtained by “gluing” together functions which are continuous over adjacent in-tervals. As long as there are only a finite number of such components, it can be shown that the


integral exists; it doesn’t matter if the function is discontinuous at the finite number of loca-tions were the functions are “glued”. But some of the properties we will be using will applyonly to continuous functions, and we may have to break a problem up into parts in order tosolve it. More about this later.

Evaluating Integrals. Among the integrals discussed in this subsection are several that re-quire the following formulæ for sums of powers of the natural numbers:

n∑

i=1

i1 =n(n + 1)

2(4)

n∑

i=1

i2 =n(n + 1)(2n + 1)

6(5)

n∑

i=1

i3 =n2(n + 1)2

4(6)

to which we could add the following trivial result31:

n∑

i=1

i0 = n . (7)

Formulæ 4, 5, 6 are proved in [1, Appendix E], but you are not expected to have read thoseproofs. The proofs given are by “Mathematical Induction”.32

Asking the Right Question. The fact that the formulæ for the sums of powers do not appear tofollow any general pattern is not because there is no pattern, but simply that we are “asking the wrongquestion”. If, instead, we had asked for the sums of what are called falling factorials, i.e., productsof an integer with successive integers immediately less than it, we would obtain the following, muchprettier results. You do not need to remember these formulæ:

n∑

i=1

i0 =n1

(8)

n∑

i=1

i =(n + 1)n

2(9)

31A definition that the product of an empty set of numbers is equal to 1 is consistent with the definition ofmultiplication of real numbers.

32Mathematical Induction was not an examination topic in MATH 140 2009 09; in the present course you arenot expected to know how to apply Mathematical Induction, but interested students are urged to read about it inthe textbook [1, pp. 77].


n∑

i=1

i(i − 1) =(n + 1)n(n − 1)

3(10)

n∑

i=1

i(i − 1)(i − 2) =(n + 1)n(n − 1)(n − 2)

4(11)

A glance at these formulæ, which are certainly “prettier” than the formulæ for the sums of the powers,shows that the first one, (8) doesn’t look as though it fits. Here again, that is because we are again“asking the wrong question”. Let’s formulate the results slightly differently, including the term i = 0 ineach of the sums; only in the case of the 0th powers does this make any difference, since 00 is definedto be 1:

n∑

i=0

i0 =n + 1

1(12)

n∑

i=0

i =(n + 1)n

2(13)

n∑

i=0

i(i − 1) =(n + 1)n(n − 1)

3(14)

n∑

i=0

i(i − 1)(i − 2) =(n + 1)n(n − 1)(n − 2)

4. (15)

Now we can see much further; we can even conjecture that there is a general result that encompassesall of these particular cases:

n∑

i=0

i(i − 1)(i − 2) · . . . · (i − r + 1) =(n + 1)n(n − 1)(n − 2) · . . . · (n − r + 1)

r. (16)

And finally, of what use are these formulæ if we need the sums of the powers of the integers, notthe sums of “falling factorials”. Any power of n can be expressed in terms of “falling factorials”, forexample

n1 = n

n2 = n(n − 1) + n

n3 = n(n − 1)(n − 2) + 3n(n − 1) + n ,

so property (16) can provide all the sums we need.The purpose of this parenthetical discussion is to illustrate that the main challenge in proofs by

induction is making the right guess, rather than in the details of the proof, which may be routine.

Linearity of the summation operator. The textbook discusses some properties of the “sigma”notation; these could be called the linearity properties of the operator Σ, and are all special


cases of the following:k+∑

i=k

(rai + sbi) = rk+∑

i=k

ai + sk+∑

i=k

bi .

I may have more to say about the sigma notation after I discuss [1, §5.5], where we shallencounter properties of the integral that have analogues for sums. For the present let it be

noted that the symbol i ink+∑i=k

ai is not a “free” variable, in that you cannot assign any values to

it: it performs a function in the symbol, but that function would be performed equally well if

we replaced i by any other symbol that is not already in use, e.g.,k+∑u=k

au,k+∑λ=k

aλ,k+∑♥=k

a♥ .

The Midpoint Rule. The “Midpoint Rule” is an approximation formula for definite inte-grals. Use of an approximation formula entails a willingness to accept an error in the cal-culation. Mathematicians normally expect to see an estimate of how good or how bad anapproximation can be before recommending their use. A partial justification of the MidpointRule is contained in [1, §7.7], a section that is to be omitted from the syllabus. For that reasonyou are asked to omit this subsection: you will not be expected to know anything about theMidpoint Rule.

Properties of the Definite Integral – Linearity Properties. The textbook lists many prop-erties of the Definite Integral, proving some of them.

1.

b∫

a

c dx = c(b − a)

2.

b∫

a

[ f (x) + g(x)] dx =

b∫

a

f (x) dx +

b∫

a

g(x) dx

3.

b∫

a

c f (x) dx = c

b∫

a

f (x) dx

4.

b∫

a

[ f (x) − g(x)] dx =

b∫

a

f (x) dx −b∫

a

g(x) dx

for any real numbers a, b, c, and any continuous functions f , g.


Some of these properties can be derived from others, or can be combined into a more generalformula. So, for example we can prove that

∫ b

a(r · f (x) + s · g(x)) dx = r

∫ b

af (x) dx + s

∫ b

ag(x) dx (17)

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx (18)

for any real numbers r, s, a, b, c, and these two equations are equivalent to the properties thatthe textbook numbers ##2, 3, 4, etc. [1, p. 387]:

∫ a

af (x) dx = 0 (19)

∫ b

a( f (x) ± g(x)) dx =

∫ b

af (x) dx ±

∫ b

ag(x) dx (20)

∫ b

ac f (x) dx = c

∫ b

af (x) dx (21)

The first property in the textbook list,∫ b

ac dx = c(b − a) (22)

states, for c ≥ 0 and b ≥ a, that the area of a rectangle of width b − a and height c is c(b − a).Note that all of these properties hold for constants a, b, c that are positive or or negative!

Here one must be careful in interpreting areas, since, in the definite integral, areas are signed— they are either positive or negative: we associate the positive sign to areas under a graphabove the x-axis, where the lower limit of the integral is not greater than the upper limit. Whenthe curve is below the x-axis, or the lower limit of the integral is not greater than the upperlimit, the area is negative. (This is the case for part of [1, Exercise 22, p. 377] which is solvedbelow: there the portion of the area that was below the x-axis cancelled part of the area abovethe x-axis; while the net result we obtained — 21 — was positive, it was not equal to thetotal of the magnitudes of the two areas above and below the x-axis, but was equal to theirdifference. The graph of the integrand crosses the x-axis at the points ±√6 − 1. The regionunder the interval [1,

√6−1] can be shown to have (negative) area −4

√6+ 28

3 ; while the regionover the interval [

√6 − 1, 4] can be shown to have (positive) area 35

3 + 4√

6.)

Properties of the Definite Integral – Additivity of the Interval. A second type or propertylisted states, in principle, that the area under a curve is the sum of the areas under any two partsinto which the curve can be decomposed:

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx (23)


for any real numbers a, b, c; here again, the constants are not necessarily positive, so one caninterpret the point c as lying outside of the interval [a, b] when a ≤ b. This property impliesanother property listed on [1, p. 373]:

∫ b

af (x) dx = −

∫ a

bf (x) dx (24)


C.4 Supplementary Notes for the Lecture of January 08th, 2010Release Date: Friday, January 08th, 2010, subject to correction

C.4.1 Summary of the last lectures

1. The objective was to define what we mean by the (signed) area under the graph of afunction; or, equivalently, the (signed) distance travelled by a moving particle, giving itsinstantaneous velocity.

2. For a function f continuous on an interval [a, b] I defined the definite integral

b∫

a

f (x) dx

as a kind of limit limmax ∆xi → 0

n→ ∞

n∑i=1

f (x∗i )∆xi , called a Riemann sum, and mentioned

that this definition and introduction was simplified for your first encounter, and some-what lacking in rigour. The treatment of the textbook, which is restricted to continuousfunctions, does not require the possibility that the width ∆xi of the ith interval be pos-sibly different from the that of other subintervals; thus we will be considering only onewidth, and denote it simply by ∆x.

3. I mentioned that it can be shown that the sum must have the same limit for all sub-divisions of [a, b] into subintervals [a = x0, x1], [x1, x2], . . ., [xi−1, xi], [xi, xi+1], . . .,[xn−1, xn = b], and all choices of x∗i in the ith subinterval (i = 1, . . . , n).

4. I illustrated computations with Riemann sums using the function f (x) = x2 over theinterval [a, b]; the variation I considered had the rectangles “hanging” from the graphof f by their upper right corners, but I suggest that students should rework the examplewith the rectangles hanging by their upper left corners. In my notes, but not discussed inclass, was the example of a similar computation for the function f (x) = x2 + 2x− 5 overthe interval [1, 4]; for part of the interval f (x) < 0, and the value of the integral includesa cancellation of “negative” and “positive” contributions to area under the graph of f .

5. I reminded you of the formulæ for summing the 0th, 1st, 2nd, and 3rd powers of the firstn natural numbers, and indicated their use in the preceding computation.

6. I indicated that computation of the integral in this way will normally not be necessary,as we will be meeting a theorem today which will enable much practical computationfor many functions. However, students are still expected to be able to carry out thecalculation of the value of a definite integral using Riemann sums.


7. I stated basic properties of the definite integral, most of which can be derived from (17),(26), (23):

∫ b

a(r · f (x) + s · g(x)) dx = r

∫ b

af (x) dx + s

∫ b

ag(x) dx

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx

C.4.2 §5.2 The Definite Integral (conclusion)

Properties of the Definite Integral – Comparison Properties. The textbook lists 3 proper-ties, which are interrelated — each of them can be used to prove the other 2.

f (x) ≥ 0 for a ≤ x ≤ b ⇒∫ b

af (x) dx ≥ 0 (25)

f (x) ≥ g(x) for a ≤ x ≤ b ⇒∫ b

af (x) dx ≥

∫ b

ag(x) dx (26)

m ≤ f (x) ≤ M for a ≤ x ≤ b ⇒ m(b − a) ≤∫ b

af (x) dx ≤ M(b − a) . (27)

While the integral∫ b

af (x) dx has, thus far, been defined only for a function f which is con-

tinuous on [a, b], we will eventually permit generalizations to that definition. Under thosegeneralizations, properties (25), (26), (27) will continue to hold wherever they “make sense”33.

Some worked examples based on problems in an earlier edition of your textbook.

Example C.2 ([7, Exercise 18, p. 391]) “Express the limit as a definite integral on the given

interval: limn→∞

n∑

i=1

exi

1 + xi∆x, on the interval [1, 5].”

Solution: This problem is not stated in perfect mathematical language, but we know what thetextbook means. We are to consider the interval 1 ≤ x ≤ 5 to be subdivided into n subintervalsof equal length ∆x, so ∆x = 5−1

n . Then we are to interpret exi

1+xias f (x∗i ), the value of a function

at a point x∗i chosen in the ith subinterval, so

1 + (i − 1)∆x ≤ x∗i ≤ 1 + i∆x .

33We will even have generalizations permitting infinite values for the integral, and the properties will holdthere, provided we don’t have to work with “values” like∞−∞ or 0 ×∞.


Since the function f (x) =ex

1 + xis continuous, by virtue of [1, Theorems 7,9, pp. 124–125],

the limit must exist:

limn→∞

n∑

i=1

exi

1 + xi∆x =

∫ 5

1

ex

1 + xdx .

(It happens that this integral is one that we will be unable to evaluate exactly, but will only beable to approximate.)

Example C.3 ([7, Exercise 42, p. 392] “Evaluate1∫

1x2 cos x dx.”

Solution: If in (18) ∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx

we set c = a and b = a, we obtain∫ a

af (x) dx =

∫ a

af (x) dx +

∫ a

af (x) dx

from which it follows that ∫ a

af (x) dx = 0 .

The present integral is of this type — the limits are equal.This result could also have been inferred from (27), by taking b = a, i.e. in

m ≤ f (x) ≤ M for a ≤ x ≤ b ⇒ m(b − a) ≤∫ b

af (x) dx ≤ M(b − a)

⇒ 0 = m · 0 ≤∫ a

af (x) d f ≤ M · 0 = 0 ,

implying that the integral is equal to 0.

Later in the course we will have a method for evaluating an integral of the formb∫

ax2 cos x dx,

where a, b are any real numbers. That method34 is not required for this “easy” problem.

Example C.4 ([7, Exercise 54, p. 392]) Use the properties of integrals to verify the inequali-ties, without evaluating the integral:

π

6≤

∫ π2

π6

sin x dx ≤ π

3.

34Integration “by parts”


Solution: The integrand, sin x, is an increasing function when x is in the 1st quadrant; thisimplies that, for

π

6≤ x ≤ π

2,

12

= sinπ

6≤ sin x ≤ sin

π

2= 1 .

The length of the interval over which the integral is being evaluated isπ

2− π

6=π

3. The value

of the integral is then bounded by the areas of two rectangles on the base[π

6,π

2

]: the lower

bound is given by the rectangle whose height is the minimum value of the function, the value atx = π

6 ; the upper bound is given by the rectangle whose height is the maximum value, attainedat x = π

2 :12· π

3≤

∫ π2

π6

sin x dx ≤ 1 · π3.

The exact value of the integral will eventually be seen to be√

32 .

Example C.5 ([7, Exercise 58, p. 392]) “Use (27) to estimate the value of the integral∫ 2

0(x3 − 3x + 3) dx .”

Solution: On the real line the given function has critical numbers at ±1; of these only x = +1is in the interval of the integral. By the 2nd Derivative Test x = 1 is a local minimum: thefunction value there is 1. At the end-points of the interval of integration, 0 and 2, the functionhas values 3 and 5. We conclude that, on the given interval, the function values are boundedas follows:

1 ≤ x3 − 3x + 3 ≤ 5 .

The length of the interval is 2−0 = 2. The value of the integral is, therefore, bounded between2 and 10. (Eventually we shall be able to evaluate this integral exactly, and shall be able to

show that its value is24

4− 3

2· 22 + 3 · 2 = 4 − 6 + 6 = 4 .

)

5.2 Exercises

[1, Exercise 22, p. 771] “Use the form of the definition of the integral...to evaluate the integral∫ 4

1(x2 + 2x − 5) dx.”

Solution: Here we are finding the area under the graph of the polynomial x2 + 2x − 5, acontinuous function, on the interval 1 ≤ x ≤ 4, i.e., the area of the region bounded by thegraph on top, the lines x = 1 and x = 4 on the two vertical sides, and the x-axis on the


bottom. We divide the interval into n parts of length 4x = 4−1n , and the end-points of the

ith subinterval are xi−1 = 1+(i−1)∆x = 1+(i−1) · 3n on the left and xi = 1+i ·4x = 1+i · 3non the right.

If we choose the sample points to be the right end-points of the subintervals, i.e., to findthe limit of the right Riemann sum, we find that

∫ 4

1(x2 + 2x − 5) dx

= limn→∞

n∑

i=1

(x2i + 2xi − 5)∆x

= limn→∞

n∑

i=1

(1 +

3in

)2

+ 2(1 +

3in

)− 5

3n

= limn→∞

n∑

i=1

((1 + 2 · 3i

n+

9i2

n2

)+ 2

(1 +

3in

)− 5

)3n

= limn→∞

n∑

i=1

1 +6n

n∑

i=1

i +9n2

n∑

i=1

i2 + 2n∑

i=1

1 +6n

n∑

i=1

i − 5n∑

i=1

1

3n

= limn→∞

n∑

i=1

1 +6n

n∑

i=1

i +9n2

n∑

i=1

i2 + 2n∑

i=1

1 +6n

n∑

i=1

i − 5n∑

i=1

1

3n

= limn→∞

(1 + 2 − 5)n∑

i=1

1 +6 + 6

n

n∑

i=1

i +9n2

n∑

i=1

i2

3n

= limn→∞

(−2n +

12n· n(n + 1)

2+

9n2 ·

n(n + 1)(2n + 1)6

)3n

= limn→∞

(−2n + 6(n + 1) +

(3n +

92

+3

2n

))3n

= 21

If we choose the sample points to be the left end-points, i.e., to find the limit of the leftRiemann sum, we obtain a very similar sum. We can write the sum, like the preceding,as a sum over the index i ranging from i = 1 to i = n, but where the function is evaluatedat the left end-points:

∫ 4

1(x2 + 2x − 5) dx

= limn→∞

n∑

i=1

(x2i−1 + 2xi−1 − 5)∆x


= limn→∞

n∑

i=1

(1 +

3(i − 1)n

)2

+ 2(1 +

3(i − 1)n

)− 5

3n

= limn→∞

n∑

i=1

((1 + 2 · 3(i − 1)

n+

9(i − 1)2

n2

)+ 2

(1 +

3(i − 1)n

)− 5

)3n

= limn→∞

n∑

i=1

1 +6n

n∑

i=1

(i − 1) +9n2

n∑

i=1

(i − 1)2 + 2n∑

i=1

1 +6n

n∑

i=1

(i − 1) − 5n∑

i=1

1

3n

At this point we can proceed in a variety of ways. One is to observe that the first termin the sums of powers is 0, so that we are summing only n− 1 non-zero powers. A moreformal way to to that is to define

j = i − 1 (28)

and to rewrite each of the sums as a sum over j, replacing the value i = 1 by j = 0 andthe value i = n by j = n − 1. We then obtain

∫ 4

1(x2 + 2x − 5) dx

= limn→∞

n−1∑

j=0

1 +6n

n−1∑

j=0

j +9n2

n−1∑

j=0

j2 + 2n−1∑

j=0

1 +6n

n−1∑

j=0

j − 5n−1∑

j=0

1

3n

= limn→∞

(1 + 2 − 5)n−1∑

j=0

1 +6 + 6

n

n−1∑

j=0

j +9n2

n−1∑

j=0

j2

3n

= limn→∞

(−2n +

12n· (n − 1)(n)

2+

9n2 ·

(n − 1)n(2n − 1)6

)3n

= limn→∞

(−2n + 6(n − 1) +

(3n − 9

2+

32n

))3n

= 21

again. The definition of j in (28) is analogous to the changes of variables that we will bemaking in definite integrals in [1, §5.5]: there, as here, we have to change the limits ofthe integral to correspond to the new values of the variable of integration. What we havehere is an example of the “finite difference calculus”, where there are results similar tothose that we will be developing in the “infinitesimal calculus”.

[1, Exercise 40, p. 378] “Evaluate the integral by interpreting it in terms of areas:10∫0|x−5| dx.”

Solution: (While we could evaluate this integral using Riemann sums, the intention isthat the student interpret the area under the curve in terms of familiar geometric objects,and use known formulæ to determine the value.)


The portion of the integral from 0 to 5 is the area of a right-angled triangle whose hy-potenuse is on the line y = −(x − 5), with base of length 5, and height 5, so its area is52

2=

252

. The portion of the integral from 5 to 10 is the mirror image of the triangledescribed above, this time with hypotenuse along the line y = x − 5; its area is the sameas the previous one, so the value of the integral is 25.

C.4.3 §5.3 The Fundamental Theorem of Calculus

In a number of areas of mathematics there are theorems that have acquired the name “TheFundamental Theorem of...”. The present section is devoted to such a theorem, also known as“The Fundamental Theorem of (the) Integral Calculus”, one part of which relates the value ofa definite integral to antiderivatives of its integrand, and provides a method for evaluating suchintegrals without the need for computing limits of complicated sums. (The formulation of thetheorem as being divided into two parts is not completely standard.)

Differentiation and Integration as Inverse Processes.

Theorem C.6 (The Fundamental Theorem of Calculus) If f is continuous on [a, b], then

1.ddx

(∫ x

af (t) dt

)= f (x).

2. F′ = f ⇒∫ b

af (x) dx = F(b) − F(a) .

Definition C.2 We may represent a difference F(b) − F(a) by

[F(x)]ba

or even more briefly byF(x)]b

a

if the latter expression is unambiguous.

I sometimes use a notation which is not standard, but is unambiguous, and write the pre-

ceding difference as

F(x)]x=bx=a


Example C.7 Following my discussion of [1, §5.2] in these notes there is35 a solution of

[1, Exercise 22, p. 377], in which the integral

4∫

1

(x2 + 2x − 5

)dx is evaluated “from first

principles”, proving that its value is 21. Let’s now verify that result using the Fundamental

Theorem. One antiderivative of x2 + 2x − 5 isx3

3+ x2 − 5x. The value of the integral is,

therefore, [x3

3+ x2 − 5x

]4

1=

(643

+ 16 − 20)−

(13

+ 1 − 5)

= 21 .

Example C.8 ([7, Exercise 12, p. 402]) “Use Part 1 of the Fundamental Theorem of Calculus

to find the derivative of the function F(x) =

∫ 10

xtan θ dθ.”

Solution: First observe that F(x) = −∫ x

10tan θ dθ. Having written the integral in the form to

which the Fundamental Theorem applies, i.e., with the variable in the upper limit, we mayapply that theorem: the derivative is minus the value of the integrand, tan θ, evaluated whereθ = x, i.e., − tan x.

(Eventually we will see that F(x) = ln cos x − ln cos(10); again students may then verifythat the Fundamental Theorem is giving us the correct derivative.)

35on pages 3016–3018


C.5 Supplementary Notes for the Lecture of January 11th, 2010Release Date: Monday, January 11th, 2010

updated on 12 January; subject to further updates and corrections

C.5.1 §5.3 The Fundamental Theorem of Calculus (conclusion)


to find the derivative of the function g(u) =

∫ u

3

1x + x2 dx.”

Solution: The derivative is the value of the integrand,1

x + x2 evaluated at the upper limit of the

integral, i.e., where x = u: g′(u) =1

u + u2 .

(Eventually we will see that g(u) = lnu + 1

u− ln

43

; students may then verify that theFundamental Theorem is giving us the correct derivative.)

5.3 Exercises

[1, Exercise 54, p. 389] “Find the derivative of the function y =

∫ 5x

cos xcos(u2) du.”

Solution: Two observations are necessary:

• The Fundamental Theorem is concerned with an integral whose upper limit is vari-able; if we wish to apply that theorem here, we shall need to transform the problemto one where only the upper limit of the integral(s) is variable.

• The Fundamental Theorem is concerned with an integral whose upper limit is theindependent variable under consideration; should we wish to permit the upper limitto vary in a more complicated way, we will need to apply the Chain Rule.

1. We shall transform the given integral into a sum of two where the lower limit ofeach is constant. We do this by splitting the interval of integration, [cos x, 5x] intotwo parts at a “convenient” point. It is not even necessary that the point we choosebe inside the interval, since the property we are applying, (18) on page 3011 ofthese notes, does not require that fact. I will choose the constant 0 to be the pointwhere the splitting occurs:

∫ 5x

cos xcos(u2) du =

∫ 0

cos xcos(u2) du +

∫ 5x

0cos(u2) du

= −∫ cos x

0cos(u2) du +

∫ 5x

0cos(u2) du


Hence

ddx

∫ 5x

cos xcos(u2) du = − d

dx

∫ cos x

0cos(u2) du +

ddx

∫ 5x

0cos(u2) du (29)

2. Each of these derivatives will be computed using the Chain Rule, with the interme-diate variable being the function appearing as the variable upper limit. In the firstcase, if we take the intermediate variable to be, say z = cos x, we have

ddx

∫ cos x

0cos(u2) du =

ddz

∫ z

0cos(u2) du · dz

dx

= cos(z2) · dzdx

= cos(cos2 x) · (− sin x)

For the second integral we take the intermediate variable to be w = 5x. Here

ddx

∫ 5x

0cos(u2) du = − d

dw

∫ w

0cos(u2) du · dw

dx

= cos(w2) · dwdx

= cos(25x2) · 5Combining the two results gives

ddx

∫ 5x

cos xcos(u2) du = − d

dx

∫ cos x

0cos(u2) du +

ddx

∫ 5x

0cos(u2) du

= − cos(cos2 x) · (− sin x) + cos(25x2) · 5= cos(cos2 x) · sin x + 5 cos(25x2)

[1, Exercise 58, p. 389] “Find the interval on which the curve y =

∫ x

0

11 + t + t2 dt is concave

upward.”

Solution: By the Fundamental Theorem, y′(x) =1

1 + x + x2 ; hence y′′(x) =−1 − 2x

(1 + x + x2)2 .

The denominator of the second derivative is a non-zero square, so it is always positive;

the function will be positive whenever −1−2x > 0, i.e., whenever x < −12

: this is wherethe graph is concave upward.

(Eventually we will be able to evaluate the integral explicitly, showing that the curve is

y =2√3

(arctan

2x + 1√3− π

6

).

Students may differentiate to check that I have found the correct second derivative.)


[1, Exercise 74, p. 390] Suppose that

b∫

0

ex dx = 3

a∫

0

ex dx . Express b in terms of a.

Solution: Substituting the values of the given integrals, from Part 2 of the FundamentalTheorem, yields

ex]b0 = 3

(ex]a

3),

which implies that

eb − e0 = 3(ea − e0

)

⇒ eb = 3ea − 2⇒ b = ln (3ea − 2) .

Problems Plus

[1, Exercise 12, p. 413] Find

d2

dx2

∫ x

0

(∫ sin t

1

√1 + u4 du

)dt .

Solution:

d2

dx2

∫ x

0

(∫ sin t

1

√1 + u4 du

)dt =

ddx

(ddx

∫ x

0

(∫ sin t

1

√1 + u4 du

)dt

)

=ddx

(∫ sin x

1

√1 + u4 du

)

=√

1 + sin4 x · cos x .

C.5.2 §5.4 Indefinite Integrals and the “Net Change” Theorem

Indefinite Integrals The traditional symbol for a “general” antiderivative F(x) of a function

f (x) (i.e., some function with the property that F′(x) = f (x)) is∫

f (x) dx, which is called the

indefinite integral of f (x). Since two antiderivatives differ by a constant (by a corollary to theMean Value Theorem), we usually write statements in the form

∫f (x) dx = F(x) + C

where F(x) is one specific antiderivative, and C is a constant of integration, intended to rangeover all real numbers. Once a particular antiderivative F has been chosen, the particular real


number C that applies in a particular situation has to be determined from additional informationthat is usually available in the problem at hand. Much of this course will be concerned withmethods for finding indefinite integrals. While the finding of the indefinite integral may be adifficult problem, the verification that a function F that is claimed to be an antiderivative of fis not, since all that needs to be done is to differentiate F and to check whether the derivativeis f . (In principle, if a function F has the property that F′ = f , then F is an antiderivative off : thus it is possible to find an antiderivative by guessing, or by simply copying the answerfrom the back of the textbook or from your neighbour’s work. The intention in the course isthat you should normally be expected to be able to show a systematic way of determining anantiderivative; there will be a very few special situations where you will be presented with anantiderivative without a convincing way of finding it.)

Note the same symbol, a stylized letter S — called the integral sign∫

— is used for

both the indefinite integral and the definite integral; while they are related by the FundamentalTheorem, they are different operations.

Even though the two operations are different, they share some similar properties. Forexample, parallel to property (17) of the definite integral on page 3011 of these notes, we canalso prove that ∫

(r f (x) + sg(x)) dx = r∫

f (x) dx + s∫

g(x) dx , (30)

where r, s are any constants. Equations like the preceding, in which an indefinite integral ap-pears on both sides of the equal sign, are normally written without any constant of integration.

The “Net Change” Theorem This is simply the author’s name for the second part of theFundamental Theorem. It is not a term in standard usage, and I am not likely to use it. Asa beginning in the development of general techniques for determining indefinite integrals, wecan reformulate results that we developed for derivatives. For example, since we know that

ddx

sin x = cos x ,

we can reformulate this result as∫

cos x dx = sin x + C .

Some reformulations require minor changes, e.g., division by an appropriate constant. Fromthe result that

ddx

xn = n · xn−1 when n , 0

we can conclude that ∫xn−1 dx =

1n

xn + C when n , 0


Function One antiderivativef (x) F(x)g(x) G(x)

f (x) + g(x) F(x) + G(x)

xn (n , −1)xn+1

n + 11x

ln |x|ex ex

cos x sin xsin x − cos xsec2 x tan x

sec x tan x sec xcsc x cot x − csc x

11 + x2 arctan x

11 + x2 −arccot x

1√1 − x2

arcsin x

1√1 − x2

− arccos x

Table 4: Some Antiderivatives

or, after the substitution of m + 1 for n,∫

xm dx =1

m + 1xm+1 + C when m , −1 .

Thus Table C.5.2 of antiderivatives on page 3025 of these notes, which I included last semesterin my notes in MATH 140 2009 09, can now be recast in the form of Table C.5.2 on page 3026below.

Applications While we have been using the 2nd part of the Fundamental Theorem to expressthe value of a definite integral in terms of the “net change” in the antiderivative, we can alsoapply the result in the opposite way: that the net change can be found by evaluating the integral.This is the spirit of the motivation that the author called “The Distance Problem” (see above,page 3005, or the solution below, on page 3029 of [1, Exercise 44, p. 397]]). We will seeanother example below in the solution of [1, Exercise 62, p. 398].


∫[r f (x) + sg(x)] dx = r

∫f (x) dx + s

∫g(x) dx

∫k dx = kx + C

∫xn dx =

xn+1

n + 1+ C (n , −1)

∫1x

dx = ln |x| + C∫

ex dx = ex + C∫

ax dx =ax

ln a+ C (a > 0)

∫sin x dx = − cos x + C

∫cos x dx = sin x + C

∫sec2 x dx = tan x + C

∫csc2 x dx = − cot x + C

∫sec x tan x dx = sec x + C

∫csc x cot x = − csc x + C∫

11 + x2 = arctan x + C

∫1

1 + x2 = −arccot x + C∫

1√1 − x2

= arcsin x + C∫

1√1 − x2

dx = − arccos x + C

Table 5: Very Short Table of Indefinite Integrals


Example C.10 ([7, Exercise 10, p. 411]) “Find the general indefinite integral of∫ (

x2 + 1 +1

x2 + 1

)dx .”

Solution: Break the integrand into two parts: the summand at the end is recognizable as thederivative of arctan x; the two terms at the beginning are multiples of powers of x, and we haveobserved earlier how to integrate them. Thus

∫ (x2 + 1 +

1x2 + 1

)dx =

∫ (x2 + 1

)dx +

∫1

x2 + 1dx

=13

x3 +11

x + arctan x + C

where the letter C represents the constant of integration. Even though there are two indefiniteintegrals on the right side of the equation, only one constant is needed: if we were to includetwo constants, as +C1 + C2, we would not gain any more freedom.

Example C.11 ([7, Exercise 11, p. 411]) Find the general indefinite integral of∫ (

2 − √x)2

dx .

Solution: One might be tempted, at first, to consider the possibility that the antiderivative ofthe given 2nd power is a multiple of the 3rd power of 2 − √x; unfortunately that temptationwill have to be resisted, as the resulting functions will not have the correct derivative. Thesimplest approach is to expand the square: since (2 − √x)2 = 4 − 4x

12 + x,

∫ (2 − √x

)2dx =

∫ (4 − 4

√x + x

)dx

= 4x − 4 · 23

x32 +

12

x2 + C


C.6 Supplementary Notes for the Lecture of January 13th, 2010Release Date: Wednesday, January 13th, 2010, subject to correction

C.6.1 §5.4 Indefinite Integrals and the “Net Change” Theorem (conclusion)

5.4 Exercises

[1, Exercise 34, p. 397] “Evaluate the integral∫ 9

1

3x − 2√x

dx.”

Solution: Simplify the integrand by dividing the denominator into the two summands ofthe numerator:∫ 9

1

3x − 2√x

dx =

∫ 9

1

(3√

x − 2√x

)dx

=

[3 · 2

3· x 3

2 − 2 · 2 · x 12

]9

1

=[2x

32 − 4

√x]9

1= (2 · 27 − 4 · 3) − (2 · 1 − 4 · 1) = (54 − 12) − (2 − 4) = 44 .

[1, Exercise 38, p. 397] “Evaluate the integral∫ π

3

0

sin θ + sin θ · tan2 θ

sec2 θdθ .”

Solution: When the integrand involves trigonometric functions, one may have to apply afamiliar identity to simplify the integration. There is often more than one way to do this.In the present example


sec2 θ= sin θ · cos2 θ + sin3 θ

= sin θ · (cos2 θ + sin2 θ) = sin θ

so∫ π

3

0


sec2 θdθ =

∫ π3

0sin θ dθ

= [− cos θ]π30 = − cos

π

3+ cos 0 = −1

2+ 1 =

12.

Eventually you will know how to evaluate the parts of this integral separately, but thepresent solution is faster than evaluating and adding the parts.


[1, Exercise 44, p. 397] Evaluate the integral

3π2∫

0

| sin x| dx.

Solution: The integrand is continuous, so we know the integral exists. However, it isnot convenient to work with an antiderivative of | sin x |. So we split the interval ofintegration into the parts [0, π] and

[π, 3π

2

]where the integrand is respectively positive

and negative, and thereby avoid working with the absolute value. For the integrands,respectively sin x and − sin x, we know antiderivatives − cos x and + cos x, so we mayapply the Fundamental Theorem to each part separately:

∫ 3π2

0| sin x| dx =

∫ π

0sin x dx +

∫ 3π2

π

(− sin x) dx

= [− cos x]π0 + [cos x]3π2π

= (−(−1) + 1) + (0 − (−1)) = 3.

[1, Exercise 58, p. 398] The velocity function is v(t) = t2 − 2t − 8 for a particle moving alonga line. Find

(a) the displacement; and

(b) the distance travelled by the particle

during the time interval 1 ≤ t ≤ 6.

Solution: Denote the position of the particle at time t by x(t); then v(t) = ddt x(t), so x is

an antiderivative of v.

(a) The displacement is simply the difference between initial and final positions of themoving particle; it is equal to the area under the graph of the velocity functionbetween the appropriate times.

displacement = x(6) − x(1) = [x(t)]61 =

∫ 6

1v(t) dt

=

∫ 6

1(t2 − 2t − 8) dt

=

[13

t3 − t2 − 8t]6

1

= (72 − 36 − 48) −(13− 1 − 8

)= −10

3


(b) The distance travelled is equal to the area under the graph of the speed functionbetween the appropriate times; the speed is the magnitude of the velocity.

distance travelled =

∫ 6

1

∣∣∣t2 − 2t − 8∣∣∣ dt =

∫ 6

1|(t + 2)(t − 4)| dt .

The function (t + 2)(t − 4) changes sign at t = −2, which is outside the interval ofintegration, and again at t = 4, which is inside the interval of integration. We canbreak the integral up into two parts at x = 4, and then we can express each of theparts without using absolute value symbols:

∫ 6

1|(t + 2)(t − 4)| dt =

∫ 4

1|(t + 2)(t − 4)| dt +

∫ 6

4|(t + 2)(t − 4)| dt

= −∫ 4

1(t + 2)(t − 4) dt +

∫ 6

4(t + 2)(t − 4) dt

= −∫ 4

1(t2 − 2t − 8) dt +

∫ 6

4(t2 − 2t − 8) dt

= −[13

t3 − t2 − 8t]4

1+

[13

t3 − t2 − 8t]6

4

= −[−18] +

[443

]=

983

Note that the factorization of the quadratic was needed in order to determine where thesplit the interval of integration, but it did not help in the actually integration operation,and had to be reversed at that stage.

[1, Exercise 62, p. 398] “Water flows from the bottom of a storage tank at a rate of r(t) =

200− 4t litres/min, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tankduring the first 10 minutes.”

Solution: Let’s denote by W(t) the amount of water (measured in litres) that has left thetank by time t. We are told that

ddt

W(t) = r(t) = 200 − 4t (0 ≤ t ≤ 50).

Then

W(10) −W(0) =

∫ 10

0

dWdt

dt

=

∫ 10

0(200 − 4t) dt =

[200t − 2t2

]10

0

= (200 · 10 − 2 · 102) − 0 = 1800 ,


so the total amount of water leaving the tank in the first 10 minutes is 1800 litres.


to evaluate the integral, or explain why it does not exist:

3∫

−2

x−5dx.”

Solution: The integrand is not defined at the point 0 in the interval [−2, 3]. This means thatwe cannot even speak of the integral at this time. (Later we will generalize our definition ofintegral to permit us to consider certain “improper” integrals where there is an infinite discon-tinuity. But even that generalization will not apply to this problem, although it is prematureto consider it here. Look at this problem again when we study [1, §7.8].) The theorem is notapplicable because the integrand is not defined at one point in the domain, and the disconti-nuity is neither removable nor a jump discontinuity: a removable discontinuity would have noeffect at all, and a jump discontinuity could be accommodated by the method of Example C.14above, i.e., by splitting the integral into two parts at the jump discontinuity.

Example C.13 ([7, Exercise 42, p. 403]) Let f (x) =

x if −π ≤ x ≤ 0

sin x if 0 < x ≤ π . Use Part 2 of

the Fundamental Theorem of Calculus to evaluate the integral

π∫

−π

f (x) dx, or explain why it

does not exist.Solution: The function f is defined piecewise, by gluing together one function defined on[−π, 0] and another defined on (0,+π]; however, it is continuous, as lim

x→0−x = 0 = lim

x→0+sin x,

and f is defined at x = 0 and equal to the common value of the two one-sided limits. We know

that one antiderivative of x is12

x2; and that one antiderivative of sin x is − cos x. Consider the

function F(x) =

x2

2 if −π ≤ x ≤ 01 − cos x if 0 < x ≤ π . This function is evidently differentiable at all

points except possibly 0; and F is an antiderivative of f on the interval −π < x < +π, with thepossible exception of the point x = 0. At x = 0 the two one-sided limits of difference quotientsare

limx→0−

F(x) − F(0)x

= limx→0−

x2

3 − 0x

= limx→0+

x3

=13· 0 = 0

limx→0+

F(x) − F(0)x

= limx→0+

(1 − cos x) − 0x

= limx→0+

1 − (1 − 2 sin2 x2 ) − 0

x


= limx→0+

(sin x

2x2

· sinx2

)

= limx→0+

sin x2

x2

· limx→0+

sinx2

= 1 · 0 = 0 ,

so the function is also differentiable at x = 0, where the derivative is equal to 0, i.e., to f (0);thus F is an antiderivative of f on the interval (−π,+π). Thus we can apply the FundamentalTheorem:

π∫

−π

f (x) dx = F(π) − F(−π) = (1 − cos π) −((−π)2

2

)= 2 − π

2

2.

But finding the antiderivative was a complicated computation, and rendered the problem moredifficult than necessary. Instead, one should proceed as follows, applying (23) in these notes,page 3011:

π∫

−π

f (x) dx =

0∫

−π

f (x) dx +

π∫

0

f (x) dx

=

0∫

−π

x dx +

π∫

0

sin x dx

=

[x2

2

]0

−π+ [− cos x]π0

=

(0 − (−π)2

2

)+ (− cos π + cos 0)

= −π2

2+ (−(−1) + 1) = 2 − π

2

2.

The lesson to be learned from this example is that there are often advantages to splitting up anintegral, even if it is theoretically possible to evaluate it without doing so.

Integration of “piecewise continuous” functions. The definition [1, Definition 2, p. 366]of a definite integral given by the textbook, is more restrictive than necessary. This definitionrequires that the integrand be continuous throughout the interval of integration. In fact, thedefinition can be weakened to apply to a broader class of functions. While we don’t requirefull generality in this course, we do wish to be able to apply the theory to functions that haveisolated “jump” discontinuities; (we can handle removable discontinuities just be “removing”them, i.e., by extending the function to the appropriate value at the points missing in the origi-nal definition). If a function f is continuous on an interval [a, c], except for a point b in (a, c)


where limx→b−

f (x) and limx→b+

f (x) both exist, but are not equal, we will define

∫ c

af (x) dx =

∫ b

af (x) dx +

∫ c

bf (x) dx (31)

that is, we will define the integral to be such as to satisfy the additivity property (23) we sawearlier in these notes, on page 3011. Note, however, that we cannot apply to the whole interval

the Fundamental Theorem to evaluate integralb∫

af (x) dx if f has a discontinuity at a point c

such that a < c < b.

Example C.14 Suppose that

f (x) =

−1 if x ≤ 01 if x > 0

Evaluate∫ 2

−1f (x) dx.

Solution: We cannot apply the Fundamental Theorem to the entire interval [−1, 2], as theintegrand is discontinuous at point 0. So we split the integral at the point x = 0, whichis the point of discontinuity. The two integrals we obtain now satisfy the conditions of theFundamental Theorem:

∫ 2

−1f (x) dx =

∫ 0

−1f (x) dx +

∫ 2

0f (x) dx

=

∫ 0

−1(−1) dx +

∫ 2

01 dx

= [−x]0−1 + [x]2

0 = (−0 + (−1)) + (2 − 0) = 1.

C.6.2 §5.5 The Substitution Rule

The “Substitution Rule” is a reformulation, in terms of the integral, of the Chain Rule.

Theorem C.15 Let u = g(x) and f (x) be respectively differentiable and continuous on a giveninterval; then

∫( f g)(x) · g′(x) dx =

∫f (g(x)) · g′(x) dx =

∫f (u) du .

In applying this theorem we usually begin with a “complicated” integral, whose form we tryto interpret like the left side of the above equation, and try to find an appropriate “substitution”of the form u = g(x) which will transform the integral into one whose integrand is one that we


are able to integrate. In practice one works with the differentials dx and du in a “mechanical”way that can be justified by the theorem.Proof: Let F and g be differentiable. Then

F′(g(x)) · g′(x) =ddx

F(g(x)) .

Integrating with respect to x, we have∫

F′(g(x)) · g′(x) dx =

∫ddx

F(g(x)) dx = F(g(x)) + C.

If we define u = g(x), f = F′, then∫

f (g(x)) · g′(x) dx =

∫ddx

F(g(x)) dx = F(u) + C =

∫f (u) du .

In practice this “substitution” is often applied by “mechanically”, substituting functions

and differentials, and such operations can be shown to be fully justifiable. The general idea inlooking for substitutions is to try to reduce the complication of the original indefinite integral.This is a subjective term, and different users may find a variety of distinct substitutions whichthey will find helpful in evaluating an indefinite integral.

Finding the appropriate substitution is one step in solving a problem. In the first prob-lems in the list of exercises the author suggests a substitution which will be helpful; eventuallystudents are expected to find an appropriate substitution on their own — often there are severalpossible choices. You should be experimenting with different substitutions and getting to knowthe types of problems each of them is useful in solving.

Example C.16 Earlier, in Example C.11 on page 3027 of these notes, I considered [7, Exercise

11, p. 411], which was concerned with the indefinite integral of∫ (

2 − √x)2

dx. We evaluated

this integral by expanding the square and then integrating the powers of x separately. Could

we use the same methods for the indefinite integral∫ (

2 − √x)10000

dx?

Solution: While we could expand the integrand in this case too, the result would have 10,001terms, each of which would have to be integrated. The result we would obtain would notbe very useful. Consider the following alternative approach, that could also have been used

when the exponent was 2. We have here an integral of the form∫

( f g)(x) · g′(x) dx, where

f (x) = x10000, and g(x) = 2 − √x. Define u = g(x) = 2 − √x, so that du = − 12√

xdx, so

dx = −2√

x du = 2(u − 2)du .


∫ (2 − √x

)10000dx = 2

∫u10000(u − 2) du

= 2∫ (

u10001 − 2u10000)

du

=2

10002u10002 − 4

10001u10001 + C

=2

10002(2 − √x)10002 − 4

10001(2 − √x)10001 + C

= (2 − √x)10001(− 2√

x10002

− 4(10001)(10002)

)+ C


C.7 Supplementary Notes for the Lecture of January 15th, 2010Release Date: Friday, January 15th, 2010 (subject to revision)

C.7.1 §5.5 The Substitution Rule (conclusion)

Review The Substitution Rule (cf., p. 3033) states that

Let u = g(x) and f (x) be respectively differentiable and continuous on a giveninterval; then

∫( f g)(x) · g′(x) dx =

∫f (g(x)) · g′(x) dx =

∫f (u) du

In Example C.16, discussed at the end of the last lecture, I showed how we could evaluate∫ (2 − √x

)10000dx using the substitution u = g(x) = 2 − √x, obtaining

dx = −2√

x du = 2(u − 2)du

⇒∫ (

2 − √x)10000

dx = 2∫

u10000(u − 2) du

= 2∫ (

u10001 − 2u10000)

du

=2

10002u10002 − 4

10001u10001 + C

=2

10002(2 − √x)10002 − 4

10001(2 − √x)10001 + C

= (2 − √x)10001(− 2√

x10002

− 4(10001)(10002)

)+ C

Example C.17 [1, Example 6, p. 403] “Calculate∫

tan x dx.”Solution: This problem does not, at first, appear to be a candidate for a substitution; but, by

expressing the tangent assin xcos x

, the textbook shows that it can be simplified by treating cos xas the new variable. That is, by defining u = cos x, which implies that du = − sin x dx, we canevaluate the integral as follows:

∫tan x dx =

∫sin x dx

cos x

= −∫

d(cos x)cos x


which has the form −∫

duu

, which we should recognize as −∫

d(ln |u|). Thus one sub-

stitution that is indicated is u = cos x: the integral becomes −∫

d(ln |u|) = − ln |u| + C =

− ln | cos x| + C. The integral could also be expressed as ln | sec x| + C or as ln |k sec x|, whereC or k are constants of integration. Three other indefinite integrals can be evaluated in similarways; perhaps you should remember36 them together:37

∫tan x dx = ln | sec x | + C = − ln | cos x | + C (32)

∫cot x dx = − ln | csc x | + C = ln | sin x | + C (33)

∫tanh x dx = ln cosh x + C (34)

∫coth x dx = ln | sinh x | + C (35)

Example C.18 ([7, Exercise 2, p. 420]) “Evaluate the integral∫

x(4 + x2)10 dx by making the

substitution u = 4 + x2.”Solution: First we should observe that we could solve this problem without any substitutionat all. We could expand the 10th power of the binomial into a polynomial of degree 20,multiply each of the terms by an additional x, and then integrate the resulting sum of 11 termsone by one. The resulting polynomial would be the correct solution. But what would youdo it the exponent were not 10, but 10000? Applying the method of substitution, we set

u = g(x) = 4 + x2, f (u) = u10, sodudx

= g′(x) = 2x. Then∫

x(4 + x2

)10dx =

∫ (4 + x2

)10 · 12

ddx

(4 + x2

)dx

=12

∫ddx

((4 + x2

)11)

dx

=1

22

(4 + x2

)11+ C .

In practice we often operate mechanically with differentials; from u = 4 + x2 we have du =

2x dx, so ∫x

(4 + x2

)10dx =

∫xu10dx =

∫u10 · x dx

=

∫u10 · 1

2du =

122

u11 + C .

36This doesn’t mean to memorize them — just to remember the “trick” needed to evaluate them.37Why are absolute signs missing in one of these cases?


But it would be bad form to stop here, since our answer has been expressed in terms of u, notthe original variable x; so we continue

=1

22

(4 + x2

)11+ C .

Example C.19 ([7, Exercise 30, p. 421]) Evaluate the indefinite integral∫

ax + b√ax2 + 2bx + c

dx .

Solution: Later in the term we will consider the integration of any function of the form

kx + `√ax2 + 2bx + c

,

but the present problem concerns a special case which is easy to handle. (Note that the textbookshould have insisted that not all of a, b, c can be 0, as then the ratio is not defined.)

Again, we begin by attempting to simplify the integrand. It would appear that the quadraticfunction whose square root appears in the denominator is the feature we should attempt tosimplify. One way to do this is to define u = ax2 + 2bx + c. This implies that

du = 2(ax + b) dx,

so we obtain∫

ax + b√ax2 + 2bx + c

dx =12

∫du√

u

= u12 + C

= (ax2 + 2bx + c)12 + C .

The problem could also be solved by the substitution v = (ax2 + 2bx + c)12 .

Example C.20 ([7, Exercise 38, p. 421], slightly changed) Evaluate the indefinite integral∫

7√x3 + 1 · x5 dx .

Solution: The method I propose to use here is based on the fact that

d(xn) = nxn−1dx

for any integer n. When an integrand is expressible in terms of a power of the variable, short byjust 1, then this type of substitution is often a good way to begin simplifying it, although further


steps might be needed. So, in this case, I begin with u = x3, which implies that du = 3x2 dx,i.e., that x2 dx = 1

3du. I then obtain∫

7√x3 + 1 · x5 dx =

13

∫7√u + 1 · u du

after which I consider further simplification. At this point I would like to simplify the 7th root.I can do this by either setting v = u+1, or w =

7√u + 1. In the first case I would obtain dv = du,while, in the second, u + 1 = w7, so du = 7w6 dw.

13

∫7√u + 1 · u du =

13

∫7√v · (v − 1) dv

=13

∫ (v

87 − v

17)

dv

=13

[7

15v

157 − 7

8v

87

]+ C

=13

[7

15(u + 1)

157 − 7

8(u + 1)

87

]+ C

=13

[7

15(x3 + 1)

157 − 7

8(x3 + 1)

87

]+ C

or13

∫7√

(u + 1) · u du =13

∫w ·

(w7 − 1

)· 7w6 dw

=73

∫ (w14 − w7

)dw

=73

[1

15w15 − 1

8w8

]+ C

=7

45(u + 1)

157 − 7

24(u + 1)

87 (u + 1)

87 + C

=7

45(x3 + 1)

157 − 7

24(x3 + 1)

87 (x3 + 1)

87 + C .

Definite Integrals Thus far I have been applying the Substitution Rule to indefinite integrals.Substitution may be applied to a definite integral

∫ b

af (g(x)) g′(x) dx

in two ways:

• Apply the Substitution Rule to the corresponding indefinite integral,∫

f (g(x))g′(x) dx ;


then apply the second part of the Fundamental Theorem to the resulting antiderivative;or

• A variant of the Substitution Rule can be formulated specifically for definite integrals.Using the same functions as described earlier, it is

∫ b

af (g(x))g′(x) dx =

∫ g(b)

g(a)f (u) du .

That is, change the limits of the new integral to the values that u has when x = a andx = b.

These two methods are equivalent — use whichever you prefer.

Symmetry The textbook reviews the definitions of even and odd functions, and considerstheir definite integrals over a finite interval centred at the origin. The author shows that thedefinite integral of an even function f over the interval −a ≤ x ≤ +a will be twice the integralover the interval 0 ≤ x ≤ a; and the integral of an odd function over the same interval will be0, because the integral to the left of 0 will cancel the integral to the right of 0. The proofs aresimple applications of the Substitution Rule.

Example C.21 Evaluate the definite integral∫ π

6

− π6tan2 θ dθ.

Solution: Normally we will evaluate an integral of this type by replacing the integrand, tan2 θ,by sec2 θ − 1. In this case we can also use the symmetry of the integrand and the interval ofintegration to further simplify the calculations:

∫ π6

− π6tan2 θ dθ =

∫ π6

− π6

(sec2 θ − 1

)dθ

= 2∫ π

6

0

(sec2 θ − 1

)dθ

since the integrand is an even function (Prove this!)

= 2 [tan θ − θ]π60

= 2(

1√3− π

6

).

5.5 Exercises

[1, Exercise 28, p. 407] “Evaluate the indefinite integral∫

tan−1 x1 + x2 dx.”


Solution: In looking for a substitution, our general intention is to try to simplify theintegrand. In the present integrand, we might be expected to consider the arctangentfactor the most complicated, so I will try to simplify by the substitution u = tan−1 x.

This implies that du =1

1 + x2 dx, which is also present in the integral. The integralbecomes ∫

tan−1 x1 + x2 dx =

∫u du =

12

u2 + C =(tan−1 x)2

2+ C .

We can verify our work by differentiating the function we claim to be an antiderivative:

ddx

(12

(tan−1 x

)2)

=12· 2 tan−1 x · 1

1 + x2 =tan−1 x1 + x2 .

[1, Exercise 24, p. 407] Evaluate the indefinite integral∫

(1 + tan θ)5 sec2 θ dθ .

Solution: In this case one should observe that there is a factor sec2 θ, which we recognizeas the derivative of tan θ. A first simplification would be obtained by the substitution

u = tan θ

since then du = sec2 θ dθ, or dθ = cos2 θ du. The integral transforms to∫

(1 + tan θ)5 sec2 θ dθ =

∫(1 + u)5 sec2 θ cos2 θ du

=

∫(1 + u)5 du .

One can observe that the integral here will be a constant multiple of (1 + u)6, and thenreplace u by tan θ. If you don’t make the observation, a second substitution would be inorder. I would observe that, in the last mentioned integral, the “most complicated” factoris 1 + u; so a substitution v = 1 + u could be attempted. This yields

dv = 0 + du

⇒∫

(1 + u)5 du =

∫v5 dv

=16· v6 + C

=16· (1 + u)6 + C

=16· (1 + tan θ)6 + C .



sin x1 + cos2 x

dx.”

Solution: First we see the cosine in the denominator; then we see in the numeratorsin x dx, which is −d(cos x). This suggests a substitution u = cos x. The integral be-

comes∫ −du

1 + u2 du = − arctan u + C = − arctan(cos x) + C.


x1 + x4 dx.”

Solution: (I am going to begin the solution by making a poor choice for a substitution,a choice that I will then “fix” by following it with a second substitution. Then I willobserve what would have been a better first choice.) Examining the integral we see thatthe powers of x present are x1 in the numerator, and x4 in the denominator. A first ideamight be to try u = x4. That would yield

∫x

1 + x4 dx =14

∫1√

u(1 + u)du

which looks more complicated than before. However, we could then undertake a secondsubstitution, v =

√u, which would correspond to an original substitution of v = x2; this

would produce∫

x1 + x4 dx =

12

∫1

1 + v2 dv

=12

arctan v + C =12

arctan(x2

)+ C

An experienced student would have been able to see the substitution v = x2 imme-diately: a substitution u = x2 is indicated when the entire integrand can be expressed insimple terms of x2, with the exception of one extra single power of x that is “left over”.

[1, Exercise 57, p. 407] Evaluate the definite integral∫ π

6

− π6tan3 θ dθ.

Solution: Eventually you will be able to integrate the cube of the tangent; but, at thispoint, you may not be able to do that. However, you can observe that the integrand is anodd function of θ. (Why?) Hence the integral from −π6 to 0 will be equal in magnitudebut opposite in sign to the integral from 0 to π

6 ; so the given integral is 0.

[1, Exercise 57, p. 407] “Evaluate the definite integral∫ π

6

− π6tan3 θ dθ.”

Solution: Eventually you will be able to integrate the cube of the tangent; but, at thispoint, you may not be able to do that. However, you can observe that the integrand is an


odd function of θ. (Why?) Hence the integral from −π6 to 0 will be equal in magnitudebut opposite in sign to the integral from 0 to π

6 ; so the given integral is 0.

[1, Exercise 65, p. 407] “Evaluate the definite integral∫ 2

1x√

x − 1 dx.”

Solution: This integral can be simplified by defining u = x − 1, so du = dx.38 Then∫

x√

x − 1 dx =

∫(u + 1)

√u du

=

∫ (u

32 + u

12)

dx

=25

u52 +

23

u32 + C

=25

(x − 1)52 +

23

(x − 1)32 + C

Taking the differences of the values of this antiderivative (i.e., with any specific valuefor C, e.g., with C = 0) at the limits yields

[25

(x − 1)52 +

23

(x − 1)32

]2

1

=

(25

(2 − 1)52 +

23

(2 − 1)32

)−

(25

(1 − 1)52 +

23

(1 − 1)32

)

=

(25

+23

)− (0 + 0) =

1615

Alternatively, the Definite Integral version of the Substitution Rule could have been used,to obtain

[25

u52 +

23

u32

]2−1

1−1=

(25

152 +

23

132

)−

(25

052 +

23

032

)etc.

The Logarithm Defined as an Integral (material in §5.6 of the 5th Edition) 39While it istheoretical, the discussion below is an essential part of the course; its purpose is to substantiatesome of the discussions in [1, §§1.5, 1.6] by replacing the weakest parts of the definitions givenat that point in the textbook. This is one of several possible ways of treating exponentials and

38An even better substitution would be v =√

x − 1, which would imply that v2 = x − 1, so dx = 2v dv, and the

integral would be equal to1∫

0(v2 + 1)v · 2v dv etc.

39The textbook material has been moved in the 6th edition to [1, Appendix G, pp. A48-A55]


logarithms “properly”; the treatment in [1, §§1.5, 1.6] was necessarily inadequate, as therewas a perceived urgency to make logarithms and exponentials available to students who didn’thave the background for a more substantial treatment.

The Natural Logarithm Previously we had “defined” the logarithm to be the inverse ofthe exponential function, whose definition was, at best, intuitive. Now we sketch very briefly amore rigorous definition of both functions, beginning with the logarithm. This theory has beendelayed until now because parts of the proofs require the concept of substitution in a definiteintegral.

Henceforth we take the following as our primary definition:

Definition C.3 Let x be any positive real number. Define

ln x =

∫ x

1

1t

dt

i.e., it is the area under the right branch of the hyperbola y =1t, between t = 1 and t = x.

Which Definition is “Correct”? The definitions in [1, §§1.5, 1.6] were all we had avail-able at that time to follow the objective of the textbook to introduce the exponential and log-arithm functions “early”. We had to rely largely on intuition in deriving properties of thesefunctions. Now that we have the integral available, we can return and replace those inadequatedefinitions by some that are more rigorous. Even the present definitions have some issues,since we haven’t rigorously proved all the properties that we are using for the definite integral.While most mathematicians today would probably follow the present development — of intro-ducing the logarithm first, as a definite integral, and the exponential as its inverse — there areother possible ways of defining the functions. Yet another way will be available to you if youfollow the theory of infinite series beyond what we will be doing in [1, Chapter 11].

Properties of the Logarithm From the first part of the Fundamental Theorem we imme-diately obtain the fact that

Theorem C.22ddx

ln x =1x. (36)

We can also prove the following, using basic properties of the integral:

Theorem C.23 Let x, y be positive real variables. Then

1. ln(xy) = ln x + ln y


2. ln 1 = 0

3. ln1x

= − ln x

4.ddx

ln |x| = 1x

5. limx→∞

ln x = ∞

6. limx→0+

ln x = −∞While I do not suggest you memorize the proofs, except possibly that of the first part below,they provide simple exercises on various aspects of the integral.Proof:

1. This is a simple application of properties of the definite integral, and in the SubstitutionRule.

ln xy =

∫ xy

1

1t

dt

=

∫ x

1

1t

dt +

∫ xy

x

1t

dt

by decomposing the interval [1, xy] into [1, x] and [x, xy]

=

∫ x

1

1t

dt +

∫ y

1

1xu

x du

using the substitution t = xu

=

∫ x

1

1t

dt +

∫ y

1

1u

du = ln x + ln y

2. ln 1 =

1∫

1

1t

dt = 0 since the integration is over an interval of length 0.

3. Using a substitution u =1t, where dt = − 1

u2 du, we obtain

ln1x

=

∫ 1x

1

1t

dt

=

∫ x

1u(− 1

u2

)du

= −∫ x

1

1u

du = − ln x


4. Recall thatddx|x| =

+1 if x > 0−1 if x < 0 .

It is convenient to write these values in the form

ddx|x| =

|x|x

if x , 0

undefined if x = 0.

Hence, by the Chain Rule and the Fundamental Theorem,

ddx

(ln |x|) =1|x| ·|x|x

=1x

when x , 0.

5. We will reconsider this proof when we study “harmonic series” in [1, §11.2, Example 7,

pp. 691-692]. Consider the Riemann sum obtained for the definite integral

2n∫

1

1t

dt. We

will hang the rectangles by their upper right-hand corners, so that the Riemann sum isclearly less than the integral. Then the area under the curve is greater than the sum

12

+13

+14

+ . . . +12n .

Let’s collect these terms into groups ending with the reciprocal of each power of 2. Weobtain

12

=12

13

+14

>14

+14

=12

15

+16

+17

+18

>18

+18

+18

+18

=12

. . . . . .1

2n−1 +1

2n−1 + 1+ . . . +

12n > 2n−1 · 1

2n =12

We see that the area of the rectangles under the curve is n2 . As we allow n → ∞ this

lower bound for the area approaches∞, implying that the full area surely→ ∞.

6. Let’s make the substitution y =1x

, so that, as x→ 0+, y→ +∞. Then

limx→0+

ln x = limy→+∞

ln1y

= − limy→+∞

ln y = −∞ .


7. I presented the preceding proof because it is classical. The definition of the logarithmas a definition integral permits a much simpler proof of the same result. Since 1

t ≤ 1 for

t ≥ 1,

x∫

1

1t

dt ≥x∫

1

1 dt = x − 1→ ∞ as x→ ∞.

The construction began with a formal definition for the (natural) logarithm. At this point wecan define the (natural) exponential function as the inverse of the logarithm. We begin bycalling the function exp(x); only after we prove that it behaves the way we expect a power tobehave, do we revert to the familiar notation.

The Natural Exponential Function We could now justify the various equations that weused intuitively in the “early transcendentals” treatment in [1, §§1.5, 1.6]. As mentioned, webegin by showing that ln, as defined above, is invertible. Then we temporarily call the inversefunction, i.e., ln−1 x, exp(x). Thus we have,

Theorem C.24

exp x = y ⇔ ln y = x (37)exp(ln x) = x (38)ln(exp x) = x (39)

exp(x + y) = (exp x) · (exp y) (40)

exp(x − y) =exp xexp y

(41)

(exp x)y = exp(xy) (42)

following which, knowing that the exponent rules are satisfied, we define

Definition C.4e = exp 1 (43)

and change our notation by recognizing that

exp x = ex . (44)

We also show that

Theorem C.25ddx

(exp x) = exp x . (45)

Using standard properties of the definite integral we can also show that

Theorem C.26

limx→−∞

exp x = 0 (46)

limx→∞

exp x = ∞ (47)


General Exponential Functions The preceding constructions have been for the naturallogarithm and the natural exponential. We may now extend these definitions to exponentialsto any positive base a , 1.

Definition C.5 For any positive real number a and any real number x we define

ax = ex ln a . (48)

This leads to the exponent rules for general (positive) bases:

ax+y = ax · ay (49)

ax−y =ax

ay (50)

(ax)y = axy (51)(ab)x = ax · bx (52)

and to the derivative propertyddx

(ax) = ax · (ln a) (53)

General Logarithmic Functions Finally the inverse function of ax is named loga, andits familiar properties are developed, e.g.,

Theorem C.27

loga x = y ⇔ ay = x (54)ddx

(loga x

)=

1x ln a

(55)

The Number e Expressed as a Limit

Theorem C.28 Let a > 0. Then limx→0

((1 + ax)

1x)

= ea

Proof:

limx→0

((1 + ax)

1x)

= limx→0

((eln(1+ax)

) 1x)

since exponential and logarithm are inverses

= limx→0

(e

1x ·ln(1+ax)

)

by the exponent rules

= limx→0

(e

ln(1+ax)x

)= lim

x→0

(e

ln(1+ax)−ln(1+0·x)x

)


= elimx→0

ln(1 + ax) − ln(1 + 0 · x)x

by the continuity of the exponential function

= e

ddx

ln(1 + ax)∣∣∣∣∣x=0

by definition of the derivative at x = 0

= e

a1 + ax

∣∣∣∣∣x=0 by the Chain Rule

= ea

C.7.2 5 Review

True-False Quiz Students tend to avoid the True-False questions because these are unlikelyto appear on quizzes or examinations. It’s true that I would not wish to have a simple True-False question in any test, since there would be a 50% chance of a correct guess. However,the True-False questions in your textbook are much more difficult than that, as they ask “De-termine whether the statement is true of false. If it is true, explain why. If it is false, explainwhy or give an example that disproves the statement.” This requirement of a proof or a coun-terexample makes these problems very challenging. The odd-numbered problems are solvedin your Student Solutions Manual [3]. I consider some of the even-numbered problems below.

[1, True-False Exercise 2, p. 409] If f and g are continuous on [a, b], then∫ b

a[ f (x) · g(x)] dx =

∫ b

af (x) dx ·

∫ b

ag(x) dx

Solution: This statement is FALSE. That is, it is not true for all constants a, b and for allfunctions f , g continuous on [a, b]. There will, of course, be functions and intervals forwhich the statement is true, but the generalized statement, quantified for all functionsand intervals is not true. While you would have know to suspect this after we study[1, §7.1], but you should be suspecting it already, and be able to construct a simplecounterexample. Here is one:

Define f (x) = g(x) = x. Then

∫ b

af (x) dx =

∫ b

ag(x) dx =

x2

2

]b

a=

b2 − a2

2.

∫ b

a[ f (x) · g(x)] dx =

∫ b

ax2 dx =

x3

3

]b

a=

b3 − a3

3


While the polynomialsb2 − a2

2· b

2 − a2

2and

b3 − a3

3look different, that does not consti-

tute a counterexample, as it could happen that we have two algebraic functions that areequal by virtue of some identity that we don’t happen to recognize at the moment. So,in order to complete the counterexample, we need to find specific values of a and b thatwill entail that ∫ b

af (x) dx ·

∫ b

ag(x) dx ,

∫ b

a[ f (x) · g(x)] dx ,

i.e., thatb2 − a2

2· b2 − a2

2,

b3 − a3

3.

One such example is a = 0, and b having any value except 0,43

:

b2

2· b2

2=

b4

4,

b3

3;

for example, take b = 1.

But the counterexample given is far from the simplest! For example, take f (x) = g(x) =

1 for all x. Then the left side of the claimed equation is equal to b − a, while the rightside is (b − a)2, which will be different from b − a except where b − a = 1 or b = a.

[1, True-False Exercise 4, p. 409] This is very similar to [1, True-False Exercise 2, p. 409];you should have no trouble constructing a counterexample.

[1, True-False Exercise 8, p. 409] If f and g are differentiable, and f (x) ≥ g(x) for a < x < b,then f ′(x) ≥ g′(x) for a < x < b.

Solution: This statement is false. Think geometrically. The condition that f (x) ≥ g(x)tells us that the graph of f is above the graph of g; the condition that f ′(x) ≥ g′(x) tells usthat the graph of f is steeper than the graph of g. So we can construct a counterexampleby finding, for a convenient function g, a function that is larger, but whose graph is notso steep. For example, take g(x) = 0 for all x; now all we need is a positive function fwhose graph is sloping downward; for example, f (x) = e−x will work, with any intervala ≤ x ≤ b. A simpler counterexample is f (x) = a + b − x, g(x) = 0.

[1, True-False Exercise 14, p. 409] All continuous functions have antiderivatives.

Solution: This problem is interesting, as it follows [1, True-False Exercise 13, p. 409],which states that “All continuous functions have derivatives.” That preceding statementis false, and you should be able to give counterexamples (e.g., |x| is continuous, but lacks


a derivative at x = 0). The present statement is TRUE, by virtue of the first part of theFundamental Theorem. If f is continuous on an interval [a, b], then

f (x) =ddx

∫ x

af (t) dt .

Exercises

[1, Exercise 70, p. 411] Evaluate limn→∞

1n

(1n

)9

+

(2n

)9

+

(3n

)9

+ . . . +(nn

)9

Solution: The form of this limit should suggest that it is a Riemann sum. There willbe more than one way to interpret this limit as such, but I will choose the interpretationthat is, in some sense, obvious. First, we see a factor 1

n that is tempting to interpret as acommon subinterval length ∆x; so let’s do that: consider the number of subintervals tobe n, and the length of the interval to be 1 — it’s convenient to take a = 0 and b = 1,

but these are not the only possible choices. Then we can interpret( in

)9

as f (x∗i ); the

most convenient choice is to think of f (x) = x9, and x∗i chosen as the right end-point ofthe subinterval [xi−1, xi]. Thus we see that the limit can be interpreted as the limit of anupper Riemann sum associated with the integral

∫ 1

0x9 dx

and its value is the value of the integral, i.e.,x10

10

]1

0=

110

.


C.8 Supplementary Notes for the Lecture of January 18th, 2010Release Date: Monday, January 18th, 2010 (subject to revision)

Textbook Chapter 6. APPLICATIONS OF INTEGRATION.

C.8.1 §6.1 Areas between Curves

If, for a ≤ x ≤ b, continuous functions f and g have the property that f (x) ≥ g(x) — i.e., ifthe graph of f is not lower than the graph of g for that interval, then the area bounded by thegraphs and the vertical lines x = a and x = b is given by the integral

∫ b

a( f (x) − g(x)) dx .

Before working some examples, I make several observations:

• When g is the 0 function — i.e., when the lower boundary of the region is the x-axis —this formula reduces to the definite integral

∫ b

af (x) dx.

• We can find the areas of certain regions by decomposing them into parts that can bedescribed as above. Sometimes there is more than one “natural” way to decompose theregion, but all decompositions should yield the same area.

• An analogous formula can be applied if we consider the “region” bounded by graphs ofthe form x = f (y), x = g(y), for a ≤ y ≤ b. In that case we say that we are integratingwith respect to y or integrating along the y-axis.

• Often we wish to find the area between two graphs determined by points where theyintersect. In these cases the vertical sides of the region have zero length.

• In solving problems it is useful to make a sketch showing an “element of area” — athin rectangle whose width is shown as dx or ∆x in the case of integration with respectto x, and analogously for the case of integration with respect to y. It is difficult for meto include such sketches in these notes, but they will be shown on the chalkboard. Thesketch is not a formal part of the solution, but you are likely to find it helps you formulateyour solution.

• Where two graphs cross in several places, so that the area between them is in severalpieces, be sure you understand what you intend if you write an integral that extendsover several pieces: if the curves interchange positions, with the upper one becomingthe lower, then the signs of the areas will change, and some of the areas will cancel. Ifthis is not what you intend, you must either write the integrand with absolute signs, as| f (x)−g(x)|, or, equivalently, find the area of each of the pieces, and add their magnitudesso as to prevent cancellation.


Example C.29 ([7, Exercise 8, p. 442]; see Figure 1 on page 3053) To find the area of the

1.0

x

0.5

0.50.0

0.25

−0.5

1.0

0.75

0.0

−1.0

Figure 1: The region(s) bounded by y = x2 and y = x4

region bounded by the curves y = x2 and y = x4.Solution: Note that the text-book used the word “region” in a general way. Not all authors usethe word in this way. For this problem there are two connected regions bounded by the curves,and the intention of the textbook was that you find the areas of both of them together.

To determine where the curves intersect, we solve their equations, obtaining (x, y) = (0, 0)and (x, y) = (±1, 1). Note that it is not enough to guess the coordinates of these points ofintersection from the graph: you must determine the coordinates by rigorous solution of theequations!


First solution: Integration with respect to x.

2∫ 1

0

(x2 − x4

)dx = 2

[x3

3− x5

5

]1

0=

23− 2

5=

415.

Second solution: Integration with respect to y. I find the area of the region in the first quad-rant. The equations of the curves are, respectively, x =

√y and x = 4

√y. The area is

∫ 1

0

(y

14 − y

12)

dy =

[45· y 5

4 − 23· y 3

2

]1

0=

45− 2

3=

215.

Now double this area. Note that the order of the curves in this integral is the reverseof that used when we integrated with respect to x, because we now order them by theirrelative distance from the y-axis.

While the limits of the two integrals look as though they are the same, the limits in the firstcase refer to the extreme values of x, while the second refer to the extreme values of y.

Example C.30 ([7, Exercise 22, p. 442]) (see Figure 2 on page 3055) To find the area of theregion bounded by the curves y = sin x and y = sin 2x between the vertical lines x = 0 andx = π

2 .Solution: To determine the points of intersection of the curves, we solve y = sin x with y =

sin 2x, and solve sin x − sin 2x = 0. Since sin 2x = 2(sin x) · (cos x), the intersections mustsatisfy sin x · (1 − 2 cos x) = 0: either

2 cos x = 1

or we must be unable to divide by sin x, because

sin x = 0 .

In the interval 0 ≤ x ≤ π2 , we must therefore have either x = π

3 (to make cos x = 12 ) or x = 0

(to make sin x = 0). This yields the points of intersection (0, 0), and(π3 ,√

32

). The

regions to be considered therefore consists of a lune-shaped region with one end at the originand the other at

(π3 ,√

32

); and a second region which begins at

(π3 ,√

32

)and ends at the vertical

line x = π2 . We will find the areas of each of the regions and add them, since that appears to

be what the textbook expects. (The wording of the question is not totally unambiguous; somereaders might be justified in assuming that the author intended signed areas to be used, ratherthan the absolute values, as I am taking.)


1.51.250.75 1.00.50.0 0.25

0.25

0.5

0.0

0.75

1.0

Figure 2: The region(s) bounded by y = sin x, y = sin 2x between x = 0 and x = π2

For the left-most region the curve y = sin 2x is above y = sin x, so the area will be

∫ π3

0(sin 2x − sin x) dx =

[−1

2cos 2x + cos x

] π3

0

=

[−1

2·(−1

2

)+

12

]−

[−1

2+ 1

]=

14

For the right region the orders of the curves are reversed, and the area is

∫ π2

π3

(sin x − sin 2x) dx =

[− cos x +

12

cos 2x] π

2

π3


=

[−0 +

12· (−1)

]−

[−1

2+

12

(−1

2

)]=

14

Thus the total area is 14 + 1

4 = 12 .

6.1 Exercises[1, Exercise 18, p. 420] (see Figure 3 on page 3056) To find the area of the region bounded

−2.5

31 20

10.0

−1

2.5

−3

0.0

5.0

−2

7.5

Figure 3: The region(s) bounded by y = 8 − x2, y = x2 between x = ±3

by the curves y = 8 − x2 and y = x2 between the vertical lines x = −3 and x = 3.

Solution: Let’s first determine where the curves intersect. Solving the equations showsthat the intersections occur when x = ±2, y = 4, i.e., in the points (±2, 4). But we areasked to find the area from x = −3 to x = 3. Thus this region has three parts:


• −3 ≤ x ≤ −2, where 8 − x2 ≤ x2;

• −2 ≤ x ≤ 2, where 8 − x2 ≥ x2;

• 2 ≤ x ≤ 3, where 8 − x2 ≤ x2.

A naive attack gives the following:

Area =

∫ 3

−3|(8 − x2) − x2| dx

=

∫ −2

−3|(8 − x2) − x2| dx +

∫ 2

−2|(8 − x2) − x2| dx +

∫ 3

2|(8 − x2) − x2| dx

= −∫ −2

−3

((8 − x2) − x2

)dx +

∫ 2

−2

((8 − x2) − x2

)dx

−∫ 3

2

((8 − x2) − x2

)dx

= −∫ −2

−3(8 − 2x2) dx +

∫ 2

−2(8 − 2x2) dx −

∫ 3

2(8 − 2x2) dx

= −[8x − 2

3x3

]−2

−3+

[8x − 2

3x3

]2

−2+

[8x − 2

3x3

]3

2

= −[(−16 +

163

)− (−24 + 18)

]+

[(16 − 16

3

)−

(−16 +

163

)]

−[(24 − 18) −

(16 − 16

3

)]

=923

This attack is naive in that we have not taken advantage of symmetry. Since the integrandis an even function, and the interval of integration is symmetric about x = 0, we can findthe integral from 0 to 3 and double it: that means evaluating 2 integrals instead of 3.

The areas could also be found by integrating with respect to y, where we would have tosum three pieces (or two if we make use of symmetry). For example, the middle regioncould be described as being double the following region: bounded by the y-axis, thecurve x =

√y, and the curve x =

√8 − y. This approach is cumbersome, as the region

has to be broken up further into two pieces: one bounded by the y-axis, x =√

y, andy = 4, and the other bounded by they-axis, y = 4, and x =

√y − 8.

[1, Exercise 32, p. 420] “Evaluate the integral and interpret it as the area of a region:∫ 4

0

∣∣∣∣√

x + 2 − x∣∣∣∣ dx .”


Solution: (see Figure 2 on page 3058) We can interpret this as the area of the region

2 40−1 1

4

1

0

2

3−2

3

Figure 4: The region(s) bounded by y =√

x + 2, y = x between x = 0 and x = 4

bounded by the curve y =√

x + 2, the line y = x, and the vertical lines x = 0 and x = 4.If we square the former equation, we obtain y2 = x+2, which can be seen to be a parabolathat is symmetric about the x-axis, with its vertex on the line x = −2 and opening to theright. Squaring the equation caused the new equation to include the lower branch of thisparabola; the original equation represents only the upper branch. The upper branch ofthe parabola and the line y = x meet both in the origin and in the point (2, 2). (The otherpoint of intersection of the parabola and the line does not lie on the upper branch of theparabola, and so is extraneous to this discussion.)


The simplest way of evaluating this integral is to break it up into two parts:∫ 4

0

∣∣∣∣√

x + 2 − x∣∣∣∣ dx =

∫ 2

0

(√x + 2 − x

)dx +

∫ 4

2

(x −√

x + 2)

dx

=

(x + 2)32

32

− x2

2

2

0

+

−(x + 2)32

32

+x2

2

4

2

=

[23· 4 3

2 − 2]−

[23· 2 3

2 − 0]

+

[−2

3· 6 3

2 + 8]−

[−2

3· 4 3

2 + 2]

=163− 2 − 2

3· 2 3

2 − 23· 6 3

2 + 18 +163− 2

=443− 4

3

√2 − 4

√6

[1, Exercise 50, p. 421] 1. “Find the number a such that the line x = a bisects the area

under the curve y =1x2 , 1 ≤ x ≤ 4

2. “Find the number b such that the line y = b bisects the area in part (a).”

Solution:

1. We solve for a the equation∫ a

1

1x2 dx =

∫ 4

a

1x2 dx

to obtain a = 85 . It follows that the area of half of the region is the common value

of the two integrals,[

1x

]a

1= 3

8 .

2. The right side of the region has an irregular boundary: the upper part has equationy = 1

x2 , i.e., x = 1√y ; the lower part is on the line x = 4. The two parts of the

boundary meet in the point(4, 1

16

). Thus the area of the rectangle bounded by the

lines x = 1, x = 4, y = 0, y = 116 is 3

16 . As this is less than half the total, we knowthat b > 1

16 . This fact is important, as it tells us that we can represent the upper halfof the area — the portion above the line y = b, by the integral

∫ 1

b

(1√y− 1

)dy ,

where the −1 in the integrand represents the lower boundary of the region — nowviewed as being “under” x = 1√

y and “over” x = 1. Setting this integral equal to 38


and solving, we obtain[2√

y − y]1

b= 3

8 , which evaluates to

1 − 2√

b + b =38

⇔ (√

b − 1)2 =38

⇔√

b = 1 ±√

38

⇔ b =118±

√32

One of the values we obtain is greater than 1, which contradicts our assumptionthat 0 ≤ b ≤ 1. The other gives the bisecting line

y =118−

√32≈ 0.150255129.

The value we rejected has a geometric significance: it is the height of a horizontalline at which the area bounded by the curve y = 1

x2 , the line x = 1 and that line isequal to 3

8 .And what would have happened if we failed to observe that b ≥ 1

16? We wouldhave obtained an equation for b that could not be solved; it would be equivalent torequiring that 5

8 − 3b = 516 when 0 ≤ b ≤ 1

16 , which are contradictory statements.

The area which I found to equal 1− 2√

b + b by integrating with respect to y couldalso be evaluated by integrating with respect to x: here the line y = b meets y = 1

x2

in the point(

1√b, b

), so the area is

∫ 1√b

1

(1x2 − b

)dx = b − 2

√b + 1 = (

√b − 1)2 ,

which we must again equate to 38 and solve.


C.9 Supplementary Notes for the Lecture of January 20th, 2010Release Date: Wednesday, January 20th, 2010 (subject to revision)

C.9.1 §6.2 Volumes

Just as we defined the area of a region as the limit of a sum of narrow rectangles, we candefine volumes as limits of sums of thin elements; these elements can be assembled in variousways. In this course you will see volumes expressed either as sums of thin slices with planarsides (called laminæ), and — for solids with rotational symmetry (called solids of revolution)— as sums of thin shells with cylindrical sides. Some problems will lend themselves onlyto one type of dissection; where a problem can be approached in more than one way, it isinstructive to try it in several ways, in order to verify your answer and also to gain experiencein choosing the method that is more efficient for different types of problems. In this section wewill be considering dissections into thin slices; where the slice has the shape of a disk with aconcentric disk cut from its centre, the author uses the term washer40.

Example C.31 1. (see Figure 1 on page 3053) ([7, Exercise 2, p. 452]) “Find the volume

1.

2.5

1.5

2

1

010.80.40.60.2

0.5

0

4.

7

5

1

6

4

0210.5 1.50

2

3

Figure 5: Regions for Example C.31

of the solid obtained by rotating the region bounded by the curves y = ex, y = 0, x = 0,x = 1 about the x-axis.” We will first solve the problem as stated, and then consider somerelated problems obtained by changing some of the data.

40Students whose native language is not English sometimes cannot understand why the author will use thename of a household appliance here; this is another meaning of the English word washer, which refers to a thindisk with a hole in the middle


Solution: (Remember to come back to this problem when we study [1, §6.3], to solve itusing the methods of that section.)

We will decompose the solid into laminæ that are thin disks. The “washers” will beobtained by rotating about the x-axis the element that we would have used for the areaif we had found the area by integrating with respect to x. For the disk whose faces arecentred at points (x, 0) and (x+∆x, 0), the volume obtained by hanging the element fromits upper left corner on the curve is πy2∆x = πe2x∆x. By the same reasoning that led usto express areas as definite integrals, we have

Volume =

∫ 1

0πe2xdx

=π

2· e2x

]1

0=π

2· (e2 − 1)

2. Now let us change the problem, asking that the solid be rotated about the y-axis. Herethe description of the cross sections will depend on the height of the cross section. Wewill be expressing everything in terms of y, not x, and integrating “with respect to y”.For y ≤ 1 (the height where the curve cuts the y-axis), the cross sections are disks ofradius 1, and the volume of that part of the solid is

∫ 1

0π12dy = πy]1

0 = π.

For y ≥ 1 the cross sections are “washers”, with outer radius 1 and inner radius x; toexpress this in terms of y we need to rewrite the equation of the curve y = ex in anequivalent way that expresses x in terms of y, as x = ln y. The washers at height y havevolume

π(12 − x2)∆y = π(1 − (ln y)2)∆y

so this part of the solid of revolution has volume

π

∫ e

1(1 − (ln y)2) dy .

where the upper limit of integration is the ordinate of the point where the line x = 1meets the curve y = ex. You are not quite ready to complete this integration. You canmake a substitution like x = ln y, under which the integral transforms to π

∫ 1

0(1−x2)ex dx,

and you know how to integrate part of this integral, as

π

∫ 1

01ex dx = πex]1

0 = π(e − 1) ,


but you are not ready to integrate π∫ 1

0x2ex dx. We will, in [1, §7.1], develop a method

to determine that∫

x2ex dx = (x2 − 2x + 2)ex + C (which is obvious by differentiation).With that we know that

Volume = π[ex − (x2 − 2x + 2)ex]10 = π[(−x2 + 2x − 1)ex]1

0 = π .

3. As another variant on this problem, consider the following: “Find the volume of thesolid obtained by rotating the region bounded by the curves y = ex, y = 0, x = 0, x = 1about the line y = −3.” The analysis is similar to what we did originally, except that thelaminæ now have a hole of radius 3 in the middle, and the outer radius is 3 units larger.This leads to an integral

Volume =

∫ 1

0π((ex + 3)2 − 32

)dx =

∫ 1

0π(e2x + 6ex

)dx

= π

[12

e2x + 6ex

]1

0

= π

[(12

e2 + 6e)−

(12

+ 6)]

=π

2

(e2 + 12e − 13

)

4. Finally, suppose that the right boundary of the region generating the solid by revolutionchanged from x = 1 to y = e2(x−1). This new right boundary also passes through (1, 0),but meets the curve in the point (x, y) =

(2, e2

). The volume will be

∫ 1

0πe2x dx +

∫ 2

1π((ex)2 −

(e2(x − 1)

)2)

dx

=π

2

(e2 − 1

)+ π

∫ 2

1e2x dx − πe4

∫ 2

1(x − 1)2 dx

=π

2

(e2 − 1

)+ π

[e2x

2

]2

1− πe4

[13

(x − 1)3]2

1

= π

(e4 − 1

2

)− πe4

3= π

(e4

6− 1

2

)

Example C.32 ([7, Exercise 8, p. 452]) Find the volume of the solid obtained by rotating theregion bounded by the curves y = sec x, y = 1, x = −1, x = 1 about the x-axis.Solution: The lines x = ±1 intersect y = sec x in the respective points (1,± sec 1). The area ofthe washer centred on the x-axis between cross sections x = X and x = X+4X is approximatelyπ(sec2 X − 12) · 4X. The volume of revolution will be

∫ 1

−1π(sec2 x − 1

)· dx = π[tan x − x]1

−1 = 2π(tan 1 − 1).


(Had the integral involved arctan 1, you would have been expected to simplify it further; butyou cannot evaluate tan 1 without calculators or techniques that you will not meet until Calcu-lus 3.)

Example C.33 ([7, Exercise 56, p. 454]) Here is a problem where the solid is not generatedby revolving a plane region about an axis. “Find the volume of the solid S: the base of S is theparabolic region (x, y)|x2 ≤ y ≤ 1 ; cross-sections perpendicular to the y-axis are equilateraltriangles.”Solution: The cross-section of the base at level y has ends with coordinates (±√y, y), so thelength of the base is 2

√y, and the area of the triangular cross-section is 1

2 ·2√

y· √3√

y =√

3·y.Integrating along the y axis we find that

Area =

∫ 1

0

√3y dy =

√

32

y2

1

0

=

√3

2(12 − 02) =

√3

2

Example C.34 ([7, Exercise 15, p. 458]) (Where this problem appeared in the cited textbook,students were asked to solve it using the method of “shells”. Let’s see if we can solve it usingthe method of “washers”.) To find the volume of the solid of revolution generated by revolvingthe region bounded by y = x2, y = 0, x = 1, x = 2 about the axis x = 1.Solution: The washers generated by elements of area parallel to the x-axis will have two kindsof descriptions, depending on whether their cross sections are above or below the point wherex = 1 meets y = x2, i.e. (1, 1): below y = 1 the cross sections are rectangles of width 2−1 = 1;above y = 1 the cross sections are rectangles obtained from such a rectangle as below y = 1by cutting away a rectangle of length x2 starting at the left end. But, in evaluating the limitof the sum of the volumes of these washers, we shall be integrating with respect to y, so weneed to express the dimensions and position of the rectangular element in terms of y. Theequation of the right branch of the parabola y = x2 is x =

√y. The hole in the washer has

radius x − 1 =√

y − 1; the washer will have area

π(12 − (√

y − 1)2) = π(2√

y − y) .

The line x = 2 meets the parabola in the point (2, 4), so we shall integrate for y ranging from0 to 4: from 0 to 1 using the constant integrand π12, and from 1 to 4 using the integrandπ(2√

y − y). The volume of revolution is

∫ 1

0π12 dy +

∫ 4

1π(2√

y − y) dy = πy]10 + π

[43

y32 − y2

2

]4

1

= π + π

[43· 8 − 8

]− π

[43· 1 − 1

2

]=

17π6


which is the same result given in the textbook for the solution that could be obtained in thenext section using the method of cylindrical shells.

6.2 Exercises

[1, Exercise 12, p. 430] Find the volume of the solid obtained by rotating the region boundedby the curves y = e−x, y = 1, x = 2 about the line y = 2.

Solution: Through the point (x, 2) on the line y = 2 the cross section will be an annulus(ring) with outer dimension 2−e−x and inner dimension 2−1 = 1; the annulus is boundedby concentric circles centred at the point (x, 2). The volume will be

∫ 2

0

(π(2 − e−2

)2 − π12)

dx

= π

∫ 2

0

(3 − 4e−x + e−2x

)dx .

The first summand of the integrand has antiderivative 3x; the second summand can beintegrated by using a substitution u = −x, which leads to an antiderivative +4e−x; thelast summand can be integrated by using a substitution u = −2x, which leads to anantiderivative − 1

2 · e−2x. Putting these three components together we find the value of theintegral to be

π

[3x + 4e−x − 1

2e−2x

]2

0= π

(6 +

4e2 −

12e4

)− π

(0 + 4 − 1

2

)

= π

(52

+4e2 −

12e4

).

[1, Exercise 39, p. 431] The textbook asks you to use a Computer Algebra System to find thevolume of the solid generated by revolving about the line y = −1 the region bounded byy = sin2 x, y = 0, for 0 ≤ x ≤ π. No Computer Algebra System is needed, although wehaven’t yet seen how to integrate this. Here is the full solution. The cross sections arewashers with centre on the line y = −1, outer radius ending on the curve y = sin2 x, andinner radius ending on the line y = 0 (the x-axis). The volume is thus

π

∫ π

0

((sin2 x − (−1))2 − 12

)dx = π

∫ π

0

(1 − cos 2x

2+ 1

)2

− 1

dx

=π

4

∫ π

0

((3 − cos 2x)2 − 4

)dx

=π

4

∫ π

0

(5 − 6 cos 2x + cos2 2x

)dx


=π

4

∫ π

0

(5 − 6 cos 2x +

1 + cos 4x2

)dx

=π

8

∫ π

0(11 − 12 cos 2x + cos 4x) dx

=π

8

[11x − 6 sin 2x +

sin 4x4

]π

0=π

8· 11π =

11π2

8

[1, Exercise 44, p. 431] Describe the solid whose volume is represented by the integral

π

∫ π2

0[(1 + cos x)2 − 1] dx .

Solution: The dx tells us that the integration is along the x-axis. That is, the planes ofthe washers are perpendicular to the x-axis. The integrand is the difference of 2 squares,multiplied by π. We may interpret this as the area of a washer whose outer radius is1 + cos x, and whose inner radius is 1. If we interpret 1 + cos x as cos x − (−1), wecan interpret this as the radius of a disk whose centre is at the point (x, y) = (x,−1),generated by a radius extending from that centre to the point (x, cos x) on the graph ofthe cosine function. The subtracted term −π12 can be interpreted as the area of a diskwhose centre is at the same point, but its radius extends from that point (x,−1) to thepoint (x, 0) above it on the x-axis. Thus the integral represents the solid of revolutionabout the line y = −1 of the region bounded by the graph of the cosine function, and thex-axis, between x = 0 and x = π

2 . (This integral is not difficult to evaluate. We will seein [1, Chapter 7] that, if we replace cos2 x by 1

2 (1 + cos 2x), the value of the integral is

π

[2 sin x +

x2

+sin 2x

4

] π2

0= 2π +

π2

4.

This is not the only way of interpreting the integral. We could also reason that it rep-resents the volume of a region rotated about the x-axis, bounded by the y-axis, the liney = 1, and the graph of y = cos x + 1.)

[1, Exercise 49, p. 431] Find the volume of a right circular cone with height h and base radiusr.

Solution: This is a standard problem that every student should be able to work.

A cone is a surface generated by joining to all points curve in a plane a fixed point (calledthe apex outside of the plane. A cone is circular if the base curve is a circle. It is rightcircular if the apex is located on the normal to the plane of the curve through the centreof that base circle.


It is convenient to set up coordinate axes so that the cone is generated by a right angledtriangle with height h and base r, i.e., with hypotenuse along the line

xr

+yh

= 1, whichtriangle is to be rotated about the y-axis. The horizontal rectangular elements of area,with height dy and length x are rotated about the y-axis to generate disk-shaped laminæ.

Expressing x as a function of y for the hypotenuse, we have x = r(1 − y

h

), so the cross-

sectional area at height y is πr2(1 − y

h

)2. Hence the volume must be

∫ h

0πr2

(1 − y

h

)2dy .

We know how to evaluate an integral of this type by expanding the square. But it is

easier to make a change of variable: u = 1 − yh

, so du = −1h

dy, dy = −h du and

Volume = −∫ 0

1πr2u2h du = πr2h

[u3

3

]0

1=

13

(πr2h) .

(cf. [9, p. 47]). If you remember this as 13π × the area of the base, you will know this

as a special case of a general theorem. In fact, it is not hard to show that the area isnot affected by the fact that the cone is a right cone: even if the apex were moved to alocation not over the centre of the circular base, the area would not change. Moreover,it can be shown that even the fact that the base is circular is not relevant! The volume ofany “cone” is as shown:

13× π × Area of base × height .

For example, the case of a square base is discussed in [1, Example 8, p. 429].

[1, Exercise 55, p. 432] Find the volume of the solid S which is a tetrahedron with three mu-tually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4cm, and 5 cm.

Solution: I find it convenient to locate the three perpendicular sides along the coordi-nate axes, with one vertex at the origin. So I locate the vertices of the tetrahedron at(0, 0, 0), (3, 0, 0), (0, 4, 0), (0, 0, 5). This tetrahedron is just a pyramid or cone on a trian-gular base; thus we know by the theory of [1, Example 8, p. 429] that the volume willbe 1

3 ×(

12 × 3 × 4

)× 5 = 10. But I will pretend we don’t know that.

Consider cross sections perpendicular to the z-axis. These are triangles whose x-dimensionat height z will be 3

5 (5− z) (by similar triangles in the xz-plane); and whose y=dimension


at height z will be 45 (5 − z), (again by similar triangles). The area of the triangle will be

12 (5 − z)2, so the volume will be

∫ 5

0

12· 12

25· (5 − z)2 dz =

625

[−1

3(5 − z)3

]5

0= − 2

25

(0 − 53

)= 10 .


C.10 Supplementary Notes for the Lecture of January 22nd, 2010Release Date: Friday, January 23nd, 2010, subject to revision

C.10.1 §6.3 Volumes by Cylindrical Shells

The integrand when using cylindrical shells The volume of a right circular cylinder (i.e.,with a disk as base, axis perpendicular to the base) is — as you can easily prove using washersor otherwise — the product of the area its base and its height. If we consider a hollowed outcylinder of radius r2, in which an inner cylinder of radius r1 is removed, then the volume willbe, if the height is h,

πr22h − πr2

1h = π(r2

2 − r21

)h .

Let’s assume that r2 = r1 + 4r, and expand this product:

π(r2

2 − r21

)h = π

((r1 + 4r)2 − r2

1

)h

= π(r2

1 + 24r · r1 + (4r)2 − r21

)h

= 2πh4r · r1 + πh(4r)2

Both of these products approach zero as we allow 4r → 0. If we wish to determine the volumeof a solid of revolution by decomposing it into cylindrical shells about the axis of revolution,it would appear that we should add elements of volume

2πh4r · r1 + πh(4r)2

and then permit the number of shells to approach infinity, and the width ∆r → 0. It can beshown that, in any such limiting process, the sum of the terms of type πh(4r)2 approaches0; that is, not only do the individual terms πh(∆r)2 approach 0, as we permit 4r → 0, buteven the sum of these terms, increasing arbitrarily in number as 4r → 0, also approaches 0.Thus the volume sought can be viewed as a Riemann sum, leading to a definite integral of theform

∫2πr1h dr. Thus, to find the volume of a solid that we can decompose into elements

which are cylindrical shells, we need only consider an integral related to terms of the first type.The integrand can be interpreted as the product of 2πr1 and h, i.e., as the area of a rectangleobtained by “cutting open” the inner surface of the cylindrical shell and “unrolling” it; then the4r can be interpreted as the thickness of a thin rectangular lamina based on that rectangle. Ofcourse, when you attempt to do that, you find that there will be a small error in that product,but, as I have mentioned, it can be shown that the totality of these errors approaches 0 as wereplace the sum of volumes of elements by the definite integral. All that remains to be done isto express h in terms of the radius, and to determine the appropriate limits for integration.

If you choose to approach these problems by substituting in formulæ, you are urged toremember how to generalize to situations where the axis of circular symmetry is parallel to


but different from one of the coordinate axes. I do not recommend memorizing these formulæ.Among the following examples are some that were not discussed in the lecture. One of these,which you should certainly read shows how to find the volume of a sphere; instead of thisexample, I worked a more difficult — but much more instructive — example which finds thevolume of a torus.

6.3 Exercises

[1, Exercise 14, p. 436] (see Figure 6 on page 3070) “Use the method of cylindrical shells to

4

2

3

1

-1

0431 20-2 -1

Figure 6: The region(s) bounded by x + y = 3 and x = 4 − (y − 1)2

find the volume of the solid obtained by rotating the region bounded by the given curvesabout the x-axis. Sketch the region and a typical shell: x + y = 3, x = 4 − (y − 1)2.”


Solution: The parabola meets the line in the points (0, 3) and (3, 0), on the coordinateaxes.

1. First we follow the instructions, using the method of cylindrical shells. The ele-ments of area that generate the shells will be narrow horizontal rectangles; at heighty the rectangle has width (4 − (y − 1)2) − (3 − y), obtained by expressing the equa-tions in the form x =function of y. The circumference of the base of the generatedcylinder is then 2πy, and the volume is

2π∫ 3

0

((4 − (y − 1)2) − (3 − y)

)y dy

= 2π∫ 3

0

(3y2 − y3

)dy

= 2π[y3 − y4

4

]3

0=

27π2

.

2. Now let us compute the volume using washers. The equation of the line may berewritten as y = 3 − x; but the parabola now splits into 2 curves — the upperbranch has equation y = 1 +

√4 − x, and the lower has equation y = 1 − √4 − x.

The description of the element of area that generates the washer will change atx = 3. To the left of x = 3 the element of area at horizontal position x has height(1 +

√4 − x) − (3 − x). We need to compute the area of the annulus the element

generates. For that purpose the length of the element is not enough, as we need toknow the distance from the axis about which it is revolving. The outer radius ofthe washer is 1 +

√4 − x, and the inner radius is 3− x, so the area of the annulus is

the difference between the areas of two disks:

π(1 +√

4 − x)2 − π(3 − x)2

and the volume of the solid generated up to x = 3 is

π

∫ 3

0

((1 +

√4 − x)2 − π(3 − x)2

)dx .

The elements of area to the right of x = 3 have inner radius 1 − √4 − x, and outerradius 1 +

√4 − x, so the area of the annular cross-section is

π(1 −√

4 − x)2 − π

(1 +√

4 − x)2

= 4π√

4 − x ,

and the volume is 4π∫ 4

3

√4 − x dx. The total volume by the method of washers is

π

∫ 3

0

((1 +

√4 − x)2 − π(3 − x)2

)dx + 4π

∫ 4

3

√4 − x dx


= π

∫ 3

0

(−(x − 1)(x − 4) + 2

√4 − x

)dx + 4π

∫ 4

3

√4 − x dx

= π

∫ 1

4(u − 3u2 + 2

√u) (−1) du + π

∫ 0

14√

u(−1) du

using the substitution u = 4 − x, du = −x, x = 4 − u

=27π

2as before.

[1, Exercise 20, p. 437] “Use the method of cylindrical shells to find the volume generatedby rotating the region bounded by the given curves about the specified axis. Sketch theregion and a typical shell: y = x2, x = y2, about y = −1.”

Solution:

1. At height y the cylinder is generated by an element of area whose horizontal di-mension is

√y − y2, and whose vertical dimension is ∆y; the circumference of the

circle generate by a point at one end of this element under revolution is 2π(y−(−1))(since the radius is the distance between a point (x, y) and the point (x,−1) belowit), so the volume is

∫ 1

0

(√y − y2

)2π(y + 1) dy

= 2π∫ 1

0

(y

32 − y3 +

√y − y2

)dy

= 2π[25

y52 − 1

4y4 +

23

y32 − 1

3y3

]1

0

=2930π .

2. It wasn’t asked for in the problem, but let’s find the same volume by the method ofwashers. The external radius of the vertical washer at x is

√x− (−1) =

√x + 1; the

internal radius of that washer is x2 − (−1) = x2 + 1. The area of the cross sectionwill be π times the difference of the squares of the radii, i.e.,

π((√

x + 1)2 − (x2 + 1)2).

The integral from x = 0 to x = 1 is again equal to 2930π.

[1, Exercise 29, p. 437] The following integral represents the volume of a solid; describe thesolid: ∫ 3

02πx5 dx .


Solution: Since the problem appears in this section, the textbook expects you to interpretthe integral as the result of application of the method of cylindrical shells. If we interpret2πx5 ·∆x as (2πx) · x4 ·∆x, we see that it is the volume of the solid generated by rotatingabout the y-axis the region bounded by the x-axis and the curve y = x4 from the point(0, 0) to the line x = 3.

But the author should not have used the definite article the, since the integral can beinterpreted in other ways. For example, we can interpret it as resulting from applicationof the method of washers also. This time we interpret 2πx5 · ∆x as π

(√2x5

)2 · ∆x: itis the area of the solid obtained by rotating about the x-axis the region bounded by thataxis and the curve y =

√2x5 between x = 0 and x = 3.

[1, Exercise 32, p. 437] The integral

π4∫

0

2π(π− x)(cos x− sin x) dx represents the volume of a

solid. Describe the solid.

Solution: The same integral could easily represent more than one solid. One interpre-tation is the following: The solid is a solid of revolution around the vertical line x = π,generated by revolving the region bounded by the y-axis and the graphs y = cos x andy = sin x up to the point where they intersect, (x, y) =

(π4 ,

π4

). But the region could be de-

formed vertically without changing the volume. For example, the region could be takento be bounded by the lines y = 0, x = 0,

y = cos x − sin x = sin( π

2 − x + x2

)· cos

π2 − x − x

2=

1√2

cos(π

4− x

)

from x = 0 to x = π4 .

[1, Exercise 43, p. 437] Use cylindrical shells to find the volume of a sphere of radius r.

Solution: Here is a case where the solid is prescribed, but not the way of generating it. Asphere can be generated by the rotation of a half-disk around its diameter. I chose to takethe diameter as the y-axis, and the equation of its boundary as x2+y2 = r2, more precisely,as x = +

√r2 − y2 where −r ≤ y ≤ r. The length of a vertical rectangular element of

area a distance of x from the axis is the distance between the points (x,√

r2 − x2) and(x,−

√r2 − x2), i.e., 2

√r2 − x2. This is the height of the cylindrical generated when the

element is rotated about the axis. The base of the shell is at a distance of x from theaxis, so it generates a circular band (annulus) of radius 2x, hence of circumference 2πx.Here I have not been precise about whether this is the inner or outer circumference ofthe annulus, since, in the limit, these distinctions have no effect on the calculations. Thethickness of the annulus is represented by the differential, dx, in the integral, and we


obtain a volume of∫ r

02πx(2

√r2 − x2) dx = −4π

3

[(r2 − x2)

32]r

0=

4πr3

3. (56)

Of course, this volume could also be evaluated by the method of washers, see [1, Exam-ple 1, pp. 423-424]. If you found the number of variables in equation (56), you couldamplify the notation, and write

∫ x=r

x=02πx(2

√r2 − x2) dx = −4π

3

[(r2 − x2)

32]x=r

x=0=

4πr3

3for all r . (57)

[1, Exercise 44, p. 437] Use cylindrical shells to find the volume of the solid torus generatedby rotating a disk of radius r around a line located a distance R from its centre.

Solution: We can take the disk to be bounded by (x − R)2 + y2 = r2, and the axis ofrotational symmetry of the torus (doughnut) to be the y-axis. The boundary of the diskis the graphs of 2 functions, y = ±

√r2 − (x − R)2, so the height of the element of area is

2√

r2 − (x − R)2. The volume is∫ R+r

R−r2πx · 2

√r2 − (x − R)2 dx = 2π

∫ r

−r(u + R)2

√r2 − u2 du

under the substitution u = x − R

= 4π∫ r

−ru√

r2 − u2 du + 4πR∫ r

−r

√r2 − u2 du

• Here the first integral can be seen to be the area under the graph of an odd functionfrom −r to the symmetrically located value +r, so the volume is 0; it may also be

integrated as[4π3· (r2 − u2)

32

]r

−r= 0.

This integral could also be evaluated “naıvely” (as I did in the lectures), by usingone of several possible substitutions, or by observation, since the integrand is ev-

idently the derivative of32·(−1

2

)· (r2 − u2)

32 , which, when evaluated between −r

and +r gives a difference of 0.

• The second integral is 2πR times the area under the curve y =√

r2 − u2 from u = −rto u = r, which can be seen to be the area of a half disk of radius r, which we know

to beπr2

2. Thus the volume is 2π2Rr2.

[1, Exercise 45, p. 437] I began the lecture by finding the volume of a right circular cone(already considered in [1, Exercise 49, p. 431], solved in the notes of the previous lecture.A solution to this variation of the problem can be found in the Student Solutions Manual[3]).


C.10.2 §6.4 Work

This section has been omitted from the syllabus because it involves physical concepts thatsome students from outside of the Faculties of Science and Engineering might not be preparedfor. If you are a Science or Engineering student, you are urged to peruse the section and trythe problems. Your instructors and TA’s will be happy to help you with any difficulties.


C.11 Supplementary Notes for the Lecture of January 25th, 2010Release Date: Monday, January 25th, 2010, subject to further revision

Review of preceding two lectures We have studied how to use a definite integral to deter-mine volumes of solids, by expressing the solid as the limit of a union of thin layers: in [1,§6.2] as the union of thin laminæ with planar sides and, in [1, §6.3], for solids of revolution, asthe union of thin circular cylindrical shells. Where the solid is obtained by revolving a planeregion about a line in that plane the solid can be called a solid of revolution about that line, andthe methods of both sections are, in principle, applicable. I do not recommend attacking theseproblems by substitution in formulæ, as there are many variants to be considered, and the blinduse of formulæ often leads to disaster when the wrong formula is applied, or the right formulais applied incorrectly. I list some formulæ below just to summarize the nature of the resultswe have found, with the intention that, in each case, you decompose the solid in one of themethods that is applicable and set up the integral by carefully examining the decomposition.

1. When a solid is decomposed into thin laminæ between parallel planes which are per-

pendicular to the x-axis: the volume is expressible in the form

b∫

a

A(x) dx, where the

decomposition ranges between values x = a and x = b, and A(u) is an expression for thecross-sectional area at x = u. (If the decomposition is along the y-axis, then the limits ofthe integral will be the limiting values of y, and we will usually express the integral interms of y; of course, the x or y that appears in the integral is irrelevant, since these arebound or “dummy” variables, that are part of our notation, and don’t actually affect thenumerical value of the integral.)

2. When a decomposition as in item 1 above has circular symmetry as a solid of revolutionabout a line x = c of a region in the xy-plane, the laminæ can be interpreted as beinggenerated by revolving a thin rectangle, and will consist of disks, possibly with a holein the middle — the textbook calls them “washers”. The integrand will be an expressionof the form π (r2(x))2 − π (r1(x))2, where r2 and r1 are the outer and inner radius of thedisk, determined by the distances of the two ends of the thin rectangle from the point ofintersection of the extended rectangle with the axis of rotational symmetry of the solid.Here we need to be prepared to work with either x or y, and possibly to consider rotationabout a line parallel to but distinct from the coordinate axes.

3. In the case of decomposition into cylindrical shells of a solid with rotational symmetryabout an axis, the cylindrical shells can also be interpreted as being generated by rotatinga thin rectangle about the axis — but here the long dimension of the rectangle is parallelto the axis. The integrand may also be interpreted as the area of a “flattened” cylindrical


shell, obtained by slitting the shell open and unrolling it; one dimension will be thecircumference of the circle generated by revolving any point of the cylinder around theaxis of symmetry — this factor will be of the form 2πx when we are at distance x fromthe axis of rotational symmetry; the other dimension will be the height of the cylindricalshell. Again, we need to adjust this formula according to the orientation of the axis ofrotational symmetry, and according as the axis is or is not a coordinate axis.

Before proceeding with the topics scheduled for today, I reminded students of the ease in whichwe can determine the volume of a sphere of radius r. A solution using cylindrical shells ([1,Exercise 43, p. 437]) can be found beginning on page 3073 of these notes. We can find thevolume using washers by taking the equation of circle to be x2 + y2 = r2, and revolving theupper semi-disk around the x-axis, obtaining

∫

−rr2π(r2 − x2) dx = 2π

[r2x − x3

3

]r

−r= 2π

(r3 − r3

3

)− 2π

(−r3 +

r3

3

)=

43· πr3 .

C.11.1 §6.5 Average value of a function

In this section the textbook defines what is meant by the term average of a continuous functionor a function “pieced” together from continuous functions over an interval a ≤ x ≤ b. Thedefinition is a generalization of the definition familiar to you of the average of a finite numberof numbers y1, y2, . . . , yn, i.e.,

average =

n∑i=1

yi

n.

If, in the finite case just mentioned, we treat each of yi as the height of a rectangle of unit width,situated so that its base is placed on the x-axis with its corners at (i − 1, 0) and (i, 0), and with

height yi, then the sumn∑

i=1

yi is the area under the graph of the function f defined by

f (x) =

y1 if 0 ≤ x ≤ 1y2 if 1 < x ≤ 2

. . .yn if n − 1 < x ≤ n

For any function that is piecewise continuous on the interval a ≤ x ≤ b, we define

average of f over [a, b] =1

b − a

∫ b

af (x) dx ,

=

∫ b

af (x) dx

∫ b

a1 dx


and so our definition is consistent with the earlier definition when the function is defined at afinite number of points x1, . . . , xn: we are simply extending the definition of that function byextending the value at any integer point to the entire interval of unit length to the left of thepoint.

The Mean Value Theorem for Integrals If f is continuous on [a, b], the Mean Value The-orem may be applied to the function g(x) =

∫ x

af (t) dt, which we know from the Funda-

mental Theorem to be differentiable. It asserts the existence of a point c in (a, b) such thatg(b) − g(a)

b − a= g′(c), i.e., such that

∫ b

af (t) dt = g(b) − g(a) = f (c) · (b − a) ,

or f (c) =

∫ b

af (x) dx

∫ b

a1 dx

=

∫ b

af (x) dx

b − a.

[1, Exercise 23, p. 445].As mentioned in connection with the Mean Value Theorem in Math 140, the spirit of this

theorem is in the existence of the point c, not in the specific values that c takes. The theoremis not constructive: it proves the existence without telling you how to find the point. (Aconstructive proof of a theorem proves existence by providing a way of finding the point; theproof we have given for the present theorem is not constructive.)

Example C.35 [1, Exercise 13, p. 445] The textbook asks you to prove that, if f is any contin-

uous function for which∫ 3

1f (x) dx = 8, then f takes on the value 4 somewhere in the interval

1 ≤ x ≤ 3. The MVT for Integrals tells you that there is a point c such that 1 ≤ x ≤ 3 and

13 − 1

∫ 3

1f (x) dx = f (c)

and the left side of this equation is exactly 82 = 4.

What if we were to ask the same question with 4 replaced by another real number, e.g.,4.001? The answer would now be negative: as a counterexample41 take the constant functionf (x) = 4. This function is continuous and has the desired area over the interval 1 ≤ x ≤ 3; butit never assumes the value 4.001.

41an example to disprove a general statement: since the statement refers to all functions with a given property,we can disprove it by exhibiting just one example.


Average velocity Suppose that the position at time t of a particle moving along the x-axis isf (t). In [1, §3.7, p. 221] the textbook has defined the Average Velocity of the particle over atime interval a ≤ t ≤ b to be

∆x∆t

=f (b) − f (a)

b − a.

You may recall being told by your instructor that the use of the word average there was alsoa generalization of the traditional meaning recalled above for the average of a finite set ofnumbers. You can now see that, since

f (b) − f (a)b − a

=

∫ b

addt f (t) dt

b − a,

that use of the word is consistent with the use we have defined here. In other words, the AverageVelocity is, in fact, the average of the velocities. So the earlier use of the word “average” was,though premature, consistent with the generalization that was planned. Previously “AverageVelocity” was a two-word name for a concept, and you would not be justified in treating thefirst word as a modifier of the second; now you may interpret that as a conventional use oflanguage, where average is an adjective. When mathematicians name concepts we try to makethe nomenclature intuitive, and consistent with earlier usage.

6.5 Exercises

[1, Exercise 4, p. 445] Find the average value of the function g(x) = x2√

1 + x3 on the interval[0, 2].

Solution:

average =1

2 − 0

∫ 2

0x2√

1 + x3 dx

We apply the substitution u = x3, so du = 3x2 dx and

average =1

2 − 0

∫ 23

0

√1 + u · 1

3du

=16· 2

3(1 + u)

32

]8

0=

19

[9

32 − 1

32]

=269

(We could also use the substitution u =√

1 + x3.) Had the problem asked us to “sketcha rectangle whose area is the same as the area under the graph of g,” we could take arectangle based on the interval 0 ≤ x ≤ 2, with height 26

9 .)


Example C.36 (taken from a problem book for students in Russian technical universities) [38,Problem 1647, p. 126] Suppose that a trough has a parabolic cross-section with equation ofthe form y = Kx2, and measures 1 meter across the top and is 1.5 meters deep. Determine theaverage depth.Solution: From the data we know the cross section passes through the point

(12 ,

32

), so 3

2 =

K(

32

)2, and K = 6. The depth of the trough at position x is 3

2 − 6x2, so

average depth =

∫ 12

− 12

(32 − 6x2

)dx

12 −

(− 1

2

)

= 2∫ 1

2

0

(32− 6x2

)dx

by symmetry, since the integrand is an even function

= 2[3x2− 2x3

] 12

0= 1

so the average depth is 1 meter, i.e., two-thirds of the way to the bottom of the trough.

6 Review

[1, Exercise 32, p. 447] “Let R1 be the region bounded by y = x2, y = 0, and x = b, whereb > 0. Let R2 be the region bounded by y = x2, x = 0, and y = b2, to the right of the linex = 0.

1. “Is there a value of b such that R1 and R2 have the same area?2. “Is there a value of b such that R1 sweeps out the same volume when rotated about

the x-axis and the y-axis?3. “Is there a value of b such that R1 and R2 sweep out the same volume when rotated

about the x-axis?4. “Is there a value of b such that R1 and R2 sweep out the same volume when rotated

about the y-axis?”

Solution: Note that the statement of the problem in the textbook is ambiguous, as thedescription of R2 applies to one region to the right of the y-axis and another to the left;for that reason I have added the italicized words. Now the two regions combine to formthe rectangle (x, y)| 0 ≤ x ≤ b, 0 ≤ y ≤ b2.

1. We are asked to investigate solutions of the equation∫ b

0x2 dx =

∫ b

0(b2 − x2) dx⇔ b3

3=

2b3

3,


which has no positive solution. (We could have evaluated either or both of theseareas by integration along the y-axis, writing the equation of the right branch of theparabola as x =

√y. With both integrations along that axis the equation would be

∫ b2

0(b − √y) dy =

∫ b2

0

√y dy .

2. I will find the volume about the x-axis using washers, and the volume about they-axis using the same elements of area, using cylinders. Equating the two areas Iobtain the equation

∫ b

0π(x2

)2dx =

∫ b

02πx · x2 dx⇔ π

[x5

5

]b

0= 2π

[x4

4

]b

0

which is equivalent to b4(2b − 5) = 0 and has one positive solution, b = 52 . (This

problem could be solved by evaluating either of the integrals in the “other” way.)

3. We have determined the volume swept out by rotating R1 about the x-axis to beπb5

5 . If we compute the volume obtained by rotating R2 about the x-axis, we find itto be

∫ b

0π((

b2)2 −

(x2

)2)

dx = π

[b4x − x5

5

]b

0=

4π5

b5 using washers, and

∫ b2

02πy · √y dy = 2π · 2

5y

52]b2

0=

4π5

b5 using cylindrical shells.

Equating the volume obtained by rotating R1 about the same axis to this, we obtainπb5

5 = 4πb5

5 , which has no positive solution.

4. The volume obtained by rotating R1 about the y-axis has been determined above tobe π

2 b4. The volume obtained by rotating R2 about the y-axis is

∫ b2

0π(√

y)2 dy = π

∫ b2

0y dy = π

[y2

2

]b2

0=π

2b4

using washers, or∫ b

02πx

(b2 − x2

)dx = 2π

[b2x2

2− x4

4

]b

0=π

2b4

using shells. Here we see that the two volumes of revolution are equal for all valuesof the parameter b.


C.12 Supplementary Notes for the Lecture of January 27th, 2010Release Date: Wednesday, January 27th, 2010,

subject to further revision

Textbook Chapter 7. TECHNIQUES OF INTEGRATION.

C.12.1 §7.1 Integration by Parts

Earlier we developed the Substitution Rule for evaluating integrals, from the Chain Rule fordifferentiation. Now we will develop another procedure for evaluation of integrals, calledIntegration by Parts, based on the Product Rule for differentiation. As with the SubstitutionRule, this rule will be applicable to both definite and indefinite integrals. If does not affect the(“independent”) variable and so there will be no change to limits in definite integrals.

Starting from the Product Rule,

ddx

[ f (x) · g(x)] =ddx

f (x) · g(x) + f (x) · ddx

g(x) ,

we integrate all 3 members with respect to x:∫

ddx

[ f (x) · g(x)] dx =

∫ [ddx

f (x) · g(x)]

dx +

∫ [f (x) · d

dxg(x)

]dx

and observe that the integral on the left is simply f (x) · g(x) + C. Moving the terms aroundgives the Rule of Integration by Parts:

f (x) · g(x) =

∫ [ddx

f (x) · g(x)]

dx +

∫ [f (x) · d

dxg(x)

]dx

⇔∫ [ddx

f (x) · g(x)]

dx = f (x) · g(x) −∫ [

f (x) · ddx

g(x)]

dx ,∫ [

f (x) · ddx

g(x)]

dx = f (x) · g(x) −∫ [

ddx

f (x) · g(x)]

dx ,

or, compactly,∫

f dg = f g −∫

g d f .

Traditionally we often name the functions u and v, or variations of these symbols42 — sincethe solution of a specific problem may require multiple applications of integration by parts.These equations are always true for differentiable functions, but we shall be applying themwhen they tend to replace a difficult integral by one that is “easier” to evaluate. Usually the

42like u1, v1, U,V , u, v, . . .


applications will be such that the integrand admits a “natural” factorization into two factors,where either one of them becomes “simpler” under differentiation, or one becomes “simpler”under integration. I begin with an example where the use of Integration by Parts is obviouslyindicated.

Example C.37 To integrate∫

xex dx.

Solution: If we factorize the integrand into u = x, v′ = ex, then u is “simplified” by differ-entiation, since u′ = 1, while v′ is not “complicated” by differentiation, where v = ex. Weobtain

∫xex dx = xex −

∫1ex dx

= xex − ex + C .

Example C.38 Sometimes the factorization of the integral is less than obvious. Consider theproblem of integrating ln x (cf. [1, Example 2, p. 454]). The “factorization” we choose is

u = ln x, dv = dx . (58)

By (58),

du =dxx

, andv = x

where we have chose one convenient antiderivative.∫

ln x dx = (ln x)x −∫

x · dxx

= (ln x)x − x + C .

This is a derivation you should remember, as we often need an antiderivative of a logarithm.43

Example C.39 Similar to the preceding example is the integration of arctan x (cf. [1, Example5, p. 456]). Here again the function does not admit a well defined factorization, but we can try(58) and obtain ∫

arctan x dx = x · arctan x −∫

x1 + x2 dx .

43Perhaps one should remember it in the “more general” form∫

ln |x| dx = x ln |x| − x + C .


Now apply a substitution to the remaining integral, either

v = x2 ⇒ dv = 2x dx orw = 1 + x2 ⇒ dw = 2x dx

Hence∫

arctan x dx = x · arctan x −∫

1w

dw

= x · arctan x − ln |w| + C= x · arctan x − ln |1 + x2| + C .

Of course, the absolute signs are not needed here because the argument of the logarithm func-tion is evidently non-negative.

There are several “standard” situations where we need to use integration by parts. Supposewe need to integrate a function of one of the following forms, where P(x) is some polynomial:

P(x) · sin x, P(x) · cos x, P(x) · ex, P(x) · cosh x, P(x) · sinh x.

In each of these cases the derivative of the polynomial is “simpler” (here meaning “of lowerdegree”), while the integral of the other factor is “not more complicated”. Repeated applica-tions cause the polynomial to disappear, leaving only an integral involving the second factor.Here again one should remember the derivation, but not memorize the formulæ, since they canbe easily reconstructed, and there are too many variations to memorize.

The rule of integration by parts need not be used in isolation: it may be necessary toprecede or follow its use by substitutions, and several applications of integration by parts couldbe needed to complete the solution to a problem.

Two applications of Integration by Parts? We saw in connection with substitutions thatwe might need to use the procedure more than once. Can that occur with Integration by Parts?It can always occur, but, if not done carefully, the second iteration will reverse the action of thefirst, and return us to the integral that we began with. Where the purpose of using Integrationby Parts is to truly simplify the integrand, then it is unlikely you will make this tactical error.But consider the first example in the next paragraph.

Solving an equation to evaluate an indefinite integral Sometimes the application of inte-gration by parts does not appear to make any progress, but a second or more applications mayeventually produce a constraint on the integral, which enables one to evaluate it. This idea willbe extended later.


Example C.40 To evaluate∫

ex cos x dx.

Solution: Set u = ex, dv = cos x, so that du = ex dx, v = sin x. Then∫

ex cos x dx = ex sin x −∫

ex sin x dx .

The new integral is of similar difficulty to the old one. We apply integration by parts again:U = ex, dV = sin x, dU = ex dx, V = − cos x:

∫ex cos x dx = ex sin x −

∫ex sin x dx

= ex sin x −(−ex cos x +

∫ex cos dx

)

= ex sin x + ex cos x −∫

ex cos dx

Could this be an instance of the pitfall that was described in the preceding paragraph? Fortu-nately not. The same integral appears on both sides of the equation, but with different coeffi-cients. If we move the integral from the right side to the left, we obtain

2∫

ex cos x dx = ex sin x + ex cos x dx + C (59)

which implies that ∫ex cos x dx =

12

(ex sin x + ex cos x) + C (60)

Several comments are appropriate:

1. If we had taken the second application of Integration by Parts as U = sin x, dV = ex dx,dU = cos x dx, V = ex, then we would have obtained

∫ex cos x dx = ex sin x −

∫ex sin x dx

= ex sin x −(ex sin x −

∫ex cos dx

)

= ex sin x − ex sin x +

∫ex cos dx =

∫ex cos x dx ,

which is a tautology. The statement obtained would not be incorrect, but we would havewasted our time to obtain an end result that could have been stated immediately, andwhich does not get us any closer to solving the problem at hand.


2. Why did the constant of integration appear in equations (59), (60) but not in the preced-ing equation? The preceding equation was of the form

∫f (x) dx = g(x) +

∫h(x) dx .

When both sides of an equation contain an indefinite integral, one understands that eachside is a set of functions which differ by a constant; no more generality is achieved if weadd the notational comment that one may add a constant to one side. But, when one sidewould consist of a single function — here it is ex sin x + ex cos x — the inclusion of aconstant changes the meaning from one specific function to all functions obtained fromit by adding any real number.

3. When we integrate the dv term, we appear to be selecting a specific antiderivative. Isthis restrictive? Should we be including a constant of integration? Try to convinceyourself that he selection of a specific antiderivative is not at all restrictive; that is, if youdo include a constant of integration, the changes to the formula will cancel each otherout. For that reason you should always choose the simplest antiderivative of v that isconvenient.

Reduction Formulæ Since integration by parts is helpful when the integrand is a product oftwo functions, one of which does not become “more complicated” under differentiation, andthe other of which does not become “more complicated” under integration, this method shouldbe able to assist in the integration of the product of a polynomial and a sine, cosine, exponen-tial, or hyperbolic function. But what happens if we have to integrate other products of thesefunctions? In some cases the differentiations and integrations do not produce major changesof simplicity, but they can lead to information that enables us to determine the integral, in theway in which the preceding example was solved. This method is particularly important whenwe wish to obtain an algorithm for evaluating certain general classes of integrals. Sometimewe will need to use the methods of the preceding paragraph in this connection, and sometimesnot.

Example C.41 Find a general formula for evaluating I(n) =

∫xne−x dx where n is a positive

integer.Solution: Let u = xn, dv = e−x dx, so du = nxn−1 dx, v = −e−x. Then

∫xne−x dx = −xne−x + n

∫xn−1e−x dx . (61)

or I(n) = −xne−x +n · I(n−1) . For any specific value of n, this formula may be used to evaluate


the integral recursively. For example, if we need to know∫

x5e−x dx, we have

∫x5e−x dx

= I(5) = −x5e−x + 5 · I(4)= −x5e−x + 5(−x4e−x + 4 · I(3))= −x5e−x + 5(−x4e−x + 4(−x3e−x + 3 · I(2)))= −x5e−x + 5(−x4e−x + 4(−x3e−x + 3(−x2e−x + 2 · I(1))))= −x5e−x + 5(−x4e−x + 4(−x3e−x + 3(−x2e−x + 2(−x1e−x + 1 · I(0)))))= −x5e−x − 5x4e−x − 5 · 4x3e−x − 5 · 4 · 3x2e−x − 5 · 4 · 3 · 2xe−x − 5!e−x + C= −

(x5 + 5x4 + 5 · 4x3 + 5 · 4 · 3x2 + 5 · 4 · 3 · 2x1 + 5 · 4 · 3 · 2 · 1x0

)e−x + C

Example C.42 [1, Exercise 50, p. 458] Let n be an integer greater than 1. Find a procedure— i.e., a reduction formula — that can be used to evaluate secn x.Solution:

∫secn x dx =

∫secn−2 x ·

(sec2 x dx

)

Applying integration by parts with u = secn−2 x, dv = sec2 x dx, we set

du = (n − 2) secn−3 x · sec x tan x dx = (n − 2) secn−2 x · tan x dx, v = tan x .

We obtain∫

secn x dx = secn−2 x · tan x − (n − 2)∫

tan2 x · secn−2 x dx

= secn−2 x · tan x − (n − 2)∫

(sec2 x − 1) · secn−2 x dx

= secn−2 x · tan x − (n − 2)∫ (

secn x − secn−2 x)

dx

which may be solved for the desired indefinite integral. First we move all copies of the sameintegral to one side of the equation:

(n − 1)∫

secn x dx = secn−2 x · tan x dx + (n − 2)∫

secn−2 x dx .

Dividing by n − 1 yields the reduction formula∫

secn x dx =1

n − 1secn−2 x · tan x +

n − 2n − 1

∫secn−2 x dx . (62)


Since we know the integrals of the 0th and 2nd powers of the secant we can now find theintegral of any even positive power. For the odd positive powers we may reduce the problemto the integration of sec x, which has not been achieved yet.

Example C.43 ([7, Exercise 36, p. 480]) Follow a substitution by integration by parts to inte-

grate∫

x5ex2dx.

Solution: The obvious substitution is u = x2, so du = 2x dx. Then∫

x5ex2dx =

12

∫u2eu du

to which we apply integration by parts withU = u2, dV = eu du, dU = 2u du,V = eu

=12

(u2eu − 2

∫u du

)

=12

u2eu −∫

ueu du

to which we apply integration by parts withu = u, dv = eu du, du = du, v = eu

=12

u2eu −(ueu −

∫eu du

)

=12

u2eu − (ueu − eu) + C

=

(12

u2 − u + 1)

eu + C

=

(12

x4 − x2 + 1)

ex2+ C

Your solution is incomplete unless you express the integral in terms of the original variable.

(This section will be discussed further at the beginning of the next lecture.)


C.13 Supplementary Notes for the Lecture of January 29th, 2010Release Date: Friday, January 27th, 2010, revised 01 February, 2010,

subject to further revision

C.13.1 §7.1 Integration by Parts (conclusion)

Recapitulation In the last lecture I introduced “Integration by Parts” — an integration tech-nique related to the Product Rule of differentiation. I discussed the routine types of applicationswe will see, as well as application to the integration of ln x and arctan x. I ended with an appli-cation which required two successive applications of the technique, followed by the solutionof an equation.

7.1 Exercises

[1, Exercise 10, p. 457] “Evaluate the integral∫

arcsin x dx.”

Solution: Since you probably don’t know an antiderivative of the inverse sine, but doknow its derivative, we can try integration by parts with u = arcsin x, dv = dx, so

du =1√

1 − x2dx and v = x. (I say try because not every attempt to apply one of the

integration rules will be successful: the rules are valid, and do convert the given integralinto another; but if you are unable to evaluate the new integral, you haven’t fully solvedthe problem, and sometimes you may have made it even more difficult to solve.)

∫arcsin x dx = x arcsin x −

∫x√

1 − x2dx

to which we apply the substitution w = 1 − x2, so dw = −2x dx

= x arcsin x −∫

1√w

(−1

2dw

)

= x arcsin x +12· 2√w + C

= x arcsin x +√

1 − x2 + C

Other substitutions that could have been used are w = x2, w =√

1 − x2.

[1, Exercise 30, p. 458] Several methods suggest themselves for evaluating∫ 1

0

r3

√4 + r2

dr.

One solution using integration by parts can be based on

u = r2, dv =r√

4 + r2dr ⇒ du = 2r dr, v =

√4 + r2 .


Hence∫ 1

0

r3

√4 + r2

dr =[rs√

4 + r2]1

0− 2

∫ 1

0r√

4 + r2 dr

=[rs√

4 + r2]1

0− 2

[12· 2

3(4 + r2)

32

]1

0

=

[r2 − 8

3·√

r2 + 4]1

0

=16 − 7

√5

3

which is approximately 0.115841, where the second integral was found by observationagain.

Some students may have difficulty observing the integral of dv, but a substitution couldmake that phase easier.

For a solution that does not use integration by parts, try the substitution u =√

4 + r2.Then u · du = r dr, and r2 = u2 − 4.

∫ 1

0

r3

√4 + r2

dr =

∫ √5

2(u2 − 4) du

=

[u3

3− 4u

]√5

2

=16 − 7

√5

3

as before.

Alternatively, one can use the substitution v = r2 + 4, so dv = 2r dr. Then∫ 1

0

r3

√4 + r2

dr =

∫ 5

4

v − 4√v

dv2

=

[13· v 3

2 − 4√

v]5

4

=

5√

53− 4√

5 −

(83− 8

),

which is the same value as obtained earlier twice.


[1, Exercise 44, p. 458] 1. Prove the reduction formula∫cosn x dx =

1n

cosn−1 x · sin x +n − 1

n

∫cosn−2 x dx . (63)

2. Use Part 1 to evaluate∫

cos2 x dx .

3. Use Parts 1, 2 to evaluate∫

cos4 x ds.

Solution: First observe that the textbook has overlooked the restriction that n ≥ 1; theequation is not meaningful when n = 0. We do know that, when n = 1,∫

cos1 x dx =11

sin x + 0. (64)

1. For n ≥ 2, we may take u = cosn−1 x, dv = cos x; then

du = (n − 1) cosn−2 x · (− sin x) dx

by the Chain Rule, and v = sin x. Let’s denote∫

cosn x dx by In. Then a first

application of Integration by Parts yields∫cosn x dx = cosn−1 x · sin x + (n − 1)

∫cosn−2 x · sin2 x dx

= cosn−1 x · sin x + (n − 1)∫

cosn−2 x · (1 − cos2 x) dx

= cosn−1 x · sin x + (n − 1) (In−2 − In) .

Collecting all terms in In to the left side of the equation, and dividing by n − 1, weobtain

In =1n· cosn−1 x · sin x +

n − 1n

In−2

for n ≥ 2, as desired.2. When n = 2, the reduction formula reduces to

I2 =12

cos x · sin x +12

∫dx

=12

cos x · sin x +x2

+ C.

3. When n = 4, a second application gives

I4 =14

cos3 x · sin x +34

I2

=14

cos3 x · sin x +34

(12

cos x · sin x +x2

)+ C

=14· cos3 x sin x +

38· cos x sin x +

38· x + C


C.13.2 §7.2 Trigonometric Integrals

This section is concerned with integrating functions that can be expressed simply in terms oftrigonometric functions. The techniques rely on heavy use of familiar trigonometric identities.In particular, integration of the following types of functions is considered:

• products of non-negative powers of sin x and cos x

• products of non-negative powers of tan x and sec x

The integration of other functions that can be reduced to functions of these two types is alsoconsidered. Strategies are developed for products of these types. However, students may wellbe able to use techniques already seen to integrate certain integrals of these types in ways otherthan those suggested here. To reiterate: YOU MAY BE ABLE TO INTEGRATE CERTAINFUNCTIONS OF THE TYPES LISTED BY USING OTHER METHODS. The objective inall of these procedures is to “simplify” the integration; where the function is a product oftrigonometric functions, this “simplification” is usually measured by a reduction in the totaldegree of the product, i.e., in the total number of trigonometric factors. Since the total is finite,repeated applications of such procedures will eventually result in successful integration.

Two ways of integrating∫

sin2 x dx and∫

cos2 x dx: One way of evaluating these integralsis to use one of the double angle formulæ from trigonometry:

sin 2θ = 2 sin θ · cos θ (65)cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1 − 2 sin2 θ (66)

which follow from the formulæ for the sine and cosine of a sum. Two formulæ involving cos 2θmay be rewritten as

sin2 θ =12

(1 − cos 2θ) (67)

cos2 θ =12

(1 + cos 2θ) (68)

From these we obtain ∫sin2 x dx =

12

∫(1 − cos 2x) dx

=12

(x − 1

2sin 2x

)+ C

∫cos2 x dx =

12

∫(1 + cos 2x) dx

=12

(x +

12

sin 2x)

+ C


As we saw earlier in connection with∫

cos2 x dx, another way to evaluate these integrals isthrough integration by parts. See [1, Exercises 43(a), 44(a)(b) p. 458] which describe howto use the reduction formula in [1, Example 6, p. 457] and an analogue for cosines for thesepurposes.

Strategy for evaluating∫

sinm x · cosn x dx

0. This is a recursive procedure: if the first 2 steps do not lead to a substitution producing anintegral that may be evaluated immediately, the last 2 steps will lead to a simplificationin the integrand, after which the procedure is begun again.

1. If n is odd, use the identity cos2 x = 1− sin2 x to convert all but one of the cosine factorsinto a function of sines. Then apply the substitution u = sin x with du = cos x dx.

2. If m is odd, proceed analogously to the preceding: use the same identity to convert all butone of the sine factors into a function of cosines. Then apply the substitution u = cos x,with du = − sin x dx.

3. If both m and n are even, use the identities cos2 x = 12 (1+cos 2x) and sin2 x = 1

2 (1−cos 2x)to express the integrand as a sum of products of sines and cosines of 2x. The degreesof these terms will be less than the degree of the integrand we started with, so that wehave simplified the problem, and can repeat the procedure until we have completed theintegration.

4. When both m and n are odd, other identities may also be used to simplify the integrand;for example, we can use sin 2x = 2 sin x · cos x combined with the two double angleformulæ mentioned immediately above.

Example C.44 To determine∫

sin4 x · cos2 x dx.

Solution:∫

sin4 x · cos2 x dx =

∫ (1 − cos 2x

2

)2

· 1 + cos 2x2

dx

=

∫ (1 − cos 2x − cos2 2x + cos3 2x

8

)dx

=18· x − 1

16sin 2x −

∫1 + cos 4x

16dx

+1

16

∫ (1 − sin2 2x

)· d

dxsin(2x) dx

etc. (Use the substitution u = sin 2x to evaluate the last integral.)


(to be continued)


C.14 Supplementary Notes for the Lecture of February 01st, 2010Release Date: Monday, February 01st, 2010,

subject to revision

C.14.1 §7.2 Trigonometric Integrals (conclusion)


sinm x · cosn x dx (Repetition from last lecture).

0. This is a recursive procedure: if the first 2 steps do not lead to a substitution producing anintegral that may be evaluated immediately, the last 2 steps will lead to a simplificationin the integrand, after which the procedure is begun again.

1. If n is odd, use the identity cos2 x = 1− sin2 x to convert all but one of the cosine factorsinto a function of sines. Then apply the substitution u = sin x with du = cos x dx.

2. If m is odd, proceed analogously to the preceding: use the same identity to convert all butone of the sine factors into a function of cosines. Then apply the substitution u = cos x,with du = − sin x dx.

3. If both m and n are even, use the identities cos2 x = 12 (1+cos 2x) and sin2 x = 1

2 (1−cos 2x)to express the integrand as a sum of products of sines and cosines of 2x. The degreesof these terms will be less than the degree of the integrand we started with, so that wehave simplified the problem, and can repeat the procedure until we have completed theintegration.

4. When both m and n are odd, other identities may also be used to simplify the integrand;for example, we can use sin 2x = 2 sin x · cos x combined with the two double angleformulæ mentioned immediately above.

Example C.45 To determine∫

sin4 x · cos2 x dx.

Solution:∫

sin4 x · cos2 x dx =

∫ (1 − cos 2x

2

)2

· 1 + cos 2x2

dx

=

∫ (1 − cos 2x − cos2 2x + cos3 2x

8

)dx

=18· x − 1

16sin 2x −

∫1 + cos 4x

16dx

+1

16

∫ (1 − sin2 2x

)· d

dxsin(2x) dx

etc. (Use the substitution u = sin 2x to evaluate the last integral.)



tanm x·secn x dx. Earlier I discussed an algorithm for integrating

products of non-negative integer powers of sines and cosines. I now consider another importantclass of products of trigonometric functions.

1. The basis of the solution I propose for most of these cases is to use the facts that

d(tan x) = sec2 x dxd(sec x) = sec x tan x dx

sec2 x = tan2 x + 1

2. If n is even and n ≥ 2, then we can replace the factor secn x by

(sec2 x

) n−22 d

dxtan x =

(tan2 x + 1

) n−22 · d

dxtan x

which, when multiplied by tanm x — where m has any value — yields an integral that issimplified by the substitution u = tan x.

3. If n = 0 we have∫

tanm x dx:

(a) When m = 0, the solution is∫

dx = x + C.

(b) When m = 1, the function integrates as∫

tan x = ln | sec x| + C = − ln | cos x| + C.

(c) When m ≥ 2, one may detach 2 powers of the tangent from the others, replacingthem by sec2 x − 1, thereby reducing the problem to the integration of a lower

power of the tangent, and an integral of the form∫

tanm−2 x · sec2 x dx, which can

be integrated following a substitution u = tan x.

Henceforth we may assume that n is odd.

4. If m is odd, we may detach one power of tan x from tanm x and one power of sec x fromsecn x, and write the integrand as

(tan2 x

)m−12 · secn−1 x · d

dxsec x

which is equal to (sec2 x − 1

)m−12 · secn−1 x · d

dxsec x

and can be integrated after a substitution u = sec x.


5. In the only remaining cases m is even and n is odd. One method would be to transformthe entire integrand into powers of sec x and then to use the reduction formula [1, Exer-cise 50, p. 458] to reduce everything to the problem of integrating sec x. Other reductionsare possible: for example, into a function expressible in terms of cos x, with one factorcos x left over — this would permit a substitution u = cos x that converts the problem tothe type we shall meet in [1, §7.3]; alternatively, one may express the integrand as a sumof powers of tan x, and develop a reduction formula for them (cf. [1, endpapers, item 75p. 9.]).

The integral∫

tan x dx. As observed above,

∫tan x dx = − ln | cos x| + C = ln | sec x| + C .

The integral∫

sec x dx. The textbook observes that

∫sec x dx = ln | sec x + tan x| + C .

Do not forget the absolute value signs when you quote these results, although you may expectthat in many of the problems you consider, where sec x+ tan x is positive, omission of the signsmay not produce any visible error.

While it is possible to prove the validity of the preceding equation simply by differentiationof the alleged antiderivative, a more direct proof requires some ingenuity.44


cotm x · cscn x dx. Analogues of the preceding techniques can

simplify integrals of these types.

44One way to derive this result is to observe that

sec x =1

cos x=

cos x

1 − sin2 x

=cos x

(1 − sin x)(1 + sin x)

=12

( cos x1 − sin x

+cos x

1 + sin x

)

using ideas of partial fractions that we shall be meeting in [1, §7.4]. Each of these summands can be integratedas a logarithm (See [33, pp. 505-506 ].)


“Other” trigonometric identities. By adding and subtracting the expansions of sin(A ± B)and cos(A ± B), one may obtain the following identities

sin A · cos B =12

(sin(A − B) + sin(A + B)) (69)

sin A · sin B =12

(cos(A − B) − cos(A + B)) (70)

cos A · cos B =12

(cos(A − B) + cos(A + B)) (71)

These identities permit the derivation of another class of useful identities. If we replace A − B

and A+ B respectively by U and V — equivalently, if we replace A byV + U

2and B by

V − U2

,we obtain

sin U + sin V = sinV + U

2· cos

V − U2

(72)

sin U − sin V = sin−V + U

2· cos

V + U2

(73)

cos U + cos V = cosV + U

2· cos

V − U2

(74)

cos U − cos V = sinV + U

2· sin

V − U2

(75)

Example C.46 Integrate∫

(sin 50x · cos 12x) dx.

Solution: (One possible solution)∫(sin 50x · cos 12x) dx =

12

∫(sin 38x + sin 62x) dx

= −12· 1

38· cos 38x − 1

2· 1

62· cos 62x + C

Example C.47 ([7, Exercise 62, p. 488]) Find the volume obtained by rotating the regionbounded by the curves y = cos x, x = 0, y = 0, x = π

2 about the axis y = 1.Solution: I give a solution, by washers. (The problem can be solved by cylindrical shells also,but that method is much more difficult.)

Volume =

∫ π2

0

(12 − (1 − cos x)2

)dx

=

∫ π2

0(2 cos x − cos2 x) dx

=

[2 sin x − x

2− sin 2x

4

] π2

0

= π(2 − π

4

).


7.2 Exercises

[1, Exercise 30, p. 466] To evaluate

π3∫

0

tan5 x sec6 x dx.

Solution: This integral can be evaluated in at least two different ways.

1. Since the exponent of the secant is even, we can use a substitution u = tan x, sodu = sec2 x dx.

π3∫

0

tan5 x sec6 x dx =

∫ tan π3

tan 0u5

(u2 + 1

)2du

=

∫ √3

0u5

(u2 + 1

)2du

=

∫ √3

0

(u9 + 2u7 + u5

)2du

=

[u10

10+

u8

4+

u6

6

]√3

0

=24310

+814

+276

=98120

2. Alternatively, since the exponent of the tangent is odd, we can use a substitutionv = sec x, so dv = sec x · tan x dx:

π3∫

0

tan5 x sec6 x dx =

∫ 2

1(v2 − 1)2v5 dv

=

[v10

10− v8

4+

v6

6

]2

1

=98120

[1, Exercise 48, p. 466] To evaluate∫

dxcos x − 1

.

Solution: This integral is not of any of the forms shown in the chapter, so some ingenuityis needed. I give more than one solution.


1. First solution, using double angle formula. When you see cos x − 1, that shouldsuggest the identity cos 2θ = 1 − 2 sin2 θ. Applying that identity here, with θ = x

2 ,yields

∫dx

cos x − 1=

∫dx

−2 sin2 x2

= −12

∫csc2 x

2dx = cot

x2

+ C

2. Second solution, using the identity sin2 x + cos2 x = 1. The idea here resemblesthe rationalization of fractions involving square roots, seen earlier.

∫dx

cos x − 1=

∫ (1

cos x − 1· cos x + 1

cos x + 1

)dx

=

∫cos x + 1− sin2 x

dx

= −∫

cos xsin2 x

dx −∫

csc2 x dx

The first integral may be evaluated in several ways, for example by using the sub-stitution u = sin x, so du = − cos x dx. That integral becomes

∫cos xsin2 x

dx =

∫duu2

= −1u

+ C1 = − csc x + C1.

The second integral can be seen immediately to be − cot x + C2. The two togethergive us ∫

dxcos x − 1

= csc x + cot x + C .

But, are the two answers equal? This can be seen through trigonometric identities. Forexample

csc x + cot x =1 + cos x

sin x=

2 cos2 x2

2 sin x2 · cos x

2

= cotx2


C.15 Supplementary Notes for the Lecture of February 03rd, 2010Release Date: Wednesday, February 03rd, 2010,

subject to revision

C.15.1 §7.3 Trigonometric Substitution

In this section we describe a type of substitutions which simplify certain commonly met inte-grals. We approach these substitutions in the reverse direction from that used earlier. Whereasearlier we investigated substitutions of the form u = g(x), this time we will usually formulateour substitutions first in the form x = h(u); that is, we will look for a substitution that willsimplify the integrand, and then try to implement it. We usually proceed “mechanically” inthese problems; but, in principle, we are postulating the existence of an inverse function, andshould be checking that there really is an inverse whenever we use the method. I will try togo through some of those steps in the first examples we consider, but, in practice, we oftenbecome careless and don’t check everything unless something indicates a problem. This isunwise; the reason that it “works” is that we usually confine the substitutions to certain wellunderstood pairs of functions/inverses, where all of the snags have already been worked out.

Example C.48 [1, Exercise 4, p. 472] To evaluate∫ 2

√3

0

x3

√16 − x2

dx.

Solution: A first look at the integrand suggests that the complication comes from the ex-pression

√16 − x2 in the denominator. We could try to simplify by giving this a new name,

u =√

16 − x2. That leads to

du = − x√16 − x2

dx⇒ x dx = −u du

and so the integral becomes∫ 2

4(u2 − 16) du =

[u3

3− 16u

]2

4=

403.

But we would like to illustrate the notion of trigonometric substitution here (in a prob-lem where it isn’t really needed!) The component

√16 − x2 suggests that we might wish

to interpret the problem geometrically, with this component arising from an application ofPythagoras’s Theorem to a right-angled triangle; equivalently, from the identity that

sin2 θ + cos2 θ = 1 .

To do this, we can first divide out the factor 16, which, when it leaves the square root, willreappear outside as 4:

√16 − x2 = 4

√1 − x2

16= 4

√1 −

( x4

)2.


This suggests a substitution

x4

= sin v or, equivalently, x = 4 sin v

that will makex4

into a sine or a cosine — either one will work. When we express the substi-tution that way we are really working with the inverse — to conform with our earlier theoryabout substitutions we should be expressing the new variable in terms of the old; so we shouldbe starting with

v = arcsinx4.

The range of values of x that interest us is 0 ≤ x ≤ 2√

3, equivalently 0 ≤ x4≤√

32

, andwe know that the inverse sine function is defined over this domain. So, beginning with thesubstitution above, we obtain

dv =1

1 −(

x4

)2 ·14

dx =dx√

16 − x2,

and the integral transforms as follows:

∫ 2√

3

0

x3

√16 − x2

dx =

∫ arcsin 2√

34

arcsin 04

64 sin3 v dv

= 64∫ π

3

0sin3 v dv

which we proceed to evaluate using the methods of the preceding section:

64∫ π

3

0sin3 v dv = 64

∫ π3

0sin v

(1 − cos2 v

)dv

= 64[− cos v +

13

cos3 v] π

3

0=

403

In practice this method works smoothly, but one must occasionally be careful about theevaluation of the inverse function, remembering our original definitions of the restricted inter-val where the inverse was taken. The method is indicated whenever we see expressions like√

1 − x2 or, more generally,√

a2 − x2, since we can transform the latter into the former by di-vision by a positive real number. The inverse cosine could be used instead of the inverse sine,and the results would be no more difficult.


Example C.49 ([7, Exercise 8, p. 494]) Evaluate∫ √

x2 − 4x4 dx.

Solution: We need a substitution that will convert x2 to 4 times the square of a secant.45 Oneway to achieve this is to make u have the property that

x = 2 sec u ;

so the actual substitution will beu = arcsec

x2,

which implies thatdx = 2 sec u tan u du .

Under the substitution what happens to√

x2 − 4? It becomes√

4 tan2 u, i.e., 2| tan u|. Do weneed the absolute signs? Recall that the inverse secant takes its values in the two intervals0 ≤ u < π

2 and π ≤ u ≤ 3π2 . In these two intervals the tangent is always positive, so the absolute

signs may be dropped.

∫ √x2 − 4x4 dx =

∫2 tan u

16 sec4 u· 2 sec u · tan u du

=14

∫sin2 u · cos u du

=1

12sin3 u + C

We can’t leave the answer in this form, as it must be expressed in terms of the original variablex. Since

sin u = tan u · cos u

=tan usec u

=

12

√x2 − 4

x2

,

112

sin3 u =112· (x2 − 4)

32

x3

Hence ∫ √x2 − 4x4 dx =

(x2 − 4)32

12x3 + C .

45Alternatively, we could make x2 four times the square of a hyperbolic cosine.


Example C.50 I have avoided computing the area of a disk until now. It is trivial with atrigonometric substitution. If the disk is the set of points (x, y) such that x2 + y2 ≤ R2, then it isthe region bounded by the 2 graphs y = ±

√R2 − x2. Its area is

∫ R

−R

(√R2 − x2 −

(−√

R2 − x2))

dx =

∫ R

−R2√

R2 − x2 dx

=

∫ R

04√

R2 − x2 dx ,

since the integrand is even, and the interval of integration is symmetric around 0. The sub-stitution u = cos−1 x

R for 0 ≤ x ≤ R implies that cos u = xR or x = R cos u, where the

interval of integration is now from u = cos−1 0 = π2 to u = cos−1 1 = 0; dx = −R sin u du;√

R2 − x2 = R| sin u| = R sin u, since the sine is positive in this interval. The integral transformsto

4∫ R

0

√R2 − x2 dx = 4

∫ 0

π2

|R sin u|(−R sin u) du

= −4R2∫ 0

π2

sin2 u du

since the sine is positive for 0 ≤ u ≤ π2

= −4R2[u2− sin

2u4

]0

π2

= −4R2[0 − π

4

]= πR2

If we had not used the evenness of the integrand to reduce the original problem to integratingover the interval 0 ≤ x ≤ R, there would have been a serious difficulty. That is because theinverse cosine function takes its values between 0 and π.

If we had naively carried through the substitution over the entire interval −R ≤ x ≤ R,we would have obtained

4∫ R

−R

√R2 − x2 dx = 4

∫ − π2π2

|R sin u|(−R sin u) du

= 4∫ 0

π2

|R sin u|(−R sin u) du + 4∫ − π2

0|R sin u|(−R sin u) du

= 4∫ 0

π2

(R sin u)(−R sin u) du + 4∫ − π2

0(−R sin u)(−R sin u) du

= −4R2∫ 0

π2

sin2 u du + 4R2∫ − π2

0sin2 u du


= −4R2∫ 0

− π2sin2 u du + 4R2

∫ π2

0sin2 u du .

But note that the integrand is an even function, so the two integrals would cancel, andthe answer would be 0. This is clearly incorrect, but what went wrong? The error wasin attempting to replace x by R cos u: u is not uniquely defined for −π2 ≤ x ≤ π

2 ! Wecould, though, have used a substitution u = sin−1 x

R over the full interval, and the correctanswer would have been obtained. We would have defined u = arcsin x

R , so sin u = xR ,

cos u du = 1R · dx.

4∫ R

−R

√R2 − x2 dx = 4

∫ π2

− π2(R cos u)R cos u du

= 4R2∫ π

2

− π2

1 + cos 2u2

du

= 2R2[u +

sin wu2

] π2

− π2

= 2R2((π

2+ 0

)−

(−π

2+ 0

))= πR2 .


C.16 Supplementary Notes for the Lecture of February 05th, 2010Release Date: Friday, February 05th, 2010

C.16.1 §7.3 Trigonometric Substitution (conclusion)

Table of trigonometric substitutions I can expand the table of substitutions of this typegiven in the textbook:

Expression Inverse Substitution Substitution Identity√

a2 − x2 θ = arcsin xa x = a sin θ 1 − sin2 θ = cos2 θ

(−a ≤ x ≤ +a)(−π2 ≤ θ ≤ π

2

)

√a2 − x2 θ = arccos x

a x = a cos θ 1 − cos2 θ = sin2 θ(−a ≤ x ≤ +a) (0 ≤ θ ≤ π)

√a2 + x2 θ = arctan x

a x = a tan θ 1 + tan2 θ = sec2 θ

(−∞ < x < +∞)(−π2 ≤ θ ≤ π

2

)

√x2 − a2 θ = arcsec x

a x = a sec θ sec2 θ − 1 = tan2 θ

(−∞ < x ≤ −a or(π ≤ θ < 3π

2 ora ≤ x < ∞) 0 ≤ θ < π

2

)

I have shown both sine and cosine versions of the first substitution, and could similarly haveproduced a cotangent version of the tangent substitution, and a cosecant version of the secantsubstitution; in practice the latter two variants are not used frequently, and usually offer noadvantages over the substitutions shown.

Another trigonometric substitution will be discussed in the next section. It is used not tosimplify a square root, but to simplify denominator terms of the form x2 + a2.

7.3 Exercises

[1, Exercise 24, p. 472] Evaluate∫

1√t2 − 6t + 13

dt.

Solution: I shall complete the square of the quadratic polynomial in the denominatorin order that, after a first substitution, this integral will be of a type that we recognize.Since

t2 − 6t + 13 = (t − 3)2 + 4 = 4(t − 3

2

)2

+ 1 ,


a first substitution u =t − 3

2, which implies that dt = 2 du, could be applied:

∫dt√

t2 − 6t + 13=

∫du√

u2 + 1.

Now I take u = tan θ, i.e., θ = arctan u. Thus −π2 < θ < +π2 .

∫du√

u2 + 1=

∫sec2 θ

| sec θ| dθ

=

∫| sec θ| dθ

=

∫sec θ dθ since − π

2< θ <

π

2= ln | sec θ + tan θ| + C

= ln∣∣∣∣∣±

(1 + tan2 θ

) 12

+ tan θ∣∣∣∣∣ + C

where the + sign is taken since |θ| < π2 and the cosine

and secant are positive in Quadrants ##1,4,

= ln∣∣∣∣∣(1 + u2

) 12

+ u∣∣∣∣∣ + C

= ln

∣∣∣∣∣∣∣∣

1 +

(t − 3

2

)212

+

(t − 3

2

)∣∣∣∣∣∣∣∣+ C

= ln∣∣∣∣∣(t2 − 6t + 13

) 12

+ (t − 3)∣∣∣∣∣ + (C − ln 2)

And I could rename the constant with a single letter, e.g., K = C − ln 2.

Hyperbolic substitutions It is possible to achieve the same sorts of simplifications by usinginverse hyperbolic functions. Since we have spent little time in becoming comfortable withthe hyperbolic functions, I will not discuss these substitutions in general, but may apply themin specific cases.

Example C.51 This is [1, Exercise 31, p. 472].

1. Use trigonometric substitution to show that∫

dx√x2 + a2

= ln(x +√

x2 + a2)

+ C


2. Use hyperbolic substitution x = a sinh t to show that∫

dx√x2 + a2

= sinh−1( xa

)+ C

Solution:

1. We can simplify the surd by making x = a tan u. But our theory of substitutions requiresthat we express u as a function of x; so we confine ourselves, for example, to −π2 < x < π

2 ,and define u = arctan x

a on that interval. Then

du =dx

1 + x2

a2

· 1a

=a dx

x2 + a2

=a dx

sec2 u

⇒ dx =1a

sec2 u du∫

dx√x2 + a2

=

∫sec2 u du| sec u|

=

∫sec u du = ln | sec u + tan u| + C

since the secant is positive in the given interval

= ln

∣∣∣∣∣∣∣xa

+

√( xa

)2+ 1

∣∣∣∣∣∣∣ + C

=

(ln

∣∣∣∣x +√

x2 + a2∣∣∣∣ − ln a

)+ C

= ln∣∣∣∣x +

√x2 + a2

∣∣∣∣ + C′

where we absorb the subtracted logarithm of a constant into a new constant of integra-tion.

2. The hyperbolic function sinh is invertible, since its derivative is positive always. If wewish to have x = a sinh u, we can define u = sinh−1 x

a . Taking differentials gives

dx = a cosh u du = a√

1 + sinh2 u du = a

√x2

a2 + 1 du =√

x2 + a2 du

Hence ∫dx√

x2 + a2=

∫du = u + C

= arcsinhxa

+ C .


Example C.52 ([7, Exercise 26, p. 494]) Evaluate∫

x2

√4x − x2

dx.

Solution: Completion of the square yields

4x − x2 = −(x2 − 4x) = 4 − (x2 − 4x + 4) = 4 − (x − 2)2 = 22

1 −(

x − 22

)2 .

Hence∫

x2

√4x − x2

dx =

∫x2

2√

1 −(

x−22

)2dx

=

∫(2u + 2)2

2√

1 − u22 du

under substitution u =x − 2

2

= 4∫ (

sin2 v + 2 sin v + 1)

dv

under substitution v = arcsin u = arcsinx − 2

2

= 4∫

1 − cos 2v2

dv − 8 cos v + 4v

= 6v − sin 2v − 8 cos v + C= 6v − 2(4 + sin v) cos v + C

Now the arcsine function takes values between −π2 and π2 , in which interval the cosine is posi-

tive; hence

cos v = +√

1 − sin2 v =√

1 − u2 =12

√4x − x2 .

We conclude that∫

x2

√4x − x2

dx = 6 arcsinx − 2

2− x + 6

2·√

4x − x2 + C .

Example C.53 ([7, Exercise 28, p. 494]) Evaluate∫

1(5 − 4x − x2) 5

2

dx.

Solution: As in the preceding example, I shall complete the square of the quadratic polynomialin the denominator in order that, after a first substitution, this integral will be of a type that werecognize. Since

5 − 4x − x2 = −(x2 + 4x − 5) = 9 − (x + 2)2 = 91 −

(x + 2

3

)2 ,


a first substitution u =x + 2

3, which implies that dx = 3 du, could be applied:

∫1

(5 − 4x − x2) 5

2

dx =181

∫du

(1 − u2)52

.

Now we can apply a second substitution u = sin φ — actually it is φ = arcsin u – where

dφ =du

(1 − u2) 1

2

. We obtain

181

∫du

(1 − u2)52

=1

81

∫dφ

cos4 φ

=1

81

∫sec4 φ dφ

=1

81

∫ (tan2 φ + 1

)sec2 φ dφ

=1

81

[13

tan3 φ + tan φ]

+ C .

The function φ was defined to be an arcsine, so its values are in the interval −π2 ≤ φ ≤ π2 , in

which the cosine is positive. Hence

tan φ =sin φcos φ

=sin φ

+

√1 − sin2 φ

=u√

1 − u2

=

x+23√

1 −(

x+23

)2=

x + 2√9 − 4x − x2

and181

[13

tan3 φ + tan φ]

=(x + 2)3

243(5 − 4x − x2) 3

2

+x + 2

81(5 − 4x − x2) 1

2


C.17 Supplementary Notes for the Lecture of February 08th, 2010Release Date: Monday, February 08th, 2010

C.17.1 §7.4 Integration of Rational Functions by Partial Fractions

In this section we shall see that an entire class of functions can be integrated by a systematicalgebraic decomposition procedure, followed by specific methods for the various componentsinto which we decompose the functions.

An example to illustrate the general procedure Before describing the general procedurelet us consider some examples that will make it easier to comprehend.

[1, Exercise 14, p. 482] Evaluate the integral∫

1(x + a)(x + b)

dx.

Solution: There will be two quite different solutions, depending on whether a = b.

Case a = b: This can be integrated directly by observation, or by using the substitutionu = x + a. ∫

1(x + a)2 dx = − 1

x + a+ C .

Case a , b: We try to decompose the integrand into the sum of fractions whose denom-inators are, respectively, x + a, and x + b. The degrees of the numerators must beless than these polynomials of degree 1, so they have to have degree 0, i.e., theyhave to be constants. So suppose that

1(x + a)(x + b)

=α

x + a+

β

x + b.

It can be shown algebraically that such a decomposition is always possible; all thatis missing is to know the values of the constants α and β. The last equation, aftermultiplication of both sides by (x + a)(x + b), yields

1 = α · (x + b) + β · (x + a) .

Here are 2 ways to find α and β:

1. Express both sides of the equation as polynomials in x, and equate the coeffi-cients of corresponding powers of x. The left side is 1x0 + 0x1. Thusdegree 0: 1 = α · b + β · adegree 1: 0 = α + β


Solving these equations gives

α =1

b − a

β =1

a − b

2. This equation must be true for all values of x. Give x “convenient” values toobtain equations in α and β, and solve those equations. Two “convenient” val-ues are x = −a and x = −b. They give, respectively, the following equations:

1 = α · (−a + b) + β · 0⇒ α =1

b − a

1 = β · (0) + β · (−b + a)⇒ β =1

a − b

We may now complete the integration:∫

1(x + a)(x + b)

dx =

∫1

a − b

(− 1

x + a+

1x + b

)dx

=1

a − b(− ln(x + a) + ln(x + b)) + C

=1

a − bln|x + b||x + a| + C

=1

a − bln

∣∣∣∣∣x + bx + a

∣∣∣∣∣ + C

This family of functions could be written in other ways. For example, since C · (a−b) = ln eC·(a−b), we could call eC·(a−b) K, and write the family as

1a − b

(ln

∣∣∣∣∣x + bx + a

∣∣∣∣∣ + ln K)

=1

a − bln

∣∣∣∣∣K ·x + bx + a

∣∣∣∣∣

where K serves as the constant of integration.

Polynomials Recall that a polynomial [1, p. 28] is a function of the form

P(x) = anxn + an−1xn−1 + · · · + a1x1 + a0x0

where a0, . . . , an are real numbers, called the coefficients of the polynomial. Except for thezero polynomial, which is the constant function 0, all polynomials will have a largest integerm such that am , 0; m is called the degree of the (non-zero) polynomial, and am is called theleading coefficient; a0 is called the constant term. We usually write x0 simply as 1, and x1

simply as x.


Rational Functions A rational function is a ratio of polynomials of the form

A(x)B(x)

where A and B are polynomial functions. The function will have as discontinuities the roots ofB, if there are any.

The goal of partial fraction decompositions In a partial fraction decomposition we expressa ratio of polynomials as a sum of “partial” fractions — fractions that have special properties.In these “partial” fractions the denominator polynomials will always be powers of one poly-nomial that is irreducible, i.e., it cannot be factored further (unless we move beyond the realnumber system to, for example, the complex number system — with which you are not ex-pected to be familiar). It is a theorem of algebra that

Theorem C.54 If a non-zero real polynomial is irreducible then it must be of one of the fol-lowing forms:

1. a non-zero constant

2. a polynomial of degree 1, of the form ax + b, where a , 0.

3. a polynomial of degree 2 without real roots, i.e., of the form ax2 + bx + c where a , 0and b2 < 4ac.

The only type of ratio to which we apply this decomposition is one where the degree of thenumerator is strictly less than the degree of the denominator; if the rational function that westart with does not have this property, then we will have to preprocess it to obtain a ratio of thistype; after the procedure of partial fraction decomposition is applied, the resulting “partial”fractions will have the same property — that the degree of their numerator will always beless than that of the denominator. We will then show that we are able to integrate all partial

fractions of these types, and so we will be able to integrate all rational polynomialsA(x)B(x)

.

The procedure we shall develop depends on having such a factorization of the denominatorof the given ratio into irreducible polynomials. In this course we shall not be concerned withfinding the factorization itself, beyond knowing

• that P(a) = 0⇒ x − a divides P(x) — the so-called “Factor Theorem”;

• how to factorize quadratic polynomials, both by using the quadratic formula, and by“completing the square”;

• that, for any positive integer n,

an − bn = (a − b)(an−1 + an−2b + . . . + abn−2 + bn−1)


• that, for any positive integer n,

a2n+1 + b2n+1 = (a + b)(a2n − a2n−1b + . . . − ab2n−1 + b2n).

Sometimes a factor x−a may divide P(x), so that P(x) = (x−a)Q(x), and again x−a may alsodivide Q(x), and possibly divide even more quotients. We may then speak of the multiplicityof the factor x − a of P(x), and say that a is a root of P(x) of that multiplicity.

“Long division” of polynomials. You should know from high school how to divide a poly-nomial B(x) into a polynomial A(x) and obtain a quotient polynomial and a remainder poly-nomial. The procedure is similar to long division of integers, and you will be reminded verybriefly of it at the lecture. (You will see an example of long division of polynomials in theleft margin of [1, p. 474], but the way you learned to write such a calculation may be slightlydifferent from that used in the textbook.)

The first step. The procedure for integratingA(x)B(x)

begins with the division of B(x) into A(x),

obtaining a quotient and a remainder:

A(x) = Q(x) · B(x) + R(x)

where the remainder has degree less than that of the divisor B(x). This is an essential step;if it is omitted, it may cause the subsequent steps to fail. However, it is not necessary if thedegree of B is greater than the degree of A, since, in that case, the quotient will be 0 and theremainder will be A(x).

An example to illustrate the general procedure.

[1, Exercise 14, p. 482] Evaluate the integral∫

1(x + a)(x + b)

dx.

Solution: I discussed an example of this type at the last lecture; the solution (whena , b), shown on pages 3111 through 3112 of these notes, may be written as

∫1

(x + a)(x + b)dx =

∫1

a − b

(− 1

x + a+

1x + b

)

=1

a − b(− ln(x + a) + ln(x + b)) + C

=1

a − bln|x + b||x + a| + C

=1

a − bln

∣∣∣∣∣x + bx + a

∣∣∣∣∣ + C


Suppose that, instead of the given integral, we wished to integrate∫

x3 + (a + b)x2 + abx + 1(x + a)(x + b)

dx .

This function cannot be expanded into partial fractions until it is arranged that the degreeof the numerator — presently 3 — be less than the degree of the denominator — 2. Ifwe divide the denominator into the numerator, we find that

x3 + (a + b)x2 + abx + 1 = x · (x + a)(x + b) + 1 .

Hence the integral may be expressed as∫ (

x +1

(x + a)(x + b)

)dx, and its value will be

x2

2+

1a − b

ln∣∣∣∣∣x + bx + a

∣∣∣∣∣ + C

What would have happened if we had attempted to expand this function into partial frac-tions? No such decomposition can exist. Using the first method (equating coefficients ofcorresponding powers) we would have obtained 4 equations which would overdeterminethe constants α and β, and which would be inconsistent — there would be no solution.But, if we used the second method, and didn’t take enough equations, we might not no-tice that there was an inconsistency, and the alleged partial fraction would be incorrect.

The general procedure. The general procedure has many facets, and some will not be ex-plicitly discussed in the lecture; you are expected to work many problems in this section of thetextbook, not only problems discussed in the lecture or appearing on WeBWorK or the quizzes.

Remember the steps we need to follow:

1. The first step in any of these problems is to ensure that the degree of thenumerator must be less than that of the denominator. If it is not, you mustdivide the denominator into the numerator, obtaining a quotient and a re-mainder. The quotient integrates as a polynomial, and we are left with theratio of the remainder to the denominator, for which the methods we arediscussing will enable a complete solution.

2. Factorize the denominator into a product of polynomials of degrees 1 and 2which are irreducible, i.e., which do not factorize further into lower degreepolynomials. The quadratic factors will be those having no real roots. Groupfactors which are exactly the same together, so that your denominator is aproduct of powers of distinct, irreducible polynomials.

3. Then the fraction must be decomposed into partial fractions using methodswe have illustrated before, and continue in this lecture.


Example C.55 [1, Exercise 28, p. 482] Evaluate the integral∫

x2 − 2x − 1(x − 1)2(x2 + 1)

dx.

Solution: The denominator has two distinct irreducible factors: x − 1, of degree 1 and multi-plicity 2, and x2 + 1, an irreducible quadratic factor of degree 2 and multiplicity 1. The degreeof the numerator is less than 4, which is the degree of the denominator, so there is no need forany long division. One type of partial fraction decomposition is of the form

x2 − 2x − 1(x − 1)2(x2 + 1)

=αx + β

(x − 1)2 +γx + δ

x2 + 1

in which we take the most general numerator in each case, of degree less than the degree of thedenominator. This is not the most useful type of partial fraction decomposition, but we willcarry this step out and then improve on it. Multiplying both sides by (x−1)2(x2 + 1), we obtaina polynomial equation

x2 − 2x − 1 = (αx + β)(x2 + 1) + (γx + δ)(x2 − 2x + 1)⇔ x2 − 2x − 1 = (α + γ)x3 + (β − 2γ + δ)x2 + (α + γ − 2δ)x + (β + δ)

Equating coefficients of corresponding powers of x yields equations corresponding to the termsof degrees 3, 2, 1, 0:

α + γ = 0β − 2γ + δ = 1α + γ − 2δ = −2

β + δ = −1

which we proceed to solve, obtaining

(α, β, γ, δ) = (1,−2,−1, 1) ,

so the decomposition is

x2 − 2x − 1(x − 1)2(x2 + 1)

=x − 2

(x − 1)2 +−x + 1x2 + 1

.

(to be continued)


C.18 Supplementary Notes for the Lecture of February 10th, 2010Release Date: Wednesday, February 10th, 2010

C.18.1 §7.4 Integration of Rational Functions by Partial Fractions (conclusion)

The general procedure (continued).

Example C.56 (continued from Example2010:31 on page 3116 of these notes, cf. [1, Exercise28, p. 482]) We shall see that we will be able to integrate the second summand immediately, bybreaking it into two parts. We can integrate the first summand if we first apply the substitutionu = x − 1:

∫x − 2

(x − 1)2 dx =

∫u − 1

u2 du =

∫ (1u− 1

u2

)du

= ln |u| + 1u

+ C

= ln |x − 1| + 1x − 1

+ C

In practice we anticipate the results of this substitution by refining the partial fraction decom-position: in place of a summand of the form

an−1xn−1 + an−2xn−2 + . . . + a0

(x − a)n

we repeatedly divide x − a into the numerator, so that we can express the numerator as a sumof powers of x−a; then we decompose the fraction and divide excess powers of x−a, to obtaina decomposition of the form

bn

(x − a)n +bn−1

(x − a)n−1 . . . +b1

x − a

in which we can integrate each of the summands at sight. This decomposition can be accom-plished as a second phase of partial fraction decomposition, or immediately, by assuming adecomposition of the form

x2 − 2x − 1(x − 1)2(x2 + 1)

=ζ

(x − 1)2 +η

x − 1+γx + δ

x2 + 1

which leads to the polynomial identity

x2 − 2x − 1 = ζ(x2 + 1) + η(x − 1)(x2 + 1) + (γx + δ)(x − 1)2

⇔ x2 − 2x − 1 = (η + γ)x3 + (ζ − η − 2γ + δ)x2 + (η + γ − 2δ)x + (ζ − η + δ)


which we proceed to solve, obtaining

(ζ, η, γ, δ) = (−1, 1,−1, 1) ,

∫x2 − 2x − 1

(x − 1)2(x2 + 1)dx

=

∫ −1(x − 1)2 dx +

∫1

(x − 1)1 dx +

∫ −xx2 + 1

dx +

∫1

x2 + 1dx

=1

x − 1+ ln |x − 1| − 1

2ln |x2 + 1| + arctan x + C

=1

x − 1+ ln |x − 1| − 1

2ln(x2 + 1) + arctan x + C

=1

x − 1+ ln

|x − 1|√x2 + 1

+ arctan x + C

7.4 Exercises

[1, Example 19, p. 482] Evaluate the integral∫

1(x + 5)2(x − 1)

dx .

Solution: Since the degree of the numerator is 0, and that of the denominator is 2 + 1 =

3 > 0, we can skip the long division step. We have to decompose the fraction into a sumof partial fractions, which we may take to be of the form

1(x + 5)2(x − 1)

=A

(x + 5)2 +B

(x + 5)1 +C

(x − 1)1 .

Multiplying both sides by (x + 5)2(x − 1), we obtain

1 = A(x − 1) + B(x − 1)(x + 5) + C(x + 5)2 . (76)

i.e.,0x2 + 0x1 + 1x0 = A(x − 1) + B(x2 + 4x − 5) + C(x2 + 10x + 25) . (77)

One way to obtain the values of A, B,C is simply to equate coefficients of like powers ofx:

0 = B + C0 = A + 4B + 10C1 = −A − 5B + 25C

which we may solve to show that (A, B,C) =(− 1

6 ,− 136 ,

136

).


Another method of solution is to assign to x “convenient” values of x, and thereby obtainmore convenient equations to solve. Two values which re “convenient” are x = 1 andx = −5, and they yield from equation (76) the following equations:

x = 1⇒ 1 = 36C

andx = −5⇒ 1 = −6A

implying that A = − 16 and C = 1

36 . To obtain the value of B we would need a thirdequation. This could be obtained by equating coefficients, as in the earlier method, or bysimply choosing another value; e.g.,

x = −1⇒ 1 = −2A − 8B + 16C =26− 8B +

1636

which yields the same values as before. Integration is straightforward:

∫1

(x + 5)2(x − 1)dx =

∫ − 1

6

(x + 5)2 +− 1

36

x + 5+

136

x − 1

dx

=16· 1

x + 5− 1

36ln |x + 5| + 1

36ln |x − 1| + K .

This indefinite integral could be further simplified, e.g., by combining the two logarith-mic terms into the logarithm of a quotient of polynomials of degree 1.

Repeated irreducible quadratic factors.

Example C.57 [7, Exercise 38, p. 504] To integratex4 + 1

x(x2 + 1)2 dx.

Solution: What distinguishes this example from those studied earlier is the multiplicity of theirreducible quadratic factor x2 + 1 in the denominator; this is the first case we have seen wherethat multiplicity exceeds 1.

Before I begin the application of the methods of this section, I will apply a substitution thatcould simplify this problem. By setting u = x2, I find that du = 2x dx, x2 = u − 1, so

∫x4 + 1

x(x2 + 1)2 dx =12

∫u2 + 1

u(u + 1)2 du .

We can seek a partial fraction decomposition:

u2 + 1u(u + 1)2 =

Au

+B

(u + 1)2 +C

u + 1


which implies identity of the numerators after the right side is taken to a common denominator:

u2 + 1 = A(u + 1)2 + Bu + Cu(u + 1)

in which I will take several “convenient” values of u to obtain equations that can be solved forthe coefficients;

u = −1 ⇒ 2 = 0 − B + 0⇒ B = −2u = 0 ⇒ 1 = Au = 1 ⇒ 2 = 4A + B + 2C ⇒ C = 0

Hence∫

x4 + 1x(x2 + 1)2 dx =

12

∫ (1u− 2

(u + 1)2

)du

=12· ln |u| + 1

u + 1+ K

= ln |x| + 1x2 + 1

+ K .

The preceding was fortuitous, a consequence of the fact that a substitution was possible. Nowlet’s use the problem as an example for the implementation of partial fraction decomposition.We proceed analogously to the last phase of the preceding example, always taking the numer-ator to be the most general polynomial whose degree is less than the degree of the irreduciblefactor in the denominator, but now taking separate summands for the powers of that irreduciblefactor. Thus we assume a decomposition of the form

x4 + 1x(x2 + 1)2 =

α

x+

βx + γ(x2 + 1

)2 +δx + η

x2 + 1

and multiply through by the denominator on the left, to obtain the polynomial identity

x4 + 1 = α(x2 + 1

)2+ (βx + γ)x + (δx + η)

(x2 + 1

)x

in which the identification of coefficients of like powers of x leads to the 5 equations

α + δ = 1η = 0

2α + β + δ = 0γ + η = 0

α = 1


implying that

(α, β, γ, δ, η) = (1,−2, 0, 0, 0) .x4 + 1

x(x2 + 1)2 =1x

+−2x

(x2 + 1

)2 .

∫x4 + 1

x(x2 + 1)2 dx =

∫1x

dx +

∫ −2x(x2 + 1

)2 dx

= ln |x| + 1x2 + 1

+ K ,

the same solution as was foruitously found earlier by using a substitution.

∫1

(x2 + 1)n dx. Should one of the partial fractions be of the formconstant(x2 + 1

)n , where n is

an integer greater than 1, we can begin the integration by a substitution u = arctan x, which

implies that du =1

x2 + 1dx, and

∫1(

x2 + 1)n dx =

∫1

sec2n−2 udu

=

∫cos2n−2 u du

=

∫ (1 + cos 2u

2

)n−1

du ,

which, in principle, we know how to integrate.

Example C.58 To evaluate the indefinite integral,∫

1(x2 + 1

)2 dx.

Let x = tan u, i.e., u = arctan x. Then du =dx

1 + x2 .

∫1

(x2 + 1

)2 dx =

∫1

sec2 udu

=

∫cos2 u du

=

∫1 + cos 2u

2du

=12

(u2

+sin 2u

4

)+ C


=arctan x

2+

sin u · cos u2

+ C

=arctan x

2+

tan u2 sec2 u

+ C

=arctan x

2+

x2(tan2 u + 1

) + C

=arctan x

2+

x2(x2 + 1

) + C

The presence of the arctangent function in the integral indicates that this could not have beenevaluated without some step equivalent to this use of the arctangent function in a substitution.

Rationalizing Substitutions. In some integrals in which the integrand is not originally arational function, it can be transformed into a rational integrand by an appropriate substitution.

Example C.59 ([7, Exercise 40, p. 504]) “Make a substitution to express the integrand as a

rational function, and then evaluate the integral∫

1

x − √x + 2dx.”

Solution: One substitution that suggests itself is u =√

x + 2. Under this substitution, u2 =

x + 2, dx = 2u du,∫

1

x − √x + 2dx =

∫2u

(u − 2)(u + 1)du

=23

∫ (2

u − 2+

1u + 1

)du

=23

(2 ln |u − 2| + ln |u + 1|) + C

=23

ln((√

x + 2 − 2)2 ∣∣∣∣√

x + 2 + 1∣∣∣∣)

+ C

=43

ln∣∣∣∣√

x + 2 − 2∣∣∣∣ +

23

ln(√

x + 2 + 1)

+ C ,

where I removed the absolute signs from the second term because√

x + 2+1 must be positive.

The arctangent substitution needs to be used when the quadratic, irreducible factors containa first degree term.

Example C.60 To integrate∫

dx4x2 + 4x + 3

.

Solution: Begin by dividing out the coefficient of the 2nd degree term, then completing thesquare:

4x2 + 4x + 3 = 4(x2 + x +

34

)= 4

(x +

12

)2

+12

= 2 + 4(x +

12

)2

.


It is clear that we can partially simplify the integral by a substitution of the form u = x +12

,

where du = dx. The integral becomes∫

du2 + 4u2 . In order to interpret the integral as an

arctangent, we need to make some type of scale change. for example, we could interpret it as∫

du2 + 4u2 =

12

∫du

1 + 2u2 =12

∫du

1 + (√

2 · u)2.

Thus an appropriate original substitution would be v =√

2 ·(x + 1

2

), where dv =

√2 dx, and

∫dx

4x2 + 4x + 3=

14

∫dx

x2 + x + 34

=1

4√

2

∫dv

v2

2 + 12

=1

2√

2

∫dv

v2 + 1

=1

2√

2arctan v + C

=1

2√

2arctan

(√2x +

1√2

)+ C ;

you should verify the correctness of this integration by differentiation.


C.19 Supplementary Notes for the Lecture of February 12th, 2010Release Date: Friday, February 12th, 2010

C.19.1 §7.5 Strategy for Integration

The textbook suggests a strategy for solving integration problems:

1. Simplify the integrand if possible.

2. Look for an obvious substitution.

3. Classify the integrand according to its form.

4. Try again. The preceding instructions are vague, depend on your experience and yourintuition, and are occasionally not appropriate, as sometimes the best way to attach aproblem will be obscured by these methods. So be prepared to try again. As for experi-ence, you need to work many problems to acquire it.

Table of integration formulæ Many of these formulæ are just recasts of familiar differenti-ation formulæ that you already know. But you should remember the integrals of tan x, cot x,sec x, csc x, even though you may know how to derive some of them.

Can We Integrate All Continuous Functions? Read this subsection. For most of the func-tions you will meet in this course it will be possible to integrate them. If that is not the case,it could be that we are asking you indirectly to do something else than to integrate. You arenot expected to be able to detect which functions cannot be integrated in terms of elementaryfunctions: this is a difficult problem even for a trained mathematician. “You may be assured,though, that the integrals in the...exercises are all elementary functions.”

7.5 Exercises

[1, Exercise 58, p. 489] Evaluate∫

x ln x√x2 − 1

dx.

Solution: A first impression is that the most complicated part of the integrand is thelogarithm, and I would like to dispose of it. One way to do that would be through asubstitution, but that could well render the denominator rather complicated. I propose totry integration by parts, and to assign u and v so that the ln x term is part of u. But howmuch of the integrand should be taken for u? I observe that if I take only the ln x term,it leaves a function that is easy to integrate, so it is an ideal first step:

u = ln x⇒ du =dxx


dv =x√

x2 − 1dx

But what is v? If you don’t see, use the substitution U = x2:∫

x√x2 − 1

dx =12

∫1√

U − 1dU .

You should be able to evaluate this last integral by sight; but, if you can’t, try the substi-tution V = U − 1:

v =12

∫1√

U − 1dU =

12

∫1√V

dV =√

V + C =√

x2 − 1 + C .

Hence ∫x ln x√x2 − 1

dx = (ln x)√

x2 − 1 −∫ √

x2 − 1x

dx .

The problem isn’t solved yet, but at least the logarithm is gone. Now I propose to try atrigonometric substitution to simplify the square root:

w = arcsecx⇒ x2 − 1 = tan2 w, x dx = sec2 w tan w dw :

∫ √x2 − 1

xdx =

∫ | tan w|sec w

· sec w tan w dw

=

∫tan2 w dw

=

∫(sec2 w − 1) dw

= tan w − w + C=√

sec2 w − 1 − arcsec x + C=√

x2 − 1 − arcsec x + C

from which we may conclude that∫

x ln x√x2 − 1

dx = (ln x)√

x2 − 1 −√

x2 − 1 + arcsec x + C1

This was not the only way to solve this problem, and it may not have been the best! ButI have written down the way I solved it first. One shorter, but less intuitive way wouldbe to use the substitution u =

√x2 − 1, which implies that du =

x√x2 − 1

dx. Then

∫x ln x√x2 − 1

dx =

∫ln√

u2 + 1 du =12

∫ln(u2 + 1) du .


Does this look simpler to you? It can be integrated by parts by taking u = ln(u2 + 1),

dv = du, so du = 2du

u2 + 1

12

∫ln(u2 + 1) du =

12

u ln(u2 + 1) −∫

u2

u2 + 1du .

The point is that the last integrand is rational, and we know how to integrate all suchfunctions! (Indeed, division of the denominator into the numerator yields a quotient of1 and a remainder of -1, so the integral can be evaluated as

∫u2

u2 + 1du =

∫1 du −

∫1

u2 + 1du

= u − arctan u + C=√

x2 + 1 − arctan√

x2 + 1 + C

Another strategy would be to use the same trigonometric inverse substitution immedi-ately: w = arcsec x⇒ dx = sec w tan w dw⇒

∫x ln x√x2 − 1

du =

∫ (sec2 w ln sec w

)dw .

In the last integral you should recognize ln sec w as being an antiderivative of tan w. Thissuggests using integration by parts with u = ln sec w and dv = sec2 w dw: du = tan w dw,v = tan w.∫ (

sec2 w ln sec w)

dw = (ln sec w)(tan w) −∫

tan2 w dw

= (ln sec w)(tan w) −∫

(sec2 w − 1) dw

= (ln sec w)(tan w) − tan w + w + C

= (ln x)√

x2 − 1 −√

x2 − 1 + arcsec x + C

In several places I have casually suppressed absolute value signs; these steps can bejustified, and are consequences of the way in which we defined the inverse functions.

Exercise C.1 To integrate∫ 2π

0

√1 + sin x dx.

Solution: This is an interesting integral which, at first glance, does not appear to fit into any ofthe families of integrals we have been studying. However, it can easily be seen that, except forthe determination of a sign, the integral is not very difficult to evaluate. But the sign questionis delicate, and even the previous edition of a well known text book overlooked this difficulty.In the Student Solution Manual to that textbook the answer was given to be −2

√1 − sin t + C,

and the following hint was given for integration:


“Multiply numerator and denominator of the integrand by√

1 − sin x.”

The hint is a good one, and does, indeed, lead to one way of solving the problem. Unfortu-nately, the answer that was given in that textbook was correct only within certain intervals.

Solution following the suggestion

∫ √1 + sin x dx =

∫ √1 + sin x ·

√1 − sin x√1 − sin x

dx =

∫ √1 − sin2 x1 − sin x

dx

=

∫ √cos2 x

1 − sin xdx =

∫ | cos x|√1 − sin x

dx

Had the absolute signs not been present, we could make the substitution u = sin x, subject towhich we would have du = cos x · dx

∫cos x√

1 − sin xdx =

∫1√

1 − udu

= −2√

1 − u + C= −2

√1 − sin x + C

Unfortunately, this function has derivative√

1 + sin x only when cos x is non-negative, i.e.,only when 4n−1

2 · π ≤ x ≤ 4n+12 · π, where n is any integer. For example, the value of the definite

integral∫ 2π

0

√1 + sin x dx is certainly not equal to

[−2√

1 − sin x]2π

0= 0 , since the integrand

is positive for most x, so the area under the curve y =√

1 + sin x must be positive; the correctvalue is

[−2√

1 − sin x] π

2

0+

[2√

1 − sin x] 3π

2

π2

+[−2√

1 − sin x]2π

3π2

=[2√

1 − sin x] 3π

2

π2

+[−2√

1 − sin x] π

2

−π2

= 2√

2 + 2√

2 = 4√

2 , 0

Other ways to find the indefinite integral While we cannot remove the sign difficultiesin this problem, we can show that the problem does, in fact, lend itself to a more systematicintegration — i.e., the hint given above is not really necessary. One way to see this is toremember that we have a trigonometric identity that expresses 1 + cos θ as a square. But, asthe trigonometric function given here is a sine, and not a cosine, one must first arrange for the


presence of a cosine. One way is as follows:

∫ √1 + sin x dx =

∫ √1 + cos

(π

2− x

)dx

=

∫ √2 cos2

(π

4− x

2

)dx

=√

2∫ ∣∣∣∣∣cos

(π

4− x

2

)∣∣∣∣∣ dx

Another approach, suggested by a student in this course several years ago, is to observe that

√1 + sin x =

√sin2 x

2+ cos2 x

2+ 2 sin

x2· cos

x2

=

∣∣∣∣∣ cosx2

+ sinx2

∣∣∣∣∣ .

To integrate this we need to determine the sign of the function inside the absolute signs. This

can be done by observing that it is equal to√

2 sin( x2

+π

4

), essentially the same function as

determined just above.A more systematic approach would have been to attempt to simplify the original integral

by the substitution u = sin x, which would imply that du = cos x dx, so

∫ √1 + sin x dx =

∫ √1 + u

cos xdu

= ±∫ √

1 + u√1 − u2

du = ±∫

1√1 − u

du

= ∓2√

1 − u + C = ∓2√

1 − sin x + C

where the sign still depends upon the interval in which x is located.

What, then, is one antiderivative of√

1 + sin x? We have found that, if we confineourselves to one interval of the form 4n+1

2 π < x < 4n+32 π, any antiderivative has the form

−2√

1 − sin x + C; and, if we confine ourselves to one interval of the form 4n+32 π < x <

4n+52 π, any antiderivative has the form 2

√1 − sin x + C. By choosing the constants to make

the function continuous (indeed, differentiable) we can patch such subfunctions together toform an antiderivative which is valid over an extended domain. The function f defined by the


following table is one such antiderivative:

x f (x)· · · · · ·

−32 π ≤ x < −1

2 π −4√

2 + 2√

1 − sin x−12 π ≤ x < 1

2π 0√

2 − 2√

1 − sin x12π ≤ x < 3

2π 0√

2 + 2√

1 − sin x32π ≤ x < 5

2π 4√

2 − 2√

1 − sin x52π ≤ x < 7

2π 4√

2 + 2√

1 − sin x72π ≤ x < 9

2π 8√

2 − 2√

1 − sin x92π ≤ x < 11

2 π 8√

2 + 2√

1 − sin x112 π ≤ x < 13

2 π 12√

2 − 2√

1 − sin x· · · · · ·

We can verify that the value of the definite integral∫ 2π

0

√1 + sin x dx is [ f (x)]2π

0 = f (2π) −f (0) =

(4√

2 − 2√

1 − sin 2π)−

(0√

2 − 2√

1 − sin 0)

= 4√

2.

C.19.2 §7.6 Integration Using Tables and Computer Algebra Systems (OMIT)

This section “is not examination material, but students are to try to solve the problems manu-ally.”

C.19.3 §7.7 Approximate Integration (OMIT)

This section is not part of the syllabus of this course.


C.20 Supplementary Notes for the Lecture of February 15th, 2010Release Date: Monday, February 15th, 2010

C.20.1 §7.8 Improper Integrals

Piecewise continuous integrands Consider a function f that is continuous everywhere in aninterval [a, b], including continuity from the right at b and from the left at a. For such functionswe have developed the theory of the definite integral, and the Fundamental Theorem applies.Now suppose that there is a point c such that a < c < b, where f has a jump discontinuity:limx→c−

f (x) and limx→c+

f (x) both exist, but are different. It is possible to define the integral of f asfollows: ∫ b

af (x) dx =

∫ c

af (x) dx +

∫ b

cf (x) dx ,

and it can be shown that the familiar properties of the definite integral hold here, with theexception of the Fundamental Theorem. We can proceed in the usual way with such integrals,until we need to actually evaluate them; then we must split them up at the jump discontinuitybefore we attempt to apply the Fundamental Theorem. More generally, this definition “works”for functions with many jump discontinuities.46 47

Other types of generalizations In this section we wish to consider other types of general-izations related to a condition of the original definition of the definite integral that fails to besatisfied. We will follow the terminology of the textbook, calling two types of “impropriety”Type 1 and Type48 2, but students should be aware that these terms are not in universal use. Thegeneral spirit of these definitions, and of the preceding generalization to jump discontinuities,is that the familiar properties of integrals proved for continuous functions should hold for thesebroader classes wherever they make sense. We shall still have to keep away from any situationthat might lead us to attempt to, for example, add +∞ to −∞, and any other undefinable oper-ations. Remember that, in this theory, you must rely on the definition, and not attempt to writedown what “makes sense”. The restrictions in some of these definitions are needed to avoidparadoxes elsewhere.

46We will not explore here what limits — if any — exist on the number of discontinuities.47We have already seen that there may be other situations where the splitting of an integral into pieces for

different parts of the domain may simplify the integration; but, where there is a jump discontinuity, the splittingis not by choice, it is by our definition of the meaning of the generalized symbol.

48Note that the terms Type 1 and Type 2 refer to a type of “impropriety” — not to a type of integral. Oneimproper integral could contain multiple instances of Type 2, and as many as 2 instances of an impropriety ofType 1.


Type 1: Infinite Intervals Our original definition of the definite integral was given for afinite interval. If we wish to speak of an integral where either or both limits are infinite, weneed to define what these are to mean. The definition we give is one that is consistent with thedefinition for finite intervals, and preserves those properties of the integral that are meaningfulwhen the limits before infinite. I repeat the boxed definition on [1, p. 509]:

Definition C.6 1.

∞∫

a

f (x) dx = limt→∞

t∫

a

f (x) dx provided the integral on the right exists for

all t ≥ a, and the limit exists as a finite number. The “improper” integral on the leftis then said to converge or to be convergent. If the limit does not exist, the improperintegral is divergent.

2. An analogous definition holds when the upper limit is −∞, or when the lower limit iseither of ±∞. Read the textbook.

3. When both limits of the definite integral are infinite, we define the value to be the sumof two integrals obtained by splitting the domain. It can be shown that it doesn’t matterwhere the line is split, the following definition will always give the same value:

∞∫

−∞

f (x) dx =

a∫

−∞

f (x) dx +

∞∫

a

f (x) dx .

Of course, the integrand f must have the entire real line R as its domain!

4. It is ESSENTIAL to understand that, in the definition of

∞∫

−∞

, TWO limits have to be de-

termined independently. IT IS NOT CORRECT TO CONSIDER ONLY limt→∞

t∫

−t

f (x) dx.

In this case one has to add two finite numbers; if either of the limits does not exist as a fi-nite number — and that includes being infinite — the sum is not finite, and the improperintegral does not exist.

Type 2: Discontinuous Integrands I have observed above how to cope with a finite, jumpdiscontinuity in the integrand. Here we are interested in other types of discontinuity, in partic-ular, discontinuities where the function has a vertical asymptote. If the discontinuity occurs atthe left end-point of the interval, then the value of the “improper” integral is defined by

b∫

a

f (x) dx = limt→a+

b∫

t

f (x) dx .


If the discontinuity is at the right end-point b, then the value of the “improper” integral isdefined by

b∫

a

f (x) dx = limt→b−

t∫

a

f (x) dx .

And, if the discontinuity occurs at a point c between a and b, then the definition is based onsplitting the integral into two parts:

b∫

a

f (x) dx =

c∫

a

f (x) dx +

b∫

c

f (x) dx ,

where both of the summands on the right are defined to be limits as above, and the two limitshave to be evaluated independently. It is important to understand that in this case it is notacceptable that the two limits be linked so that one may affect the other.

Example C.61 The improper integral

2∫

0

1x − 1

dx does not converge, since at least one of the

limits

limt→1−

t∫

0

1x − 1

dx , limt→1+

2∫

t

1x − 1

dx

does not exist; indeed, neither of them exists! We say that this improper integral diverges.

Without this severe definition we might find that some of the properties we wish the integralto possess might not be present.

Example C.62 ([1, Example 4, p. 511]) “For what values of p is the integral∫ ∞

1

1xp dx con-

vergent?”Solution:

∫ ∞

1

1xp dx = lim

t→∞

∫ t

1

1xp dx

=

limt→∞

[ln x]t1 when p = 1

limt→∞

11 − p

[x1−p

]t

1when p , 1

=

limt→∞

[ln t − 0] = ∞ when p = 11

1 − plimt→∞

[1

tp−1 − 1]

=1

p − 1when p > 1

11 − p

limt→∞

[t1−p − 1

]= ∞ when p < 1


This result will be needed in Chapter 11 in connection with the “p-series test” [1, p. 700]. In

the special case p = 1, it states that the area under the hyperbola y =1x

from x = 1 indefinitelyto the right is unbounded. By symmetry, that region should have the same area as the areabetween x = 0 and x = 1 under the same curve, and over the line y = 1. That area is the valueof the improper integral

∫ 1

0

1x

dx = limt→0+

∫ 1

t

1x

dx

= limt→0+

[ln x]1t

= − limt→0+

ln t = +∞

Example C.63 Consider the improper integrals∫ +∞

−∞

1x2 + 1

dx ,∫ +∞

−∞

xx2 + 1

dx .

In both of these cases we must split the interval of integration at a finite point, and considertwo improper integrals of Type 1 independently. In the first case an antiderivative is arctan x,so

∫ +∞

−∞

1x2 + 1

dx =

∫ 0

−∞

1x2 + 1

dx +

∫ +∞

0

1x2 + 1

dx

= lima→−∞

∫ 0

a

1x2 + 1

dx + limb→+∞

∫ b

0

1x2 + 1

dx

= lima→−∞

(0 − arctan a) + limb→+∞

(arctan b − 0)

= −(−π

2

)+π

2= π ,

and the improper integral is convergent. (We could have split the interval at any convenientpoint other than 0.)

But, in the second case, we have∫ +∞

−∞

xx2 + 1

dx =

∫ 0

−∞

xx2 + 1

dx +

∫ +∞

0

xx2 + 1

dx

= lima→−∞

∫ 0

a

xx2 + 1

dx + limb→+∞

∫ b

0

xx2 + 1

dx

= lima→−∞

(0 − ln(x2 + 1)

2

)+ lim

b→+∞

(ln(x2 + 1)

2− 0

).


Here we find that each of the limits is infinite: the integral is said to be divergent. In as-signing a meaning to a doubly infinite integral we do not permit +∞ and −∞ to be ap-proached at the same rate. That is, it is not permitted to interpret this integral as being equal

to lima→∞

a∫

−a

xx2 + 1

dx. (Here the finite, symmetric integral is 0, so the limit is 0. But we do not

give that meaning to the original, improper integral, which we insist diverges.

Extended Definition of the Integral The original definition in the textbook — in terms ofRiemann sums — was for continuous functions f , where the interval of integration was a finiteinterval [a.b]. This definition has been extended in several different ways.

1. First extension: to functions with a finite number of finite jump discontinuities. Thesame notation, viz.

b∫

a

f (x) dx

is used; but it now means the sum of integrals over the various (disjoint) subintervalswhere the function is continuous. Remember that continuity over a closed interval [a, b]means

• continuity at every point x such that a < x < b,

• continuity from the right at x = a; and

• continuity from the left at x = b. (When a = b the integral is defined to be 0, withno continuity restrictions.)

There is a need for an explicit definition here, since, where a function f defined ona ≤ x ≤ b has a jump discontinuity at c, where a < c < b, then the function will usuallybe continuous on one side of c, so, for the subinterval on the other side of c, there is stillthe need for a limiting operation of some kind, since our original definition did requirecontinuity of the integrand over the full closed interval of the integral. If it happens thatthe jump discontinuity is removable, then it can simply be ignored.)

2. Type 1 Improper Integral with one limit of integration equal to +∞. We define

+∞∫

a

f (x) dx = limb→+∞

b∫

a

f (x) dx ; and

a∫

+∞

f (x) dx = limb→+∞

a∫

b

f (x) dx .


3. Type 1 Improper Integral with one limit of integration equal to −∞. We define

b∫

−∞

f (x) dx = lima→−∞

b∫

a

f (x) dx ; and

−∞∫

b

f (x) dx = lima→−∞

a∫

b

f (x) dx .

4. Type 1 Improper Integral with one limit of integration equal to −∞ and one limitof integration equal to +∞. We define

+∞∫

−∞

f (x) dx =

a∫

−∞

f (x) dx +

+∞∫

a

f (x) dx and,

∫ −∞

+∞f (x) dx =

∫ −∞

af (x) dx +

∫ a

+∞f (x) dx ,

where a is any convenient point chosen to break the line into two semi-infinite rays. Inpractice one often chooses a = 0, but that is not required. It is not permitted to computethe two limits together — each of them must exist as a finite limit, independent of theother.

5. Type 2 Improper Integral where end-point a is a point of infinite discontinuity. Herewe “excise” the point a and define

b∫

a

f (x) dx = limc→a+

b∫

c

f (x) dx .

6. Type 2 Improper Integral where end-point b is a point of infinite discontinuity. Herewe “excise” the point b and define

b∫

a

f (x) dx = limc→b−

c∫

a

f (x) dx .

7. Type 2 Improper Integral where a point c such that a < c < b is a point of infinitediscontinuity. Here we “excise” the point c and define

b∫

a

f (x) dx =

c∫

a

f (x) dx +

b∫

c

f (x) dx .


This means that the integral is the sum of two separate limits, where the “bad” pointhas been excised at one end of each of the smaller intervals of integration. The twolimits must be computed separately, and both of them must exist (as finite limits) for theintegral to be said to converge. It is not permitted to compute the two limits at the sametime in a symmetric way.

With these extended definitions we can show that the familiar rules for definite integrals arestill operational, where they make sense. Because the definitions for improper integrals in-volve limits, we use the terminology of convergence and divergence, the same terminology wecould use in connection with other limits, and which we will see again when we study infinitesequences and series.

Comparison Test for Improper Integrals The definitions I have sketched concern limits ofthe values of certain integrals. Where we are unable to evaluate certain integrals directly, wecan still justify a comparison theorem similar to that we saw in connection with finite integrals:

Theorem C.64 (Comparison “Test” for Improper Integrals) Suppose that f and g are con-tinuous functions such that, on the interval a ≤ x ≤ ∞, 0 ≤ g(x) ≤ f (x). Then

∞∫

a

f (x) dx converges ⇒∞∫

a

g(x) dx converges

∞∫

a

g(x) dx diverges ⇒∞∫

a

f (x) dx diverges

Example C.65 Evaluate the definite integral

e3∫

1

dx

x√

ln x. Show that the integral is a conver-

gent, improper integral, and find its value.Solution: The integrand is not defined at x = 1, since one factor in the denominator is expressedin terms of ln x. For x > 1 the substitution u = ln x is valid. Hence

e3∫

1

dx

x√

ln x= lim

t→1+

e3∫

t

dx

x√

ln x

= limt→1+

3∫

ln t

1√u

du

= limt→1+

[2√

u]3

ln t

= limt→1+

[2√

3 − 2√

ln t]

= 2√

3 − 2 limt→1+

[√ln t

]= 2√

3 ,


since, as t → 1+, ln t → 0, so√

ln t → √0 = 0, by the continuity of the function√

t from theright at t = 1.

7.8 Exercises

[1, Exercise 18, p. 515] Determine whether the integral

∞∫

0

dzz2 + 3z + 2

is convergent or di-

vergent. Evaluate it if it is convergent.

Solution: Using standard methods of Partial Fractions we can show that∫

dzz2 + 3z + 2

=

∫ (1

z + 1− 1

z + 2

)dz

= ln |z + 1| − ln |z + 2| + C

= ln∣∣∣∣∣z + 1z + 2

∣∣∣∣∣ + C .

Hence∞∫

0

dzz2 + 3z + 2

= lima→∞

[ln

∣∣∣∣∣z + 1z + 2

∣∣∣∣∣]a

0

= lima→∞

∣∣∣∣∣∣ln

1 + 1a

1 + 2a

∣∣∣∣∣∣ − ln12

= ln 1 − ln12

= ln 2

Note that we could not have expressed

∞∫

0

dzz2 + 3z + 2

as the difference of the improper

integrals

∞∫

0

dzz + 1

and

∞∫

0

dzz + 2

, as both of these improper integrals are divergent and the

difference of their “values” is not defined.

[7, Exercise 52, p. 516] “Use the Comparison Theorem to determine whether the integral∞∫

1

x√1 + x6

dx is convergent or divergent.”

Solution: At first glance, this integral suggests a substitution u = x2. While that wouldsimplify its form, it would not enable us to integrate it immediately, and it is not nec-essary, since we can prove the convergence without this step. A simpler attack is to


observe that 1 + x6 > x6, so, for positive x,√

1 + x6 > x3,1√

1 + x6<

1x3 . Hence we can

consider the limit limt→∞

t∫

1

xx3 dx = lim

t→∞

[−1

x

]t

1= 1 − lim

t→∞1t1 = 1. The convergence of the

larger, improper integral implies the convergence of the given one.

[1, Exercise 75, p. 517] “Show that∞∫

0

x2e−x2dx =

12

∞∫

0

e−x2dx .”

Solution: This is an interesting question, because we cannot express an antiderivativeof e−x2

in terms of elementary functions. So, at first glance, one wonders how it will bepossible to work with these integrals. Before doing that, I must prove that the integralsare convergent — otherwise we don’t have any right to include them as numbers in anequation.

I note that

x ≥ 1 ⇒ −x2 ≤ −x (multiplying the inequality by a negative number)⇒ e−x2 ≤ e−x (exponential function is increasing)

⇒ for t ≥ 1

t∫

1

e−x2dx ≤

t∫

1

e−x dx =1e− 1

et

Hence, as t → ∞, the integral on the right approaches1e

, i.e., the improper integral

∞∫

1

e−xdx =1e.

By the [1, Comparison Theorem, p. 514], the improper integral

∞∫

1

e−x2dx is also con-

vergent, and is less than1e

. But we were considering the integral from 0, not from 1!The integral from 0 to 1 can be bounded in another way, since the reasoning given aboveis valid only for x ≥ 1. For example,

−x2 < 0 ⇒ e−x2< e0 = 1

⇒1∫

0

e−x2dx <

1∫

0

1 dx = 1


Hence∞∫

0

e−x2dx =

1∫

0

e−x2dx +

∞∫

1

e−x2dx < 1 +

1e.

This is not the exact value of the integral; in fact, it can be shown that

∞∫

0

e−x2dx =

√π

2,

but you are not expected to know this fact, nor how to prove it.

Now to prove the desired equality. Let us apply integration by parts with dv = xe−x2dx

and u = x, so v = −12e−x2

, and du = dx:

t∫

0

x2e−x2dx = −1

2xe−x2

]t

0+

12

t∫

0

e−x2dx

=t

2et2+

12

t∫

0

e−x2dx.

By l’Hospital’s Rule,

limt→∞

tet2

= limt→∞

12tet2

= 0 .

Hence

∞∫

0

x2e−x2dx = lim

t→∞

t∫

0

x2e−x2dx

=12

limt→∞

∫

y0te−x2

dx

=12

∞∫

0

e−x2dx .


C.21 Supplementary Notes for the Lecture of February 17th, 2010Release Date: Wednesday, February 17th, 2010

Textbook Chapter 8. FURTHER APPLICATIONS OF INTEGRATION.

C.21.1 §8.1 Arc Length

Just as with the earlier concepts of area, volume, and average, we are faced first with adoptinga definition that appears to have the properties that we associate with the concept, and, at thesame time, is workable in practice. The length of an arc will be defined to be the limit — ifthere is a limit — of the sum of the lengths of the sides of an approximating polygon formedby choosing points closer and closer together on the curve, and joining them by line segments.Note that we haven’t even defined what we mean in general by a curve, so the definition wegive will apply at first only to the graph of a function.

Suppose that we wish to find the length of the arc of the graph of y = f (x) between thepoints (a, f (a)) and (b, f (b)). We can subdivide the interval [a, b] on the x-axis by intermediatevertices, so that we have a sequence a = x0, x1, x2, . . . , xn = b of points on the x-axis. If wedefine

∆xi−1 = xi − xi−1, and

∆ f (xi−1) = ∆yi−1 = yi − yi−1 = f (xi) − f (xi−1)

then the distance between successive points Pi−1 = (xi−1, f (xi−1)) and Pi = (xi, f (xi)) is

|Pi−1Pi| =√

(xi − xi−1)2 + (yi − yi−1)2 =√

(∆xi)2 + (∆yi)2 ,

which square root can be expressed as either of the following:√

1 +

(∆ f (xi)

∆xi

)2

· ∆xi =

√1 +

(∆xi

∆yi

)2

· ∆yi .

Note that the orientation of the increments in x and y is not relevant, as the increments appearin these formulæ only as magnitudes. When we pass to the limit, as the “mesh” of pointsselected on the x-axis become closer and closer together, the first of these expressions, givesrise in the limit to the following integral representing the length of the arc:

b∫

a

√1 + ( f ′(x))2 dx.


If the curve is given by an equation in the form x = g(y), then we find the arc length from thepoint (g(k), k) to (g(`), `) to be

`∫

k

√1 + (g′(y))2 dy.

When the function whose graph is y = f (x) is invertible, both formulæ are applicable, and theygive the same length.49

Evaluation of these integrals can often require an approximation method, as the integrandstend to be of types for which a function expressible in terms of elementary functions is un-available. For that reason the problems that one meets in calculus books are often confined toa small set of functions for which antiderivatives can be found.

The Arc Length Function. If we fix a point on a curve, we can then define a functionthat expresses distance along the curve from the fixed point. This distance is expressed as anintegral with a variable upper limit, and is signed, so that, in effect, we have parameterized thecurve with a variable — usually denoted by the symbol s — uniquely denoting the position ofa point on a path along the curve. This practice differs from that employed when we evaluatethe length of the arc between two points, where only the magnitude is of interest. I have written“...on a path along the curve” rather than “...on the curve”, because we shall be generalizingthese ideas in [1, Chapter 10], to consider curves that are not the graphs of functions; in thosegeneralizations a curve may cross itself, and the same point could be traversed more than onceby a point whose path we are studying: in that case it is the length of the path that will begiven by the arc length function.

Example C.66 Circumference of a circle. What is the circumference of the circle x2+y2 = R2

(where R > 0)?Solution: Since the equation given is not the graph of a function (because the curve crossessome vertical lines more than once), let’s find the length of the upper arc from x = −R to

x = +R, and double it. This is given by the function f (x) =√

R2 − x2 = R

√1 −

( xR

)2.

y′ = −xR√

1 −(

xR

)2

1 +(y′)2

=1

1 −(

xR

)2

49Passage between the two forms can be seen to result from the change of variable given by y = f (x).


Circumference = 2

R∫

−R

1√1 −

(xR

)2dx

= 4

R∫

0

1√1 −

(xR

)2dx

since the integrand is even, and the interval symmetric around 0

= 4 limt→R−

t∫

0

1√1 −

(xR

)2dx

since the integral has a Type 2 impropriety at x = R

= 4R limt→R−

tR∫

0

1√1 − u2

du

under the substitution u =xR

= 4R lima→1−

a∫

0

1√1 − u2

du

under the substitution a =tR

= 4R lima→1−

arcsin a∫

arcsin 0

cos v| cos v| dv

under the substitution v = arcsin u

= 4R lima→1−

arcsin a∫

0

1 dv

since cos v is non-negative for −π2 ≤ v ≤ π2

= 4R lima→1−

[v]arcsin a0

= 4R arcsin 1 by continuity of arcsin

= 4R · π2

= 2πR .

Of course, we didn’t need to apply this last substitution v = arcsin u because we know two


antiderivatives of1√

1 − u2:

4R limt→R−

tR∫

0

1√1 − u2

du = 4R limt→R−

[arcsin u]tR0

= 4R limt→R−

[arcsin

tR− 0

]

= 4R [arcsin 1 − 0] by continuity of arcsin

= 4R[π

2− 0

]= 2πR .

Example C.67 ([7, Exercise 8, p. 552]) Find the length of the arc y =x2

2− ln x

4, (2 ≤ x ≤ 4).

Solution: Note the way in which the information is presented: we need a description of theunderlying curve, here given by an equation, and specifications of the portion of the curvewhose length is to be determined, here given by an interval 2 ≤ x ≤ 4 or in the alternativenotation x ∈ [2, 4]. Only the absolute value of the length is of interest, so we need not becareful about which of the end-points 2, 4 is in which limit of the integral; alternatively, it isthe absolute value of the integral that we seek.

y =x2

2− ln x

4⇒ dy

dx= x − 1

4x

⇒√

1 + (y′)2 =

√1 +

(x2 − 1

2+

116x2

)

⇒√

1 + (y′)2 =

√x2 +

12

+1

16x2 =

∣∣∣∣∣x +14x

∣∣∣∣∣

Hence the arc length between x = 2 and x = 4 (where the function∣∣∣∣∣x +

14x

∣∣∣∣∣ is equal to x +14x

)

is

4∫

2

∣∣∣∣∣x +14x

∣∣∣∣∣ dx =

4∫

2

(x +

14x

)dx

=

[x2

2+

ln x4

]4

2

= 8 +2 ln 2

4− 2 − ln 2

4= 6 +

ln 24

= 6 + ln4√2 .


Example C.68 ([7, Exercise 12, p. 552]) Find the length of the curve y = ln x for 1 ≤ x ≤√

3.Solution: Let’s attack this problem by integrating along either axis.

Integrating along the x-axis:

y = ln x ⇒ dydx

=1x

⇒√

1 +

(dydx

)2

=

√1 +

1x2

Hence

arc length =

√3∫

1

√1 +

1x2 dx .

To complete the integration, one approach is to try the substitution u =√

x2 + 1 tosimplify the integral. Then

x dx = u du√3∫

1

√1 +

1x2 dx =

2∫

√2

u2

u2 − 1du

=

2∫

√2

1 +

12

u − 1−

12

u + 1

du

=

u + ln

√u − 1u + 1

2

√2

= 2 −√

2 + ln1√3− ln

√ √2 − 1√2 + 1

= 2 −√

2 − ln 32− ln

1√2 + 1

= 2 −√

2 − ln 32

+ ln(√

2 + 1) , etc.

Integrating along the y-axis: Here the curve can be rewritten as x = ey, whose length is tobe found for 0 ≤ y ≤ ln 3

2 .

arc length =

ln 32∫

0

√1 + e2y dy


=

2∫

√2

v2

v2 − 1dv

under substitution v =√

1 + e2y

=

2∫

√2

1 +

12

v − 1−

12

v + 1

dv

=

[v +

12

ln∣∣∣∣∣v − 1v + 1

∣∣∣∣∣]2

√2

= 2 −√

2 − ln 32− 1

2ln

√2 − 1√2 + 1

= 2 −√

2 − ln 32− 1

2ln

√

2 − 1√2 + 1

·√

2 − 1√2 − 1

= 2 −√

2 − ln 32− 1

2ln

(√2 − 1

)2

= 2 −√

2 − ln 32

+ ln(√

2 + 1) .

Example C.69 ([7, Exercise 14, p. 552]) Find the length of the curve y2 = 4x, 0 ≤ y ≤ 2.Solution: Since x =

y2

4 , dxdy =

y2 , and the arc length is the integral

2∫

0

√1 +

y2

4dy =

π4∫

0

sec v · 2 sec2 v dv

under the substitution y = 2 tan v, i.e., v = arctan y2

= [tan v · sec v + ln | sec v + tan v|] π40=√

2 + ln∣∣∣∣√

2 + 1∣∣∣∣ .

(I have applied the reduction formula for∫

sec3 v dv determined in [1, p. 458, Exercise 50,§7.1].)

8.1 Exercises

[1, Exercise 12, p. 530] Find the length of the curve y = ln(cos x) for 0 ≤ x ≤ π

3.


Solution: y′ = − tan x⇒√

1 + (y′)2 = | sec x|.

Arc length =

π3∫

0

| sec x| dx

=

π3∫

0

sec x dx

= [ln | sec x + tan x|]π30

= ln |2 +√

3| − ln |1 + 0| = ln(2 +√

3) .

C.21.2 §8.2 Area of a Surface of Revolution

I develop a formula for the area of a surface of revolution by observing that an element of arcof length ∆s will, when rotated about an axis whose distance from the element is r, generatean element of surface whose area is approximately 2πr · ∆s. Remembering that

∆s =

√1 +

(∆ f (x)

∆x

)2

· ∆x =

√1 +

(∆x∆y

)2

· ∆y .

we can integrate with respect to either x or y as the conditions of the problem demand (providedthe derivative exists).

(to be continued)


C.22 Supplementary Notes for the Lecture of February 19th, 2010Release Date: Friday, February 19th, 2010; corrected 10 March, 2010

subject to further correction

C.22.1 §8.2 Area of a Surface of Revolution (conclusion)

8.2 Exercises

[1, Exercise 12, p. 537] Find the area of the surface obtained by rotating the curve x = 1+2y2

(1 ≤ y ≤ 2) about the x-axis.

Solution:

x = 1 + 2y2 ⇒ dxdy

= 4y

⇒√

1 +

(dxdy

)2

=√

1 + (4y)2

⇒ Surface Area = 2π

2∫

1

y√

1 + (4y)2 dy

=

[π

24

(1 + 16y2

) 32]2

1

=π

24·(65√

65 − 17√

17)

[1, Exercise 22, p. 537] ...Find the area of the surface obtained by rotating the curve y =√x2 + 1 (0 ≤ x ≤ 3) about the x-axis.

Solution:

y =√

x2 + 1

⇒ y′ =x√

x2 + 1

⇒√

1 + (y′)2 =

√1 + 2x2

√1 + x2

⇒ Area of revolution =

3∫

0

2π√

1 + x2 ·√

1 + 2x2

√1 + x2

dx

= 2π

3∫

0

√1 + 2x2 dx



=2π√

2

arctan 3√

2∫

0

sec3 θ dθ

under the substitution θ = arctan(x√

2)

=π√2

arctan 3√

2∫

0

sec θ dθ +π√2

[tan θ · sec θ]arctan 3√

20

by the reduction formula [1, Exercise 50, p. 458]

=π√2

[tan θ · sec θ + ln | sec θ + tan θ|]arctan 3√

20

= π

[3√

19 +1√2

ln(√

19 + 3√

2)].

Note that the textbook suggested the use of either a table of integrals or a computeralgebra system, but that neither was needed, as the solution of this problem is wellwithin the abilities of a student in this course (if she has time to do the calculations).

[1, Exercise 24, p. 537] Find the area of the surface obtained by rotating the curve y = ln(x+1)(0 ≤ x ≤ 1) about the y-axis.

Solution:

y = ln(x + 1) ⇒ dy =dx

x + 1

⇒√

1 + (y′)2 =

√1 +

1(x + 1)2

Area of surface = 2π

1∫

0

x

√1 +

1(x + 1)2 dx

= 2π

2∫

1

(u − 1)

√1 +

1u2 du

under the substitution u = x + 1

= 2π

2∫

1

√1 + u2 du − 2π

2∫

1

√1 + u2

udu

2∫

1

√1 + u2 du =

arctan 2∫

π4

sec3 θ dθ


under the substitution θ = arctan u

=12

[tan θ · sec θ + ln | sec θ + tan θ|]arctan 2π4

by the reduction formula [1, Exercise 50, p. 458]

=12

[(2√

5 + ln(√

5 + 2))−

(√2 + ln(

√2 + 1)

)]

2∫

1

√1 + u2

udu =

arctan 2∫

π4

(sec θ · tan θ + csc θ) dθ

under the substitution θ = arctan u= [sec θ + ln | csc θ − cot θ|]arctan 2

π4

=

√

5 + ln√

5 − 12

−

(√2 + ln(

√2 − 1)

)etc.

Here again the integral does not require special software or tables, just patience.

[1, Exercise 4, p. 562] 1. The curve y = x2, (0 ≤ x ≤ 1), is rotated about the y-axis. Findthe area of the resulting surface.

2. Find the area of the surface obtained by rotating the curve in part 1 about the x-axis.

Solution:

1. Sincedydx

= 2x,

Area = 2π∫ 1

0x√

1 + 4x2 dx

= 2π[23· 1

8·(1 + 4x2

) 32

]1

0

=π

6

(5

32 − 1

).

2. Here Area = 2π

1∫

0

x2√

1 + 4x2 dx . To simplify this integral I begin with the sub-

stitution 2x = tan θ, i.e., θ = arctan 2x. Then 2 dx = sec2 θ dθ, so

Area = 2π

arctan 2∫

0

tan2 θ

4· sec θ · sec2 θ

2dθ


=π

4

arctan 2∫

0

tan2 θ · sec3 θ dθ

=π

4

arctan 2∫

0

(sec5 θ − sec3 θ

)dθ .

Earlier, in Example C.42 on 3087 I had solved [1, Exercise 50, p. 458], provinga reduction formula (Equation (62))for integrals of positive integer powers of thesecant:

∫secn x dx =

1n − 1

secn−2 x · tan x +n − 2n − 1

∫secn−2 x dx .

From this formula we see that∫

sec3 θ dθ =12

sec θ · tan θ +12

∫sec θ dθ

=12


ln | sec θ + tan θ| + C∫

sec5 θ dθ =14

sec3 · tan θ +34

∫sec3 θ dθ

=14

sec3 · tan θ +34

(12


ln | sec θ + tan θ|)

+ C

=14

sec3 · tan θ +38


ln | sec θ + tan θ| + C

Note that the value of these antiderivatives (with C = 0) at θ = 0 is 0, and thatsec arctan 2 =

√5. Hence

Area =π

4

[14

sec3 θ · tan θ − 18

sec θ · tan θ − 18

ln | sec θ + tan θ|]arctan 2

0

=π

32

(2 · 5 3

2 · 2 − 512 · 2 − ln

∣∣∣∣√

5 + 2∣∣∣∣)

= π

9

16

√5 −

ln(√

5 + 2)

32

.

C.22.2 §8.3 Applications to Physics and Engineering (OMIT)

Omit this section. (But read it if you are majoring in Physics or Engineering!)


C.22.3 §8.4 Applications to Economics and Biology (OMIT)

Omit this section. (But read it if you are majoring into economics or biology.)

C.22.4 §8.5 Probability (OMIT)

Omit this section.

Textbook Chapter 9. DIFFERENTIAL EQUATIONS. (OMIT)

No parts of this chapter are included in the syllabus.


C.23 Supplementary Notes for the Lecture of March 01st, 2010Release Date: Monday, March 01st, 2010

Important Announcement

In earlier versions of the timetable for Section 1 of MATH 140 2010 01, thetimes shown for Quizzes Q3, and Q4 were not correct. Quiz Q3 will be ad-ministered during the week March 08-12, and Quiz Q3 will be administeredduring the week March 22–26. The timetable has now been corrected.

Textbook Chapter 10. PARAMETRIC EQUATIONS AND POLARCOORDINATES.

C.23.1 §10.1 Curves Defined by Parametric Equations

A parameter is just a variable. When we call a variable by this term we usually are thinkingof a function or set of functions involving the variable as representing a family of objects. Inthis first contact, the family will be a set of points. We will be taking the variable to be areal variable, and so it is natural to consider not only the family of points on the graphs of afunction, but also the way in which the points are generated by the assignment of real numbersto the parameter as representing a point moving along the curve. We can name the parameterwith any available symbol. Often we use the letter t; and a common use for this representationis to treat t as time, so that the curve can be thought of as the trajectory of (i.e., the pathtraced out by) a moving point. If we adopt this point of view, and if the parameter values arechosen from an interval a ≤ t ≤ b, then we can speak of the curve (x(t), y(t)), and can thinkof (x(a), y(a)) and (x(b), y(b)) as, respectively, the initial and terminal points. Curves givenparametrically in this way need not be graphs of functions: a curve may cross vertical linesmore than once, and may even cross itself, possibly more than once.

Graphing Devices This subsection may be omitted, as, in this course, we shall not be con-cerned with the use of graphing devices.

Can the graph of a function, given non-parametrically, be expressed in parametric form?The curve y = f (x) can be expressed in parametric form as, for example, x = t, y = f (t). Butthere are infinitely other ways of expressing it parametrically, for example

x = t3, y = f (t3),


but not necessarilyx = t2, y = f (t2),

since the latter would include only the points with non-negative abscissæ.

Parametric vs. nonparametric representation of a curve When we represent a curve inparametric form, the parametrization sometimes contains information beyond what is availablein a non-parametric representation. Often we can see a dynamic way of actually tracing outthe curve by allowing the parameter to range through its domain. So, for example, the curve

x = cos ty = sin t

can be thought of as being traced out by a point that moves around the unit circle centred atthe origin, starting at the point (1, 0) at time t = 0, counterclockwise at a rate of 1 radian perunit time as t increases (and clockwise at a rate of 1 radian per unit time as t decreases). If wecompare the curve with

x = cos 2ty = sin 2t

we see that both trace out the same circle, covering the curve infinitely often, but the secondcurve moves twice as fast. If we wanted to eliminate the multiple covering, we could includethe inequalities 0 ≤ t < 0 in the first case, or 0 ≤ t < π

2 in the second.The standard way to transform from parametric to non-parametric equations is to eliminate

the parameter “between” the equations, which can be interpreted as solving one equation forthe parameter, and substituting that value into the second equation. However, this operationsometimes “loses information”.

“Cartesian” representation of curves Where the textbook speaks of a Cartesian represen-tation, I prefer to speak of a non-parametric representation. All the representations for curvesconsidered in these two sections are Cartesian, since they all refer to the system of represen-tation associated with Rene Descartes. In the following two sections we will be consideringcurves represented in another way — in the so called polar representation.

Example C.70 [7, Exercise 13, p. 656] asks you to find a Cartesian equation of the curvex = cos2 θ, y = sin2 θ,

(−π2 < θ < π

2

). We can eliminate θ by solving one equation for θ and

substituting into the second; or, more elegantly, by adding the two equations together, obtainingan equation that does not involve θ:

x + y = 1


But this equation does not convey all of the information we started with. One type of infor-mation that has been lost is the fact that both x and y, being squares, are non-negative. Thusit is not the whole line x + y = 1 that is equivalent, but only the line segment joining thepoints (0, 1) and (1, 0). This feature is essential, since the doubly-infinite line is not the curverepresented by the parametric equations. Another type of information that is contained in theparametrization is the way in which this line segment is traversed. The point is moving backand forth between the point (1, 0) and the point (0, 1), covering the open segment in a parame-ter interval of length π

4 . Distance is not covered at a constant speed — the point moves fastestin the middle of the segment; but, so far, we are not interested in how fast the point is moving.

There is more than one correct way to describe this curve, but giving the equation x +y = 1is not enough.

Example C.71 [7, Exercise 17, p. 656]

(a) Eliminate the parameter to find a Cartesian equation of the curve.

(b) Sketch the curve and indicate with an arrow the direction in which the curve is traced asthe parameter increases:

x = cosh t y = sinh t

For every value of t a point with these coordinates can be seen to lie on the curve x2 − y2 = 1,which is a hyperbola with two branches, one opening to the right, and passing through the point(1, 0), the other opening to the left, and passing through the point (−1, 0); the curves are bothasymptotic to the lines y = ±x. But, since the hyperbolic cosine is positive, the parametrizationapplies only to the right branch of the hyperbola: the curve comes in from −∞ from below,passes through (1, 0) at t = 0, and the moves off along the upper half as t → +∞. So one wayto describe the curve non-parametrically is

x2 − y2 = 1x ≥ 0.

This curve may be parameterized in other ways. For example, we could represent it as

x = sec ty = tan t

(cf. [7, Exercise 14, p. 656]) but this time we need to restrict the values the parameter maytake, for example by

−π2< θ <

π

2.


The Cycloid In this subsection the textbook describes the construction of this interestingcurve. You are not expected to remember specific properties of this curve, nor its history.However, you should be able to work with this curve if the parametric equations are givento you, in the same way as you would be expected to be able to work with any reasonablecurve given parametrically. The particular parametrization developed in the textbook is x =

r(θ − sin θ), y = r(1 − cos θ).

Families of Parametric Curves Here the author considers a family of curves given para-metrically, in which the various members of the family all have a similar equation obtainedby assigning different values to a variable in the parametric equations. That variable is alsocalled a parameter, and the family of curves could be called a parametric family of curvesrepresented by parametric equations. Of course, the parameter that represents the family ofcurves is not the same as the parameter that represents the family of points on any specificcurve. For example, x2 + y2 = a2 could be considered a parametric family of curves (circles)given in non-parametric form; we could represent the same family in parametric form by

x = a cos ty = a sin t

where the parameter t represents position on a specific curve, and the parameter a representsthe different curves in the family. We could also interpret the equation by switching the rolesof the parameters: the curve

x = a cos t y = a sin t (78)

with t constant and a variable represents the points on the line through the origin inclined tothe positive x-axis by an angle of t radians. This point of view will become important in [1,§10.3].

Here again, you are expected to be able to work with reasonable families of parametriccurves, but not to know specific properties of those families (with the exception of the obviousparametric equations for familiar curves, like the circle).

Example C.72 Equation of a line in the plane. The line in the plane through the point(x0, y0) and with slope m has non-parametric equation y = y0 + m(x− x0). It can be representedparametrically in infinitely many ways. If we choose to relate the parameter to distance alongthe line, one can show that the following equations represent the line:

x = x0 + ty = y0 + mt

Check that the line segment joining any two points on this line has slopem1

= m. (Take twopoints with coordinates (x0 + a, y0 + ma) and (x0 + b, y0 + mb).)


C.23.2 §10.2 Calculus with Parametric Curves

Tangents If a curve is given parametrically by (x, y) = (x(t), y(t)), then (subject to certainconditions) we can differentiate y with respect to x by passing through an intermediate variablet. Recalling that

dxdt· dt

dx=

dxdx

= 1 ,

we have

dydx

=dydt· dt

dx

=dydt· 1

dxdt

so we can determine the first derivativedydx

in terms of the derivatives of the functions f and

g. This calculation can be extended to the second derivatived2ydx2 , although the expressions are

not as pretty:

d2ydx2 =

ddx

(dydx

)

=ddt

(dydx

)· dt

dx

=ddt

dydtdxdt

· 1

dxdt

=

d2ydt2 ·

dxdt− d2x

dt2 ·dydt(

dxdt

)2

· 1

dxdt

=

d2ydt2 ·

dxdt− d2x

dt2 ·dydt(

dxdt

)3

Rather than substituting in this formula, memorized, you are advised to be able to carry outthis computation for a specific parameterized curve. It enables us to study the concavity of


a parameterized curve, and to apply the 2nd derivative test if necessary in an optimizationproblem.

At this point in the lecture I began to discuss the curve which is featured in [1, Exercise 6,p. 636]. The discussion will be continued at the next lecture.


C.24 Supplementary Notes for the Lecture of March 03rd, 2010Release Date: Wednesday, March 03rd, 2010

C.24.1 §10.2 Calculus with Parametric Curves (continued)

Example C.73 (cf. [1, Exercise 6, p. 636]) (see Figure 7 on page 3158) Let’s investigate the

1

0

-2

0.5

-0.5

-1.5

-1

1.510.50-1 -0.5-1.5

Figure 7: The curve x = cos θ + sin 2θ, y = sin θ + cos 2θ

curve given parametrically by

x = cos θ + sin 2θ (79)y = sin θ + cos 2θ (80)


Differentiating the parameterizing functions, we obtain

dxdθ

= − sin θ + 2 cos 2θ

= −4 sin2 θ − sin θ + 2

= −4(sin2 θ +

14

sin θ − 12

)

= −4(sin θ +

18

)2

− 3364

= −4sin θ +

1 +√

338

sin θ +

1 − √338

dydθ

= cos θ − 2 sin 2θ

= −4 cos θ(sin θ − 1

4

)

The textbook asks us to find an equation for the tangent to the curve at the point with parameter

value θ = 0. We find that, when θ = 0,(dxdθ,

dydθ

)= (2, 1), so the slope of the tangent at the

point (x(0), y(0)) = (1, 1) is 12 , and an equation for the tangent is

y = 1 +12

(x − 1)

or x − 2y + 1 = 0.We can now use the same curve to illustrate some of the other theory of this section. For

example, we can determine where the curve is horizontal, by solving the equationdydx

= 0,

which implies thatdydθ

= 0. We find that this happens when

cos θ = 0 or sin θ =14,

which we could proceed to solve. We can also determine where the curve is vertical, by

determining wheredxdθ

= 0; this happens when

sin θ =−1 ± √33

8

All of these equations could be solved. Since the curve is expressed entirely in terms ofsin θ, cos θ, sin 2θ, cos 2θ, and these functions are periodic, repeating themselves after θ passes


through an interval of length 2π, we can see the whole curve by confining θ to the interval

0 ≤ θ ≤ 2π. In that interval the cosine vanishes when θ =π

2,

3π2

, and sin θ =14

when

θ = arcsin14

and π − arcsin14

. These are the 4 points where the curve is horizontal; one ofthose point is the origin, because the curve passes through the origin several times, one of thosetimes with a horizontal tangent. The curve has the shape of a “3-leafed rose” or a “trefoil” (3-leafed clover) centred at the origin, with one petal bisected by the y-axis. We will see othercurves with this shape when we study polar coordinates in [1, §10.3].

For the same curve given parametrically by equations (79), (80), let’s consider the follow-ing question, similar to [1, Exercise 25, p. 636]: “Show that the curve..has (several)...tangentsat (x, y) = (0, 0), and find their equations.”Solution: If we set the coordinates equal to zero and solve, we can simplify the resultingequations, to obtain:

cos θ · (1 + 2 sin θ) = 0(sin θ − 1) · (1 + 2 sin θ) = 0 .

The curve passes through the origin whenever both of the equations are satisfied. This meansthat either

sin θ = −12, (81)

or both50 of the following equations must hold:

cos θ = 0 (82)sin θ = 1 . (83)

For θ between 0 and 2π, these last equations are satisfied when θ =π

2; equation (81) is satisfied

when θ =7π6,

11π6

. The curve, which has the shape of a 3-leafed clover or a “3-leafed rose”,passes through the origin 3 times, and we can find the slopes of the tangents in the usual way,

by taking the ratio

dydθdxdθ

.

50Note that the second equation implies the first, but the first does not imply the second; we could thus haveshown only the second equation.


Areas. Consider the arc of the curve x = f (t), y = g(t) determined by points with parametervalue t between α ≤ t ≤ β. Since

∫ β

α

g(t) · f ′(t) dt =

∫ f (β)

f (α)y · dx ,

the first integral represents the area between the arc and the x-axis. Remember, though, thatthe curve need no longer be the graph of a function, so it could cross vertical lines more thanonce. This means that the area could be “folded”, and there might be portions that could becounted negatively and canceling the portions that you are interested in. For that reason youshould use this integral only where you are clear about the shape of the region whose area youare finding, and there might be situations where the region should be broken up into parts andthe areas of the parts found separately.

Arc Length. In a similar way to the preceding, we can argue that the length of the arcα ≤ t ≤ β of the curve x = u(t), y = v(t) between α ≤ t ≤ β is given by the integral

∫ β

α

√(dxdt

)2

+

(dydt

)2

dt .

Note that, when we consider the parametrization

x = t y = f (t) (a ≤ x ≤ b)

of the graph of the function y = f (x) between the points (a, f (a)) and (b, f (b)), this reducesto the formula we derived earlier. Here again, be careful that you are finding the length of thecurve that you intend. In this case there cannot be any cancellation, since the integrand is asquare root, which cannot be negative. If you obtain a negative length, it could simply be aconsequence of the direction in which you have parameterized the curve, which is harmless;or of an error you have made in you calculations, which is serious.

Surface Area. We can also adapt, in the obvious ways, our previous formulæ for area ofsurfaces of revolution.

Example C.74 Let’s determine a formula for the surface area of a doughnut. Suppose that thedoughnut is generated by the curve

x = R + r cos θy = 0 + r sin θ


where 0 ≤ θ ≤ 2π, where this circle, centred at the point (x, y) = (R, 0), is revolved around they-axis. Assume R ≥ r.

Area of revolution =

∫ 2π

02π(R + r cos θ)

√(dxdθ

)2

+

(dydθ

)2

dθ

=

∫ 2π

02π(R + r cos θ)

√r2 dθ

= 2πr [Rθ + r sin θ]2π0 = 2πr[R · 2π − R · 0] = 4π2rR.

Volumes. The textbook appears to say little about this, but here again the earlier formulæ canbe adapted, to determine, for example, the volume of revolution generated by a given curveabout a given axis.

10.2 Exercises

[1, Exercise 54, p. 637] “Find the total length of the astroid x = a cos3 t, y = a sin3 t.”

Solution: This curve is generated over an interval of length 2π: we can take 0 ≤ t ≤ 2π.the curve looks like a deformed circle, that has been pinched towards the centre at thepoints away from where it crosses the coordinate axes.

Total length =

∫ 2π

0

√(dxdt

)2

+

(dydt

)2

dt

=

∫ 2π

0

√(−3a cos2 · sin t

)2+

(3a sin2 t · cos t

)2dt

=

∫ 2π

03|a| · | cos t · sin t| dt

By symmetry we can find the length of one quarter of the curve:

Total length = 4∫ π

2

03|a| · | cos t · sin t| dt

= 4|a|∫ π

2

03 cos t · sin t dt

= 6|a|[sin2 t

] π2

0= 6|a| .

[1, Exercise 48, p. 637] Find the length of the loop of the curve x = 3t − t3, y = 3t2.


Solution: Where does this loop begin and end? We need to find all parameter valuest1, t2 where t1 , t2 but

3t1 − t31 = 3t2 − t3

2

3t21 = 3t2

2

The equations can be simplified to

(t1 − t2)(3 − t21 − t1t2 − t2

2) = 0(t1 − t2)(t1 + t2) = 0

When we know that t1 , t2, the equations further simplify to the system

3 − t21 − t1t2 − t2

2 = 0t1 + t2 = 0

which imply that t1 = −t2 = ±√3. Thus these two curves intersect in the points thathave parameter values ±√3.

Since√(

dxdt

)2

+

(dydt

)2

=

√(3 − 3t2)2

+ (6t)2

= 3(1 + t2) ,

the length of the arc of the loop is

∫ √3

−√33(1 + t2) dt = 2

[3t + t3

]√3

0= 12

√3 ,

since the integrand is an odd function, and the limits of integration are symmetricallylocated around t = 0.


C.25 Supplementary Notes for the Lecture of March 05th, 2010Release Date: Friday, March 05th, 2010

C.25.1 §10.2 Calculus with Parametric Curves (conclusion)

One topic remaining, not discussed in the previous lectures is summarized in the followingsubsection, which was included earlier in these notes on page 3160:

Areas. Consider the arc of the curve x = f (t), y = g(t) determined by pointswith parameter value t between α ≤ t ≤ β. Since

β∫

α

g(t) · f ′(t) dt =

f (β)∫

f (α)

y · dx ,

the first integral represents the area between the arc and the x-axis. Remember,though, that the curve need no longer be the graph of a function, so it could crossvertical lines more than once. This means that the area could be “folded”, and theremight be portions that could be counted negatively and canceling the portions thatyou are interested in. For that reason you should use this integral only where youare clear about the shape of the region whose area you are finding, and there mightbe situations where the region should be broken up into parts and the areas of theparts found separately.

I illustrate this theory with the following example:

10.2 Exercises (conclusion)

[1, Exercise 34, p. 637] “Find the area of the region enclosed by the astroid x = a cos3 θ,y = a sin3 θ.” This is the same curve considered above in [1, Exercise 54, p. 637]. Let’sconsider the arch in the first quadrant. The area will be 4 times the area in the firstquadrant, which is

a∫

0

y dx =

∣∣∣∣∣∣∣∣∣

π2∫

0

y(θ) · dx(θ)dθ

dθ

∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣∣

π2∫

0

a sin3 θ ·(−3a cos2 θ · (− sin θ)

)dθ

∣∣∣∣∣∣∣∣∣.

You know one way to evaluate an integral of this type — by replacing sin2 θ and cos2 θ

respectively by1 − cos 2θ

2and

1 + cos 2θ2

. A variant of that method is as follows:∣∣∣∣∣∣∣∣∣

π2∫

0

a sin3 θ ·(−3a cos2 θ · (− sin θ)

)dθ

∣∣∣∣∣∣∣∣∣= 3a2

π2∫

0

a sin4 θ · cos2 θ dθ


= 3a2

π2∫

0

1 − cos 2θ2

·(sin 2θ

2

)2

dθ

=3a2

8

π2∫

0

(sin2 2θ − sin2 2θ · cos 2θ

)dθ

=3a2

8

π2∫

0

(1 − cos 4θ

2− sin2 2θ · cos 2θ

)dθ

=3a2

8

[θ

2− sin 4θ

8− sin3 2θ

6

] π2

0

=3πa2

32,

so the area of the interior of the astroid is3πa2

8.

Laboratory Project: Bezier Curves Omit this subsection.

C.25.2 §10.3 Polar Coordinates

Polar Curves. In the polar coordinate system we locate points in the plane by taking a specialpoint, called the pole, a special half-line or ray which emanates from the pole, called the polaraxis and a direction for measuring positive angles — usually taken to be the counter-clockwisedirection. Any point can be located if we know its distance from the pole, usually denoted byr, and the angle that the line joining them makes with the polar axis in the positive direction,usually denoted by θ. However, the angle is not unique, since angles of θ and θ + 2nπ willgive the same point for any integer n. So here we have one of the essential differences betweenpolar and cartesian coordinates:

Theorem C.75 The polar coordinates of a point are never unique.

In the case of the pole itself, the angle θ is totally undetermined: once we know that r = 0, anyangle θ will give the same point.

Convention permitting negative r. It is convenient to broaden the multiplicity of coordi-nates by permitting the distance from the pole to be negative. We do this by agreeing that (r, θ)and (−r, θ+π) represent the same point. This convention permits for continuous representationof certain curves, but causes complications at various stages: occasionally additional care isrequired.


Relations between polar and cartesian coordinates. In theory we can set up polar andcartesian systems independently in the plane, placing the pole at any convenient place. Inpractice we often place the pole at the origin of a cartesian system, with the polar axis along thepositive x-axis. When the author of the textbook suggests that you are to consider two systemsat the same time, and gives no other information, this is what he expects you to do.51 Whenthe polar and cartesian systems are placed in this “standard” way, the following relationshipshold:

x = r · cos θy = r · sin θ

x2 + y2 = r2

yx

= tan θ

Note that, while it is possible to transform from polar to cartesian coordinates without ambi-guity, it is not always possible to move painlessly in the other direction; this is because of thenon-uniqueness of polar coordinates, about which much more will be said. If you are given anequation in cartesian coordinates, you can transform it to polar by substituting the appropriateformulæ for x and y and simplifying — try [1, Problems 21-26, p. 648]; if you are given thecoordinates of a point in polar coordinates, you can transform to cartesian in the same way —try [1, Exercises 3-4, p. 647].

Consequences of Non-uniqueness of Polar Coordinates. This is a difficult topic, and willrequire considerable practice before you will become comfortable with it! Some of the prob-lems you will have are related to understanding what is meant by an equation for a curve:you need to understand that any curve can be represented in multiple ways, even when we usecartesian coordinates. But, when the representation of the points themselves is not unique, theresults can be confusing.

We saw with parametric equations that the same point on a curve can appear more than onceon a “curve”. In that context there was a “natural” way of tracing out the curve, by followingincreasing values of the parameter. When we come to study polar coordinates, the situation ismuch more complicated, because there is no “natural” way of following the generation of thecurve, and no one set of coordinates for a point has preferential status with respect to others.Polar curves can be expressed by any relationship between r and θ, although more often thannot the equation will be in the form r = f (θ); however, you will see some equations in the formθ = f (r) — for example, the equation θ =

π

4represents the line through the pole inclined at an

angle ofπ

4to the polar axis.

51However, there is an important application, involving conic sections — ellipses, parabolæ, hyperbolæ, wherewe place the origin at a different point; this topic is not on the syllabus of the present course, but you may readabout it in [1, §10.6].


Symmetry When we studied symmetry in the context of cartesian coordinates, we couldhave considered [1, pp. 19-20, 308]

• reflective symmetry in a vertical line, particularly in the y-axis;

• reflective symmetry in a horizontal line, particularly in the x-axis;

• rotational symmetry about a point, particularly about an angle of π around the origin(which is equivalent to reflective symmetry in both the x-axis and the y-axis); and

• periodicity of a graph under a horizontal shift.

These are not the only types of symmetries that a graph can possess, so the study in the textbookis selective, and we haven’t investigated thoroughly what principles were involved.

Similarly, when we consider symmetry in the context of polar coordinates, we will notattempt a thorough study, but will consider types of symmetries which the polar system isparticularly able to accommodate. Here are the types of symmetry that the textbook mentions:

• reflective symmetry in the polar axis — exhibited by the invariance of the equation of acurve under the substitution θ 7→ −θ.52

• rotational symmetry under a rotation through an angle of π around the pole — exhibitedby the invariance of the equation under the transformation r 7→ −r or under θ 7→ θ + π

• reflective symmetry under reflection in the line θ =π

2, exhibited by invariance under the

transformation θ 7→ π − θ.Again, these are not the full range of symmetries that can occur in the plane. Because ofthe non-uniqueness of coordinates, curves can have the symmetries listed without exhibitinginvariance under the transformations listed. For example, the curve θ = π is the line thatextends the polar axis. It certainly has symmetry in the polar axis, but the equation is notunchanged when we replace θ by −θ. You are not expected to be an expert in the subject ofsymmetries.

Graphing Polar Curves with Graphing Devices. Omit this section — this is a device-freecourse.

52This description is incomplete. See the discussion below on page 3177 of Example C.78 ([7, Exercise 37, p.678]).


Where is the point with polar coordinates (r, θ)? The ambiguity in polar coordinates is notin locating a point with given coordinates — it is only that all points possess multiple sets ofcoordinates. If you are given the coordinates (r, θ), you may locate the point by:

• First, locating the ray obtained by turning the polar axis (the distinguished ray that em-anates from the pole, relative to which we refer all coordinate angles) through an angleof θ in the positive direction; and

• then

– if r ≥ 0, proceeding along that ray a distance of r; or

– if r ≤ 0, proceeding along the extension of the polar axis beyond the pole a distanceof −r.

Example C.76 ([7, Exercise 2, p. 677]) Plot the point whose polar coordinates are(−1,−π2

).

Then find two other pairs of polar coordinates of this point, one with r > 0, and one with r < 0.Solution: The second coordinate tells us that the point is on the line inclined to the polar axisat an angle of −π2 radians measured in the positive direction: this takes us to the ray which isobtained by turning the polar axis in the clockwise direction through a right angle. But thenthe negative first polar coordinate tells us to proceed along the opposite ray for 1 unit. If thepolar system is superimposed on a cartesian system in the usual way, the point is the unit pointon the positive y-axis. This point has the following other sets of polar coordinates:

with positive r:(1, π2 + 2nπ

)

with negative r:(−1, −π2 + 2mπ

)or

(−1, 3π

2 + 2`π)

where n, m, and ` are any integers.

Example C.77 ([7, Exercise 12, p. 677]) “Sketch the region in the plane consisting of pointswhose polar coordinates satisfy the conditions −1 ≤ r ≤ 1, π

4 ≤ θ ≤ 3π4 .”

Solution: Let’s separate the portion of the region described by positive r, zero r, and negativer:

positive r. 0 < r ≤ 1, π4 ≤ θ ≤ 3π

4 describes a sector of the unit disk centred at the pole — allpoints in the region bounded by two perpendicular radii bisecting the first and secondquadrants. The points on the radii and the bounding circle are included.

zero r. The region described by r = 0, π4 ≤ θ ≤ 3π

4 consists of the pole alone: the θ coordinateis irrelevant for the pole.

negative r. −1 ≤ r < 0, π4 ≤ θ ≤ 3π

4 describes the region antipodal to the one described forpositive r — between radii at angles π + π

4 and π + 3π4 .


10.3 Exercises

[1, Exercise 14, p. 648] “Find a formula for the distance between the points with polar coor-dinates (r1, θ1) and (r2, θ2).”

Solution: One way to solve this problem is to transform to cartesian coordinates, thenapply the usual distance formula, after which cancellations will occur. A more directway is to apply the Law of Cosines to a triangle whose vertices are the pole and the twogiven points. That is easy if we assume that r1 and r2 are non-negative, so they representthe lengths of the sides of the triangle. To solve the problem in this way we need toconsider several cases. The result is the same: the distance is

√r2

1 + r22 − 2r1r2 cos(θ1 − θ2) .

[1, Exercise 18, p. 648] Identify the curve r = 2 cos θ + 2 sin θ.

Solution: The textbook suggests finding a cartesian equation first. This is not alwayseasy, but is not impractical in the present problem. If we multiply the given equation byr, we obtain the equation

r2 − 2r cos θ − 2r sin θ = 0

which we see is equivalent to

x2 + y2 − 2x − 2y = 0

i.e., to(x − 1)2 + (y − 2)2 = 2

which is a circle with radius√

2 centred at the point (1, 1). However, we must note that,in multiplying an equation by a factor — equivalently in multiplying the two equations

r = 2 cos θ + 2 sin θ (84)r = 0 (85)

we were, in the second equation, possibly permitting new points to be included in the“curve”. We must carefully analyze whether the equation r = 0 did that. But we knowthat r = 0 represents only the pole! And the pole was already on the given curve: itappears there as

(r, θ) =

(0,

3π4

).

So multiplying by r = 0 does not introduce any new points.53

53But what would happen if you were to multiply an equation like r = 1 by r = 0?


[1, Exercise 30, p. 648] Describe the curve with the equation r2 − 3r + 2 = 0.

Solution: The equation can be rewritten as

(r − 1)(r − 2) = 0 ,

which is satisfied by all points whose r-coordinate is either 1 or 2, i.e., by all points oneither of two concentric circles around the pole.

[1, Exercise 33, p. 648] Describe the curve with the equation r = 2(1 − sin θ) (θ ≥ 0).

Solution: (see Figure 8 on page 3170) Under the transformation θ 7→ π − θ the equation

0

-2

-1

-3

-4

210-1-2

Figure 8: The cardioid with equation r = 2(1 − sin θ)

is unchanged, so the curve has symmetry in the y-axis. This is a “heart-shaped” curve,


called a cardioid. The textbook describes another example in [1, Example 9, pp. 645-646], where the author shows, as we could show here, that the curve is tangent to they-axis. We call the point of tangency (here, the pole) a cusp. (The condition θ ≥ 0 doesnot restrict the curve in any way, since the function 2(1 − sin θ) is periodic with period2π: any part of the curve that might require a negative value θ0 of θ to represent it, canalso be represented by the θ0 + 2πn, where n is any integer; taking n sufficiently largewill make this value positive.)


C.26 Supplementary Notes for the Lecture of March 08th, 2010Release Date: Monday, March 08th, 2010

C.26.1 §10.3 Polar Coordinates (continued)

10.3 Exercises (conclusion)[1, Exercise 34, p. 648] Describe the curve with equation r = 1 − 3 cos θ. (see Figure 9 on

page 3172)

2

0

1

-1

-2

0-1-2-3-4

Figure 9: The limacon r = 1 − 3 cos θ,

Solution: It is not practical to represent this curve in cartesian coordinates. We notethat, when θ is replaced by −θ, the equation is unchanged. This tells us that the curve is


symmetric about the polar axis. Try tracing it out by starting with the point with θ = 0. Ifwe superimpose a cartesian system in the usual way, that point is the point with cartesiancoordinates (−2, 0). The curve passes through the pole first when θ = arccos 1

3 , about70 degrees. It crosses the y-axis at the point with cartesian coordinates (x, y) = (0, 1),and then moves to its maximum distance of 1 + 3 = 4 from the pole when θ = π. Thenit returns, crossing the negative y-axis at (x, y) = (0,−1), and passing again through thepole. The curve is one of the family called limacons, see them sketched in [1, Example11, p. 647].

Tangents to Polar Curves I discuss how to determine the tangents to curves of the formr = f (θ). The discussion is based on the observation that, if we superimpose polar and cartesiansystems in the most usual way, we can express x and y in terms of θ by

x = f (θ) · cos θy = f (θ) · sin θ

and thereby interpret θ as a parameter in the parametric representation of a curve. We find that

dydx

=

dydθdxdθ

=

drdθ· sin θ + r · cos θ

drdθ· cos θ − r · sin θ

. (86)

In the case of the cardioid r = 2(1 − sin θ), we find that

dydx

=2 sin θ − 1

4 cos θ

when cos θ , 0. When θ = 0, at the unit point on the initial ray, the tangent has slope − 14 . As θ

increases to π6 the tangent becomes vertical; then, as θ > π

6 and θ → π2−, the tangent approaches

the vertical, with positive slope. Similarly, as θ → π2

+, the tangent approaches the vertical, withnegative slope. We say that this curve has a cusp at the pole.

[1, Exercise 35, p. 648] Describe the curve with equation r = θ, with θ ≥ 0.

Solution: This curve is a spiral, turning around the pole. (see Figure 10 on page 3174)If, however, we were to ask about the curve with equation r = θ, with θ ≤ 0, it is also aspiral around the pole. (see Figure 11 on page 3175) The superposition of the two curvesis shown in Figure 12 on page 3176.


30

10

-30

30

20

20-20-30

-20

10

-10

-1000

Figure 10: The spiral with equation r = θ, (θ ≥ 0)


30

10

-30

30

20

20-20-30

-20

10

-10

-100

0

Figure 11: The spiral with equation r = θ, (θ ≤ 0)


30

10

-30

20

03010-30

-20

-10

-20 -10 200

Figure 12: The full spiral with equation r = θ, −∞ < θ < +∞


Example C.78 ([7, Exercises 37, 40, p. 678]) Describe the curves with equations r = sin 2θ,r = sin 5θ.Solution: First let’s consider the curve r = sin 2θ. (see Figure 13 on page 3177) While the

0.6

0.2

-0.6

0.4

0

-0.4

-0.2

0.60.40.20-0.2-0.6 -0.4

Figure 13: The “4-leafed rose” with equation r = sin 2θ,

equation does not remain unchanged when we replace θ by −θ, it changes to r = − sin 2θ,which contains the same points, since it can be rewritten as −r = sin 2(θ + π), which can beobtained from the original equation by the transformation

(r, θ) 7→ (−r, θ + π) .

Thus this curve is symmetric about the polar axis. It is also symmetric about the y-axis, sincethe replacement of θ by π

2 − θ leaves the equation unchanged. The curve is a “4-leafed rose”,where each petal is tangent to the cartesian axes, if located in the usual way.


But, when the multiplier of θ is an odd integer, the situation changes. (see Figure 14 onpage 3178) The curve r = sin 5θ is again a “rose”, passing through the pole every 36 degrees.

0.8

0

0.4

-0.4

-0.8

0.80 0.4-0.8 -0.4

Figure 14: The “5-leafed rose” with equation r = sin 5θ,

It is not symmetric about the y-axis, nor about the x-axis. (It is symmetric under reflections inother axes, and also under rotation through certain angles around the pole.)

Example C.79 ([7, Exercises 42, p. 678]) Sketch the curve with the polar equation r2 = sin 2θ.Note that there are no points on this curve for π

2 < θ <π2 since, in that interval, sin 2θ < 0. The

entire curve is traced out for 0 ≤ θ ≤ π2 : every value of θ gives rise to two points on the graph.

(see Figure 15 on page 3179)

Example C.80 Where is the curve r = sin 2θ horizontal?


0.8

0

0.4

-0.4

0.80.40-0.8 -0.4

-0.8

0.3

0.1

-0.3

0.2

0

-0.2

-1 0 0.5 1

-0.1

-0.5

Figure 15: The lemniscates r2 = sin 2θ, r2 = cos 2θ

Solution: We setdydx

= 0 in (86):

drdθ· sin θ + r · cos θ = 0

⇔ 2 cos 2θ · sin θ + sin 2θ · cos θ = 0⇔ 2(1 − 2 sin2 θ) · sin θ + 2 sin θ · cos2 θ = 0⇔ 2 sin θ(2 − 3 sin2 θ) = 0

so the tangents will be horizontal when

sin θ = 0,±√

23,

i.e., when

θ = 0,± arcsin

√23, π ± arcsin

√23.

C.26.2 §10.4 Areas and Lengths in Polar Coordinates

Areas To find the area bounded by a curve given in polar coordinates we express the area asthe limit of a sum of narrow triangles whose bases are along the bounding curve, and whoseupper vertex is at the pole. It was shown in the lecture that the area subtended by the arc of the


curve r = f (θ) between θ = a and θ = b is then

b∫

a

12· ( f (θ))2 dθ .


C.27 Supplementary Notes for the Lecture of March 10th, 2010Release Date: Wednesday, March 10th, 2010

C.27.1 §10.4 Areas and Lengths in Polar Coordinates (continued)

Areas To find the area bounded by a curve given in polar coordinates we express the area asthe limit of a sum of narrow triangles whose bases are along the bounding curve, and whoseupper vertex is at the pole. It was shown in the lecture that the area subtended by the arc of thecurve r = f (θ) between θ = a and θ = b is then

b∫

a

12· ( f (θ))2 dθ .

(This could be shown in two different ways: by treating the element of area as a sector of a cir-cle, or as a narrow, isosceles triangle.) As with all other formulæ involving polar coordinates,one must use this formula with care. Be sure that you know precisely what the region lookslike; in case of doubt, break the region up into parts, and find the areas of the parts separately.The limits must be chosen carefully, to be sure that, for example, you are not computing morearea than you intend. For example, the curve r = cos θ is a circle of radius 1

2 centred at thepoint

(12 , 0

), and passing through the pole. The curve is swept out as θ ranges from 0 to π, so

the area of the disk is

12

π∫

0

cos2 θ dθ =14

π∫

0

(1 + cos 2θ) dθ =π

4.

Compare this with the area of the disk r = 1, centred at the pole, where the curve is swept outas θ ranges from 0 to 2π;

Area =12

2π∫

0

1 dθ = π .

In that case, if we were to stop at π, we would obtain only the area of the upper half-disk.Rather than attempting to memorize rules about limits for integrals, I suggest you carefullyanalyze each problem individually.

Finding the intersections of curves in polar coordinates The textbook [1, p. 651-652]discusses the difficulties that of finding intersections which are a consequence of the multiplesets of coordinates for points. Some textbooks suggest that finding the intersections can onlybe done visually, and that is not true. In fact, intersections can be found algebraically, but one


must be careful and thorough. I attach below a discussion I prepared for a class some yearsago, where another textbook had made a false statement that I felt obliged to correct; the erroris continued in the current edition of the same textbook.

Example C.81 In an example in another textbook [31, Example 8, p. 579], [30, Example 8, p.635] the objective is to find the points where the curves

r = 1 + sin θ (87)r2 = 4 sin θ (88)

intersect. It was stated in the textbook solution that only one of the points of intersectioncan be found algebraically, and that the others can be found only “when the equations aregraphed”. We show here all intersection points can be found algebraically! We never resort tocalculations on a sketch: all procedures can be justified theoretically — the sketch serves onlyto help visualize a situation that can be adequately described verbally and/or with mathematicalformulæ.

(cf. Figure 16, page 3182) Had the curves been given in cartesian coordinates, we could

–2

–1

0

1

2

–1 –0.5 0.5 1

Figure 16: Intersecting polar curves r = 1 + sin θ, r2 = 4 sin θ

have found all intersections by solving the equations simultaneously. Why can’t we solve the


polar equations in the same way? The difficulty derives from the fact that any point has in-finitely many different polar representations. More precisely, a point that can be representedby polar coordinates (r, θ) also has coordinates ((−1)nr, θ + nπ), where n is any integer — pos-itive or negative; moreover, the pole can be represented by (0, θ), where θ is any real number.To determine the points of intersection, one must consider the possibility that the same pointappears with different coordinates.

Solve the given equations algebraically: By eliminating sin θ between the two equations, weobtain r2 = 4(r−1), which implies that (r−2)2 = 0, so r = 2, and sin θ = 2−1 = 1. Hence

θ =π

2+2mπ, where m is any integer, and the points of intersection are

(2,π

2+ 2mπ

): but,

by the convention described above, these are representations of the same point, whose

“simplest” representation is(2,π

2

).

Transform the equations in all possible ways and solve again: Apply to one of the equa-tions the transformation

(r, θ) 7→ (−r, θ + π) (89)

and solve it with the original form of the other equation. Repeat this process until theequations transform to a pair already solved. Equation (87) transforms to

−r = 1 + sin(θ + π) (90)

which is equivalent tor = −1 + sin θ (91)

which equation we solve with (88). Eliminating sin θ yields r = 2 ± 2√

2, so sin θ =

3± 2√

2. The upper sign is inadmissible, as a sine cannot exceed 1 in magnitude. Hencer = 2 − 2

√2 and

sin θ = 3 − 2√

2 . (92)

The solutions to (92) are θ = sin−1(3 − 2

√2)+ 2mπ and θ = π− sin−1

(3 − 2

√2)+ 2mπ;

we may take m = 0, as all other values of m give the same two points:(2 − 2

√2, sin−1

(3 − 2

√2))

and(2 − 2

√2, π − sin−1

(3 − 2

√2)).

As the first coordinate in these cases is negative, we could equally well represent thepoints as (

2√

2 − 2, sin−1(3 − 2

√2)

+ π)

and (2√

2 − 2,− sin−1(3 − 2

√2)).

A second application of (89), to (90), restores the original equation; hence there are noother intersection points, except possibly the pole.


Check whether the pole is on both curves: On (87) the pole appears as(0,

3π2

), etc.; on (88)

it appears as (0, 0), etc. Thus the pole is also a point of intersection. The reason we didnot find it when we solved pairs of equations is that it appears on the two curves onlywith different sets of coordinates, no two related by (89).

Example C.82 [7, Exercise 29, p., 683] “Find the area of the region that lies inside bothcurves: r = sin θ, r = cos θ.”Solution: (see Figure 17 on page 3184) The curves are circles through the pole. To see this,

0.75

0.5

0.0

0.25

−0.5

1.0

1.0

0.75

0.25

0.5

−0.25

0.0−0.25−0.5

Figure 17: Curves r = sin θ, r = cos θ

multiply the first by r = 0 (which, of course, could be bringing the pole into the curve). Theresulting equation is r2 = r sin θ, which, in cartesian coordinates, would be x2 + y2 = y, a


circle with centre (in cartesian coordinates)(0, 1

2

)and radius 1

2 . The operation of multiplyingby r = 0 did not, however, alter the curve, since the pole was already on the curve, with polarcoordinates (r, θ) = (0, 0). In the same way we can show that the second curve is a circle ofthe same radius centred at the point (r, θ) =

(12 , 0

); polar coordinates for the centre of the first

circle are, for example,(

12 ,

π4

).

Where do the curves meet? (This was [7, Exercise 37, p. 683], which should have pre-ceded the present problem in the exercises!) We can begin this investigation by solving theirequations, which imply that tan θ = 1, so θ = π

4 + nπ, where n is any integer; that implies thatr = ± 1√

2: the + sign is associated with the cases where n is even, the − sign with those where

n is odd. But (r, θ) =(

1√2, π4 + 2mπ

)and (r, θ) =

(− 1√

2, π4 + (2m + 1)π

)are all the same point —

having cartesian coordinates(

12 ,

12

).

But the curves also meet at the pole, even though this did not show up when we solved theequations. This is because the pole appears on the first curve with θ = nπ, and on the secondcurve with θ = (n + 1

2 )π. The only practical way to determine whether curves intersect at thepole is to examine each of them separately for that point.

Could it be that the curves meet in any other points? While a sketch does not suggest that,we should never rely on a sketch! If we transform the first equation under the transformation(r, θ) 7→ (−r, θ + π), we find that the equation does not change (except for a sign change onboth sides); the same applies to the second equation. Thus, in the present example, there are nointersections other than the pole in which a point will appear on the two curves with differentcoordinates. We can comfortably proceed to the integration part of the problem, confident thatwe have not missed any points of intersection.

We can see that the region whose area is sought is symmetric about the line θ = π4 , so it

suffices to find half of it and to double it. The lower half of the area is subtended at the pole bythe arc 0 ≤ θ ≤ π

4 of the circle r = sin θ; hence

Area = 2 · 12

π4∫

0

sin2 θ dθ

=12

π4∫

0

(1 − cos 2θ) dθ

=12

[θ − sin 2θ

2

] π4

0=π

8− 1

4.

(You could have solved this problem using Cartesian coordinates.)

10.4 Exercises


[1, Exercise 22, p. 653] “Find the area enclosed by the loop of the strophoid r = 2 cos θ −sec θ.”

Solution: (see Figure 18 on page 3186) (The textbook did not expect a rigorous proof

0.2

-0.2

-0.6

10.80.60.40.2-0.2-0.4

0.6

0.4

0

-0.4

0

Figure 18: The strophoid r = 2 cos θ − sec θ

that the curve crosses itself only at the pole.) First we observe that the curve is entirelytraced out when θ passes through an interval of length 2π; so, without limiting generality,we may confine ourselves to such an interval, say −π2 < θ < π

2 , and thereby remove thesome possible ambiguity of coordinates. We must exclude the points where sec θ isundefined. For convenience, let’s confine θ to the union of intervals

−π2< θ <

π

2and

π

2< θ <

3π2.


Consider a point on the second branch, say with θ = φ + π, where −π2 < φ < π2 . We find

that, for this point,

r = 2 cos(φ + π) − sec(φ + π)= −2 cos φ + sec φ= −(2 cos φ − sec φ)

So we see that the point is the same as the point (−r, φ). Thus, in order to see the wholecurve, it suffices to investigate angles θ between −π2 and π

2 ; some of the points will,however, appear with negative r-values.

We observe that the replacement of θ by −θ does not change the equation: this tells usthat the curve is symmetric about the polar axis and its extension.

For the problem to be meaningful, there should be just one loop. The curve passesthrough the pole when 0 = 2 cos θ−sec θ, equivalently, when cos θ = ± 1√

2. In the interval

to which we have confined θ, this will occur when θ = ±π4 . If we follow the curve as θcomes from −π2 , we find that r is negative, and the point is in the 2nd quadrant. It passesthrough the pole first when θ = −π4 , and is then in the fourth quadrant, striking the polaraxis when θ = 0, at the point (r, θ) = (1, 0) and it then moves into the first quadrant,following the mirror image of the portion in the fourth quadrant, passing through thepole again when θ = π

4 , and moving into the 3rd quadrant. If we consider the cartesiancoordinates of the curve in parametric form, we have

x = 2 cos2 θ − 1y = sin 2θ − tan θ

and see that, as cos θ → 0, x → −1, and y → ±∞, so the curve is asymptotic to thevertical line x = −1. Our present problem is to determine the area of the loop fromθ = −π4 to θ = π

4 , or, by symmetry,

2(12

) π4∫

0

(2 cos θ − sec θ)2 dθ =

π4∫

0

(4 cos2 θ + sec2 θ − 4

)dθ

=

π4∫

0

(2 cos 2θ + sec2 θ − 2

)dθ

= [sin 2θ + tan θ − 2θ]π40

=

(1 + 1 − π

2

)− 0 = 2 − π

2.


[1, Exercise 26, p. 653] Find the area of the region that lies inside the curve

r = 2 + sin θ (93)

and outside the curver = 3 sin θ . (94)

Solution: (see Figure 19 on page 3188) The first step is to determine where the curves

3

1

2

1 20-2 -1

-1

0

Figure 19: Curves r = 2 + sin θ, r = 3 sin θ

intersect. We can start by solving equations (93) and (94) algebraically. We find thatr = 3 and θ = π

2 + 2nπ, where n is any integer. The second curve is a circle with centre atthe point (r, θ) =

(32 ,

π2

). The first curve appears to be some sort of oval, and touches the


circle at its topmost point, which we shall call A, with coordinates (r, θ) =(3, π2

). Could

there be any other points of intersection? There don’t appear to be any, from the sketch,but we can resolve this question algebraically. The pole does not lie on first curve, sincethat would entail that 0 = 2 + sin θ, i.e., that the sine of an angle exceeds 1 in magnitude.Since that is not possible, the pole cannot lie on the curve. If there were to be any otherintersections, they would have to have different sets of coordinates on the 2 curves. Weinvestigate what happens to (94) under the transformation

(r, θ) 7→ (−r, θ + π)

and find that the equation does not change significantly: it becomes −r = sin(θ + π) =

− sin θ, which is evidently equivalent to the original equation. Equation (93) changes to−r = 2 + sin(θ + π) = 2 − sin θ, or

r = −2 + sin θ. (95)

When we solve this equation with (94), we find that r = −3, sin θ = −1: in the interval0 ≤ θ ≤ 2π we find only the point (r, θ) =

(−3, 3π

2

), which is the same point as (r, θ) =(

3, π2), so we don’t find any new intersections. Thus there are no other intersections than

A.

A direct way to solve this problem is to find the area of the region bounded by the outercurve, and then subtract from it the area of the disk inside. The outer curve is traced outas θ ranges from 0 to 2π, so the area of the region it bounds will be

12

2π∫

0

(2 + sin θ)2 dθ =12

2π∫

0

(4 + 4 sin θ + sin2 θ) dθ

=12

2π∫

0

(4 + 4 sin θ +

1 − cos 2θ2

)dθ

=12

[9θ2− 4 cos θ − 1

4sin 2θ

]2π

0=

9π2

The inner curve bounds an area of

12

π∫

0

9 sin2 θ dθ =94

π∫

0

(1 − cos 2θ) dθ

=94

[θ − 1

2sin 2θ

]π

0=

94· π .


What would have happened if we had taken the upper limit of integration to be 2π? Wewould have found twice the area, because the entire circle is swept out as θ ranges overan interval of length π.

Thus we see that the area of the region between the curves is

9π2− 9π

4=

9π4.

Could we find the area by taking the difference of two squares under the integral sign,as is suggested in [1, Example 2 and Figures 5 and 6, p. 651]? We could do this for theportion of the area above the polar axis and its extension, where we would obtain

12

π∫

0

((2 + sin θ)2 − (3 sin θ)2

)dθ

=12

π∫

0

(4 + 4 sin θ − 4(1 − cos 2θ)) dθ

=12

[−4 cos θ + 2 sin 2θ]π0 = (2 + 0) − (−2 + 0) = 4

For the remainder of the area only curve (93) and the extended polar axis serve as aboundary:

12

2π∫

π

(2 + sin θ)2 dθ =12

[9θ2− 4 cos θ − 1

4sin 2θ

]2π

π

=12

[(9π − 4 − 0) −

(9π2− (−4) − 0

)]=

9π2− 4

which, when added to the area of the upper portion, gives the same area as before.

What would have happened if we had subtracted the square of 3 sin θ over the full rangeof the integration? We would have been subtracting the area of the inside disk twice!

And what would have happened if we had taken (95) as the equation of the outer curve?If we used it only to find the area of the (larger) region bounded, we would obtain thecorrect value there. But, if we had taken the difference of squares under the integral sign,we would obtain values for regions that do not correspond to the one whose are we aretrying to find.

Before finding an area in polar coordinates, (or, indeed, in cartesian coordinates as well),you are urged to make a sketch and study the element of area that you are summing inthe limit, to ensure that the integral represents the area that you are seeking.


Example C.83 Another example of finding the intersections of curves.

Find the intersections of the limacon r = 1 − 2 cos θ and the circle r = 1. [29]

Solution: (see Figure 20 on page 3191) When we solve the given equations, we find the inter-

1.5

0

1

0.5

-0.5

-1.5

10-1-2-3

-1

Figure 20: Intersections of the limacon r = 1 − 2 cos θ with the circle r = 1

section points (r, θ) for which cos θ = 0, i.e.,(1,±π2

). But a glance at the graphs shows that

there appears to be a third point of intersection. Under the transformation (r, θ) → (−r, θ + π),the given equation for the circle r = 1 does not change in a significant way. However, theequation of the limacon changes to r = −1 − 2 cos θ; when we solve this latter equation withthe equation r = 1, we find that cos θ = −1, so the point (r, θ) = (1, π) is another point ofintersection: i.e., (1, π) on the circle, (−1, 0) on the original equation for the limacon.


To finish this investigation we should investigate whether the pole is on both curves, sincethat may not be detected by this method of transforming the equations. We note, however, thatthe curve r = 1 cannot possibly contain the pole, since all of its points are 1 unit from thepole. Thus the investigation is complete: there are three point of intersection, and we havenow found them all.

Example C.84 A problem from a recent examination:

[12 MARKS] Use polar coordinates — no other method will be accepted — to

find the area of the region bounded by the curve r = 2 and the line r =1

cos θ, and

containing the pole.

Solution: (see Figure 21 on page 3193) The first step is to determine where the two curvesintersect. While the points of intersection can all be found algebraically, that procedure istedious, so it is best to make a sketch first to see whether the elaborate procedure is required.You should be able to see immediately that the curve r = 2 is a circle of radius 2 centred atthe pole. But what about r = sec θ? Since the secant is never less than 1 in magnitude, youknow this curve does not pass through the pole. You also know that, as θ ranges between −π2and π

2 , the function ranges from −∞ to +∞ monotonely, so the curve will be closest to the polewhen θ = 0 — when it is 1 unit from the pole. As θ increases to +π

2 , r increases steadily —monotonely — the curve crossing the circle r = 2 and moving a greater and greater distancefrom the pole. What happens when r < 0? A point with coordinates (−a, θ), where a > 0 willstill be a distance a from the pole! (It’s located on the ray obtained by turning the polar axisthrough an angle of θ + π.) So, as r → −∞, the curve again crosses each of the circles centredat the pole and moves a greater and distance from the pole. This is all you need to know inorder to solve the problem, but you should be able to see that this is a very simple curve: just

multiply both sides of r =1

cos θby cos θ, and you see that the equation is r cos θ = 1, or, in

cartesian coordinates, x = 1 — the curve is a straight line perpendicular to the polar axis!We have seen that there will be just 2 points of intersection of the curves. Solving their

equations by eliminating r we obtain cos θ = 12 , so θ = ±π3 : the points of intersection are

(r, θ) =

(2,±π

3

).

We saw that the element of area in the form of a thin triangle with apex at the pole (or a thinsector of a disk with centre at the pole) has area of the form 1

2r2 dθ. In this problem such anelement would have to be described in two different ways:

−π3≤ θ ≤ π

3: The area of this triangle is

12

π3∫

− π3

1cos2 θ

dθ =12

[tan θ]π3− π3

=√

3 .


2

0

1

-1

20 1-2

-2

-1

Figure 21: Intersections of the curve r = sec θ with the circle r = 1

π

3≤ θ ≤ 5π

3: The area of this portion of the disk is

12

5π3∫

π3

22 dθ =12

[4θ]5π3π3

=8π3.

Thus the area of the region determined by the two curves, and containing the pole, is√

3+8π3.


This problem could also have been solved by subtracting from the area of the disk, π22, thearea given by

12

π3∫

− π3

(22 − sec2 θ) dθ =12

[4θ − tan θ]π3− π3

=4π3−√

3 .


C.28 Supplementary Notes for the Lecture of March 12th, 2010Release Date: Friday, March 12th, 2010

C.28.1 §10.4 Areas and Lengths in Polar Coordinates (conclusion)

Arc Length To develop a formula for the arc length of a curve given in polar coordinates asr = f (θ), we can apply the theory of curves given in parametric form to the curve

x = f (θ) · cos θy = f (θ) · sin θ

We find that (dxdθ

)2

+

(dydθ

)2

=

(drdθ

)2

+ r2

so the length is given by the integral

L =

b∫

a

√(drdθ

)2

+ r2 dθ

where the limits θ = a and θ = b need to be determined from the parametrization of the curve.As usual, you should be careful to determine the appropriate values of θ to define the portionof the curve that interest you.

10.4 Exercises (continued)

[1, Exercise 46, p. 654] Find the exact length of the arc of the polar curve r = e2θ from θ = 0to θ = 2π.

Solution:

Length =

2π∫

0

√(drdθ

)2

+ r2 dθ

=

2π∫

0

√(2e2θ)2

+(e2θ) r2 dθ

=√

5

2π∫

0

e2θ dθ

=

√5

2

[e2θ

]2π

0=

√5(e4π − 1

)

2


[1, Exercise 54, p. 654] Graph the curve r = cos2 θ2 , and find its length.

Solution: The equation can be simplified to read

r =1 + cos

(2 · θ2

)

2

which we recognize to be a cardioid. As θ ranges over the interval 0 ≤ θ ≤ 2π the entirecurve is traced out once.

Length =

2π∫

0

√(drdθ

)2

+ r2 dθ

=12

2π∫

0

√(− sin θ)2 + (1 + cos θ)2 dθ

=12

2π∫

0

√2(1 + cos θ) dθ

=12

2π∫

0

√2(2 cos2 θ

2

)dθ

=

2π∫

0

∣∣∣∣∣cos(θ

2

)∣∣∣∣∣ dθ

=

[2 sin

θ

2

]π0−

[2 sin

θ

2

]2π

π

= 2 − (−2) = 4 .

C.28.2 §10.5 Conic Sections

Omit this section.

Textbook Chapter 11. INFINITE SEQUENCES AND SERIES.


C.28.3 §11.1 Sequences

A sequence is an ordered set of objects, usually labelled with either the non-negative integers0, 1, . . . , n, . . . or the positive integers 1, 2, . . . , n, . . . . In this chapter we shall be interested insequences of real numbers or of real functions. Technically, such a sequence is a function thatmaps either the non-negative integers or the positive integers on to the set of real numbers (orto the set of real functions). We usually denote a sequence by a single letter, e.g. a, and showthe labelling by either a parenthesized value, as a(n), or a subscripted values, as an; when wewish to talk about a sequence with specific terms, we may just list the terms with a generalterm that shows the pattern we are describing, if there is one, as in

1, 1, 2, 3, 5, 8, 13, 21, 24, 55, 89, . . .

which is the Fibonacci sequence, in which every entry past the second is the sum of the twoentries immediately preceding it. Unless the general term is described unambiguously (as here,where Fn+2 = Fn+1 + Fn for all n ≥ 3), there will be more than one way to generalize a patternthat appears to hold between a few terms at the beginning of the sequence.

Notation When a sequence consists of terms a0, a1, a2, . . . we may speak of the sequencean, or perhaps the sequence ann=0,1,..., or sometimes simply the sequence an. Other notationsare also possible, and should be understandable from the context.

The limit of a sequence We define limn→∞

an = L in a way analogous to the definition oflimx→∞

f (x) = L; another way of writing the same definition is

an → L as n→ ∞.The precise definition is to be found as [1, Definition 2, p. 677], but is not on the syllabus —you are not expected to be able to work with the precise definition of the limit of a sequence.Note that we usually do not write the braces and when we speak of the limit of a sequence.

Since you bring to this course some understanding of limits of functions, we will occasion-ally appeal to the following

11.1 Exercises

[1, Exercise 32, p. 685] Determine whether the sequence

ln nln 2n

converges.

Solution:

ln nln 2n

=ln n + ln 2 − ln 2

ln n + ln 2

= 1 − ln 2ln 2n


The subtracted fraction has a constant numerator, but the denominator is increasing, sothe fraction is decreasing as n increases; hence 1 minus the fraction is increasing; infact, we can see that the subtracted fraction is approaching 0, so the given fraction isapproaching 1.

C.28.4 Sketch of Solutions to Problems on the Final Examination in MATH 141 200501

(not for discussion at the lecture)

The best way to prepare for the examination in MATH 141 2010 01 is to study the textbook,working problems, and verifying your solutions using the Student Solutions Manual: for sev-eral reasons I don’t recommend your studying exclusively from old examinations. Howeverthe old examinations do have some uses, and I am including one of them here, with solutions,as a timely reminder that less than 1 month remains between now and the final examination inthis course. I am hoping to discuss last April’s examination briefly in a final lecture; there willnot be time for me to discuss the following examination in the lectures.

1. SHOW ALL YOUR WORK!

(a) [4 MARKS] Evaluate∫ 3

0|x − 1| dx .

Solution:

∫ 3

0|x − 1| dx =

∫ 1

0|x − 1| dx +

∫ 3

1|x − 1| dx

=

∫ 1

0(1 − x) dx +

∫ 3

1(x − 1) dx

=

[x − x2

2

]1

0+

[x2

2− x

]3

1

=

(12− 0

)+

(32−

(−1

2

))=

52

(b) [3 MARKS] Evaluateddx

∫ 5

x

√4 + t2 dt .

Solution:ddx

∫ 5

x

√4 + t2 dt = − d

dx

∫ x

5

√4 + t2 dt = −

√4 + x2 .


(c) [3 MARKS] Evaluateddx

∫ x2

2πsec t dt .

Solution: Let u = x2. By the Chain Rule

ddx

∫ x2

2πsec t dt =

ddu

∫ u

2πsec t dt · du

dx= sec u · 2x = sec(x2) · 2x .

(d) [3 MARKS] Evaluate∫

x5( 3√

x3 + 1)

dx .

Solution: Let u =(x3 + 1

) 13 , so u3 = x3 + 1, 3u2 du = 3x2 dx.

∫x5

( 3√x3 + 1

)dx =

∫ (u3 − 1

)·u·u2 du =

u7

7−u4

4+C =

(x3 + 1

) 73

7−

(x3 + 1

) 43

4+C


For each of the following series you are expected to apply one or more tests for conver-gence or divergence and determine whether the series is convergent. In each case youmust answer 3 questions:

• Name the test(s) that you are using.

• Explain why the test(s) you have chosen is/are applicable to the given series.

• Use the test(s) to conclude whether or not the series is convergent.

(a) [4 MARKS]∞∑

n=2

2 − cos nn

Solution: This is a positive series. Since 0 <1n≤ 2 − cos n

n, for all n, the terms

are bounded below by the terms of the harmonic series, a positive series knownto diverge. We may apply the Comparison Test to such pairs of series, and mayconclude that the given series also diverges.

(b) [4 MARKS]∞∑

n=0

n(−3)n

4n

Solution: As formulated in your textbook, we may apply the Ratio Test to thesequence of absolute values of ratios of terms to their predecessors. Here

limn→∞

∣∣∣∣∣∣∣(n+1)(−3)n+1

4n+1

n(−3)n

4n

∣∣∣∣∣∣∣ = limn→∞

(1 +

1n

)· 3

4=

34< 1 ,

from which we may conclude that the given series is (absolutely) convergent.


(c) [4 MARKS]∞∑

n=2

1n ln n

Solution: Define f (x) = 1x ln x . Then f is a positive, continuous function, taking

on the values of the given sequence at the positive integer points. By the IntegralTest, the series and the following improper integral will either both converge orboth diverge.

∫ ∞

2

dxx ln x

= limb→∞

∫ b

2

dxx ln x

= limb→∞

[ln ln x]b2 = lim

b→∞ln ln b − ln ln 2

which approaches +∞ as n→ ∞. Hence the series diverges (to∞).

3. BRIEF SOLUTIONS Express the value of each of the following as a definite integralor a sum, product, or quotient of several definite integrals, but do not evaluate the inte-gral(s). It is not enough to quote a general formula: your integrals must have integrandand limits specific to the given problems:

(a) [6 MARKS] The area of the region bounded by the parabola y = x2, the x-axis, andthe tangent to the parabola at the point (1, 1).Solution: The tangent to the parabola at the point

(x, x2

)has slope 2x; at (1, 1) the

slope is 2. The equation of the tangent is y = 1 + 2(x − 1) = 2x − 1, which lineintercepts the x-axis in the point

(12 , 0

).

If we evaluate the area by integration with respect to x, we have

∫ 12

0x2 dx +

∫ 1

12

(x2 − (2x − 1)

)dx =

[x3

3

] 12

0+

[(x − 1)3

3

]1

12

=1

24+

124

=1

12

We can also integrate with respect to y:∫ 1

0

(−√y +

y + 12

)dy =

[−2

3y

32 +

y2

4+

y2

]1

0= −2

3+

14

+12

=1

12.

(Note that you were not expected to evaluate the integrals: I did so in order to usethe opportunity of having 2 methods in order to verify my work.)

(b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 4 theregion bounded by x = 0 and the curve x =

√sin y (0 ≤ y ≤ π).

Solution: This problem is from your textbook [7, Exercise 25, p. 459], and is solvedin the Student Solution Manual, [9, p. 269]; it is one of the problems for which theCD-Roms accompanying the textbook provide extra help.


Integrating by the method of Cylindrical Shells centred on the line y = 4, we obtain

the volume to be 2π∫ π

0(4 − y)

√sin y dy.

(c) [3 MARKS] The area of the surface obtained by revolving about the y-axis thecurve y = ex, 1 ≤ y ≤ 2.

Solution: Integrating with respect to x gives the integral∫ ln 2

02πx√

1 + e2x dx. Inte-

grating with respect to y gives the integral∫ 2

12π(ln y)

√1 +

1y2 dy.

(d) [2 MARKS] The average value of the function2x

(1 + x2)2 over the interval 0 ≤ x ≤2.

Solution: The answer expected was either 12

∫ 2

02x

(1+x2)2 dx or

∫ 2

02x

(1+x2)2 dx∫ 2

01 dx

. The inte-

gral can be evaluated by using substitutions like u = x2, or u = x2 + 1.


[12 MARKS] Evaluate the indefinite integral∫

2x3 + 3x2 + 3x2 + x − 12

dx .

Solution: Since the degree of the numerator of the rational function which is the inte-grand is not less than the degree of the denominator, the first step is to divide denominatorinto numerator, obtaining

2x3 + 3x2 + 3x2 + x − 12

= (2x + 1) +23x + 15

x2 + x − 12.

The polynomial quotient may be integrated to∫

(2x + 1) dx = x2 + x + C. The excessmust be expanded using the method of Partial Fractions. Assuming an expansion of theform

23x + 15(x + 4)(x − 3)

=A

x + 4+

Bx − 3

,

we multiply both sides by (x+4)(x−3) to obtain the identity 23x+15 = A(x−3)+B(x+4).We may obtain 2 equations for A, B by assigning to x the successive values x = 3 andx = 4, obtaining B = 12 and A = 11. Hence

∫2x3 + 3x2 + 3x2 + x − 12

dx =

∫ (2x + 1 +

11x + 4

+12

x − 3

)

= x2 + x + 11 ln |x + 4| − 12 ln |x − 3| + C ,


which I then checked by differentiation.


(a) [9 MARKS] Use integration by parts to prove that, for integersm ≥ 2, ∫

cosm x dx =1m

cosm−1 x · sin x +m − 1

m

∫cosm−2 x dx

Solution: (The instructions asked that the problem be solved using Integration byParts. Otherwise, the reduction formula could also have been proved by differenti-ating both sides and showing that the same derivative was obtained.)Let u = cosm−1 x, dv = cos(x) · dx, so du = (m − 1) cosm−2 x · (− sin x) dx andv = sin x. Then

∫cosm x dx = cosm−1 x · sin x +

∫sin2 x(m − 1) cosm−2 x dx

= cosm−1 x · sin x + (m − 1)∫ (

cosm−2 x − cosm x)

dx

⇒ m∫ m

x dx = cosm−1 x · sin x + (m − 1)∫

cosm−2 x dx

⇒∫

cosm x dx =1m


m

∫cosm−2 x dx .

(b) [3 MARKS] Showing all your work, use the formula you have proved to evaluate∫ π2

0cos6 x dx.

Solution:

cos6 x dx =16

cos5 x · sin x +56

∫cos4 x dx

=16

cos5 x · sin x +56

(14

cos3 x · sin x +34

∫cos2 x dx

)

=

(16

cos5 x +5

24cos3 x

)sin x +

58

(12

cos x · sin x +12

∫dx

)

=

(16

cos5 x +5

24cos3 x +

516

cos x)

sin x +5

16x + C .

Evaluating between 0 andπ

2we obtain 0 +

5π32

=5π32

.



Consider the curve C defined by

x = 2 cos t − cos 2ty = 2 sin t − sin 2t .

(a) [8 MARKS] Determine the points where the arc of the curve given by

π

4≤ t ≤ 7π

4

has a vertical tangent.Solution:

dxdt

= −2 sin t + 2 sin 2t

dydt

= 2 cos t − 2 cos 2t

Tangents are vertical where dydx is infinite, i.e., where dx

dt = 0 but dydt , 0.

dxdt

= 0 ⇔ −2 sin t + 4 sin t · cos t = 0

⇔ sin t(cos t − 1

2

)= 0

⇔ sin t = 0 or cos t =12

⇔ t = π,π

3,

5π3

for t restricted to lie in the intervalπ

4≤ t ≤ 7π

4. We check the values of dy

dt at thesethree points, and find that the respective values are −4, 2, 2, none of which is 0.Hence the tangents will be vertical at the following points:

t = π (x, y) = (−3, 0)

t =π

3(x, y) =

32,

√3

2

t =5π3

(x, y) =

32,−−

√3

2


(b) [4 MARKS] Determine the length of the arc of the curve given by

0 ≤ t ≤ 2π .

Solution:

(dxdt

)2

+

(dydt

)2

= (−2 sin t + 2 sin 2t)2 + (2 cos t − 2 cos 2t)2

= 4 + 4 − 8(sin t · sin 2t + cos t · cos 2t)= 8 − 8 cos(2t − t) = 8(1 − cos t) .

The length of the arc is

√8∫ 2π

0

√1 − cos t dt =

√8∫ 2π

0

√2 sin2 t

2dt

= 4∫ 2π

0

∣∣∣∣∣sint2

∣∣∣∣∣ dt

= 4∫ 2π

0

(sin

t2

)dt

= −8[cos

t2

]2π

0= −8(−1 − 1) = 16 .


(a) [5 MARKS] Determine whether the following integral is convergent; if it is con-vergent, determine its value: ∫ 1

−1

dx√1 − x2

Solution: The integral is improper because the integrand is not defined — becomesinfinite — at both ends of the interval of integration. According to the definition,we must integrate between points away from ±1, and allow the limit to be takenindependently. The safest way to do that is to split the improper integral into two,and then to evaluate the two of them separately.

∫ 1

−1

dx√1 − x2

= lima→−1+

∫ 0

a

dx√1 − x2

+ limb→+1−

∫ b

0

dx√1 − x2

= lima→−1+

(arcsin 0 − arcsin a) + limb→+1−

(arcsin b − arcsin 0)

= 0 −(−π

2

)+

(π

2

)− 0 = π .


(b) [5 MARKS] Determine whether the following series is conditionally convergent,absolutely convergent, or divergent.

∞∑

n=1

(−1)n n!nn

Solution: I will apply the Ratio Test to the sequence of absolute values of the ratios

of terms to their predecessors:

∣∣∣∣∣∣∣(−1)n+1 (n+1)!

(n+1)n+1

(−1)n n!nn

∣∣∣∣∣∣∣ =(n + 1)!

(n + 1)n+1 ·nn

n!=

(1 +

1n

)−n

→ 1e

as n → ∞. As this limit is less than 1, the given series is absolutely convergent,hence divergent.

(c) [3 MARKS] Determine whether the sequence an = ln(n + 1) − ln n is convergent;if it is convergent, carefully determine its limit.Solution: This is not a question about telescoping series! It is a problem aboutsequences!

ln(n + 1) − ln n = ln(1 +

1n

). As n → ∞, 1 + 1

n → 1, so ln(1 + 1

n

)→ ln 1 = 0 (by

continuity of the logarithm function from the right at the point 1).


[12 MARKS] Find the area of the region bounded by the curves

r = 4 + 4 sin θr sin θ = 3

which does not contain the pole.

Solution: The problem is to determine the area cut off from the cardioid r = 4 + 4 sin θby the line y = 3. First we must determine where the curves cross. Solving the polarequations by eliminating sin θ between them we obtain r2 − 4r − 12 = 0, equivalentto (r − 6)(r + 2) = 0. The values r = 6,−2 yield corresponding values sin θ = 1

2 ,− 32 .

Of these the second is impossible, as a sine cannot be less than 1. We conclude thatsin θ = 1

2 and the values of θ in the interval 0 ≤ θ ≤ 2π are θ = π6 ,

5π6 : the curves intersect

in (r, θ) =(6, π6

)and (r, θ) =

(6, 5π

6

). A naive way to solve the problem is to integrate

directly,

Area =12

∫ 5π6

π6

((4 + 4 sin θ)2 − (3 csc θ)2

)dθ

=12

∫ 5π6

π6

(16

(1 − cos 2θ

2+ 2 sin θ + 1

)− 9 csc2 θ

)dθ


=12

[24θ − 32 cos θ − 4 sin 2θ + 9 cot θ]5π6π6

= 8π + 12√

3 .

Another way to find the area is to observe that the triangle with base on y = 3 betweenthe points of intersection and its third vertex at the pole has area 1

2 (6√

3) ·3 = 9√

3. Then

one can simply subtract this area from12

∫ 5π6

π6

(4 + 4 sin θ)2 dθ.


C.29 Supplementary Notes for the Lecture of March 15th, 2010Release Date: Monday, March 15th, 2010

C.29.1 §11.1 Sequences (conclusion)

Theorem C.85 [1, Theorem 3, p. 678] If f is a function, and an is such that f (n) = an for alln, and if lim

x→∞f (x) = L, then lim

n→∞an = L.

We also generalize our definition of the limit of a sequence to permit limits to be ±∞; thedefinition is again analogous to the corresponding definition for the meaning of lim

x→∞f (x) =

±∞.A sequence that has limit L is said to converge to L; if there is no limit, the sequence is

said to diverge. We do not permit L to be ±∞ in this usage, so a sequence whose limit is ±∞is said to diverge.

Limit Laws We can prove limit laws for sequences analogous to the limit laws we saw inthe previous course for functions. I shall not give the details in these notes. Remember theusual restriction that we cannot divide by 0, so there must be a restriction on the quotient law.The limit laws can be extended to infinite limits whenever the operations can be justified; so,we can think of∞+∞ = ∞, but we cannot attach a meaning to∞+ (−∞), nor of 0 · ∞, as it isnot possible to assign to these expressions a meaning that will be consistent with the algebraicoperations on real numbers.

Increasing and Decreasing Sequences We defined the concepts of increasing and decreas-ing in connection with real functions of a real variable (cf. [1, p. 20]. The identical defini-tions apply for sequences, where we consider the domain of the function to be the positiveor non-negative integers. When a sequence is either increasing or decreasing we say that it ismonotonic or monotone. Sometimes we find it convenient to work with a slightly weaker prop-erty than increasing, and may speak of a sequence as being non-decreasing, which permits thefunction to remain constant for a while.

A function is bounded above if there exists a number M which is greater than all values ofthe function. For example, the sine function is bounded above, since sin x ≤ 1 for all x in itsdomain. We could also have observed that sin x < 10000, and conclude that such a statementjustifies the conclusion that the function is bounded above: we don’t care how high above thegraph of the function the bounding line appears, only that such a line exists. But the functiontan x is not bounded above. In the same way we can define bounded below and bounded —meaning bounded both above and below. An important theorem we shall need in this chapteris


Theorem C.86 1. A sequence that is bounded above and monotonely increasing (or evenmonotonely non-decreasing) is convergent.

2. A sequence that is bounded below and monotonely decreasing (or even monotonely non-increasing) is convergent.

(Note that the theorem stated here is stronger than [1, Monotonic Sequence Theorem, p., 683].)

Example C.87 [1, Exercise 36, p. 685] Determine whether the sequence an = ln(n + 1) − ln nconverges.Solution: We might be tempted to analyze a difference by using the difference law for limits.But we find that each of the terms approaches +∞, and we cannot give a meaning to ∞ − ∞.So that approach will not work.

However, if we observe that the difference of logarithms is

lnn + 1

n= ln

(1 +

1n

)

we can reason as follows. As n → ∞, 1n → 0, 1 + 1

n → 1 + 0 = 1. By continuity of thelogarithm, the sequence will approach ln 1 = 0. (I am using Theorem C.85 above.)

Sequences of powers For a fixed real number a we know the behavior of a function ax asx→ ∞, and how it depends on a; this permits us to study the behavior of an when the exponentis restricted to integer values:

limn→∞

an = +∞ if a > 1

limn→∞

an = 1 if a = 1

limn→∞

an = 0 if −1 < a < 1

limn→∞

an does not exist if a ≤ −1

(Note that, when a limit is ±∞, as in the first case above, we still say that the limit does notexist!)

Example C.88 Consider the sequence an, where an =n!nn . Does it converge? If so, to what

value?Solution: Consider the limit of the ratio of an+1 to an. It is

an+1

an=

(n + 1)!n!

· nn

(n + 1)n+1 =1(

1 + 1n

)n →1e

as n→ ∞. Thus, for sufficiently large n, every term is less than half of the one before it, whichis positive. Compared to the nth term, the (n + 10)th will be less than 1

1000 of the size; then + 20th will be less than one-millionth the size, etc. The sequence is thus approaching 0.


Example C.89 ([7, Exercise 57, p. 711]) Determine whether the sequence is increasing, de-creasing, or not monotonic. Is the sequence bounded?

an = cosnπ2

Solution: The values of this sequence are (starting at n = 0) 1, 0,−1, 0, 1, 0,−1, 0, . . .. Thesequence is neither increasing nor decreasing, so it is not monotonic, and the [1, MonotonicSequence Theorem, p. 683] is not applicable. The sequence is bounded. It does not converge— but that is not a consequence of the fact that the sequence does not satisfy the conditions of

the theorem. (Compare with the sequencecos nπ

2

nwhich converges to 0.)


[1, Exercise 52, p. 685] Investigate the convergence or divergence of the sequence

an =1 · 3 · 5 · . . . · (2n − 1)

n!

Solution:

an =1 · 3 · 5 · . . . · (2n − 1)

n!

≥ 1 · 2 · 4 · . . . · (2n − 2)n!

=2n−1

n

We can determine the limit of this ratio by L’Hospital’s Rule:

limn→∞

2n−1

n= lim

n→∞2n−1 · ln 2

1= lim

n→∞ln 22· 2n = +∞

since it is a non-zero multiple of the sequence of powers of a constant whose magnitudeexceeds 1; thus the original sequence diverges.

Example C.90 For each of the following sequences, determine whether it converges; if itdoes, find the limit.

1. an =sinh xcosh x

2. an =

√3 + sin n

n


Solution:

1.

limn→∞

sinh xcosh x

= limn→∞

ex − e−x

ex + e−x

= limn→∞

1 − e−2x

1 + e−2x

=

limn→∞

(1 − e−2x

)

limn→∞

(1 + e−2x

)

=11

= 1

2. Consider the fraction under the radical sign: the numerator is bounded, while the de-nominator approaches infinity, so the fraction approaches 0. By a familiar result aboutthe composition of continuous functions, the square root approaches 0.

C.29.2 §11.2 Series

The sum of the terms in a sequence. Having defined what we mean by convergence of asequence an — analogous to the existence, for a function a(x), of lim

x→∞a(x) — we proceed

to apply that concept to generalize the concept of addition. To a mathematician, the operationof addition is a binary operation on numbers, that is, it maps two real numbers on to one realnumber. This operation has a number of properties that mathematicians take as primitive:

• it is commutative: x + y = y + x for all real numbers x and y;

• it is associative: (x + y) + z = (x + y) + z for all real numbers x, y, z;

• it has an “additive identity”, which we call zero and denote by 0, with the property thatx + 0 = x for every real number x;

and other properties that we will not discuss here. The second property permits us to talk ofmore general sums than of just 2 numbers. If we have three real numbers x, y, z, then thesecond property permits us to define x + y + z to be the common value of (x + y) + z and(x + y) + z; we can generalize this idea to the sum of x1 + x2 + . . . + xn, although we will notconsider the details of this definition in this calculus course.54 But the definition we obtain inthis way applies only to a finite sequence of numbers. In this section we consider a broader

54This is a topic that would likely appear in a first Abstract Algebra course, like MATH 235.


definition, to permit us to speak of the sum of all the terms in a sequence. In this context weno longer write the sequence with commas, as

a0, a1, . . . , an, . . . ,

but, instead, place plus signs between the terms, writing

a0 + a1 + . . . + an + . . . ,

or, more compactly, as∞∑

n=0

an ,

and we speak of a series, rather than a sequence. Note that the word series is both singularand plural in English: there is no word “serie” in English55. Our definition proceeds from thepartial sums, which we define to be the terms in the sequence

a0, a0 + a1, a0 + a1 + a2, . . . , a0 + a1 + . . . + an, . . .

or, more compactly, as

0∑

n=0

an,

1∑

n=0

an, . . . ,

m∑

n=0

an, . . . ,

all of which are finite sums, and therefore well defined. (Note that we had to use a differentletter — m — for the general term in this sequence, because the letter n was “busy”56.) We saythat ∞∑

n=0

an = L

if L is the limit of the sequence of partial sums of the series. The series is then said to converge,or to be convergent; if it does not converge, it diverges or is divergent. Here, as with seriesand functions, we generalize to write that a sum = ∞ or = −∞, but still describe such series asdivergent. We will often consider series which we will know to be convergent, without beingable to specify the precise value of the limit.

“Geometric” series. A geometric series is one, each of whose terms after the first is a con-stant multiple of its predecessor. We will often represent such series as

a + ar + ar2 + . . . + arn + . . .

55But the singular word for “series” in French is serie, and the plural is series.56Technically, we call n a bound variable in this case.


where a , 0 and the “common ratio” is the ratio of each term to its predecessor, here denotedby r. By “standard” methods, usually seen in high school, one can prove that the value of thepartial sum of m + 1 terms is

a + ar + . . . + arm =

m∑

n=0

arn =

a (1 − rm)

1 − rif r , 1

(m + 1)a if r = 1.

This ratio can be seen to have the following limit properties:

a + ar + . . . + arm + . . . = limm→∞

m∑

n=0

arn

=

=a

1 − rif −1 < r < 1

= +∞ if r ≥ 1 and a > 0= −∞ if r ≥ 1 and a < 0diverges if r ≤ −1 and a , 0

.


C.30 Supplementary Notes for the Lecture of March 17th, 2010Release Date: Wednesday, March 17th, 2010

C.30.1 §11.2 Series (conclusion)

Divergence of the “harmonic” series. The harmonic series is

1 +12

+13

+ . . . +1n

+ . . . .

In the next section we will see a proof, using an improper integral, that this series diverges.Here we shall see a simpler proof without integrals. We need only observe the followinginequalities:

12 ≥ 1

2

13 + 1

4 >14 + 1

4 = 24 = 1

2

15 + 1

6 + 17 + 1

8 >18 + 1

8 + 18 + 1

8 = 48 = 1

2

. . .1

2n−1 + 1+

12n−1 + 2

+ . . . +1

2n − 1+

12n > 2n−1 · 1

2n =12

Thus we can make the partial sums as large as we like by proceeding out sufficiently far in theseries. It follows that ∞∑

n=1

1n

= +∞ ,

so the harmonic series diverges.

“Necessary” and “sufficient” conditions. Suppose that A and B are sentences that may betrue or false. If A cannot be true except when B is true, we say that the truth of A entails thetruth of B, or that

A implies B

and write this symbolically asA⇒ B .

We call A a sufficient condition for B. An example of such an implication where x and y aremembers of the “universe” of real numbers is

x = y2 − 2y + 4⇒ x ≥ 3


since y2 − 2y + 4 = (y − 1)2 + 3, and a square cannot be negative.We can interpret the statement

A⇒ B

in another way, by observing that, when A is true, B cannot be false. We can say that B is anecessary condition for A, since A cannot be true unless B is true.

Mathematicians usually search for conditions that are both necessary and sufficient, sincethese can characterize a situation. But often we have to be satisfied with conditions that areeither of one type or the other.

A “necessary” condition for convergence of a series The implication that interests use hereis in the “universe” of series of real numbers. It states that

∞∑

n=0

an is convergent ⇒ limn→∞

an = 0 .

This — once it has been proved — is a test that can be applied to a series to see whetherthe series is convergent. It can never prove that a series is convergent! What it can prove,sometimes, is that a certain series is not convergent, since no series that “fails the test” canbe convergent. Many textbooks call this test the “nth term test”; your textbook calls it “‘The’test for divergence”. This is not a standard term, and you might wish to avoid using it, sincea listener who has not read Stewart’s books might not know what you are referring to.57 Inpractice you should “internalize” this test and always apply it, since you might otherwisewaste time trying to prove that a divergent series is convergent.

“Telescoping” series. Sometimes the partial sums can be interpreted as sums where thereis heavy cancellation of intermediate terms, leaving only a few at each end. An example is∞∑

n=1

1n(n + 1)

=

∞∑

n=1

(1n− 1

n + 1

)where the partial sum of the first N terms is simply 1 − 1

N + 1.

(cf. [1, Example 6, p. 691]).

Changes of variable in a sum. We have already seen the concept of changing a variable inconnection with definite integrals. We can carry out similar changes in sums — both finite andinfinite, although we will not investigate all the details. So, for example, we can change the

57You will meet other tests that could be given such a name, although this is the most likely one to distinguishin this way. Stewart’s name for the test may be no more objectionable than the name in general use: it is not goodform to use the symbol n in the name of a test, since the test does not depend on giving that particular name tothe variable that indexes the members of the sequence; my objections to the name Stewart assigns are that (1) it isnot universally accepted; and (2) the definite article suggests uniqueness, and this is not the only test that exists.


name of the index of summation, and writem∑

r=0

ar in place of the summ∑

n=0

an, by a “change of

variable” given by r = n. More generally, we could define s = n + 4, say, and changem∑

n=0

an

intom+4∑

s=4

as−4. Notice how the “limits of summation” have to be changed in the same way that

we changed the limits of integration in a definite integral.

Operations on series. Convergent series may be added term by term, or multiplied by aconstant: ∞∑

n=0

(an + bn) =

∞∑

n=0

an +

∞∑

n=0

bn

∞∑

n=0

can = c∞∑

n=0

an

But note well: while you may add convergent series term by term, you may not rearrange aseries; we shall see later that rearrangement of the terms can result in a series having a differentsum, or even in a convergent series being rendered divergent.

11.2 Exercises

[1, Exercise 36, p. 695] Determine whether the series∞∑

n=1

2n2 + 4n + 3

is convergent or diver-

gent by expressing the partial sum as a telescoping sum (as in [1, Example 6, p. 691]).If it is convergent, find its sum.

Solution: Since

2n2 + 4n + 3

=2

(n + 1)(n + 3)=

1n + 1

− 1n + 3

,

the Nth partial sum is equal to

N∑

n=1

2n2 + 4n + 3

=

N∑

n=1

(1

n + 1− 1

n + 3

)

=

N∑

n=1

1n + 1

−N∑

n=1

1n + 3

=

N∑

n=1

1n + 1

−N+2∑

m=3

1m + 1


setting m = n + 2 in the second sum

=

N∑

n=1

1n + 1

−N+2∑

n=3

1n + 1

replacing the variable m by n

=

2∑

n=1

1n + 1

+

N∑

n=3

1n + 1

−N∑

n=3

1n + 1

−N+2∑

n=N+1

1n + 1

=

2∑

n=1

1n + 1

−N+2∑

n=N+1

1n + 1

=12

+13− 1

N + 2− 1

N + 3.

As N → ∞, this partial sum approaches 12 + 1

3 = 56 . (In this type of problem we don’t

expect students to be write the proof out formally in this way; I have included thissolution so that you can see how the problem can be solved “properly”, without the useof “dots”.


n=1

lnn

n + 1is convergent or divergent

by expressing sn is a telescoping sum (as in [1, Example 6, p. 691]). If it is convergent,find its sum.

Solution: In a similar way to that shown above in the solution to [1, Exercise 36, p. 695],we can show that the partial sum of the first N terms of the series is ln 1 − ln N + 1. AsN → ∞ the partial sum approaches −∞: the series diverges.

[1, Exercise 44, p. 695] Express the “repeating decimal expansion” 6.254 as a ratio of inte-gers.

Solution: By a “repeating decimal expansion”, we intend that the digits under the hori-zontal line are to be repeated as a subsequence of the expansion indefinitely; thus

6.254 = 6.25454545454... ,

which we can interpret as the sum of a series

6 +2

10+

541000

+54

100000+

5410000000

+ . . .

which can be seen as a constant added to a geometric series:(6 +

210

)+

541000

1 +1

100+

(1

100

)2

+

(1

100

)3

+ . . .


whose sum is6 +

210

+54

1000· 1

1 − 1100

= 6 +15

+3

55=

34455

.

(In this way we can show that any “repeating decimal” is a rational number.)

[1, Exercise 50, p. 695] Find the values of x for which the series∞∑

n=0

(x + 3)n

2n converges. Find

the sum of the series for those values of x.

Solution: The given series can be rewritten as∞∑

n=0

((x + 3)

2

)n

: it can be viewed as a

geometric series with initial term 1, and common ratiox + 3

2. We know that a geometric

series converges if and only if its common ratio is less than 1 in magnitude. Thus the

given series converges iff∣∣∣∣∣x + 3

2

∣∣∣∣∣ < 1. Solving this inequality for x we obtain

∣∣∣∣∣x + 3

2

∣∣∣∣∣ < 1 ⇔ −1 <x + 3

2< +1

⇔ −2 < x + 3 < 2⇔ −5 < x < −1 .

Thus the given series converges for and only for x ∈ (−5,−1). For x in this interval thesum of the series is

first term1 − common ratio

=1

1 − x+32

= − 21 + x

.

C.30.2 §11.3 The Integral Test and Estimates of Sums

Thus far we have met only one “test” that can be applied to series to investigate whether theseries converges. That is the test Stewart calls “The Test for Divergence”, and it gives negativeinformation: it states that, if lim

n→∞an does not exist, or if the limit does exist but its value is

not 0, then the series∞∑

n=0

an is divergent; this test cannot be used to confirm a suspicion that a

series does converge. This is the only test we shall have that can be applied to “general” series,where the signs of the terms do not follow a specific pattern.

The Integral Test. The idea of this test is to interpret the terms of a series∞∑

n=1

an as a sum

of areas. We can do this by interpreting an as the area of a rectangle whose width is 1, and


whose height is an. The test will be restricted to sequences where the terms are monotonelydecreasing. Suppose that we know of a function f defined on the interval [1,∞), whose graphpasses through the left upper end-points of the rectangles. Then the area under the curve willbe less than the sum of the areas of the rectangles. If we can show that the area under the curveis infinite, we will be able to conclude that the sum of the series is divergent. In a similar way,if we can pass the graph of a function f through the right upper endpoints of the rectangles,then the area under the curve will exceed the sum of the areas of the rectangles: if the areaunder the curve is convergent, then the same can be said about the series. In these ways wecan shown that the series converges precisely when the integral representing the area underthe curve converges! We call this result The “Integral” Test.


C.31 Supplementary Notes for the Lecture of March 19th, 2010Release Date: Friday, March 19th, 2010; corrected on 08 April, 2010; subject to further

correction

Review of the preceding lecture

• Theorem:∞∑

i=0

ai converges⇒ limn→∞

an = 0

• The contrapositive of this theorem is “The” Test for Divergence: If limn→∞

an either does

not exist, or does exist but does not equal 0, then the series∞∑

i=0

ai diverges.

• The harmonic series diverges — proved by grouping successive terms.

• “telescoping” series

• evaluating a “repeating” decimal expansion as a rational number

• The Integral Test — requiring the association with a positive series of a continuous,decreasing, positive function f taking, at the positive integer points, values equal to theterms of the series

C.31.1 §11.3 The Integral Test and Estimates of Sums (conclusion)

Theorem C.91 (The Integral Test) If f is a continuous, positive, decreasing58 function de-fined on the interval [1,∞), and if the sequence an has the property that an = f (n). Then

∞∑

n=1

an is convergent ⇔∞∫

1

f (x) dx is convergent .

The convergence or divergence of both the series and the improper integral do not depend onthe lower limits of the sum and the definite integral. (However, when, below, we investigatebounds for the actual value of the sum of the series, then our bounds will depend on the specificvalues we choose for the limits of sum and integral.)

58Actually, it suffices for the function to be non-increasing: we can permit the function values to stay at thesame level so long as they eventually decrease again and eventually approach 0.


Tests of Positive Series for convergence. Most of the tests we shall meet require a specificarrangement of signs for the terms. In fact, all but one of them will require that all the signsare the same. We call such series “positive” series, thinking of all the signs as being +; butsimilar results hold when all signs are −. The first test of positive series will be discussed inthis section. I do not think of this as the most elementary of the tests, and would have preferredto discuss the material of the next sections first; but I will grudgingly follow the order of topicsin the textbook.

“Ultimate” satisfaction of properties imposed on the terms of a series. In the statementgiven for the a sum beginning with the 1st term, and compared it with the integral whoselower bound was x = 1. Both of these lower bounds can be replaced by any integers youlike, provided the series and function are defined whenever you refer to them. Changing theselower bounds does not change the truth of the logical equivalence. It could, however, changethe value of the sum of the series or the definite integral.

Application: the “p-series”. The “p-series” are the series∑ 1

np , where p is a positiveconstant. The integral test shows that

∑ 1np is convergent if and only if p > 1 .

The case p = 1 is the harmonic series, which we have already shown to be divergent, usinga different proof. (This application is sometimes called the “p-series test”.) Since the testswe will meet in [1, §11.4] involve comparing a given series with series that we know to beconvergent and divergent, the p-series are particularly important as they provide us with afamily of series that can be used for comparison purposes.

Estimating the Sum of a Series In the proof of the Integral Test we compared the sum of aseries of decreasing positive terms with the area under the graph of a function that is positiveand decreasing. This comparison can be refined, and gives rise to inequalities that bound thevalue of the sum from below and above.

By associating with a decreasing, positive series∑

an a decreasing positive function f , we

are able to “trap” the value of the partial sumsN∑

n=1

an between the definite integrals giving the

areas for regions respectively contained in and containing the region represented by the partialsum, when interpreted as the area of a region formed by rectangles of width 1 and respectiveheights 1, 2, . . ., N. From the inequalities

N∑

n=1

an ≤ a1 +

∫ N

1f (x) dx


N∑

n=1

an ≥∫ N+1

1f (x) dx

we are able to relate the convergence of the infinite sum to that of the improper integral∫ ∞

1f (x) dx.

Let us, extending the notation used in the enunciation of the theorem, define

Rn =

∞∑

m=n+1

am;

that is, Rn is the sum of the “tail” or Remainder of the series, starting with the (n + 1)st term.Then ∫ ∞

n+1f (x) dx ≤ Rn ≤

∫ ∞

nf (x) dx ,

or, alternatively, ∫ ∞

n+1f (x) dx ≤ Rn ≤ an+1 +

∫ ∞

n+1f (x) dx ,

or0 ≤ Rn −

∫ ∞

n+1f (x) dx ≤ an+1 .

In particular, when n = 0, this gives the inequalities

0 ≤∞∑

n=1

an −∫ ∞

1f (x) dx ≤ a1 .

While, for the purposes of testing convergence, it is sufficient to demand “ultimate” sat-isfaction of the conditions of the Integral Test, this is not sufficient if we wish to determinebounds for the sum. In that case the estimations discussed above depend on the comparison ofthe area under a curve and the sum of the areas of the step function that represents the series;these comparisons are valid only when the condition the the function be decreasing is satisfied.It is possible to refine this type of bounding of partial sums and remainders, as in [1, Examples5,6, pp. 701-702], but we will not go further in this course.

Example C.92 We proved earlier that the telescoping series∞∑

n=1

1n(n + 1)

converges, and we

found the sum. The integral test could be used to prove that it converges, but the test wouldnot give the exact value of the sum, although it would provide bounds which compare the sum


to the improper integral∫ ∞

1

1x(x + 1)

dx = lima→∞

((ln a − ln 1) − (ln(a + 1) − ln 2))

= lima→∞

(ln

aa + 1

+ ln 2)

= ln 1 + ln 2 = 0 + ln 2 = ln 2 ,

so the sum of the series is bounded between ln 2 = 0.6931471806 and ln 2+ 12 = 1.1931471806,

which is a weaker result than the fact we already know that the sum is equal to 1.

11.3 Exercises

[1, Exercise 6, p. 703] Use the Integral Test to determine whether the series∞∑

n=1

1√n + 4

is

convergent or divergent.

Solution: We begin by verifying that the Integral Test is applicable to this series:

• Since the terms are reciprocals of square roots, they are positive.

• The function f (x) =1√

x + 1is decreasing. This can be shown either by showing

that f ′(x) = − 12 (x + 1)−

32 < 0, or by the following reasoning:

x is an increasing function of x⇒ x + 1 is an increasing function of x⇒ √

x + 1 is an increasing function of x since√

preserves order

⇒ 1√x + 1

is a decreasing function of x.

The Integral Test tells us that the series will converge iff the improper integral

∞∫

1

dx√x + 1

converges. But

∞∫

1

dx√x + 1

= lima→∞

a∫

1

dx√x + 1

= lima→∞

[2√

x + 1]a

1= +∞ .

so the series must also diverge.

[1, Exercise 20, p. 704] To investigate the convergence of∞∑

n=1

1n2 − 4n + 5

.


Solution: We know that∫

1x2 − 4x + 5

dx =

∫1

(x − 2)2 + 1dx = arctan(x − 2) + C ,

and might be tempted to compare the integral∫ ∞

1

1x2 − 4x + 5

dx with the series. This,

however, is not directly possible from the Integral Test, because the function f (x) =1

x2 − 4x + 5is decreasing only for x ≥ 2: from x = 1 to x = 2 the function is increasing.

Thus we can apply the integral test to the series∞∑

n=2

1x2 − 4x + 5

, provided we first verify

that the necessary conditions are satisfied by the function f whose values at the positiveintegers coincide with the terms of the series:

1. Is f decreasing?

ddx

f (x) =−(2x − 4)(

x2 − 4x + 5)

=−2(x − 2)

((x − 2)2 + 1

)2 .

The numerator is negative for x > 2; the denominator is a non-zero square, alwayspositive. Thus the ratio is negative for x > 2, and so f is decreasing for x > 2.

2. Is f continuous? f is a rational function, and [1, Theorem 5(b), p. 122] all rationalfunctions are continuous on their domain. (The domain of f is all of R.)

Bounding the sum for n ≥ 2 from below yields

∞∑

n=2

1n2 − 4n + 5

≥∫ ∞

2

dxx2 − 4x + 5

= lima→∞

arctan(a − 2) − arctan 0

=π

2− 0 =

π

2.

Bounding the sum for n ≥ 3 from above yields

∞∑

n=3

1n2 − 4n + 5

≤∫ ∞

2

dxx2 − 4x + 5

=π

2,


which implies that

∞∑

n=2

1n2 − 4n + 5

≤ 122 − 4(2) + 5

+

∫ ∞

2

dxx2 − 4x + 5

= 1 +π

2.

Thus the series∞∑

n=1

1n2 − 4n + 5

converges, and its sum is bounded between1 + π

2and

3 + π

2. (If we were to add more of the terms at the beginning of the series manually, we

could determine much “tighter” bounds for the sum of the series.)

[1, Exercise 28, p. 704] Find the values of p for which the series∞∑

n=3

1n ln n[ln(ln n)]p is con-

vergent.

Solution: The denominators of the fractions being summed are products of n, ln n, anda power of ln ln n, all of which are increasing functions of n. Hence we may apply the

Integral Test with the function f (x) =1

x(ln x) ([ln(ln x)]p), confident that it is decreasing

and continuous. When x ≥ 3, ln x > 1.098 > 1, so ln ln x > 0, and f (x) > 0. Whenp = 1,

a∫

3

dxx ln x[ln(ln x)]1 = [ln ln ln x]a

3 = ln ln ln a − ln ln ln 3→ ∞

as a→ ∞; hence, for the series to converge, we must certainly have p > 1; (any positivepower of p less than 1 will yield an integral which is larger than the integral we obtainwhen p = 1). But, when p > 1,

a∫

3

dxx ln x[ln(ln x)]p = − 1

p − 1·[

1(ln ln ln x)p−1

]a

3

→ 1p − 1

1ln ln ln 3

as a → ∞; since the improper integral is divergent for 0 < p < 1, the positive series isalso divergent. Thus the given series diverges for 0 < p ≤ 1, and converges for p > 1 —the same values for the exponent as for p-series.



C.32 Supplementary Notes for the Lecture of March 22nd, 2010Distribution Date: Monday, March 22nd, 2010, subject to further revision

Review of the last lecture

• After reviewing the previous lecture, I completed my treatment of the Integral Test.

• It time permits, I may return to further discussion of the use of integrals to bound series.

• Discussion included [1, Exercise 28, p. 704],∞∑

n=3

1n(ln n) (ln ln n)p , which can be shown

to behave “similarly” to p-series.

• There are often many, significantly different choices of series with which to compare.You need to work problems and check your solutions afterward in the manual to developthe necessary experience.

C.32.1 §11.4 The Comparison Tests

We continue to study series in a “reverse-logical” order. That is, as we proceed, we will beencountering properties that are more and more basic. In this section we will meet anothertheorem that — like the Integral Test — applies only to series whose terms all have the samesign, the so-called “positive series”, or “series of positive terms”. Finally, in [1, §11.6] we willsee a theorem [1, Theorem 3, p., 715] that will justify our continued investigation of serieswith positive terms. When we have covered all the sections on the syllabus, I hope to return tothe various tests.

The comparison tests are designed so that we can infer the convergence or divergence ofa given positive series by comparison of its terms with those of another series that we knowto be convergent or divergent. So, in order to use this test, we need to have available as largeas possible a family of positive series whose convergence or divergence are known. Let’sremember for which series we know such facts; these include

• positive geometric series of common ratio less than 1 (convergent)

• positive geometric series of common ration 1 or more (divergent)

• harmonic series (divergent)

• p-series (convergent if and only if p > 1)

• positive series whose terms do not approach 0 (divergent)


• series obtained from a given series by deleting any finite number of terms at the begin-ning (have the same convergence properties as the original series)

• series obtained from a given series by multiplying by a positive constant (have the sameconvergence properties as the original series)

Theorem C.93 (Comparison Test) Let∑

an be a series with positive terms.

1. If∑

bn is a series of positive terms such that bn ≤ an for all n, then∑

an converges ⇒∑

bn converges .

2. If∑

bn is a series of positive terms such that an ≤ bn for all n, then∑

an diverges ⇒∑

bn diverges .

3. In either of the preceding results, the words “for all n” may be replaced by “for all nsufficiently large”, which means, in mathematical jargon, that all we require is that theinequality holds after some integer, which could be as large as you like; if the inequalityfails a finite number of times, even an enormously large finite number of times, that issufficient for the purposes of the hypotheses of the theorem.

Prior to applying this test we may first adjust the series∑

an that we are using for comparisonpurposes by

• deleting or otherwise changing a finite number of terms at the beginning;

• multiplying by a positive constant

The following related test is sometimes easier to use:

Theorem C.94 (Limit Comparison Test) Let∑

an be a series with positive terms. If∑

bn

is a series of positive terms such that

limn→∞

an

bn= c > 0

then ∑an diverges ⇔

∑bn diverges .

∑an converges ⇔

∑bn converges .

(There is a sharper version of this theorem that permits the limit to be either 0 or ∞; but it isdelicate, and difficult for students to remember, so we usually do not discuss it in this course,cf. [1, Exercises 40-42, p. 709].)


Estimating Sums This question will be illustrated by the solution below of part of [1, Exer-cise 32, p. 709].

Example C.95 ([7, Exercise 20, p. 735]) Determine whether the series∞∑

n=1

1 + 2n

1 + 3n converges

or diverges.

Solution: We could not compare the terms of this series naively with those of2n

3n , which is aconvergent geometric series because the terms here are slightly larger. However a number ofother possibilities suggest themselves. For example, we have

1 + 2n

1 + 3n ≤1 + 2n

3n =

(13

)n

+

(23

)n

< 2(23

)n

which is the general term of a geometric series with common ratio 23 < 1, which is conver-

gent. Other comparisons are possible, some of them complicated to prove. For the purpose ofproving only convergence (and not bounding the series) one tries to find the simplest possiblecomparison.

This problem can also be solved by applying the Limit Comparison Test, again in compar-ison with

∑(23

)n:

limn→∞

1 + 2n

1 + 3n(23

)n = limn→∞

(1 + 2n

2n · 3n

1 + 3n

)

= limn→∞

((2−n + 1

) · (3−n + 1))

= 1 × 1 = 1.

Since the limit is a positive real number, and since the geometric series∑(

23

)nconverges, the

given series must also converge.

Example C.96 ([7, Exercise 34, p. 735]) Estimate the error when the sum of the first 10 terms

of the following sum is used to approximate the sum of the series:∞∑

n=1

1 + cos nn5 .

Solution: The “tail” of the series will be estimated from above. Observe that

1 + cos nn5 ≤ 2

n5 .

We know that1n5 ≤

1x5 for n − 1 ≤ x ≤ n .


Comparing the sum of the terms on the left — interpreted as rectangles of width 1 lying underthe graph of f (x) = 1

x5 — we have

∞∑

n=11

1n5 ≤

∫ ∞

10

dxx5 =

1104 .

It follows that the “tail” of the series — the sum beginning with term n = 11 — is no greaterthan 2 × 10−4. This gives an upper bound for the error if the series is truncated at term n = 10.This method will not give a useful lower bound for the error, however, since we know only that

1 + cos nn5 ≥ 0

n5 .

This is certainly not the best possible lower bound, but we can’t do better in this course.

11.4 Exercises In some of the discussions below I give only hints about the solution.

[1, Exercise 12, p. 709] Determine whether the series converges or diverges:∞∑

n=0

1 + sin n10n .

Solution: The Limit Comparison Test cannot be used, since the terms behave erraticallybecause of the sine function, and there is not likely to exist a limit, no matter what seriesyou choose to compare with. But

1 + sin n10n <

210n ,

so the terms of the series are smaller than the terms of a geometric series with commonratio 1

10 , which is less than 1, and must converge.


n=0

√n

n − 1.

Solution: The terms can be made smaller by making the denominator larger, into n. Thatyields a divergent p-series.


n=0

12n + 3

.

Solution: The terms can be made smaller by making the denominator into 2n + 4. Thenew series is a multiple

(12

)of a harmonic series. The harmonic series diverges, and that

property doesn’t change if we discard 2 terms at the beginning, nor if we then multiplyby 1

2 .



n=1

n!nn converges or diverges.

Solution: The general term is

n!nn =

1n· 2

n· 3

n· . . . · n − 1

n· n

n

≤ 1n· 2

n· 1 =

2n2

for n ≥ 3, which is the general term in a convergent p-series. Hence the given seriesis convergent. It is impractical to use the limit comparison test here, because the limitwould be difficult to compute, and because it would be zero or infinite, depending onwhich series is in the numerator; our version of this test does not permit the limit to beeither 0 or∞.


n=1

sin1n

converges or diverges.

Solution: We know the value of limn→∞

sin 1n

1n

to be 1; so the Limit Comparison Test shows

this series diverges because the harmonic series diverges. Could we have used the Com-parison Test? Unfortunately, the geometric argument that we used to prove the resultquoted earlier showed [1, p. 190-191] that

0 < sin1n<

1n.

In its present form this inequality is not useful for proving the divergence of the givenseries, since the series known to be divergent occupies the outer position rather thanlying between the terms of the given series and 0.


n=1

1

n1+ 1n

converges or diverges.

Solution: It appears that the series “resembles” the Harmonic Series, and we will try tocompare it in the limit. The guess is a fortunate one, since the limit does exist.

limn→∞

1n1+ 1

n

1n

= limn→∞

1

n1n

= limn→∞

1

eln nn

=1

e limn→∞

ln nn

by continuity of exponential


=1e0 = 1 by L’Hospital’s Rule

We apply the Limit Comparison Test: since the Harmonic Series diverges and the ratioof the terms of the given series to those of this divergent series approaches a non-zeropositive real number, the given series also diverges.

The same argument could have been used to prove that the series∞∑

n=1

1

n1+ 1ln n

is also divergent; the terms in this last series are smaller than those of the series we weregiven.

C.32.2 Sketch of Solutions to Problems on the Final Examination in MATH 141 200601

(not for discussion at the lecture)

The best way to prepare for the examination in MATH 141 2009 01 is to study the textbook,working problems, and verifying your solutions using the Student Solutions Manual: for sev-eral reasons I don’t recommend your studying exclusively from old examinations. Howeverthe old examinations do have some uses, and I am including a second one of them here, withsolutions. I am hoping to discuss last April’s examination briefly in a final lecture; there willnot be time for me to discuss the following examination in the lectures.


For each of the following series you are expected to apply one or more tests for conver-gence or divergence and determine whether the series is convergent. In each case youmust answer 3 questions:




(a) [4 MARKS]∞∑

n=1

1(tanh n)2 + 1

Solution: As n → ∞, tanh n = en−e−n

en+e−n = 1−e−2n

1+e−2n → 11 = 1 , so the terms summed in

this series have limit 112+1 = 1

2 , 0. By the “Test for Divergence” this series cannotconverge.


(b) [4 MARKS]∞∑

n=1

n2ne−n2

Solution:

nth root of nth term =n√

n2ne−n2= n2e−n =

n2

en .

By l’Hospital’s Rule (∞∞ case) the limit of this nth root as n→ ∞ is

limn→∞

n2

en = limn→∞

2nen = lim

n→∞2en = 0 < 1 .

By the Root Test, the series is convergent.

(c) [4 MARKS]∞∑

n=1

n2 − 85n + 12n(n + 6)2

Solution: The ratio of the nth term of this series to the nth term of the Harmonic

series∞∑

n=1

1n is

n2 − 85n + 12(n + 6)2 =

1 − 85n + 12

n2(1 + 6

n

)2 → 1 , 0 .

By the Limit Comparison Test the given series is divergent, since the Harmonicseries is divergent.



−1|x| dx .

Solution:∫ 2

−1|x| dx =

∫ 0

−1|x| dx +

∫ 2

0|x| dx

=

∫ 0

−1(−x) dx +

∫ 2

0x dx

= −[

x2

2

]0

−1+

[x2

2

]2

0

=12

+ 2 =52

(b) [3 MARKS] Evaluate

e3∫

1

dt

t√

1 + ln t.


Solution: Under the substitution u = ln t, du = dtt and

e3∫

1

dt

t√

1 + ln t=

∫ 3

0

du√1 + u

= 2(1 + u)12]3

0= 2(√

4 −√

1) = 2(2 − 1) = 2


∫ x2

0et2 dt .

Solution: With u = x2,

ddx

∫ x2

0et2 dt =

ddx

∫ u

0et2 dt = eu2

2x = e(x2)2

2x = 2xex4,

by the Fundamental Theorem of Calculus.

(d) [4 MARKS] Evaluate

limn→∞

1n

(0n

)7

+

(1n

)7

+

(2n

)7

+ . . . +

(n − 1

n

)7 .

Solution: You are asked to evaluate the limit of a Riemann sum for∫ 1

0x7 dx =[

x8

8

]1

0= 1

8 , (where the values of the function are taken at the left end-point of eachsubinterval). Accordingly the value is 1

8 .

SHOW ALL YOUR WORK!

3. For each of the following series you are expected to apply one or more tests for conver-gence or divergence and determine whether the series is convergent. In each case youmust answer 3 questions:




(a) [4 MARKS]∞∑

n=1

1(tanh n)2 + 1

Solution: As n → ∞, tanh n = en−e−n

en+e−n = 1−e−2n

1+e−2n → 11 = 1 , so the terms summed in

this series have limit 112+1 = 1

2 , 0. By the “Test for Divergence” this series cannotconverge.


(b) [4 MARKS]∞∑

n=1

n2ne−n2

Solution:

nth root of nth term =n√

n2ne−n2= n2e−n =

n2

en .

By l’Hospital’s Rule (∞∞ case) the limit of this nth root as n→ ∞ is

limn→∞

n2

en = limn→∞

2nen = lim

n→∞2en = 0 < 1 .

By the Root Test, the series is convergent.

(c) [4 MARKS]∞∑

n=1

n2 − 85n + 12n(n + 6)2

Solution: The ratio of the nth term of this series to the nth term of the Harmonic

series∞∑

n=1

1n is

n2 − 85n + 12(n + 6)2 =

1 − 85n + 12

n2(1 + 6

n

)2 → 1 , 0 .

By the Limit Comparison Test the given series is divergent, since the Harmonicseries is divergent.

4. BRIEF SOLUTIONS Express the value of each of the following as a definite integralor a sum, product, or quotient of several definite integrals, but do not evaluate the inte-gral(s). It is not enough to quote a general formula: your integrals must have integrandand limits specific to the given problems, and should be simplified as much as possible,except that you are not expected to evaluate the integrals.

(a) [3 MARKS] Expressed as integral(s) along the x-axis only, the area of the regionbounded by the parabola y2 = 2x + 6 and the line y = x − 1. An answer involvingintegration along the y-axis will not be accepted.

DEFINITE INTEGRAL(S) ONLY (DO NOT EVALUATE)


Solution: ∫ −1

−32√

2(x + 3) dx +

∫ 5

−1

( √2(x + 3) − (x − 1)

)dx

(b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 theregion bounded by the curves y = x3 and y = x2. For this question you are to useonly the method of “washers”.


Solution: Only the first integral was required.

π

∫ 1

0

((1 − x3

)2 −(1 − x2

)2)

dx

= π

∫ 1

0

(−2x3 + x6 + 2x2 − x4

)dx

= π

[−2x4

4+

x7

7+

2x3

3− x5

5

]1

0

= π

(−2

4+

17

+23− 1

5

)=

23π210

.

(c) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 theregion bounded by the curves y = x3 and y = x2. For this question you are to useonly the method of “cylindrical shells”.


Solution: Only the first integral was required.

2π∫ 1

0(1 − y)

(y

13 − y

12)

dy


= 2π∫ 1

0

(y

13 − y

43 − y

12 + y

32)

dy

= 2π(34− 3

7− 2

3+

25

)=

23π210

(d) [3 MARKS] The length of the curve whose equation is

x2

4+

y2

9= 1 .


Solution: This is a special case of [7, Exercise 20, p. 553].

x2

4+

y2

9= 1 ⇔ 9x2 + 4y2 = 36

⇒ dydx

= −9x4y

⇒√

1 +

(dydx

)2

=

√81x2 + 16y2

4|y|

⇒ arc length = 2∫ 2

−2

12

√16 + 5x2

4 − x2 dx

where the factor 2 is needed because the integral is the length of either the top orthe bottom arc of the ellipse; this integral is improper. A cleaner solution is foundby parameterizing the curve as x = 2 cos θ, y = 3 sin θ (0 ≤ θ ≤ 2π). Then the arclength is ∫ 2π

0

√4 sin2 θ + 9 cos2 θ dθ .



x5 + xx4 − 16

dx .


Solution: The first step is to divide denominator into numerator:

x5 + x = x(x4 − 16) + 17x⇒ x5 + xx4 − 16

= x +17x

x4 − 16.

Next we could separate the resulting remainder fraction, whose numerator has degreeless than that of its denominator, into partial fractions:

17xx4 − 16

=Ax + Bx2 − 4

+Cx + Dx2 + 4

=E

x − 2+

Fx + 2

+Cx + Dx2 + 4

⇒ 17x = E(x + 2)(x2 + 4) + F(x − 2)(x2 + 4)+Cx(x − 2)(x + 2) + D(x − 2)(x + 2)

We can determine the constants by a combination of the standard methods: giving xconvenient values, and equating coefficients of like powers of x:

x = 2 ⇒ 34 = 32E ⇒ E =1716

x = −2 ⇒ −34 = −32F ⇒ F =1716

x = 0 ⇒ 0 = 6E − 6F − 8D⇒ D = 0

coeff x3 : ⇒ 0 = E + F + C ⇒ C = −178

Hencex5 + x

x4 − 16= x +

1716

(1

x − 2+

1x + 2

− 2xx2 + 4

)

Hence∫

x5 + xx4 − 16

dx =x2

2+

1716

ln

∣∣∣∣∣∣x2 − 4x2 + 4

∣∣∣∣∣∣ + C

A better tactic would have been to make a substitution u = x2 in the integral prior toseparation into partial fractions:∫

17xx4 − 16

dx =172

∫du

(u − 4)(u + 4)=

1716

∫ (1

u − 4− 1

u + 4

)du =

1716

ln∣∣∣∣∣u − 4u + 4

∣∣∣∣∣ + C


Showing all your work, evaluate each of the following:


(a) [4 MARKS]∫

cos x · cosh x dx

Solution: Using integration by parts,

u = cos x ⇒ du = − sin x dxdv = cosh x dx ⇒ v = sinh x

⇒∫

cos x · cosh x dx = (cos x)(sinh x) +

∫sinh x · sin x dx .

A second integration by parts yields

U = sin x ⇒ dU = cos x dxdV = sinh x dx ⇒ V = cosh x

⇒∫

sin x · sinh x dx = (sin x)(cosh x) −∫

cosh x · cos x dx

⇒∫

cos x · cosh x dx = (cos x)(sinh x) + (sin x)(cosh x) −∫

cos x · cosh x dx.

Now we can move the integral from the right side of the equation to join the sameintegral with opposite sign on the left side, yielding

2∫

cos x · cosh x dx = (cos x)(sinh x) + (sin x)(cosh x) + C

or ∫cos x · cosh x dx =

(cos x)(sinh x) + (sin x)(cosh x)2

+ C′ .

(b) [5 MARKS]

1∫

−3

√x2 + 2x + 5 dx

Solution: We begin by completing the square:

1∫

−3

√x2 + 2x + 5 dx =

1∫

−3

√(x + 1)2 + 4 dx .

Then the integrand can be changed to the form of√

X2 + 1:

1∫

−3

√(x + 1)2 + 4 dx = 2

1∫

−3

√(x + 1

2

)2

+ 1 dx .


Here an appropriate substitution would be to set x+12 equal to either tan θ or sinh θ;

I have chosen to use the trigonometric substitution:

x + 12

= tan θ ⇒ θ = arctanx + 1

2⇒ dx = 2 sec2 θ dθ

⇒1∫

−3

√x2 + 2x + 5 dx = 2

∫ π4

− π4| sec θ| · 2 sec2 θ dθ

⇒1∫

−3

√x2 + 2x + 5 dx = 8

∫ π4

0sec3 θ dθ

Finally we have to find an antiderivative of sec3 θ, which can be integrated by“standard” methods, e.g., by integration by parts with u = sec θ, dv = sec2 θ dθ:

du = sec θ tan θ dθv = tan θ∫

sec3 θ dθ = sec θ tan θ −∫

sec θ tan2 θ dθ

= sec θ tan θ −∫

sec θ(sec2 θ − 1) dθ


sec3 θ dθ +

∫sec θ dθ


sec3 θ dθ + ln | sec θ + tan θ| + C

⇒ 2∫

sec3 θ dθ = sec θ tan θ + ln | sec θ + tan θ| + C

⇒∫

sec3 θ dθ =sec θ tan θ + ln | sec θ + tan θ|

2+ C′ .

Here1∫

−3

√x2 + 2x + 5 dx = 8

∫ π4

0sec3 θ dθ = 4

√2 + 4 ln(

√2 + 1) .

(c) [4 MARKS]∫

sin2 x · cos2 x dx

Solution: One way to simplify this integral is to use the identity

sin x cos x =sin 2x

2:


∫sin2 x · cos2 x dx =

14

∫sin2 2x dx

=18

∫(1 − cos 4x) dx =

18

(x − sin 4x

4

)+ C

=x8− sin 4x

32+ C .

This could also have been simplified using the algorithm proposed in the textbook:∫

sin2 x · cos2 x dx =14

∫(1 − cos 2x)(1 + cos 2x) dx

=14

∫(1 − cos2 2x) dx

=14

∫ (1 − 1 + cos 4x

2

)dx

=18

∫(1 − cos 4x) dx etc.



x = x(t) = 10 − 3t2

y = y(t) = t3 − 3t ,

where −∞ < t < +∞.

(a) [8 MARKS] Determine the value ofd2ydx2 at the points where the tangent is horizon-

tal.Solution:

dxdt

= −6t

dydt

= 3t2 − 3

dydx

=

dydtdxdt

=t2 − 1−2t

= −12

(t − 1

t

)

d2ydx2 =

ddx

(dydx

)


=ddt

(dydx

)· dt

dx

=

ddt

(dydx

)

dxdt

=

ddt

(−1

2

(t − 1

t

))

dxdt

=

−12

ddt

(t − 1

t

)

−6t

= −12

1 +1t2

−6t=

t2 + 112t3 .

The tangent is horizontal whendydx

= 0, i.e., when t2 − 1 = 0 or t = ±1. At suchparameter values

d2ydx2 =

2±12

= ±16.

(b) [4 MARKS] Determine the area of the surface of revolution about the x-axis of thearc

(x(t), y(t)) : −√

3 ≤ t ≤ 0.

Solution: The area of the surface of revolution about the x-axis is

2π∫ 0

−√3y(t)

√(dxdt

)2

+

(dydt

)2

dt = 2π∫ 0

−√3(t3 − 3t)

√36t2 + 9(t2 − 1)2 dt

= 2π∫ 0

−√3(t3 − 3t)(3t2 + 3) dt

= 2π∫ 0

−√3(t5 − 2t3 − 3t) dt

= 6π[t6

6− 2t4

4− 3t2

2

]0

−√3= 27π .



(a) [5 MARKS] Showing detailed work, determine whether the following integral isconvergent; if it is convergent, determine its value:

∫ 0

−1

dx

x23

.

Solution: This integral is improper because the integrand is undefined, hence dis-continuous at the right endpoint.

∫ 0

−1

dx

x23

= limb→0−

∫ b

−1

dx

x23

by definition

= limb→0−

[3x

13]b

−1= 3 lim

b→0−b

13 − (−3) = 0 + 3 = 3.

Since the limit exists, the improper integral is convergent.


∞∑

n=1

(−1)n

n − ln n

Solution: Let f (x) = 1x − ln x . then

f ′(x) = − 1 − 1x

(x − ln x)2 .

For x > 1, 1− 1x > 0, so f ′(x) < 0. Thus the terms of the given series are monotonely

decreasing in magnitude, and alternating in sign.We observe that

limx→∞

ln xx

= limx→∞

1x

1= 0;

it follows that the limit of the ratio of corresponding terms of the given series andthe series

∑ 1n is

limn→∞

1n−ln n

1n

= limn→∞

11 − ln n

n

=1

1 − limn→∞

ln nn

= 1 .

Since the Harmonic Series diverges, this implies (by the Limit Ratio Test) that theseries of absolute values of the given series is divergent; and the given series iseither conditionally convergent or divergent. The limit of the terms of the series is

limn→∞

1n − ln n

= limn→∞

1n· lim

n→∞1

1 − ln nn

= 0 · 1 = 0 .


This, combined with the property that the terms of the given series are alternatingin sign and monotonely decreasing in magnitude, permits us to conclude by theLeibniz Test that the series converges. Thus this alternating series is convergent,but not absolutely convergent: it is conditionally convergent.

(c) [3 MARKS] Give an example of a sequence an with the property that limn→∞

an = 0

but∞∑

n=1

an = +∞. You are expected to give a formula for the general term an of

your sequence.Solution: The Harmonic Series an = 1

n is one example of a series which has thedesired property.


[12 MARKS] The arcr = 1 − cos θ (0 ≤ θ ≤ π)

divides the area bounded by the curve

r = 1 + sin θ (0 ≤ θ ≤ 2π)

into two parts. Showing all your work, carefully find the area of the part that containsthe point (r, θ) =

(12 ,

π2

).

Solution: Since the functions defining the curves are both periodic with period 2π, wecan confine our attention to any parameter interval of length 2π. We need to determinewhere the curves intersect. Solving the two given equations algebraically gives tan θ =

−1, implying that θ = 3π4 ,

7π4 in the interval 0 ≤ θ ≤ 2π. The only point whose parameter

value is in the intervals given for the two arcs is θ = 3π4 . However, we always need to be

alert to the possibility that there are intersections of the curves in which the parametervalues could be different on the two curves. One place where precisely that occurs forthese two curves is the pole: on r = 1−cos θ the pole appears as (r, θ) = (0, 0), which is inthe restricted interval 0 ≤ θ ≤ π; on the cardioid r = 1 + sin θ the pole appears as

(0, 3π

2

).

Could there be any other points of intersection? Any others could be found by solvingthe variants of the equations obtainable by applying the transformation (r, θ) = (−r, θ+π)repeatedly. We can show that there are no such points that appear in this way.

It follows that the region whose area we seek is bounded by two arcs:

• The arc of r = 1 − cos θ bounded by 0 ≤ θ ≤ 3π4 .

• One arc of r = 1 + sin θ: either −π2 ≤ θ ≤ 3π4 or 3π

4 ≤ θ ≤ 3π2 . (In the first of these

I have, for convenience, used the negative value θ = −π2 . You might wish to objectthat the prescribed interval for θ was 0 ≤ θ ≤ 2π. I could, instead have described


2

1

1.5

0.5

01-1.5 -1 0.5-0.5-2 0

Figure 22: The curves with equations r = 1 − cos θ, (θ ≤ 0), and r = 1 + sin θ, and the point(12 ,

π2

)

the first of these arcs as being made up of two pieces: one being 3π2 ≤ θ ≤ 2π, and

the second being 0 ≤ θ ≤ 3π4 .)

But the point(

12 ,

π2

)is only in the region whose boundary contains the arc 3π

4 ≤ θ ≤ 3π2 of

the curve r = 1 − cos θ.

To find the area, think of a line segment being drawn from the pole to(1 + 1√

2, 3π

4

). The

area is then the sum of the area below the line segment and the area above it, i.e.,

12

∫ 3π4

0(1 − cos θ)2 dθ +

12

∫ 3π2

3π4

(1 + sin θ)2 dθ

=12

∫ 3π4

0

(1 − 2 cos θ +

1 + cos 2θ2

)dθ +

12

∫ 3π2

3π4

(1 + 2 sin θ +

1 − cos 2θ2

)dθ


=12

[3θ2− 2 sin θ +

sin 2θ4

] 3π4

0+

12

[3θ2− 2 cos θ − sin 2θ

4

] 3π2

3π4

=9π8−√

2 − 14.


C.33 Supplementary Notes for the Lecture of March 24th, 2010Distribution Date: Wednesday, March 24th, 2010, subject to further revision

C.33.1 §11.5 Alternating Series

I continue to follow the syllabus in the order of topics in the textbook. This order is the reverseof that which I would prefer, in that the most basic results for this topic will appear in the nextsection, where the textbook considers theorems about positive series. As with the precedingsection, we are investigating a specialized topic concerning series prior to meeting a moregeneral theorem in [1, §11.6] at the next lecture. The present section will broader the class ofseries that we can work with, albeit to a very special class of series.

An alternating series is one in which the signs alternate. Our main interest, as in thepreceding sections will be in three questions, listed in the order of our priorities:

1. Does the series converge?

2. If the series converges, what is its sum?

3. If we truncate the series at a specific partial sum, can we obtain a “good” upper boundfor the error — i.e., for the difference between the sum of the series and the given partialsum.

Our interest in alternating series derives partly from some specific series that will arise later in[1, Chapter 11], in the portion of the chapter assigned to Calculus III. There we would meetseveral important power series which represent some familiar functions; and many of thoseseries have alternating signs.

By the “Test for Divergence”, an alternating series cannot converge unless the limit of thesequence of its terms is 0. For these series, and these series alone, we have a partial converseof that test:

Theorem C.97 (Leibniz’s Alternating Series Test) An alternating series∞∑

n=1

(−1)n−1bn for which:

(i) 0 ≤ bn+1 ≤ bn for all n; and

(ii) limn→∞

bn = 0

converges.


Estimating Sums While the estimation of errors can be complicated for other series, thefollowing very simple bounds hold for alternating series:

Corollary C.98 (to the Leibniz Alternating Series Test) Continuing with the same notation

as the theorem, the remainder upon truncation of an alternating series∞∑

n=1

(−1)n−1bn (bn ≥ 0; n =

1, 2, . . .) at the Nth term cannot exceed the very next term; more precisely,∣∣∣∣∣∣∣∞∑

n=N+1

(−1)n−1bn

∣∣∣∣∣∣∣ < |bN+1| .

Example C.99 Determine whether the series∞∑

n=0

(−1)n 12n + 1

converges.

Solution: This is an alternating series. Since the denominators are increasing, while the nu-merators stay fixed, the terms are decreasing; the limit of the denominators is ∞, so the termsare approaching 0. By the Leibniz Theorem, the series converges. (It is beyond this course,but it can be shown that this series converges to arctan 1, i.e., to π

4 , and thereby provides a wayof approximating π to any desired accuracy. It is, however, not a good series for that purpose,since it converges rather slowly.)

Example C.100 Determine whether the series∞∑

n=0

(−1)n 1n + 1

converges.

Solution: This is an alternating series. Since the denominators are increasing, while the nu-merators stay fixed, the terms are decreasing; the limit of the denominators is ∞, so the termsare approaching 0. By the Leibniz Theorem, the series converges. (It is beyond this course,but this series can be proved to converge to ln(1 + 1), i.e., to ln 2.)

Example C.101 ([7, Exercise 26, p. 740]) How many terms of the series∞∑

n=1

(−1)nn4n do we

need to add in order to find the sum to within an error of 0.002?Solution: As n→ ∞ lim n

4n = lim 14n ln n = 0. The terms are decreasing, since

n4n >

n + 14n+1 ⇔ 4n > n + 1⇔ n >

13.

Thus the Leibniz Theorem is applicable, and we apply the error estimate associated with thatTheorem. If we truncate the series at the Nth term, then we know that the error will be less

thanN + 14N+1 . If we can find N satisfying the condition that

N + 14N+1 < 0.002 , (96)


we will have the desired accuracy. Since

ddx

x4x =

1 · 4x − x · 4x · ln 4(4x)2 =

1 − x ln 44x ,

which is negative provided x > 1ln 4 , which is certainly true if x ≥ 1. Thus we need only find the

first value of N for which inequality (96) holds. By repeated calculations we find the first valueis N = 5. That is, the series may be truncated by summing so that the last term we include is(−1)55

45 and the error will be within the prescribed tolerance.59

The student should understand that the Leibniz estimate for error is not necessarily as sharpas it might be. It is often possible to obtain much better approximations than the Leibniz esti-mate might guarantee; such investigations are beyond this course. The Leibniz theorem statesthat an alternating series whose terms are non-decreasing in magnitude and are approaching0 is convergent. However, an alternating series whose terms are not decreasing, but whoseterms do approach 0, may still converge! For example, suppose that we modify the “alternat-

ing harmonic series”∞∑

n=1(−1)n+1 1

n in the following way: term number 2n is the negative fraction

(−1)2n+1 12n . Let’s divide just these terms by 2. The resulting alternating series can be shown

to still be convergent; but it will converge to a sum which is less than the sum of the originalseries; in fact, the reduction converges to a sum of

12·

12

1 − 12

=12.

But the condition of the Leibniz Theorem that requires that the terms be monotonely decreasingin magnitude is not satisfied, so the convergence cannot be inferred from the Leibniz Theorem.Nonetheless, the series does converge (since the partial sums are equal to the difference of thepartial sums of two series known to converge).

11.5 Exercises

[1, Exercise 14, p. 713] Test for convergence or divergence the series∞∑

n=1

(−1)n−1 ln nn

.

Solution: The series is alternating. The limit of the terms can be seen by L’Hospital’sRule to be lim

n→∞1n = 0. To determine whether the terms are decreasing, we can consider

ddx

(ln x

x

)=

1 − ln xx2 < 0

59We can actually sum this infinite series, using methods analogous to the method used to sum a geometricprogression. The sum can be shown to be − 4

25 . The sum of the first 5 terms is approximately -0.161133, so theerror is approximately 0.001133, which is within the desired tolerance.


for x > e. Hence the Leibniz Theorem applies, and the series must converge. It’s notimportant where the terms start to decrease — only that they ultimately decrease.60


n=1

(−1)n√

n1 + 2

√n

.

Solution: The general term of the given series is,

(−1)n√

n1 + 2

√n

= (−1)n 11√n + 2

whose magnitude → 12

as n → ∞. Thus the sequence of terms of this series does nothave a limit as n→ ∞, and the series cannot converge, by the “Test for Divergence”. Infact, the other condition of the Leibniz Theorem, that the terms decrease in magnitude,also fails to hold! But here we can say more: not only can we state that the Leibniz Testyields no information, but we can state that the Test for Divergence shows that the seriesactually diverges.


n=1

(−1)n−1 ln nn

.

Solution: The series is alternating. The limit of the terms can be seen by L’Hospital’sRule to be lim

n→∞1n = 0. We have determined this limit for 2 reasons:

• as an application of “The” Test for Divergence — our result is inconclusive, andgives no information about divergence;

• as a condition of the Leibniz Alternating Series Test — here our result is positive,and, when combined with the yet unproved fact that the terms are decreasing inmagnitude, will show that the series does, indeed, converge.

To determine whether the terms are decreasing, we can consider

ddx

(ln x

x

)=

1 − ln xx2 < 0

for x > e. Hence the Leibniz Theorem applies, and the series must converge. It’s notimportant where the terms start to decrease — only that they ultimately decrease.61

60However, the Leibniz estimate for the error would not be valid until the terms have begun to decrease.61However, the Leibniz estimate for the error would not be valid until the terms have begun to decrease.



n=1

(−1)n sinπ

n.

Solution: This problem should be compared with [1, Exercise 31, p. 709] (cf. thesenotes, p. 3229). In the earlier problem we proved the divergence of a series whose termsresemble those of the present series, except that it was not alternating. But the presentseries satisfies the conditions of the alternating series test, as the terms alternate in sign,are monotonely decreasing (because the sine function is an increasing function to theright of 0), and the limit of the terms is 0. Thus the series converges. When we havecovered [1, §11.6] we will be able to describe the present series as being ConditionallyConvergent — it is convergent, but loses that property if all terms are replaced by theirabsolute value.


n=1

(−1)n cosπ

n.

Solution: Since the magnitudes of the terms of this series approach 1 as n → ∞, theterms of the series have no limit! By “The” Test for Divergence, this series diverges!


n=1

(−1)n nn

n!(cf. [1, Ex-

ercise 30, p. 709][1, Exercise 18, p. 720])

Solution: The terms may be factorized as

(−1)n nn

n!= (−1)n n

1· n

2· . . . n

n − 1· n

n.

While this is an alternating series, the limit of the terms cannot be 0, since the fractionsinto which I have factored the terms are, in magnitude, all greater than or equal to 1;thus the series must diverge, by “The” Test for Divergence; this also shows that not allconditions of the Leibniz Alternating Series Test are satisfied, but from that we can onlyconclude that that test does not apply, not that the series diverges.


n=1

(−1)n(n5

)n.

Solution: This series also diverges, because the terms to not have limit 0.

C.33.2 Solutions to Problems on the Final Examination in MATH 141 2007 01

Among the instructions for this examination was the following:

There are two kinds of problems on this examination, each clearly marked asto its type.


• Most of the questions on this paper require that you

SHOW ALL YOUR WORK!

Their solutions are to be written in the space provided on the page wherethe question is printed. When that space is exhausted, you may write on thefacing page. Any solution may be continued on the last pages, or the backcover of the booklet, but you must indicate any continuation clearly on thepage where the question is printed!

• Some of the questions on this paper require only

BRIEF SOLUTIONS

for these you are expected to write the correct answer in the box provided;you are not asked to show your work, and you should not expect partial marksfor solutions that are not correct.

You are expected to simplify your answers wherever possible.You are advised to spend the first few minutes scanning the problems. (Please

inform the invigilator if you find that your booklet is defective.)


Your answers must be simplified as much as possible.


−1|x|2 dx.

Solution: ∫ 2

−1|x|2 dx =

∫ 2

−1x2 dx =

[x3

3

]2

−1=

8 − (−1)3

= 3 .


0∫

1

t4dt√t5 + 1

.

Solution: Either by observation, or by applying a substitution like u = t5, or u =

t5 + 1, or u =√

t5 + 1, we can show that

0∫

1

t4dt√t5 + 1

=25

√t5 + 1

]0

1=

25

(1 −√

2) = −25

(√

2 − 1) .

(c) [3 MARKS] Determine the value of

1n

(0n

)3

+

(1n

)3

+

(2n

)3

+ . . . +

(n − 1

n

)3 .


Solution: Note that the question did not ask for the value of the limit as n→ ∞—it asked simply for the value of the sum (as a function of n).

1n

(0n

)3

+

(1n

)3

+

(2n

)3

+ . . . +

(n − 1

n

)3

=

n−1∑

i=0

i3

n4

=

((n − 1)n

2

)2

n4 =(n − 1)2

4n2 .

(Had the instructions been to evaluate the limit as n → ∞, then the given productcould have been interpreted as a left Riemann sum with n intervals of equal length

∆n = 1n . The value would have been

1∫

0

x3 dx =

[x4

4

]1

0=

14

.)

(d) [3 MARKS] Suppose it is known that f ′(x) = 4 cosh x for all x. Showing all yourwork, determine the value of f (1)− f (−1), expressed in terms of the values of eitherexponentials or hyperbolic functions.Solution: Integrating the given equation yields f (x) = 4 sinh x + C, where C issome (fixed) real constant C. Since f ′ exists, f is continuous everywhere, so it willbe the same constant C that is involved throughout the interval [−1,+1]; hence

f (1) − f (−1) = 4 sinh 1 + C − (sinh(−1) + C)= 4(sinh 1 + sinh 1)= 8 sinh 1 .

Alternatively one could work with exponentials, beginning with f ′(x) = 2 (ex + e−x),

to show that f (1) − f (−1) = 4(e − 1

e

).

(e) [4 MARKS] Evaluateddx

∫ x2

12

ett dt when x = 1 .

Solution:

ddx

∫ x2

12

ett dt =d

du

∫ u

12

ett dt · du

dxwhere u = x2

= euu · dudx

= e(x2)(x2) · 2x


which simplifies to 2x · ex(2x2); when x = 1, the value is 2e.


For each of the following series you are expected to apply one or more tests for conver-gence or divergence to determine whether the series is absolutely convergent, condition-ally convergent, or divergent. All tests used must be named, and all statements must becarefully justified.

(a) [4 MARKS]∞∑

n=1

(−n − 2)n(n − 2)n

(2n2 + 1)n

Solution:∞∑

n=1

(−n − 2)n(n − 2)n

(2n2 + 1)n =

∞∑

n=1

(−1)n (n2 − 4)n

(2n2 + 1)n .

This is an alternating series. Applying the Root Test to the series of absolute valuesof the terms of the given series, we have

n

√(n2 − 4)n

(2n2 + 1)n =

∣∣∣∣∣∣n2 − 4

2n2 + 1

∣∣∣∣∣∣ =1 − 4

n2

2 + 1n2

for n ≥ 2. As n → ∞, this ratio→ 12 < 1 . Since the limit of the ratio is less

than 1, the original alternating series is absolutely convergent.

(b) [4 MARKS]∞∑

n=1

(−1)n+1 n!n22n

Solution: If we define an = (−1)n+1 n!n22n , we find, using the Ratio Test, that

∣∣∣∣∣an+1

an

∣∣∣∣∣ =(n + 1)n2

(n + 1)21

=n2

2(n + 1)→ ∞ > 1 .

Hence the original series diverges.

(c) [4 MARKS]∞∑

n=1

(−1)n sin1n

Solution: The series of absolute values may be compared, using the Limit Compar-ison Test, with the Harmonic series, a positive series known to be divergent. Since


limn→∞

sin 1n

1n

= 1 , 0, the series of absolute values is also divergent, and the original

series is either (1) conditionally convergent, or (2) divergent. But

ddx

(sin

1x

)=

(cos

1x

)·(− 1

x2

)< 0

as x > 1. Hence the given series of alternating terms are monotonely decreasingin magnitude. As n → ∞, sin 1

n → sin 0 = 0, by the continuity of the sine func-tion. Thus the conditions of the Alternating Series Test are satisfied, and the seriesconverges. We conclude that the given series is conditionally convergent.

3. BRIEF SOLUTIONS Express each of the following as a definite integral or a sum,product, or quotient of several definite integrals, simplified as much as possible; you arenot expected to evaluate the integrals.

R is defined to be the region enclosed by the curves x + y = 6 and y = x2; C is the arcy = 3x (−1 ≤ x ≤ 2).

(a) [3 MARKS] The region R is rotated about the x-axis. Give an integral or sum ofintegrals whose value is the volume of the resulting solid.


Solution: By solving the equations of the line x + y = 6 and the parabola y = x2,we find the points of intersection to be (−3, 9) and (2, 4). Either of the followingintegrals or sums of integrals was acceptable. (Students were not expected to eval-uate the integrals; I have done so here in order to verify my own work, and to showyou that the integrals weren’t really hard to evaluate.)

i. Using the method of “Washers”, we find the volume to be

π

∫ 2

−3

((6 − x)2 − (x2)2

)dx = π

∫ 2

−3

(−x4 + x2 − 12x + 36

)dx

which can be shown to be equal to500π

3.

ii. Using the method of cylindrical shells, we find the volume to be∫ 4

0(2πy) · (2√y) dy +

∫ 9

4(2πy) · (6 − y +

√y) dy


whose value can again be shown to equal500π

3.

(b) [3 MARKS] The region R is rotated about the line x = 5. Give an integral or sumof integrals whose value is the volume of the resulting solid.


Solution: Here again, students were expected only to state the integrals, but not toevaluate them.

i. Using the method of cylindrical shells: We will be rotating about x = 5vertical elements of area; the width can be taken to be ∆x, as integration willbe with respect to x. The element extends from a point (x, x2) to (x, 6 − x), soits height is 6 − x − x2. The volume is∫ 2

−32π(5 − x) · (6 − x − x2) dx = 2π

[x4

4− 4x3

3− 11x2

2+ 30x

]2

−3=

1375π6

.

ii. Using the method of washers:

π

∫ 4

0

((5 +

√y)2 − (−5 +

√r)2

)dy + π

∫ 9

4((5 +

√y)2 − (5 − (6 − y))2) dy

= 20π∫ 4

0

√y dy + π

∫ 9

4(24 + 10

√y + 3y − y2) dy =

1375π6

.

(c) [3 MARKS] Express in terms of integrals — which you need not evaluate — theaverage length that R cuts off from the vertical lines which it meets.


Solution: The length of the segment cut off from the line x = a has been shownabove to be 6 − a − a2. The average of this function over the interval −3 ≤ a ≤ 2 isthe ratio ∫ 2

−3(6 − a − a2) da

2 − (−3)=

256.


Students weren’t asked to evaluate the integral. I have done so here to verify mywork; in this case I can’t verify by comparing answers obtained in two differentways, but I can, at least, examine the magnitude of my answer, and decide whetherit is reasonable for the given data. For example, if my answer had been 25, Iwould know something was wrong, by considering the sizes of the numbers beingaveraged, and the maximum value obtained by the function on the given interval.

(d) [2 MARKS] Give an integral whose value is the length of C; you need not evaluatethe integral.


Solution: When y = 3x, y′ = 3x ln 3. The arc length is∫ 2

−1

√1 + (3x ln 3)2 dx .

(e) [3 MARKS] Given an integral whose value is the area of the surface generated byrotating C about the line y = −1; you need not evaluate the integral.


Solution:

2π∫ 2

−1(3x + 1)

√1 + 32x(ln 3)2 dx


[12 MARKS] Evaluate the indefinite integral

∫ x(x2 − 4

)(x − 2) + 4

(x2 − 4

)(x − 2)

dx .

Solution: The numerator of the integrand has degree not less than that of the denomina-tor. Accordingly, the first step is to divide denominator into numerator and to integrate


the quotient separately, and to factorize the denominator and group like factors together:

∫ x(x2 − 4

)(x − 2) + 4

(x2 − 4

)(x − 2)

dx =

∫ (x +

4(x + 2)(x − 2)2

)dx

=x2

2+

∫4

(x + 2)(x − 2)2 dx .

Next we apply to the new integrand the method of partial fractions:

4(x + 2)(x − 2)2 =

Ax + 2

+B

x − 2+

C(x − 2)2

⇒ 4 = A(x − 2)2 + B(x + 2)(x − 2) + C(x + 2)⇒ 4 = 4C when x = 2; and

4 = 16A when x = 2

implying that A = 14 and C = 1. Comparing coefficients of x2 on both sides of the

identity yields 0 = A + B, implying that B = − 14 . Now we can continue integration of

the original function:

∫ x(x2 − 4

)(x − 2) + 4

(x2 − 4

)(x − 2)

dx =x2

2+

∫4

(x + 2)(x − 2)2 dx

=x2

2+

14

∫ (1

x + 2− 1

x − 2

)dx +

∫dx

(x − 2)2

=x2

2+

14

ln∣∣∣∣∣x + 2x − 2

∣∣∣∣∣ −1

x − 2+ C .



(a) [4 MARKS]∫

e−x · cos x dx

Solution: I will apply Integration by Parts twice. First, with u = e−x, dv = cos x dx(so du = −e−x dx, v = sin x) yields

∫e−x cos x dx = e−x sin x +

∫e−x sin x dx ,

and the second (applied to the integral resulting from the first application), withU = e−x, dV = sin x dx (so dU = −e−x dx, V = − cos x) yields

∫e−x cos x dx = e−x (sin x − cos x) −

∫e−x cos x dx .


By collecting both integral terms on one side, and dividing by 2, we obtain∫

e−x cos x dx =12

e−x(sin x − cos x) + C .

Another way to solve this problem is to assume — from experience — that theindefinite integral is equal to a sum of the form

∫e−x · cos x dx = e−x(A sin x + B cos x) + C ,

where A and B are “undetermined” constants, to be determined. One could thendifferentiate this equation, top obtain

e−x cos x = −e−x(A sin x + B cos x)e−x(A cos x − B sin x) .

This is an identity – true for all values of x. If we assign “convenient” values, forexample x = 0 and x =

π

2, we obtain equations

A − B = 1 ,A + B = 0 ,

which we can solved, to obtain A = 12 , B = −1

2 , yielding the same solution as foundearlier.

(b) [5 MARKS]

52∫

− 12

x√8 + 2x − x2

dx

Solution:

52∫

− 12

x√8 + 2x − x2

dx =

52∫

− 12

x√9 − (x − 1)2

dx

=13

52∫

− 12

x√1 − ( x−1

3 )2dx .

I will use the substitution u =x − 1

3, so du =

dx3

.

13

52∫

− 12

x√1 − ( x−1

3 )2dx =

13

∫ 12

− 12

3u + 1√1 − u2

.3 du


=[−3√

1 − u2 + arcsin u] 1

2

12

=

−3

√34

+ arcsin12

−−3

√34− arcsin

12

= 2 arcsin12

= 2 · π6

=π

3.

(c) [4 MARKS]∫ (

cos2 x +1

cos2 x

)· tan2 x dx

Solution:∫ (

cos2 x +1

cos2 x

)· tan2 x dx =

∫ (sin2 x + tan2 x · sec2 x

)dx

=

∫ (1 − cos 2x

2+ tan2 x · sec2 x

)dx

=x2− sin 2x

4+

tan3 x3

+ C .


Consider the arc C defined by

x = x(t) = cos t + t sin ty = y(t) = sin t − t cos t ,

where 0 ≤ t ≤ π2 .

(a) [6 MARKS] Determine as a function of t the value ofd2ydx2 .

Solution:

dxdt

= − sin t + sin t + t cos t = t cos t

dydt

= cos t − cos t + t sin t = t sin t

dydx

=

dydtdxdt

=t sin tt cos t

= tan t when t , 0, π2 ,

d2ydx2 =

ddx

(dydx

)=

ddt

(dydx

)· dt

dx

=

ddt

(dydx

)

dxdt

=

ddt tan t

t cos t=

sec2 tt cos t

=1

t cos3 t,


when t , 0, π2 .

(b) [6 MARKS] Determine the area of the surface generated by revolving C about they-axis.Solution: The area of the surface generated is

2π∫ π

2

0x(t)

√(t cos t)2 + (t sin t)2 dt

= 2π∫ π

2

0(cos t + t sin t)|t| dt = 2π

∫ π2

0(t cos t + t2 sin t) dt .

At this point I interrupt the calculations to integrate by parts. First, taking u = t,dv = cos t dt (so du = dt, v = sin t),

∫t cos t dt = t sin t −

∫sin t dt = t sin t + cos t + C1 .

Then, taking u = t2, dv = sin t dt (so du = 2t dt and v = − cos t),∫

t2 sin t dt = −t2 cos t + 2∫

t cos t dt = −t2 cos t + 2(t sin t + cos t) + C2 .

Incorporating these results into the earlier integral yields

2π∫ π

2

0(t cos t + t2 sin t) dt

= 2π[(t sin t + cos t) + (−t2 cos t + 2t sin t + 2 cos t)

] π2

0

= 6π(π

2− 1

).



∫ π

π2

sec x dx .

Solution: The integrand has an infinite discontinuity at x =π

2.

∫ π

π2

sec x dx = lima→( π2 )+

∫ π

asec x dx


= lima→( π2 )+

[ln | sec x + tan x|]πa= lim

a→( π2 )+(ln | − 1 + 0| − ln | sec a + tan a|)

= 0 − lima→( π2 )+

ln | sec a + tan a| .

Here both sec a and tan a are negative, as the argument is in the second quadrant,so they both approach −∞. This implies that their sum also approaches −∞, andthat ln | sec a + tan a| approaches +∞; hence the improper integral approaches −∞,and is not convergent.

(b) [5 MARKS] Showing all your work, carefully determine whether the series∞∑

n=3

4n ln n

is convergent.Solution: Since both factors in the denominator are increasing, the terms of thispositive series are decreasing monotonely. We can apply the Integral Test with thefunction f (x) = 4

x ln x . But∫ ∞

3

dxx ln x

= limb→∞

[ln ln x]b3 = +∞

is a divergent, improper integral, so the series is also divergent.

(c) [3 MARKS] Showing all your work, determine whether the following sequenceconverges; if it converges, find its limit:

a1 = 1.a2 = 1.23a3 = 1.2345a4 = 1.234545a5 = 1.23454545a6 = 1.2345454545

etc., where each term after a2 is obtained from its predecessor by the addition onthe right of the decimal digits 45.Solution:

1.23 + 45(0.0001 + 0.000001 + . . .) = 1.23 +45

10000

(1 +

1100

+1

1002 + . . .

)

= 1.23 +45104 ·

11 − 1

100


= 1.23 +45

9900= 1.23 +

51100

=13581100

=679550

.


[10 MARKS] The polar curves

r = 2 + 2 sin θ (0 ≤ θ ≤ 2π)and

r = 6 − 6 sin θ (0 ≤ θ ≤ 2π)

divide the plane into several regions. Showing all your work, carefully find the area ofthe region bounded by these curves which contains the point (r, θ) = (1, 0).

Solution: (see Figure 23 on page 3262)) Both of these curves are cardioids, but their sizesand orientations are different. The first step is to determine the points of intersection ofthe curves. Solving the equations, by eliminating r “between” them, we obtain sin θ = 1

2 ,so θ = π

6 + 2nπ, 5π6 + 2nπ, etc. For θ in the given interval, then, we have obtained the

points of intersection(3, π6

),(3, 5π

6

).

(Are these the only points of intersection? We didn’t find the pole to be a point of intersection— but it is! It lies on the first cardioid as the point

(0, 3π

2

), and on the second cardioid as the

point(0, π2

)— the same point, but appearing on the two curves with different coordinates! Could

there be yet other points of intersection? These could be only at points which, like the pole,appeared with different sets of coordinates. The only other way in which different coordinatescan occur is because of the convention that (r, θ) and (−r, θ + π) are the same point. If we applythe transformation (r, θ) → (−r, θ + π) to the cardioid r = 2 + 2 sin θ (0 ≤ θ ≤ 2π), weobtain the equation −r = 2 − 2 sin θ (0 ≤ θ ≤ 2π); and, when we solve this equation withthe original equation of the second cardioid, i.e., r = 6 − 6 sin θ (0 ≤ θ ≤ 2π), we obtainθ = π

2 and r = 0, i.e., the pole. Similarly, if we apply the transformation (r, θ) → (−r, θ + π)to the cardioid r = 6 − 6 sin θ (0 ≤ θ ≤ 2π), we obtain the equation −r = 6 + 6 sin θ (0 ≤θ ≤ 2π); and, when we solve this equation with the original equation of the first cardioid, i.e.,r = 2+2 sin θ (0 ≤ θ ≤ 2π), we obtain θ = 3π

2 and r = 0, i.e., again the pole. Finally, if we applythe transformation to both of the original equations, we obtain sin θ = −1

2 , r = −3, which yieldsthe same 2 points we found in the original solving of equations. A second application of thetransformation (r, θ)→ (−r, θ + π) takes either equation back to its original form. To summarize,we now know all the possible points of intersection: the pole, and the two points

(3, π6

),(3, 5π

6

).)

The region that interests us is bounded by the two arcs:

r = 2 + 2 sin θ(3π2≤ θ ≤ 13π

6

)(97)


4

-4

04 80-8 -4

-12

-8

Figure 23: The cardioids with equations r = 2 + 2 sin θ, r = 6 − 6 sin θ

and (98)

r = 6 − 6 sin θ(π

6≤ θ ≤ π

2

)(99)

Technically the first of these curves is described incorrectly, since the values shown forthe angle are not in the given interval; so we should write, instead,

r = 2 + 2 sin θ(3π2≤ θ ≤ 2π

), (100)

r = 2 + 2 sin θ(0 ≤ θ ≤ π

6

), (101)

and (102)


r = 6 − 6 sin θ(π

6≤ θ ≤ π

2

). (103)

There are various ways in which the area can be calculated (see Figure 24 on page3263)). One method is to first draw a line joining (0, 0) and

(3, π6

). The area of the

1.5

0.5

1

20

2.51.50.5 10

-0.5

Figure 24: The region bounded by cardioids r = 2 + 2 sin θ, r = 6 − 6 sin θ and containing thepoint (r, θ) = (1, 0)

subregion above the line is12

π2∫

π6

(6 − 6 sin θ)2 dt; the area of the region below the line is

12

π6∫

− π2

(2 + 2 sin θ)2 dθ. To evaluate these definite integrals, observe that

∫(1 − sin θ)2 dθ =

∫(1 − 2 sin θ + sin2 θ) dθ


=

∫ (1 − 2 sin θ +

1 − cos 2θ2

)dθ

=3θ2

+ 2 cos θ − sin 2θ4

+ C .

In a similar fashion we can prove that∫

(1 + sin θ)2 dθ =3θ2− 2 cos θ − sin 2θ

4+ C .

Hence

12

π2∫

π6

(6 − 6 sin θ)2 dt =

[27θ + 36 cos θ − 9

sin 2θ2

] π2

π6

=27π

2−

9π + 36 ·√

32− 9

2·√

32

= 9π − 63√

34

12

π6∫

− π2

(2 + 2 sin θ)2 dt =

[3θ + 4 cos θ − sin 2θ

2

] π6

− π2

=

π2 + 4 ·√

32−√

32

−(−−3π

4

)

= 2π +7√

34

,

so the total area is 11π − 18√

3. This area could have been computed in other ways. Forexample, half the area of the smaller cardiod can be seen to be 3π, and from this onecould subtract the integral

12

π2∫

π6

((2 + 2 sin θ)2 − (6 − 6 sin θ)2

)2dθ ,

which can be shown to equal 18√

3 − 8π. The integration of this difference of squaresdepends on the fact that the two curves are “traced out” in the same direction; this shouldbe verified before using this “combined” method.


C.34 Supplementary Notes for the Lecture of March 26th, 2010Distribution Date: Friday, March 26th, 2010, subject to further revision

C.34.1 §11.6 Absolute Convergence and the Ratio and Root Tests

Finally, after meeting, in [1, §§11.2–11.4] several tests for the convergence of positive series,we see a reason for our interest in this type of series. To state the theorem we require twodefinitions:

Definition C.7 1. A series∑

an is absolutely convergent if the series∑ |an| is convergent.

2. A series∑

an is conditionally convergent if∑

an is convergent but not absolutely con-vergent, i.e., if

∑an converges but

∑ |an| diverges.

Note that, thus far, you must not interpret the word absolutely as an adverb modifying theparticiple convergent; until the following theorem is available, you should treat absolutelyconvergent as a two-word name for a property, nothing more. Now we state the theorem thatwill permit a broader interpretation of the name.

Theorem C.102 A series which is absolutely convergent is convergent.

(I shall not prove the theorem in the lectures, but a short proof can be found in your textbook.)Now that we have this theorem, we can interpret the word absolutely as a modifier —

if we drop the word from a statement, the resulting statement is still true; we didn’t needsuch a reservation with the term conditionally convergent, since the definition stated that aconditionally convergent series is convergent and . . . .

While investigation of the convergence of general series, i.e., series whose terms have bothplus and minus signs, can be difficult, we can begin by considering the series of absolutevalues: if the resulting series converges, we can infer that the original series is convergent.But, if the series of absolute values diverges, or if we are unable to say anything about it, thenwe can make no inference at all.

The Ratio Test. The Ratio Test is a test for positive series which considers the limit of theratio of a term to its predecessor. The textbook states the test in terms of the absolute values ofterms of a general series, so we will present it in that variant:

Theorem C.103 (Ratio Test) 1. If limn→∞|an+1||an| = L < 1, then the series

∑an is absolutely

convergent.

2. If limn→∞|an+1||an| = L > 1, then the series

∑an is divergent.


3. If limn→∞|an+1||an| = 1, or if the limit does not exist, then this test provides no information

concerning the possible convergence of the series∑

an.

The Root Test. The Root Test is another test for positive series; here also the textbookpresents it in a form that applies to all series.

Theorem C.104 (Root Test) 1. If limn→∞

n√|an| = L < 1, then the series

∑an is absolutely

convergent.

2. If limn→∞

n√|an| = L > 1, then the series

∑an is divergent.

3. If limn→∞

n√|an| = 1, or if the limit does not exist, then this test provides no information

concerning the possible convergence of the series∑

an.

You will need to practice on many problems in order to become comfortable with all the testsyou have met, and to know the limitations of each of them. You should not be surprised ifsome problems are amenable to the use of more than one test.

Rearrangements The textbook reports on the result of Bernhardt Riemann that the termsof a conditionally convergent series may be written in another order to create series that willconverge to any given real number, and even to diverge.

Example C.105 ([7, Exercise 20, p. 746]) Determine whether the series

∞∑

n=2

(−1)n

(ln n)n

is absolutely convergent, conditionally convergent, or divergent.Solution: Consider the series of absolute values. The ratio of the (n+1)st term to its predecessoris complicated, so we will look at the nth root of the nth term, obtaining

1ln n→ 0

as n → ∞. The given series is absolutely convergent. If we were interested only in checkingthe convergence of the given series, then the Leibniz Alternating Series Test would have beensufficient, and easy to use; but it would have given no information about absolute convergence.

Example C.106 ([7, Exercise 24, p. 746]) Determine whether the series

∞∑

n=1

(−1)n

(arctan n)n


is absolutely convergent, conditionally convergent, or divergent.Solution: I will apply the Root Test. The nth root of the absolute value of the nth term is

1arctan n

, which approaches2π< 1. Hence the series is absolutely convergent.

11.6 Exercises

[1, Exercise 3, p. 719] I modify the problem: Determine whether the series∞∑

n=0

(−1)n 1, 000, 000n

n!

is absolutely convergent, conditionally convergent, or divergent.

Solution: Apply the Ratio Test to the series of absolute values. The ratio of the (n + 1)th

term to the nth is100000n + 1

, and100000n + 1

→ 0 as n → ∞. Thus, by the Ratio Test, theseries is absolutely convergent. This conclusion could also be proved in other ways.


n=0

sin 4n4n


Solution: Neither the Root Test nor the Ratio Test is useful here, because the limits donot exist. Compare the series of absolute values with the geometric series

∑ 14n . Since

the latter converges, so does the former. Thus the given series is absolutely convergent.


n=1

3 − cos n

n23 − 2


Solution: For n > 1000 — (we don’t need the best bound here) — this is a positiveseries; the numerator behaves erratically, and we will not be able to calculate a limitusing the Ratio or Root Tests. But since, for large n, the denominator is larger thann

23 , we would expect this series to diverge by comparison with the appropriate p-series.

More precisely, the divergence of the p-series with p = 23 implies the divergence of the

given series, since3 − cos n

n23 − 2

>3 − 1

n23 − 2

>3 − 1

n23

=2

n23

.


C.35 Supplementary Notes for the Lecture of March 29th, 2010Distribution Date: Monday, March 29th, 2010, subject to further revision

C.35.1 §11.6 Absolute Convergence and the Ratio and Root Tests (conclusion)

Review of the last lecture After reviewing the Leibniz Alternating Series Test, I workedexamples on these series, and mentioned the concept of conditional convergence, to be definedin the next section. In §11.6 we finally saw some justification for our development of tests forthe convergence of positive series.

• definitions of absolute and conditional convergence

• Theorem: Absolute convergence implies convergence.

• The Ratio Test and Root Test are essentially tests for the convergence of positive series,but they are formulated in your textbook for any series, since the limits involved are eachapplied to the absolute values of roots and/or ratios.

• In each of these tests a limit is investigated, and that limit must be different from 1 forthe test to yield any useful information. If the limit does not exist, or if the limit existsand is equal to 1, the test fails to give any information.

• When the limit exceeds 1, that implies that the terms do not approach 0, and so diver-gence follows from “The” Test for Divergence.



n=1

(−1)n−1 2n

n4 is absolutely convergent,

conditionally convergent, or divergent.

Solution: This is an alternating series: the only test we have to apply directly to suchseries is the Leibniz Alternating Series Test (and the Test for Divergence, which forms a

part of the Leibniz Test). To determine limn→∞

2n

n4 , where both numerator and denominatorbecome infinite, we can use L’Hospital’s Rule. This will need to be applied 4 times, after

which we have limn→∞

2n(ln 2)4

24= +∞ . The terms of the original series do not approach

+∞, since they are alternating in sign; but the non-existence of a limit shows that, inparticular, the limit of terms is not 0, so the Test for Divergence tells us that this seriesmust diverge. We don’t even need to check the second condition of the Leibniz Test(which could be shown to fail). Thus, even though we thought we were applying the


Leibniz Test, the failure of convergence derives from the Test for Divergence, which isone part of the Leibniz Test.

The preceding was correct, but naive. It would have been better to apply the Ratio Test,since

limn→∞

2n+1

(n + 1)4

2n

n4

= 2 limn→∞

( nn + 1

)4= 2

(limn→∞

nn + 1

)4= 2 ,

implying that the given series diverges.


n=1

(−1)n−1 n22n

n!is absolutely conver-

gent, conditionally convergent, or divergent.

Solution:

1. This is an Alternating Series, and we could have begun by applying the LeibnizTest. We would find that the terms are decreasing only if

(n + 1)22n+1

(n + 1)!· n!

n22n < 1

which is equivalent, after reduction, to n − 1 >√

3; thus the condition of be-ing monotonely decreasing is ultimately satisfied (here, for n ≥ 4). But how do

we determine limn→∞

(−1)n+1 n2

2n ? Simply observe that the (n + 1)st term is obtained

from the nth by multiplying by2n + 2

n2 . This factor will be less than, say, 12 , when

(n − 2)2 > 8, i.e., when n ≥ 5. So, for such n, each term is less than 12 of its prede-

cessor; this shows that the nth term is bounded by a constant multiple of 12n , which

approaches 0 as n → ∞. Thus the terms approach 0, so the Test for Divergencereveals nothing in itself! However, as we have now proved that both conditionsof the Leibniz Test apply to this Alternating Series, we can infer that the series isconvergent.

2. But, to answer the given question, we need to know whether the series is absolutelyconvergent or conditionally convergent, so more information is required: we haveproved only that it is at least conditionally convergent.

3. To check for absolute convergence we need to apply one of our tests to the positiveseries obtained by changing all the signs to +. Of the tests we have, the Integral Testis not useful, since we don’t have any “nice” function available to take on the valuesof n! at the integer points n; there does exist such a function, but its development


is far beyond this course. The obvious test to try is the Ratio Test, since the termsare products, and the ratio of successive terms has many factors which cancel. We

find that the ratio of the (n + 1)st term to the nth is2(n + 1)

n2 = 2(1n

+1n2

)which

approaches 0 as n → ∞. Since this limit exists, and is less than 1, the Ratio Testtells us that the series of absolute values is convergent — i.e., that the original seriesis absolutely convergent.

(cf. [1, Exercise 17, p. 720]) Determine whether the series

∞∑

n=0

(−1)n

n ln n


Solution: Here the ratio of the absolute value of the (n + 1)th term to that of the nth is(1 +

1n

)· ln(n + 1)

ln n

in which the first factor approaches 1, and the second (by L’Hospital’s Rule) also ap-proaches 1. Thus the Ratio Test fails to give useful information about this series. We arenot helpless, however. Let’s first consider the series of absolute values. The terms aredecreasing and approaching 0, so we may use the Integral Test to show that it is diver-gent. On the other hand, the original series is alternating, and the terms are decreasingand approaching 0; hence, by the Leibniz Test, that series converges; consequently it isconditionally convergent.

[1, Exercise 17, p. 720] is concerned with the series∞∑

n=0

(−1)n

n ln n. The preceding discussion

shows that a series with terms which are smaller in magnitude is not absolutely conver-gent; hence, by the Comparison Test, the series of this exercise is surely not absolutelyconvergent; we could also probe the absence of absolute convergence by comparing withthe harmonic series. But it is conditionally convergent, since the conditions of the Leib-niz test are easily shown to be satisfied. (We could also probe the absence of absoluteconvergence by comparing with the harmonic series.)

[1, Exercise 26, p. 720] Determine whether the series

25

+2 · 65 · 8 +

2 · 6 · 105 · 8 · 11

+2 · 6 · 10 · 145 · 8 · 11 · 14

+ . . .



Solution: The ratio of the (n + 1)st term to the nth is4n + 23n + 5

, which approaches 43 > 1.

By the Ratio test, this series diverges.

[1, Exercise 30, p. 720] The terms of a series∑

an are defined “recursively” by a1 = 1, an+1 =2 + cos n√

n· an (n ≥ 1). Determine whether

∑an converges or diverges.

Solution: The ratio of the (n + 1)st to the nth term is 2 + cos n√

n (n ≥ 1). As n→ ∞ thenumerator is bounded between 1 and 3, while the denominator becomes infinitely large;by the Squeeze Theorem, applied to

0 ≤ an+1

an≤ 3√

n,

we see that limn→∞

an+1

an= 0. By the Ratio Test, this shows that the series converges abso-

lutely.

[1, Exercise 32, p. 720] For which positive integers k is the series∞∑

n=1

(n!)2

(kn)!convergent?

Solution: The ratio of the (n + 1)st term to the nth is

((n + 1)!)2

(k(n + 1))!(n!)2

(kn)!

=(n + 1)!(n + 1)!(kn)!

n!n!(kn + k)!=

(n + 1)2

(kn + k)(kn + k − 1) · . . . · (kn + 1)

in which the factors in the denominator are k in number, and all of them lie betweenn + 1 and kn + k. As long as k > 2, this fraction has more 1st degree factors in thedenominator than in the numerator, so it will approach 0 as n → ∞, and we note that

0 < 1; when k = 2 the fraction is(n + 1)2

(2n + 2)(2n + 1), which approaches 1

4 < 1 as n → ∞;

when k = 1 the fraction is equal to n + 1, and approaches +∞. By the Ratio Test theseries will converge for k ≥ 2, and will diverge when k = 1.


C.36 Supplementary Notes for the Lecture of Wednesday, March 31st,2010

Distribution Date: Wednesday, March 31st, 2010subject to correction

C.36.1 §11.7 Strategy for Testing Series

This section is concerned with improving your methods for attacking series problems. Plan-ning strategies really needs some experience: after you have worked a substantial number ofquestions from earlier sections in the chapter, you might wish to read the 8 points that thetextbook gives on [1, p. 721]. These points will be of much more use to you after you haveaccumulated some experience in applying the individual tests.

Exercise C.2 ([7, Problems 2 and 4, p. 748]) You are asked to test for convergence or diver-

gence the series∞∑

n=1

n − 1n2 + n

and∞∑

n=1

(−1)n−1 n − 1n2 + n

.

Solution:

1. Consider the given positive series.

(a) It is useful, where possible, to formulate a good conjecture (guess) about the out-come of the investigation, in order to know whether to concentrate efforts on at-tempting to prove convergence or divergence. I suggest that a valuable first strate-gic step is to replace the general term by a “simpler” term: in this case I wouldlook at

∑

n

nn2 . Since this gives the harmonic series, a first guess in this case would

be that the series diverges. I would thus try to use some test that relates the givenseries to the harmonic series.

(b) The simplest method would be to use the Limit Comparison Test:

limn→∞

n − 1n2 + n

1n

= limn→∞

n − 1n + 1

= 1 , 0 .

Hence the given series and the Harmonic Series converge or diverge together. Weknow that the Harmonic Series diverges; hence the given series also diverges.

(c) We could also have attacked this problem naively by using the Integral Test, with

f (x) =x − 1x2 + x

. Since

f ′(x) =−x2 + 3x(x2 + x

)2 = − x(x − 3)(x2 + x

)2 ,


the function is decreasing for n > 3; and

limn→∞

n − 1n2 + n

= limn→∞

1 − 1n

1 + n= 0 .

Then the series will converge or diverge according as the following improper inte-gral converges or diverges:

∫ ∞

3

x − 1x2 + x

dx =

∫ ∞

3

(−1

x+

1x + 1

)dx

expanding by partial fractions

= limb→∞

∫ b

3

(−1

x+

2x + 1

)dx

= limb→∞

[− ln x + 2 ln(x + 1)]b3

= limb→∞

[ln

(x + 1)2

x

]b

3

= limb→∞

(ln

(b + 1)2

b− ln

163

)

= limb→∞

(ln

(b + 2 +

1b

)− ln

163

)

= ∞ .

Thus this integral diverges, and the given series must also diverge.

2. Now consider the series∞∑

n=1

(−1)n−1 n − 1n2 + n

. We know from the preceding discussion

that the given alternating series does not converge absolutely. However, the AlternatingSeries Test may be applied to show that the alternating series does converge. Thus theseries converges conditionally. So the answer to the question as stated is that the givenseries does converge.

11.7 Exercises


n=1n2e−n3

.

Solution: The terms suggest a function with an obvious antiderivative. More precisely,if f (x) = x2e−x3

, then∫

f (x) dx = − 13e−x3

+ C. Thus it would appear that we could usethe Integral Test here. But let’s check that the conditions for that test are satisfied:

• Is the given series positive: YES


• Does the function f match the series values at integer terms: YES

• Is f (continuous and) positive (not just at the integer points): YES

• Is f decreasing? NOT CHECKED YET!

f ′(x) = x(2 − 3x3

)e−x3

, a product of 3 factors. The first factor is positive for x > 0; thelast factor is an exponential, so it is always positive; and the middle factor is negative

for x >(

23

) 13 , so it is certainly negative for x ≥ 1. Thus f ′ < 0, and f is decreasing. The

Integral Test tells us that the series will converge or diverge according as the improperintegral

∫ ∞1

f (x) dx converges or diverges. But

∫ a

0x2e−x3

dx = −13

(e−a − e−1

)→ 1

3eas a→ ∞ .

From the convergence of the integral we infer the convergence of the given series.

Another approach would have been to use the Ratio Test:

limn→∞

(n + 1)2e−(n+1)3

n2e−n3 =

(1 +

1n

)2

· 1e3n2+3n+1

→ 12 · 0 = 0 < 1 as n→ ∞.

The Root Test could also be used, although the calculation is more difficult:

n√

n2 · e−n3= e

2 ln nn · e−n2 → e0 · 0 = 1 · 0 = 0 < 1 as n→ ∞ ,

by l’Hospital’s Rule. In either of these cases the limit being less than 1 implies that theseries is absolutely convergent.


n=1sin n.

Solution: Here is a case where the Test for Divergence is needed. None of the othertests you know will be helpful. As n → ∞, sin n does not approach a limit. (However,students in this course could not be expected to prove that fact rigorously without somestrong hint.) Thus the series cannot converge.


k=1

k + 55k .

Solution: The series looks at first like a geometric series; but the numerators are increas-ing, so we can’t compare with a geometric series; the Limit Comparison Test will also

not be helpful, at least not in comparing with the Geometric Series∑ 1

5k . But severalother possibilities suggest themselves.


Compare with a geometric series which converges more slowly: e.g., with∑(

25

)k. Here

limk→∞

k + 55k

(25

)k = limk→∞

k + 52k = lim

k→∞1

2k ln 2= 0 .

Unfortunately, this limit being 0, we cannot use the form of the Limit ComparisonTest given in your book. We could use the version given in [1, Exercise 40(a), p.709], but that was not studied in our course.

Ratio Test:limk→∞

ak+1

ak=

15

limk→∞

k + 6k + 5

=15< 1

hence the given positive series converges.

Root Test:

limk→∞

k√

ak =15

limk→∞

k√k + 5 =

15

limk→∞

eln(k + 5)

k =15

elimk→∞

ln(k + 5)k .

It can be shown by l’Hospital’s Rule that the exponent approaches 0, so the kth rootapproaches 1

5 , which is less than 1.

In fact one can determine the actual sum here, but it involves techniques slightly beyondthe course.


n=1tan

(1n

).

Solution: As n → ∞, 1n → 0, and tan 1

n → 0 (since it is the ratio of sin 1n to cos 1

n , andcos 1

n → cos 0 = 1, while sin 1n → 0). Thus the Test for Divergence does not eliminate

the possibility that the series may converge, just as the same test did not eliminate thepossibility that the Harmonic Series might converge. If you remember that

limn→∞

tan 1n

1n

= limn→∞

sin 1n

1n · cos 1

n

= limn→∞

sin 1n

1n

· limn→∞

sec1n

= 1 · 1 = 1

you can apply the Limit Comparison Test, comparing the given series with the HarmonicSeries. Since the limit is a non-zero real number, and since the Harmonic Series is knownto diverge, the given series must also diverge.



n=2

1(ln n)ln n .

Solution: Observe that

1(ln n)ln n =

1(eln ln n)ln n =

1e(ln ln n)(ln n) =

1(eln n)ln ln n =

1nln ln n .

We can arrange for the exponent, i.e., ln ln n, to be greater than, for example, 2, by takingn > ee2

= 1618.17799.... Thus we may compare the given series with the p-series∑ 1

n2 ,which is known to converge, and infer that the given series also is convergent.


n=1

( n√2 − 1

).

Solution: I start with the observation that the following factorization holds:

an − bn = (a − b)(an−1 + an−2b + . . . + abn−2 + an−1

),

for any real numbers a and b, Taking a =n√2 and b = 1, we have

n√2 − 1 =

(n√2

)n − 1(

n√2)n−1

+(

n√2)n−2

+ . . . +n√2 + 1

.

The numerator is equal to 1. All of the n terms in the denominator are smaller than(n√2

)n ≤ 2, so the denominator is smaller than 2n. Thus the terms are greater than termsin a multiple of the harmonic series, so the series must diverge, by the Comparison Test.

We could have used the Limit Comparison Test also, since we know that

limn→∞

n√2 = lim

n→∞e

ln 2n = elimn→∞ ln 2

n e0 = 1 .

Now apply l’Hospital’s Rule:

limn→∞

n√2 − 11n

= limn→∞

− ln 2n2

n√2

− 1n2

= limn→∞

(ln 2)n√2 = ln 2 ,

sincelimn→∞

n√2 = lim

n→∞

(eln 2

) 1n

= limn→∞

eln 2n = e lim

n→∞ln 2n = e0 = 1 .


By the Limit Comparison Test, the divergence of the harmonic series implies the diver-gence of the given series. (Note that, if we alternate the signs in this problem, we obtaina series whose terms are decreasing and approaching 0, so the Alternating Series Testtells us that that series converges, i.e., that it is conditionally convergent.)

C.37 Supplementary Notes for the Lecture of Wednesday, April 7th, 2010Distribution Date: Wednesday, April 07th, 2010; corrected April 17, 2010;

subject to further correction

The main topic of today’s lecture will be a discussion of one version of the final examinationfrom April, 2009. However, before beginning a report of that discussion in these notes, I amincluding a set of draft solutions for the final examination from April, 2008. As always, I urgeyou not to base your studying on final examinations from previous years.

C.37.1 Final Examination in MATH 141 2008 01 (one version)

This examination was written during a labour disruption, when the services of Teaching As-sistants were not available for grading purposes. The following additional instructions weredistributed with the examination.

VERSION nMcGILL UNIVERSITYFACULTY OF SCIENCEFINAL EXAMINATION

IMPORTANT ADDITIONAL INSTRUCTIONS

MATHEMATICS 141 2008 01 CALCULUS 2EXAMINER: Professor W. G. Brown DATE: Monday, April 14th, 2008ASSOCIATE EXAMINER: Mr. S. Shahabi TIME: 09:00 – 12:00 hours

A. Part marks will not be awarded for any part of any question worth [4 MARKS] or less.

B. To be awarded part marks on a part of a question whose maximum value is 5 marks ormore, a student’s answer must be deemed to be more than 75% correct.

C. While there are 100 marks available on this examination 80 MARKS CONSTITUTE APERFECT PAPER. You may attempt as many problems as you wish.

All other instructions remain valid. Where a problem requires that all work be shown, thatremains the requirement; where a problem requires only that an answer be written in a boxwithout work being graded, that also remains the requirement.


Students are advised to spend time checking their work; for that purpose you could verifyyour answers by solving problems in more than one way. Remember that indefinite integralscan be checked by differentiation.

W. G. Brown, Examiner.

Instructions

1. Fill in the above clearly.

2. Do not tear pages from this book; all your writing — even rough work — must be handed in.You may do rough work for this paper anywhere in the booklet.

3. Calculators are not permitted. This is a closed book examination. Regular and translationdictionaries are permitted.

4. This examination booklet consists of this cover, Pages 1 through 9 containing questions; andPages 10, 11, and 12, which are blank. Your neighbour’s version of this test may be differentfrom yours.

5. There are two kinds of problems on this examination, each clearly marked as to its type.

• Most of the questions on this paper require that you SHOW ALL YOUR WORK!Their solutions are to be written in the space provided on the page where the questionis printed; in some of these problems you are instructed to write the answer in a box,but a correct answer alone will not be sufficient unless it is substantiated by your work,clearly displayed outside the box. When space provided for that work is exhausted, youmay write on the facing page. Any solution may be continued on the last pages, or theback cover of the booklet, but you must indicate any continuation clearly on the pagewhere the question is printed!

• Some of the questions on this paper require only BRIEF SOLUTIONS ; for these youare expected to write the correct answer in the box provided; you are not asked to showyour work, and you should not expect partial marks for solutions that are not correct.

You are expected to simplify your answers wherever possible.

You are advised to spend the first few minutes scanning the problems. (Please inform theinvigilator if you find that your booklet is defective.)

6. A TOTAL OF 100 MARKS ARE AVAILABLE ON THIS EXAMINATION.




(a) [2 MARKS] Evaluate∫

4 − 6x1 + x2 dx .

Solution:∫

4 − 6x1 + x2 dx = 4

∫dx

1 + x2 − 3∫

2x1 + x2 dx

= 4 arctan x − 3 ln(1 + x2) + C


2∫

0

y2√

y3 + 1 dy .

Solution: I apply the change of variable u = y3 (or v =√

y3 + 1), under whichdu = 3y2 dy:

2∫

0

y2√

y3 + 1 dy =

∫ 8

0

√u + 1 · 1

3du

=13· 2

3

[(u + 1)

32]8

0

=29

(9

32 − 1

32)

=529.

(c) [3 MARKS] Evaluate∫

sin(18 θ) · cos(30 θ) dθ .

Solution:

i. The easiest way to solve this problem is by using a trigonometric identitywhich relates products to sums:

∫sin(18 θ) cos(30 θ) dθ =

12

∫(sin 48θ + sin(−12θ)) dθ

= − 196

cos 48θ +1

24cos 12θ + C .

ii. The problem could also be solved laboriously by two applications of inte-gration by parts: Take u = sin 18θ, dv = cos 30θ dθ, implying that du =

18 cos 18θ, v = 130 sin 30θ:

∫sin(18 θ) · cos(30 θ) dθ =

130

sin 18θ · sin 30θ − 1830

∫cos 18θ · sin 30θ dθ .


A second application of integration by parts, with U = cos 18θ, dV = sin 30θ dθ,implying that dU = −18 sin 18θ dθ, V = − 1

30 cos 30θ, yields∫

sin(18 θ) · cos(30 θ) dθ

=1

30sin 18θ · sin 30θ − 18

30

(− 1

30cos 18θ · cos 30θ − 18

30

∫sin 18θ · cos 30θ dθ

).

This equation may be solved for the desired indefinite integral, yielding∫

sin(18 θ) cos(30 θ) dθ

=1

1 −(

1830

)2

(1

30sin 18θ · sin 30θ +

18302 cos 18θ · cos 30θ

)+ C

=196

(5 sin 18θ · sin 30θ + 3 cos 18θ · cos 30θ) + C .

The validity of this solution can be demonstrated by differentiation.


(a) [3 MARKS] Simplifying your answer as much as possible, evaluateddx

∫ √3

−xearcsin z dz .

Solution:

ddx

∫ √3

−xearcsin z dz = − d

dx

∫ −x

√3

earcsin z dz

= −earcsin(−x) · ddx

(−x)

= earcsin(−x) = e− arcsin x

(The last simplification was not required.)

(b) [4 MARKS] For the interval 2 ≤ x ≤ 5 write down the Riemann sum for thefunction f (x) = 3 − x, where the sample points are the left end-point of each of nsubintervals of equal length.

Solution: The intervals have length5 − 2

n=

3n

; f (2) = 1, xi = 2 +3in, f (xi) =

3 −(2 +

3in

)= 1 − 3i

n. The Riemann sum is

3n

n−1∑

i=0

(1 − 3i

n

)or

3n

n∑

i=1

(1 − 3(i − 1)

n

)


(c) [4 MARKS] Determine the value of the preceding Riemann sum as a function ofn, simplifying your work as much as possible. (NOTE: You are being asked todetermine the value of the sum as a function of n, not the limit as n→ ∞.)Solution:

3n

n−1∑

i=0

1 − 3n

n−1∑

i=0

i

=3n

(n − 3

n· n(n − 1)

2

)

= −32

+9

2n.


For each of the following series determine whether the series diverges, converges condi-tionally, or converges absolutely. All of your work must be justified; prior to using anytest you are expected to demonstrate that the test is applicable to the problem.

(a) [4 MARKS]∞∑

n=3

(1

n√

ln n

)

Solution: Since√

x, and ln x are both increasing functions of x, so is their compo-sition

√ln x; since x is also an increasing function of x, so is the product x

√ln x.

Thus the denominators of these fractions are increasing, and the fractions must bedecreasing; (this is because the function 1

x is a decreasing function of x). We ob-serve also that the limit of n

√n as n→ ∞ is infinite; hence the limit of the terms of

the series is 0. As the function 1x ln x is continuous, we have proved that the condi-

tions of the integral test have been satisfied: the series will converge iff the integral∞∫

3

1x ln x

dx converges. (The fact that the function is decreasing could also have

been shown by differentiation.)∫ a

3

dx

x√

ln x= 2√

ln x]a

3= 2√

ln a − 2√

ln 3→ ∞ as a→ ∞ .

We conclude, by the Integral Test that the given series diverges. (The proof that theterms were the values of a function which satisfied the conditions of the theoremwas an essential part of the solution.)

(b) [4 MARKS]∞∑

n=1

(−1)n+1

√4n + 5

3n + 10Solution: We may begin by considering the corresponding series of absolute val-

ues,∞∑

n=1

√4n + 5

3n + 10. If we attempt to apply the Ratio Test, we find that the limit of


the ratios of successive terms is 1, so that test is useless here. But we can apply theLimit Comparison Test, comparing with the divergent p-series

∑n−

12 :

limn→∞

√4n + 5

3n + 101

n12

= limn→∞

√4 + 5

n

3 + 10n

=23> 0 .

Since this limit is positive, and since the p-series to which we have compared isdivergent, we may conclude that the series of absolute values is also divergent; andhence that the original series we were given is not absolutely convergent. But thatstill leaves open the question of whether that series is conditionally convergent ordivergent. I will check whether the conditions of the Leibniz Alternating SeriesTest are satisfied:

limn→∞

√4n + 5

3n + 10= lim

n→∞

1√n·

√4 + 5

n

3

= 0

ddx

√

4x + 53x + 10

=

12· 2 · 1√

4x + 5· (3x + 10)) − (

√4x + 5 · 3)

(3x + 10)2

=−9x − 5

(3x + 10)2√

4x + 5< 0 ,

hence we may apply the Leibniz Test, and conclude that the given series is conver-gent. Since it has been shown to not be absolutely convergent, it is conditionallyconvergent.

(c) [4 MARKS]∞∑

n=1

(cot−1

(1

n + 1

)− cot−1

(1n

))

Solution: This is a telescoping series:

N∑

n=0

(cot−1

(1

n + 1

)− cot−1

(1n

))= − cot−1 1 + cot−1 1

N + 2.

As N → ∞ this partial sum approaches − cot−1 1 + cot−1 0 = −π4 + π2 = π

4 , so this is

the value to which the series converges. Since the derivative of cot−1 x is − 11 + x2 ,

the function is monotonely decreasing. Hence the terms of the given series are allpositive: a positive series which is convergent is absolutely convergent.


(The problem did not require the student to find the precise value of the sum. Ifone did not notice that the given series was telescopic, one could still have appliedthe integral test to prove that it is absolutely convergent. It can be seen that

∫ ∞

0

(cot−1 1

x + 1− cot−1 1

x

)dx

=

∫ ∞

0(arctan(x + 1) − arctan(x)) dx

= lima→∞

[(arctan(x + 1) − ln(1 + (x + 1)2)

2

)−

(arctan x − ln(1 + x2)

2

)]a

1

= lima→∞

[arctan(x + 1) − arctan x +

12

ln

∣∣∣∣∣∣1 + x2

1 + (x + 1)2

∣∣∣∣∣∣]a

1

= 0 − 12

ln25.

)

4. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by thecurves 2y = x, y = 2x, and x2 + y2 = 5.

(a) [4 MARKS] The region R is rotated about the line x = −1. Give an integral or sumof integrals whose value is the volume of the resulting solid.Solution:

Using “Washers”:

π

∫ 1

0

((2y + 1)2 −

(y2

+ 1)2)

dy + π

∫ 2

1

(( √5 − y2 + 1

)2 −(y2

+ 1)2)

dy

Using Cylindrical Shells:

2π∫ 1

0(x + 1)

(2x − x

2

)dx + 2π

∫ 2

1(x + 1)

(√5 − x2 − x

2

)dx

(b) [4 MARKS] Let L(a) denote the length of the portion of line y = a which lies insideR. Express in terms of integrals — which you need not evaluate — the average ofthe positive lengths L(a).Solution:

12

(∫ 1

0

(2y − y

2

)dy +

∫ 2

1

( √5 − y2 − y

2

)dy

)

(c) [4 MARKS] Let C1 be the curve x(t) = t, y(t) = cosh t (0 ≤ t ≤ ln 2). Simplifyingyour answer as much as possible, find the length of C1.



Solution:

dxdt

= 1

dydt

= sinh t√(

dxdt

)2

+

(dydt

)2

=√

1 + sinh2 t = | cosh t| = cosh t

Length =

∫ ln 2

0cosh t dt

= [sinh t]ln 20

= sinh ln 2 − sinh 0

=3ln 2 − e− ln 2

2− 0

=

2 − 12

2=

34.


(a) [8 MARKS] Evaluate the indefinite integral∫

36(x + 4)(x − 2)2 dx .

Solution: We observe that the degree of the numerator of the integrand is less thanthe degree of the denominator; thus we do not need to divide denominator intonumerator and obtain a quotient and remainder. Next we need to find a PartialFraction decomposition of the integrand. Assuming a decomposition of the form

36(x + 4)(x − 2)2 =

Ax + 4

+B

x − 2+

C(x − 2)2

we obtain, by taking the right side to a common denominator,

36 = A(x − 2)2 + B(x + 4)(x − 2) + C(x + 4) .

Students should know two methods to find the values of these constants: either,by assigning “convenient” values to x and thereby obtaining enough equations thatcan be solved for the coefficients, or by comparing coefficients of powers of x on


the two sides of the equation. These methods, or a combination of them, yieldA = 1, B = −1, C = 6. Hence

∫36

(x + 4)(x − 2)2 dx =

∫ (1

x + 4− 1

x − 2+

6(x − 2)2

)dx

= ln∣∣∣∣∣x + 4x − 2

∣∣∣∣∣ −6

x − 2+ C .

(b) [4 MARKS] Determine whether

∞∫

3

36(x + 4)(x − 2)2 dx converges. If it converges,

find its value.

∫ a

3

36(x + 4)(x − 2)2 dx =

ln

∣∣∣∣∣∣∣∣∣∣

1 +4a

1 − 2a

∣∣∣∣∣∣∣∣∣∣− 6

a − 2

− (ln 7 − 6)

→ 0 − 0 − (ln 7 − 6) = 6 − ln 7 as a→ ∞ .

Since the limit exists, the integral converges to the limiting value, 6 − ln 7.



(a) [4 MARKS]∫ √

e√

x dx

Solution: This integral can be solved by a series of substitutions; for example, the

substitution u =

√x

2, which implies that x = 4u2, dx = 8u du yields

∫ √e√

t dt =

∫e√

x2 dx = 8

∫u · eu du ,

which can then be integrated by parts, as follows: with u = u, dv = eu du, du = du,v = eu, so

∫ √e√

t dt = 8∫

u · eu du

= 8(u · eu −

∫eu du

)

= 8 ((u − 1)eu) + C

=(4√

x − 8)

e√

x2 + C .


(b) [5 MARKS]

0∫

− 12

x√3 − 4x − 4x2

dx

Solution:√

3 − 4x − 4x2 =√

4 − (1 + 2x)2

= 2

√1 −

(1 + 2x

2

)2

.

0∫

− 12

x√3 − 4x − 4x2

dx =12

∫ 0

− 12

x√1 −

(1+2x

2

)2dx

=12

∫ 12

0

u − 12√

1 − u2du

=12

[−√

1 − u2 − 12· arcsin u

] 12

0

=12

−√

32− 1

2arcsin

12

− (−1 − 0)

=12

−√

32− 1

2· π

6+ 1

=12− π

24−√

34.

(c) [4 MARKS]∫ π

0sin2 t cos4 t dt .

Solution: To evaluate integrals of this type the method of the textbook is to replacethe square powers of sines and cosines by functions of the cosine of twice the angle.The following is a variant of that method, using both the sine and the cosine of thedouble angle.

∫ π

0sin2 t cos4 t dt =

∫ π

0

(2 sin t · cos t)2

4· cos2 t dt

=

∫ π

0

(sin2 2t

4

)·(1 + cos 2t

2

)dt

=

∫ π

0

(sin2 2t

8+

sin2 2t · cos 2t8

)dt

=

∫ π

0

(1 − cos 4t

16+

3 · sin2 2t · cos 2t · 232

)dt



=

[t

16− sin 4t

64+

sin3 2t48

]π

0

=π

16.


Consider the curve C2 defined by x = x(t) = 1 + e−t , y = y(t) = t + t2 .

(a) [2 MARKS] Determine the coordinates of all points where C2 intersects the x-axis.Solution:

y = 0 ⇒ t + t2 = 0⇒ t = 0,−1t = 0 ⇒ (x, y) = (2, 0)

t = −1 ⇒ (x, y) + (1 + e, 0) ,

so the points of intersection with the x axis are (2, 0) and (1 + e, 0).

(b) [2 MARKS] Determine the coordinates of all points of C2 where the tangent ishorizontal.Solution:

dxdt

= −e−t

dydt

= 1 + 2t

dydx

= 0 ⇒dydtdxdt

= 0⇒ dydt

= 0⇒ t = −12

t = −12⇒ (x, y) =

(1 +√

e,−14

).

(c) [6 MARKS] Determine the area of the finite region bounded by C2 and the x-axis.

Solution: The area is

2∫

1+ 1e

y dx =

∫ 0

−1y(t) · dx

dt(t) dt =

∫ 0

−1

(t + t2

) (−e−t) dt

=[e−t(t + 1) + e−t

(t2 + 2t + 2

)]0

−1

=[e−t ·

(t2 + 3t + 3

)]0

−1= 3 − e(1 − 3 + 3) = 3 − e .


The preceding evaluation required two indefinite integrals that could be obtainedby integration by parts:

• Taking u = t, v′ = e−t, which imply that u′ = 1, v = −e−t,∫

t · e−t dt = −te−t +

∫e−t dt = −e−t(t + 1) + C

• Taking U = t2, V ′ = e−t, which imply that U′ = 2t, V = −e−t,∫

t2 · e−t dt = −t2e−t +

∫2t · e−t dt = −(t2 + 2t + 2)e−t + C .


(a) [5 MARKS] Showing all your work, determine whether the series∞∑

n=2

√n(√

n + 2 −√

n − 2)

is convergent or divergent.Solution: This is a telescoping series.

N∑

n=2

√n(√

n + 2 −√

n − 2)

=

N∑

n=2

√n ·√

n + 2 −N−2∑

m=0

√m ·√

m + 2

= −√

3 +√

N(N + 2) +√

(N − 1)(N + 1)→ ∞ as N → ∞ .

As the partial sums are approaching infinity, they do not have a finite limit, and theseries is, by definition, divergent.Alternatively, one could argue that

√n(√

n + 2 −√

n − 2)

=

√n (n + 2 − n + 2)√n + 2 +

√n − 2

=4√

n√n + 2 +

√n − 2

=4√

1 + 2n +

√1 − 2

n

→ 2 , 0 as n→ ∞.

By “The” Test for Divergence, the series diverges.


(b) [5 MARKS] Showing all your work, determine whether the following sequenceconverges; if it converges, find its limit:

a1 = 3.a2 = 3.14a3 = 3.1414a4 = 3.141414a5 = 3.14141414a6 = 3.1414141414

etc., where each term after a2 is obtained from its predecessor by the addition onthe right of the decimal digits 14.

an = 3 + 0.14(1 +

1100

+1

1002 + . . . +1

100n−1

)

= 3 +0.14

(1 − 1

100n

)

1 − 1100

→ 3 +0.140.99

=31199

as n→ ∞ .


Curves C3 and C4, respectively represented by polar equations

r = 4 + 2 cos θ (0 ≤ θ ≤ 2π) (104)and

r = 4 cos θ + 5 (0 ≤ θ ≤ 2π) , (105)

divide the plane into several regions.

(a) [8 MARKS] Showing all your work, carefully find the area of the one region whichis bounded by C3 and C4 and contains the pole.Solution: (cf. Figure 25 on page 3291 of these notes) Solving the given equations

yields θ =2π3,

4π3

and r = 3. The points of intersection are(3,

2π3

),(3,

4π3

).

(Strictly speaking, the student should also solve using equations −r = 4 + cos(−θ)and −r = 4 cos(−θ)+5, taking all 2×2 combinations of the equations, in case a pointof intersection appeared on the two given curves with different sets of coordinates;that would not have yielded any new points of intersection in this problem. It couldalso have been necessary to check for the curves’ possibly passing through the pole— but that does not happen in this problem, since, when we set r = 0, we obtainan equation for θ which cannot be solved.)There are several different ways of finding the area in question.


6

2

-6

4

0

-4

-2

8420-2 6

Figure 25: The curves with equations r = 4 + 2 cos θ, r = 4 cos θ + 5

i. Find the area of the small oval (a limacon) and subtract that of the regionon the left. The area of the limacon is

12

∫ 2π

0(2 cos θ + 4)2 dθ =

12

∫ 2π

0(2(1 + cos 2θ) + 16 cos θ + 16) dθ

=12

[18θ + sin 2θ + 16 sin θ]2π0 = 18π .

The area of the region to be subtracted is

2 · 12

∫ π

2π3

((4 + 2 cos θ)2 − (4 cos θ + 5)2

)dθ


=

∫ π

2π3

((16 + 16 cos θ + 2(1 + cos 2θ)) − (8(1 + cos 2θ) + 40 cos θ + 25)) dθ

= [−15θ − 3 sin 2θ − 24 sin θ]π2π3

= (−15π − 0 − 0) −−10π + 3 ·

√3

2− 24 ·

√3

2

=21√

32− 5π ,

so the area of the region in question is 18π − 21√

32

+ 5π = 23π − 21√

32

.

ii. Find the area of the larger outer oval and subtract the area of the regionto the right of the desired region. The area bounded by the larger oval is

12

∫ 2π

0(4 cos θ + 5)2 dθ =

12

∫ 2π

0(8(1 + cos 2θ) + 40 cos θ + 25) dθ

=12

[33θ + 4 sin 2θ + 40 sin θ]2π0 = 33π .

The area to be subtracted is

2 · 12

∫ 2π3

0

((4 cos θ + 5)2 − (4 + 2 cos θ)2

)dθ = [15θ + 3 sin 2θ + 24 sin θ]

2π3

0

=

10π − 3√

32

+24√

32

− 0

= 10π +21√

32

.

Thus the net area is equal to

33π −10π +

21√

32

= 23π − 21√

32

.

iii. Join the pole by line segments to the points of intersection of the twocurves. Then compute the areas of the two subregions of the desired regionbounded by these line segments. The area to the right of the line segments is

212

∫ 2π3

0(2 cos θ + 4)2 dθ =

∫ 2π3

0(2(1 + cos 2θ) + 16 cos θ + 16) dθ

= [18θ + sin 2θ + 16 sin θ]2π3

0

=

12π −√

32

+16√

32

− 0 = 12π +15√

32

.


The area to the left of the line segments is

212

∫ π

2π3

(4 cos θ + 5)2 dθ

= [33θ + 4 sin 2θ + 40 sin θ]π2π3

= 11π − 18√

3 .

Summing these two areas yields 23π − 21√

32

.

(b) [4 MARKS] Find another equation — call it (105*) — that also represents C4, andhas the property that there do not exist coordinates (r, θ) which satisfy equations(104) and (105*) simultaneously. You are expected to show that equations (104)and (105*) have no simultaneous solutions.Solution: In my discussion of the preceding part of the problem I have shown (byreplacing (r, θ) by (−r, θ + π)) that an alternative equation for the second curve is−r = 4 cos(−θ)+5, equivalently r = −4 cos θ−5. When this equation is solved with

r = 4 + 2 cos θ, we obtain, as a consequence, that cos θ = −32

, which is impossible.

C.37.2 Draft Solutions to the Final Examination in MATH 141 2009 01 (Version 4)

Instructions


2. Calculators are not permitted. This is a closed book examination. Regular and translation dictio-naries are permitted.

3. . . . A TOTAL OF 75 MARKS ARE AVAILABLE ON THIS EXAMINATION.

4. You are expected to simplify all answers wherever possible.

• Most questions on this paper require that you SHOW ALL YOUR WORK!

. . . To be awarded partial marks on a part of a question a student’s answer for that partmust be deemed to be more than 50% correct.

• Some questions on this paper require only BRIEF SOLUTIONS ; . . . you are not asked toshow your work, and you should not expect partial marks for solutions that are not correct.




t3 cos t2 dt .

Solution: One substitution is u = t2, which implies that du = 2t dt,∫

t3 cos t2 dt =12

∫u cos u du .

Now we can integrate by parts, taking U = u, dV = cos u du, so dU = du, V =

sin u.∫

t3 cos t2 dt =12

∫u cos u du

=12

(u sin u −

∫sin u du

)

=12

(u sin u + cos u) + C

=12

(t2 sin

(t2)

+ cos(t2))

+ C .

This problem could also have been attacked by an immediate application of inte-gration by parts: u = t2, dv = t cos t2 dt ⇒ du = 2t dt, v = 1

2 sin t2:∫

t3 cos t2 dt = t2 · 12

sin t2 −∫ (

sin t2)· 2t dt

=12

(t2 · sin t2 + cos t2

)+ C .

Of course, whichever method you used, you should check your answer by differ-entiation.

(b) [4 MARKS] Simplifying your answer as much as possible, evaluate the derivative

ddt

t2∫

0

tanh x2 dx .

Solution: Let u = t2.

ddt

∫ t2

0tanh x2 dx =

ddt

∫ u

0tanh x2 dx

=

(d

du

∫ u

0tanh x2 dx

)· du

dt

=(tanh u2

)· du

dt=

(tanh t4

)· 2t .




(a) [4 MARKS] Evaluate

12∫

1√2

dx√1 − x2 · arcsin x

.

Solution: Let u = arcsin x, so du =dx√

1 − x2. Then

12∫

1√2


=

arcsin 12∫

arcsin 1√2

duu

=

π6∫

π4

duu

= ln |u|]π6π4

= lnπ

6− ln

π

4= ln

23.

It should be no surprise that this answer is negative, since1√2>

12

, and the inte-

grand is positive in the interval[

12 ,

1√2

].

(b) [4 MARKS] Evaluate∫

2y√y2 − y + 1

dy

Solution: By completion of the square we find that

y2 − y + 1 =

(y − 1

2

)2

+34

=34

1 +

(2√3

(y − 1

2

))2

=34

1 +

(2y − 1√

3

)2 .

We may thus substitute u =2y − 1√

3, obtaining du =

2√3· dy, y =

u√

3 + 12

.

∫2y√

y2 − y + 1dy =

∫ u√

3 + 1√

32

√1 + u2

·√

32

du


=√

3∫

u√1 + u2

du +

∫du√

1 + u2

=√

3 ·√

1 + u2 +

∫du√

1 + u2

= 2√

y2 − y + 1 +

∫du√

1 + u2.

TO BE CONTINUED


C.38 Supplementary Notes for the Lecture of Friday, April 09th, 2010Distribution Date: Friday, April 09th, 2010, subject to correction

C.38.1 Final Examination in MATH 141 2009 01 (Version 4, continued)

2. (b) [4 MARKS] Evaluate∫

2y√y2 − y + 1

dy

Solution: By completion of the square we find that

y2 − y + 1 =

(y − 1

2

)2

+34

=34

1 +

(2√3

(y − 1

2

))2

=34

1 +

(2y − 1√

3

)2 .

We may thus substitute u =2y − 1√

3, obtaining du =

2√3· dy, y =

u√

3 + 12

.

∫2y√

y2 − y + 1dy =

∫ u√

3 + 1√

32

√1 + u2

·√

32

du

=√

3∫

u√1 + u2

du +

∫du√

1 + u2

=√

3 ·√

1 + u2 +

∫du√

1 + u2

= 2√

y2 − y + 1 +

∫du√

1 + u2.

Now we can apply a trigonometric (or hyperbolic) substitution, like tan θ = u =2y − 1√

3, i.e., θ = arctan

2y − 1√3

, so du = sec2 θ dθ. Thus

∫2y√

y2 − y + 1dy = 2

√y2 − y + 1 +

∫sec θ dθ

= 2√

y2 − y + 1 + ln | sec θ + tan θ| + C

= 2√

y2 − y + 1 + ln

∣∣∣∣∣∣∣∣

√1 +

(2y − 1√

3

)2

+2y − 1√

3

∣∣∣∣∣∣∣∣+ C


= 2√

y2 − y + 1 + ln

∣∣∣∣∣∣∣2√

y2 − y + 1 + 2y − 1√3

∣∣∣∣∣∣∣ + C

= 2√

y2 − y + 1 + ln∣∣∣∣2

√y2 − y + 1 + 2y − 1

∣∣∣∣ + C′ .

SHOW ALL YOUR WORK!

3. For each of the following series determine whether the series diverges, converges condi-tionally, or converges absolutely. All of your work must be justified; prior to using anytest you are expected to demonstrate that the test is applicable to the problem.

(a) [4 MARKS]∞∑

n=1

(−1)n+1(cos n

2

)n

Solution: We can’t use the nth root test here, as the sequencecos n

2

n=1,2,...

has no

limit as n → ∞. But, if we first consider the positive series of the magnitudes ofthe terms of the given series, we have

0 ≤∣∣∣∣∣(cos n

2

)n∣∣∣∣∣ ≤12n .

Since the series whose general term is12n is a convergent (geometric) series, we

may apply the Comparison Test to conclude that the series of absolute values ofthe terms of the original series is convergent; hence the original series is absolutelyconvergent.

(b) [4 MARKS]∞∑

n=2

(−1)n 1

n√

ln n.

Solution: As x → ∞, x, ln x, and√

x are all increasing. Hence f (x) =1

x√

ln xis

decreasing. As it is a continuous function for sufficiently large x, we may apply theIntegral Test. The divergence of the improper integral

∞∫

2

dx

x√

ln x= lim

a→∞

[2√

ln x]a

2= +∞ ,

implies the divergence of the series of absolute values. Thus the series∞∑

n=2

(−1)n 1

n√

ln nis at most conditionally convergent. Since x, ln x, and

√x are increasing functions


of x, 1x ln x is a decreasing function; and, as x → ∞, this last function approaches

0. Since the conditions of the Leibniz Alternating Series Theorem are satisfies, theoriginal series is convergent; but, as it is not absolutely convergent, is is thereforeconditionally convergent.

(c) [4 MARKS]∞∑

n=4

(−1)n 1n

ln (3n + 1)

Solution: I apply the “ Test for Divergence”:

limn→∞

ln (3n + 1)n

= limn→∞

3n · ln 33n + 1

1

= (ln 3) · limn→∞

11 + 1

3n

= ln 3 > ln e = 1 .

Hence the limit limn→∞

((−1)n ln (3n + 1)

n

)does not exist, and the given series diverges.

4. [9 MARKS] SHOW ALL YOUR WORK!

(a) [3 MARKS] Evaluate limn→∞

1n·

n∑

r=1

cos2(rπ

n

) . (Hint: This could be a Riemann

sum.)Solution: First consider the area under the graph of f (x) = cos2 x between x = 0and x = π. Divide the interval into n parts of equal width ∆x =

π

n. If we hang

rectangular elements of area from the curve by their upper right-hand corners, then

the area of the ith rectangle will beπ

n· cos2 iπ

n. In order to be able to interpret the

given sum as a Riemann sum, we will need to scale the function by a constant. So

let’s use the functioncos2 xπ

. Thus the area will be

π∫

0

cos2 xπ

dx =

π∫

0

1 + cos 2x2π

dx =1

2π

[x +

12

sin 2x]π

0=

12.

(b) [3 MARKS] Showing all your work, prove divergence, or find the limit of an =

arctan(−2n) as n→ ∞.

Solution: As n→ ∞, 2n→ ∞, and arctan(−2n)→ −π2

.


(c) [3 MARKS] Showing all your work, prove divergence, or find the value of∞∑

n=1

∞∑

i=1

15i

n

.

Solution: Since∞∑

i=1

15i = lim

n→∞

15

(1 −

(15

)n)

1 − 15

=

1545

=14,

∞∑

n=1

∞∑

i=1

15i

n

=

∞∑

n=1

(14

)n

=

14

1 − 14

=13.

TO BE CONTINUED


C.39 Supplementary Notes for the Lecture of Monday, April 12th, 2010Distribution Date: Monday, April 12th, 2010, subject to correction

C.39.1 Final Examination in MATH 141 2009 01 (Version 4, continued)



t + 12t2 − t − 1

dt .

Solution: We observe that

2t2 − t − 1 = (2t + 1)(t − 1) = 2(t +

12

)(t − 1) ,

a product of two distinct linear factors. There must therefore exist a partial fractiondecomposition of the form

t + 12t2 − t − 1

=A

2t + 1+

Bt − 1

=A(t − 1) + B(2t + 1)

(2t + 1)(t − 1).

The constants can be determined by comparing coefficients of corresponding pow-ers of t in the numerator polynomial, or as follows, by assigning to t “convenient”values of t in the identity t + 1 = A(t − 1) + B(2t + 1) (or by a combination of thetwo methods):

t = 1 ⇒ 2 = 0 + 3B⇒ B =23

t = −12⇒ 1

2= A

(−3

2

)+ B · 0⇒ A = −1

3.

Hence∫

t + 12t2 − t − 1

dt =

∫ (−1

3· 1

2t + 1+

23· 1

t − 1

)dt

= −16

ln |2t + 1| + 23

ln |t − 1| + C

=16

ln(t − 1)4

|2t + 1| + C .


(b) [2 MARKS] Determine whether the following improper integral converges or di-verges; if it converges, find its value:

∞∫

4

t + 12t2 − t − 1

dt .

Solution: We have found an antiderivative in the preceding part. Thus

∞∫

4

t + 12t2 − t − 1

dt = lima→∞

[16

ln(t − 1)4

|2t + 1|]a

4

=16

lima→∞

ln(a − 1)4

|2a + 1| −16

ln819.

In the factorization

(a − 1)4

2a + 1=

(a3

)·(1 − 1

a

)4

2 + 1a

the last factor has limit 12 as a→ ∞. Thus lim

a→∞(a − 1)4

2a + 1= +∞; this implies that its

logarithm also approaches ∞ as a → ∞ (cf. [1, Result 4, p. A50]). Thus the givenimproper integral diverges.


Consider the curve C1 defined by x = 3t2 , y = 2t3 , (t ≥ 0).

(a) [7 MARKS] Showing all your work, determine the area of the surface generatedwhen the arc 0 ≤ t ≤ 1 of C1 is rotated about the y-axis.Solution:

dxdt

= 6t

dydt

= 6t2

√(dxdt

)2

+

(dydt

)2

= 6|t|√

1 + t2

Surface Area = 2π

1∫

0

x · dxdt

dt


= 2π

1∫

0

3t2 · 6t√

1 + t2 dt

= 36π

1∫

0

t3√

1 + t2 dt

= 36π

√2∫

0

(u2 − 1

)· u2 du

using substitution u =√

1 + t2, so u du = t dt

= 36π[u5

5− u3

3

]√2

1

=24π

5

(√2 + 1

).

(b) [2 MARKS] Showing all your work, determine all points — if any — where thenormal to the curve is parallel to the line x + y = 8.Solution: The normal has slope

−dxdtdydt

= − 6t6t2 = −1

t.

Line x + y = 8 has slope −1, so we may impose the condition that −1t

= −1, whichtells us that the normal to the curve has the desired property only at the point withparameter value t = 1. That is the point (x, y) = (3 · 12, 2 · 13) = (3, 2).


Curves C3 and C4, are respectively represented by polar equations

r = 3 + 3 cos θ (0 ≤ θ ≤ 2π) (106)and

r = 9 cos θ (0 ≤ θ ≤ π) . (107)

(a) [7 MARKS] Showing all your work, carefully find the area of the region lyinginside both of the curves.Solution: (cf. Figure 26 on page 3304 of these notes) When we solve the two given


Figure 26: The curves with equations r = 3+3 cos θ, (0 ≤ θ ≤ 2π), and r = 9 cos θ, (0 ≤ θ ≤ π)

equations simultaneously (for variables r and cos θ), we find that

cos θ =12

(108)

and r = 92 . The only solution θ to equation (108) which satisfy the condition

0 ≤ θ ≤ π is θ =π

3: thus one of the points of intersection of the two curves

is the point (r, θ) =

(92,π

3

). Another point of intersection could be identified

by symmetry, or by solving with equation (107) the equation −r = 3 − 3 cos θ(0 ≤ θ ≤ 2π), which we can obtain from equation (106) by replacing (r, θ) re-spectively by (−r, θ + π); this is another equation for the same cardioid; (there isno distinct second equation for the circle, as the change described here leaves the

equation invariant). Simultaneous solution of these equations yields cos θ = −12

,


and yields the point (r, θ) =

(−9

2,

2π3

), which point can also be represented as

(r, θ) =

(92,−1π

3

). The curves also intersect at the pole, which appears on curve C3

with coordinates (r, θ) = (0, π), and on C4 with coordinates (r, θ) =

(0,π

2

).

I will find the portion of the desired area above the line θ = 0, and double it. AndI will divide this upper area into two parts by the line segment joining the pole to

the upper point of intersection, (r, θ) =

(92,π

3

). The upper area to the right of this

line segment has magnitude

12

π3∫

0

(3 + 3 cos θ)2 dθ =92

π3∫

0

(1 + 2 cos θ +

1 + cos 2θ2

)dθ

=92

[3θ2

+ 2 sin θ +sin 2θ

4

] π3

0

=92

π2 +9√

38

.

The area to the left of the line segment has area

12

π2∫

π3

(9 cos θ)2 dθ =812

π2∫

π3

1 + cos 2θ2

dθ

=812

[θ

2+

sin 2θ4

] π2

π3

=27π

4− 81

√3

16.

Hence the area of the region interior to both closed curves is 2(9π) = 18π.Suppose that you wished to solve the problem without appealing to symmetry.Then the area could be decomposed into 3:

i. the sector subtended by the pole on the cardioid; andii. the two sectors subtended by the pole on the circle from θ = π

3 to θ = 2π3 .

The area of the sector subtended on the cardioid can be represented by an integralfrom θ = −π3 to θ = +π

3 . To the student who might object that this appears to involvevalues of θ outside of the given interval of values, one could respond that the single


integral could also be given by two separate integrals, with the same integrand, thefirst from 0 to π

3 , and the second from 5π3 to 2π:

12

π3∫

− π3

(3 + 3 cos θ)2 dθ +12

2π3∫

π3

(9 cos θ)2 dθ

i.e.,

12

2π∫

5π3

(3 + 3 cos θ)2 dθ +12

π3∫

0

(3 + 3 cos θ)2 dθ +12

2π3∫

π3

(9 cos θ)2 dθ .

(b) [3 MARKS] Determine the length of the curves which form the boundary of theregion whose area you have found.Solution: The region has a boundary formed by four arcs. I will find the lengthsof the two arcs above θ = 0, and double the sum. For the arc of the circle C4,√

r2 + (r′)2 = 9, a constant; for the arc of the cardioid C3,√

r2 + (r′)2 =

√9(1 + cos θ)2 + 9 sin2 θ = 3

√2 ·√

1 + cos θ = 6∣∣∣∣∣cos

θ

2

∣∣∣∣∣ .

Hence the length of the arc of the cardioid is

6

π3∫

0

cosθ

2dθ = 12

[sin

θ

2

] π3

0= 6 .

The length of the arc of the circle is

9

π2∫

π3

dθ =3π2.

The boundary of the region is therefore of length

2(6 +

3π2

)= 12 + 3π .

Suppose that you wished to solve the problem without appealing to symmetry. Thecircumference could be represented by 4 separate integrals: 1 for the circle, from


π3 to 2π3, and 1 for the cardioid, from θ = −π3 to θ = +π3 . Again, to the student who

might object that this appears to involve values of θ outside of the given intervalof values, one could respond that the single integral could also be given by twoseparate integrals, with the same integrand, the first from 0 to π

3 , and the secondfrom 5π

3 to 2π:

6

2π∫

5π3

cosθ

2dθ + 6

π3∫

0

cosθ

2dθ + 9

∫ 2π3

π3

dθ .


C.40 Supplementary Notes for the Lecture of Wednesday, April 14th,2010

Distribution Date: Wednesday, April 14th, 2010, subject to correction

C.40.1 Final Examination in MATH 141 2009 01 (Version 4, concclusion)

8. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the

curves y =8x2 , x = y, x = 1.

(a) [3 MARKS] The region R is rotated about the line x = −1 to create a 3-dimensionalsolid, S 1. Give an integral or sum of integrals whose value is the volume of S 1;you are not asked to evaluate the integral(s).Solution: Solving the given equations, we find that the vertices of the region R are(1, 8), (1,1), and (2,2). I will solve the problem in two ways. Students were notasked to evaluate the integrals, but I will do so:

Using Cylindrical Shells:

Volume =

2∫

1

(x + 1)(

8x2 − x

)dx

= 2π

2∫

1

(−x2 − x +

8x

+8x2

)dx

= 2π[−1

3· x3 − 1

2· x2 + 8 ln |x| − 8

x

]2

1= π

(16 ln 2 +

13

).

Using “Washers”: We need to express the first equation as x =

√8y

. There will

be two types of “washers”, so the volume will have to be expressed as thesum of two integrals; the change in description of the right boundary of the“washer” occurs at height y = 2.

Volume = π

2∫

1

((y − (−1))2 − (1 − (−1))2

)dy

+π

8∫

2

√8y− (−1)

2

− (1 − (−1))2

dy


= π

2∫

1

(y2 + 2y − 3

)dy + π

8∫

2

8y

+4√

2√y− 3

dy

= π

[13· y3 + y2 − 3y

]2

1+ π

[8 ln y + 8

√2y − 3y

]8

2

=7π3

+ π(16 ln 2 − 2) = π

(16 ln 2 +

13

).

(b) [3 MARKS] The region R is rotated about the x- axis to create a 3-dimensionalsolid, S 2. Give an integral or sum of integrals whose value is the volume of S 2

obtained only by the Method of Cylindrical Shells; you are not asked to evaluatethe integral(s).Solution: Students were asked to express this volume of revolution only by usingthe method of cylindrical shells, and not to evaluate the integral(s). I will solve theproblem both ways.

Using Cylindrical Shells: I will again need to express the non-linear boundary in

the form x =

√8y

.

Volume = 2π

2∫

1

y · (y − 1) dy + 2π

8∫

2

y

√8y− 1

dy

= 2π[y3

3− y2

2

]2

1+ 2π

[√8 · 2

3· y 2

3 − y2

2

]8

2

=49π

3.

Using “Washers”:

Volume =

2∫

1

π(

8x2

)2

− πx2

dx

= π

2∫

1

(64x4 − x2

)dx

= π

[− 64

3x3 −x3

3

]2

1=

49π3

.


(c) [3 MARKS] Calculate the area of R.Solution: The area is

2∫

1

(8x2 − x

)dx =

[−8

x− x2

2

]2

1=

52

;

or, integrating along the y-axis:

2∫

1

(y − 1) dy +

8∫

2

√8y− 1

dy

=

[y2

2− y

]2

1+

[4√

2y − y]8

2

=12

+ (8 − 6) =52.


D Problem Assignments from Previous Years

D.1 1998/1999The problem numbers listed below refer to the textbook in use at that time, [31], [33]. Formany of the problems there are answers in the textbook or in the Student Solution Manual[34].

D.1.1 Assignment 1

§5.2: 5, 11, 15, 21, 29

§5.3: 3, 9, 15, 35, 47

§5.4: none

§5.5: 17, 27, 33, 41

§5.6: 47, 55, 59, 65

§5.7: 21, 27, 33, 39, 45, 51, 57

§5.8: 33, 39, 45, 51, 57

D.1.2 Assignment 2

§6.1: none

§6.2: 3, 9, 15, 21, 27, 31, 35, 41

§6.3: 3, 9, 15, 21, 27, 31, 39, 43

§6.4: 3, 9, 15, 21, 27, 31, 35, 41

§3.8: none

Chapter 7: none


D.1.3 Assignment 3

§8.2: 5, 13, 21, 29, 39, 45, 53

§9.2: 5, 13, 21, 29, 39

§9.3: 5, 13, 21, 29, 39, 41

§9.4: 5, 13, 21, 29, 39

§9.5: 5, 9, 17, 21, 29, 33

§9.6: 5, 9, 17, 21, 29, 33

D.1.4 Assignment 4

§9.7: 13, 17, 21, 25, 29, 33

§9.8: 21, 23, 29, 33, 39

§10.2: 39, 41, 43, 45, 47, 49, 51, 53, 57

§10.3: 9, 13, 17, 21, 23, 29, 33, 35

§10.4: 3, 5, 9, 13

D.1.5 Assignment 5

§11.2: 9, 17, 23, 33, 39

§11.3: 3, 9, 15, 21, 29, 35, 47

§11.4: 3, 9, 15, 21, 29, 35, 45, 47

§11.5: 3, 9, 15, 21, 23

§11.6: 3, 9, 15, 21, 29

§11.7: 3, 9, 15, 21, 27, 33


D.2 1999/2000(Students had access to brief solutions that were mounted on the web.)

D.2.1 Assignment 1

Before attempting problems on this assignment you are advised to try some “easy” problemsin the textbook. In most of the following problems there is a reference to a “similar” problemin the textbook. You should always endeavour to show as much of your work as possible, andto reduce your solution to “simplest terms”. Remember that the main reason for submittingthis assignment is to have an opportunity for your tutor to grade your work; the actual gradeobtained should be of lesser significance.

In Exercises 1-5 below, evaluate the indefinite integral, and verify by differentiation:

1. (cf. [31, Exercise 5.2.5, p. 294])∫ (

3x4 − 5x

12 − x2 + 4x−3

)dx

2.∫ (

3x− 2

1 + x2

)dx

3. (cf. [31, Exercise 5.2.13, p. 294])∫ (

xex2 − e4x)

dx

4. (cf. [31, Exercise 5.2.19, p. 294])∫

(1 − √x)(2x + 3)2 dx

5. (cf. [31, Exercise 5.2.27, p. 294])∫

(4 cos 8x − 2 sin πx + cos 2πx − (sin 2π)x) dx

6. (cf. [31, Example 5.2.8, p. 289]) Determine the differentiable function y(x) such thatdydx

=1√

1 − x2and y

(2−

12

)=π

2.

7. (This is [31, Exercise 5.2.51, p. 295] written in purely mathematical terminology.) Solve

the initial value problem:ddx

(dydx

)= sin x, where y = 0 and

dydx

= 0 when x = 0. [Hint:

First use one of the initial values to determine the general value ofdydx

from the given“differential equation”; then use the second initial value to determine y(x) completely.]

8. ([31, Exercise 5.3.4, p. 306]) Write the following in “expanded notation”, i.e. without

using the symbol∑

:6∑

j=1

(2 j − 1).


9. (cf. [31, Exercise 5.3.18, p. 306]) Write the following sum in “summation notation”:

x − x3

3+

x5

5− x7

7+ ... ± x999

999where the signs are alternating +, −, +, −, ... The sign of the last term has not been given— you should determine it.

10. (cf. [31, Example 5.3.6, p. 302]) Given thatn∑

i=1

i =n(n + 1)

2,

n∑

i=1

i2 =n(n + 1)(2n + 1)

6,

n∑

i=1

i3 =n2(n + 1)2

4,

determine limn→∞

(n + 1)3 + (n + 2)3 + ... + (2n)3

n4 .

D.2.2 Assignment 2

1. Evaluate the following integrals:

(a)∫ 3

1(x − 1)4 dx

(b)∫ 1

0(2ex − 1)2 dx

(c)∫ π

0sin 4x dx.

2. Interpreting the following integral as the area of a region, evaluate it using known areaformulas: ∫ 6

0

√36 − x2 dx.

3. Use properties of integrals to establish the following inequality without evaluating theintegral: ∫ 1

0

11 +√

xdx ≤

∫ 1

0

11 + x3 dx.

4. Deduce the Second Comparison Property of integrals from the First Comparison Prop-erty [31, p. 325, §5.5].

5. Apply the Fundamental Theorem of Calculus [31, p. 331, §5.6] to find the derivative ofthe given function: ∫ x

−1(t2 + 2)15 dt.


6. Differentiate the functions

(a)∫ x3

0cos t dt

(b)∫ 3x

1sin t2 dt.

7. Solve the initial value problemdydx

=√

1 + x2 , y(1) = 5 . Express your answer interms of a definite integral (which you need not attempt to evaluate). This problem canbe solved using the methods of [31, Chapter 5].

8. Evaluate the indefinite integrals:

(a)∫

2x√

3 − 2x2 dx

(b)∫

x2 sin(3x3) dx

(c)∫

x + 3x2 + 6x + 3

dx

9. Evaluate the definite integrals:

(a)∫ 8

0t√

t + 2 dt

(b)∫ π/2

0(1 + 3 sin η)3/2 cos η dη

(c)∫ π

0sin2 2t dt.

10. Sketch the region bounded by the given curves, then find its area:

(a) x = 4y2, x + 12y + 5 = 0

(b) y = cos x, y = sin x, 0 ≤ x ≤ π

4.

11. Prove that the area of the ellipsex2

a2 +y2

b2 = 1 is A = πab. This problem can besolved using the methods of [31, Chapter 5]. It is not necessary to use methods of [31,Chapter 9].



D.2.3 Assignment 3

In all of these problems you are expected to show all your work neatly. (This assignment isonly a sampling. Your are advised to try other problems from your textbook; solutions to somecan be found in the Student Solution Manual [32].)

1. [31, Exercise 6.1.6, p. 382] As n → ∞, the interval [2, 4] is to be subdivided into nsubintervals of equal length ∆x by n − 1 equally spaced points x1, x2, ..., xn−1 (where

x0 = 2, xn = 4). Evaluate limn→∞

n∑

i=1

1xi

∆x by computing the value of the appropriate

related integral.

2. (a) [31, Exercise 6.2.6, p. 391] Use the method of cross-sections to find the volume ofthe solid that is generated by rotating the plane region bounded by y = 9 − x2 andy = 0 about the x-axis.

(b) (cf. Problem 2a) Use the method of cylindrical shells to find the volume of the solidthat is generated by rotating the plane region bounded by y = 9−x2 and y = 0 aboutthe x-axis.

(c) Use the method of cross-sections to find the volume of the solid that is generatedby rotating the plane region bounded by y = 9 − x2 and y = 0 about the y-axis.

(d) (cf. Problem 2c) Use the method of cylindrical shells to find the volume of the solidthat is generated by rotating the plane region bounded by y = 9−x2 and y = 0 aboutthe y-axis.

3. (a) [31, Exercise 6.2.24, p. 392] Find the volume of the solid that is generated byrotating around the line y = −1 the region bounded by y = 2e−x, y = 2, and x = 1.

(b) (cf. Problem 3a) Set up an integral that would be obtained if the method of cylin-drical shells were used to represent the volume of the solid that is generated byrotating around the line y = −1 the region bounded by y = 2e−x, y = 2, and x = 1.YOU ARE NOT EXPECTED TO EVALUATE THE INTEGRAL.

4. (cf. [31, Exercise 6.2.40, p. 392]) The base of a certain solid is a circular disk with di-ameter AB of length 2a. Find the volume of the solid if each cross section perpendicularto AB is an equilateral triangle.

5. (a) [31, Exercise 6.3.26, p. 401] Use the method of cylindrical shells to find the volumeof the solid generated by rotating around the y-axis the region bounded by the

curves y =1

1 + x2 , y = 0, x = 0, x = 2.


(b) (cf. Problem 5a) Use the method of cross sections to find the volume of the solidgenerated by rotating around the y-axis the region bounded by the curves y =

11 + x2 , y = 0, x = 0, x = 2.

6. (cf. [31, Exercise 7.3.69, p. 450]) Find the length of the arc of the curve y =ex + e−x

2between the points (0, 1) and (ln 2, 2).

7. (a) [31, Exercise 6.4.30, p. 411] Find the area of the surface of revolution generatedby revolving the arc of the curve y = x3 from x = 1 to x = 2 around the x-axis.

(b) (cf. 7a) Set up an integral for, BUT DO NOT EVALUATE, the area of the surfaceof revolution generated by revolving the arc of the curve y = x3 from x = 1 to x = 2around the y-axis.

8. [31, Exercise 7.2.44, p. 442] Evaluate the indefinite integral∫

x + 1x2 + 2x + 3

dx

9. (cf. [31, Exercise 7.2.36, p. 442]) Determine the value of the function f (x) =

∫ x

−1

t2

8 − t3 dt

for any point x < 2.

10. (cf. [31, Exercise 7.3.70, p. 450]) Find the area of the surface generated by revolvingaround the x-axis the curve of Problem 6.

D.2.4 Assignment 4

1. Differentiate the functions:

(a) sin−1(x50)

(b) arcsin(tan x)

(c) cot−1 ex + tan−1 e−x

2. Showing all your work, evaluate the integrals:

(a)∫

dx√1 − 4x2

(b)∫

dx2√

x(1 + x)


(c)∫

ex

1 + e2x dx

(d)∫ cot

√y csc

√y√

ydy

(e)∫

(ln t)8

tdt

(f)∫

tan4 2x sec2 2x dx

(g) THIS PROBLEM SHOULD BE OMITTED. IT MAY BE INCLUDED IN AS-

SIGNMENT 5.∫

x2

√16x2 + 9

dx

3. Use integration by parts to compute the following integrals. Show all your work.

(a)∫

t cos t dt

(b)∫ √

y ln y dy

(c) THIS PROBLEM SHOULD BE OMITTED. IT MAY BE INCLUDED IN AS-

SIGNMENT 5.∫

x2 arctan x dx

(d)∫

csc3 x dx

(e)∫

ln(1 + x2) dx

4. Showing all your work, evaluate the following integrals:

(a)∫

cos2 7x dx

(b)∫

cos2 x sin3 x dx

(c)∫

sin3 2xcos2 2x

dx

(d)∫

sec6 2t dt


D.2.5 Assignment 5

1. [31, Exercise 9.5.6, p. 540] Find∫

x3

x2 + x − 6dx. (Your solution should be valid for x

in any one of the intervals x < 3, −3 < x < 2, x > 2.)

2. [31, Exercise 9.5.8, p. 540] Find∫

1(x + 1)(x2 + 1)

dx.

3. (a) [31, Exercise 9.5.23] Find∫

x2

(x + 2)3 dx.

(b) Find the volume of the solid of revolution generated by the region bounded byy =

x

(x + 2)32

, y = 0, x = 1, and x = 2 about the x-axis.

(c) Find the volume of the solid of revolution generated by the region bounded byy =

x

(x + 2)32

, y = 0, x = 1, and x = 2 about the y-axis.

4. [31, Exercise 9.5.38, p. 540] Make a preliminary substitution before using the methodof partial fractions: ∫

cos θsin2 θ(sin θ − 6)

dθ

5. [31, Exercise 9.6.6, p. 547] Use trigonometric substitutions to evaluate the integral∫x2

√9 − 4x2

dx.

6. [31, Exercise 9.6.26, p. 547] Use trigonometric substitutions to evaluate the integral∫1

9 + 4x2 dx.

7. [31, Exercise 9.6.35, p. 547] Use trigonometric substitutions to evaluate the integral∫ √x2 − 5x2 dx.

8. [31, Exercise 9.7.14, p. 553] Evaluate the integral∫

x√

8 + 2x − x2 dx.

9. [31, Exercise 9.8.17, p. 561] Determine whether the following improper integral con-

verges; if it does converge, evaluate it:∫ ∞

−∞

xx2 + 4

dx.

10. [31, Exercise 9.8.27, p. 561] Determine whether the following improper integral con-

verges; if it does converge, evaluate it:∫ ∞

0cos x dx.


11. (cf. [31, Exercise 9.8.14, p. 561]) Determine whether the following improper integral

converges; if it does converge, evaluate it:∫ +8

−8

1

(x + 4)23

dx.

12. [31, Exercise 10.2.2, p. 580] Find two polar coordinate representations, one with r ≥ 0,and the other with r ≤ 0 for the points with the following rectangular coordinates:

(a) (−1,−1),

(b) (√

3,−1),

(c) (2, 2),

(d) (−1,√

3),

(e) (√

2,−√2),

(f) (−3,√

3).

13. For each of the following curves, determine — showing all your work — equations inboth rectangular and polar coordinates:

(a) [31, Exercise 10.2.20, p. 580] The horizontal line through (1, 3).

(b) [31, Exercise 10.2.26, p. 580] The circle with centre (3, 4) and radius 5.

14. (a) [31, Exercise 10.2.56, p. 581] Showing all your work, find all points of intersectionof the curves with polar equations r = 1 + cos θ and r = 10 sin θ.

(b) Showing all your work, find all points of intersection of the curves with polar equa-tions r2 = 4 sin θ and r2 = −4 sin θ.

[Note: The procedure sketched in the solution of [31, Example 10.2.8, p. 579] for findingpoints of intersection is incomplete. Your instructor will discuss a systematic procedurein the lectures.]

D.2.6 Assignment 6

1. Find the area bounded by each of the following curves.

(a) r = 2 cos θ,

(b) r = 1 + cos θ.

2. Find the area bounded by one loop of the given curve.

(a) r = 2 cos 2θ,

(b) r2 = 4 sin θ.


3. Find the area of the region described.

(a) Inside both r = cos θ and r =√

3 sin θ.

(b) Inside both r = 2 cos θ and r = 2 sin θ .

4. Eliminate the parameter and then sketch the curve.

(a) x = t + 1, y = 2t2 − t − 1.

(b) x = et, y = 4e2t.

(c) x = sin 2πt, y = cos 2πt; 0 ≤ t ≤ 1. Describe the motion of the point (x(t), y(t)) ast varies in the given interval.

5. Find the area of the region that lies between the parametric curve x = cos t, y =

sin2 t, 0 ≤ t ≤ π, and the x-axis.

6. Find the arc length of the curve x = sin t − cos t, y = sin t + cos t; π/4 ≤ t ≤ π/2.7. Determine whether the sequence an converges, and find its limit if it does converge.

(a) an =n2 − n + 72n3 + n2 ,

(b) an =1 + (−1)n√n

(3/2)n ,

(c) an = n sin πn,

(d) an =

(n − 1n + 1

)n

.

8. Determine, for each of the following infinite series, whether it converges or diverges. Ifit converges, find its sum.

(a) 1 + 3 + 5 + 7 + . . . + (2n − 1) + . . . ,

(b) 4 + 43 + . . . + 4

3n + . . . ,

(c)∞∑

n=1

(5−n − 7−n),

(d)∞∑

n=1

( eπ

)n.

9. Find the set of all those values of x for which the series∞∑

n=1

( x3

)nis a convergent geomet-

ric series, then express the sum of the series as a function of x.


10. Find the Taylor polynomial in powers of x − a with remainder by using the given valuesof a and n.

(a) f (x) = sin x; a = π/6, n = 3.

(b) f (x) =1

(x − 4)2 ; a = 5, n = 5 .

11. Find the Maclaurin series of the function e−3x by substitution in the series for ex.

12. Find the Taylor series for f (x) = ln x at the point a = 1.

13. Use comparison tests to determine whether each of the following infinite series convergeor diverge.

(a)∞∑

n=1

11 + 3n ,

(b)∞∑

n=1

√n

n2 + n,

(c)∞∑

n=1

sin2(1/n)n2 .

D.3 2000/2001(In the winter of the year 2001 Assignments based on WeBWorK were used, although theexperiment had to be terminated in mid-term because of technical problems.)

D.4 2001/2002This was the first time WeBWorK assignments were used exclusively in this course.

D.5 MATH 141 2003 01WeBWorK assignments were used exclusively for assignments. The questions are not avail-able for publication.

D.6 MATH 141 2004 01WeBWorK assignments were used exclusively for assignments. The questions are not avail-able for publication.


D.7 MATH 141 2005 01WeBWorK assignments were used for online assignments; the questions are not available forpublication. In addition, these written assignments were intended to provide students with in-dividualized opportunities to work problems for which the textbook often provided examples;at the time, appropriate materials available from WeBWorK for this purpose were limited. Theindividualization was often based on the student number.

D.7.1 Written Assignment W1

Your written assignments will usually be mounted on the WeBWorK site, and will usually beindividualized, that is, your problems will not be exactly the same as those of other students.This first written assignment is based on your WeBWorK assignment R1. Subsequent writtenassignments may be designed in other ways. Because of its general form, it was possible torelease this assignment in the document Information for Students in MATH 141 2005 01; someof the other assignments may appear only on your WeBWorK site.

Your completed assignment must be submitted together with your solutions to quiz Q1,inside your answer sheet for that quiz. No other method of submission is acceptable.

Purpose of the written assignments These assignments are designed to help you learn howto write full solutions to problems. While they carry a very small weight in the computation ofyour final grade, conscientious completion of the assignments should help you substantially inlearning the calculus, and help prepare you for your final examination.

Certificate Your assignment will not be graded unless you attach or include the followingcompleted certificate of originality:

I have read the information on the web page

http://www.mcgill.ca/integrity/studentguide/,

and assert that my work submitted for W1 and R1 does not violate McGill’sregulations concerning plagiarism.

Signature(required) Date(required)

The assignment questions Your assignment consists of the following problems on your ver-sion of R1:


##3, 4, 5, 6, 7, 8

(Teaching Assistants are not primarily checking for plagiarism; but, if they detect it, they maybe obliged to report any apparent violations to the Associate Deans.)

Complete solutions are required It is not enough to give the correct answer; in fact, thenumerical answer alone may be worth 0 marks. You should submit a full solution, similar tosolutions in Stewart’s textbook or the Student Solutions Manual, which are where you shouldlook if you have doubts about the amount of detail required in a solution.

Use of calculators You are expected to complete the entire assignment without the use ofa calculator. In particular, you are expected to be familiar with the values of trigonometricfunctions at “simple” multiples and submultiples of π.

Not all problems may be graded On all of the written assignments it is possible that theTeaching Assistant will grade only a small number of the solutions you submit. The numbersof the questions that will be graded will not be announced in advance, even to the tutor. Forthat reason you are advised to devote equal attention to all of the problems.


Written assignment W2 is based on your WeBWorK assignment R3, but some problems arebeing modified. Your completed assignment must be submitted together with your solutions toquiz Q2, inside your answer sheet for that quiz. No other method of submission is acceptable.

Purpose of the written assignments These assignments are designed to help you learn howto write full solutions to problems. While they carry a very small weight in the computation ofyour final grade, conscientious completion of the assignments should help you substantially inlearning the calculus, and help prepare you for your final examination.

Certificate Your assignment will not be graded unless you attach or include the followingcompleted certificate of originality, signed in ink:




and assert that my work submitted for W2 and R3 does not violate McGill’sregulations concerning plagiarism.


The assignment questions Your assignment consists of the following problems on your ver-sion of R1:

1. Problem 1 of R3, solved by integration with respect to x. Include in your solution a roughsketch of the region, showing a typical element of area.

2. Problem 1 of R3 solved by integration with respect to y. This will require rewriting theequations of the curves appropriately. You may assume without proof that

∫ln x dx = x(ln x − 1) + C ,

a fact which you will see derived later in the course. Include in your solution a roughsketch of the region, showing a typical element of area.

3. Problem 6 of R3, evaluated using the Method of Washers. Include in your solution arough sketch of the plane region which generates the solid, showing a typical element ofarea which will generate a typical washer.

4. Problem 6 of R3, evaluated using the Method of Cylindrical Shells. Include in yoursolution a rough sketch of the plane region which generates the solid, showing a typicalelement of area which will generate a typical cylindrical shell.

Complete solutions are required It is not enough to give the correct answer; in fact, thenumerical answer alone may be worth 0 marks. You should submit a full solution, similar tosolutions in Stewart’s textbook or the Student Solutions Manual, which are where you shouldlook if you have doubts about the amount of detail required in a solution.

Use of calculators You are expected to complete the entire assignment without the use of acalculator.


Not all problems may be graded On all of the written assignments it is possible that theTeaching Assistant will grade only a small number of the solutions you submit. The numbersof the questions that will be graded will not be announced in advance, even to the tutor. Forthat reason you are advised to devote equal attention to all of the problems.


Unlike the preceding written assignments Written Assignment W3 is not directly based on yourWeBWorK assignments, although some problems will be similar to WeBWorK assignmentproblems. Your completed assignment must be submitted together with your solutions to quizQ3, inside your answer sheet for that quiz. No other method of submission is acceptable.

Certificate Your assignment will not be graded unless you attach or include the followingcompleted statement of originality, signed in ink:



and assert that my work submitted for W3 does not violate McGill’s regula-tions concerning plagiarism.


The assignment questions The parameters in these problems are based on the digits of your9-digit McGill student number, according to the following table:

Parameter name: A B C D E F G H JYour student number:

Before starting to solve the problems below, determine the values of each of these integerconstants; then substitute them into the descriptions of the problems before you begin yoursolution.

1. Showing all your work, systematically determine∫ (

Ax2 + Bx + C)

e−x dx by repeated

integration by parts: no other method of solution will be accepted. Verify by differenti-ation that your answer is correct.


2. Showing all your work, use trigonometric or hyperbolic substitutions to evaluate

each of∫

du√u2 − (F + 1)2

,

∫du√

u2 + (F + 1)2,

∫du√

−u2 + (F + 1)2.

Verify that your answers are correct by differentiation.

3. Let K =⌊

J+42

⌋=

[[J+4

2

]].62 Showing all your work, develop for this integer K a reduction

formula of the following type that can be used to evaluate

In(x) =

∫xn (sin(Kx)) dx in terms of In−2(x):

∫xn (sin Kx) dx = L · xn cos Kx + M · xn−1 sin Kx + N

∫xn−2 (sin Kx) dx

where n ≥ 2 and L, M,N are constants that you are expected to determine only byintegration by parts. Again showing all your work, use the reduction formula you havejust determined to evaluate

∫x2 sin Kx dx, and test by differentiation the answer that it

gives — you should recover x2 sin Kx.

4. Showing all your work, evaluate both of the integrals∫

sinF+1 x · cos2 x dx and∫

sinF+2 x · cos2 x dx

Complete solutions are required It is not enough to give the correct answer; in fact, thefinal answer alone may be worth 0 marks. You should submit a full solution, similar to solu-tions in Stewart’s textbook or the Student Solutions Manual, which are where you should lookif you have doubts about the amount of detail required in a solution.

Not all problems may be graded On all of the written assignments it is possible that theTeaching Assistant will grade only a small number of the solutions you submit. The numbersof the questions that will be graded will not be announced in advance, even to the tutor. If forno other reason, you are advised to devote equal attention to all of the problems.


Your completed assignment must be submitted together with your solutions to quiz Q4, insideyour answer sheet for that quiz. No other method of submission is acceptable.

62Determine K from J using the greatest integer function, defined in your textbook, page 110. Your textbookuses the notation

[[J+4

2

]], but some authors write the function as

⌊J+4

2

⌋.







The assignment questions Some of the parameters in these problems are based on the digitsof your 9-digit McGill student number, according to the following table:

Parameter name: A B M D E F G H JYour student number:

Before starting to solve the problems below, determine the values of each of these integerconstants; then substitute them into the descriptions of the problems before you begin yoursolution. It is not enough to give the correct answer; in fact, the final answer alone could beworth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook orthe Student Solutions Manual, which are where you should look if you have doubts about theamount of detail required in a solution. Not all problems may be graded.

1. Showing all your work, systematically determine the value of the following integral:∫x

(x + J)2(x − H − 1)1 dx . Verify your work by differentiation of your answer: you

should recover the integrand. (Systematically means that you to use the methods youlearned in this course for the treatment of problems of this type, even if you happen tosee some other method that could be used in this particular example.)

2. Showing all your work, use a substitution to transform the integrand into a rationalfunction, then integrate the particular integral that is assigned for your particular valueof G:

If G = 1,4,or 7:∫

1

x − √x + 2dx

If G = 0, 2, 5, or 8:∫

cos 2xsin2 2x + sin 2x

dx


If G = 3, 6, or 9:∫

1

x − 2 − 2√

x + 1dx

3. Problem 55, pages 504-505 of your textbook, describes a substitution discovered by KarlWeierstrass (1815-1897) for the evaluation of rational functions of sin x and cos x intoan ordinary rational function. It states that if, for x such that −π < x < π, we define

t = tan x2 , then cos x =

1 − t2

1 + t2 and sin x =2t

1 + t2 and that, at a consequence, dx =

21 + t2 dt . You are not asked to verify these facts. You are asked to use the substitutionto transform the following integral and then to evaluate it:

∫ π2

π3

11 + sin x − cos x

dx

4. Use the trigonometric identities given in your textbook on page 487 to evaluate the fol-lowing integral:

∫cos Mx ·cos Dx ·sin x dx where M and D are the digits of your student

number, defined above.

5. Showing all your work, determine whether each of the following integrals is convergentor divergent. If it is convergent, determine its value (again showing all your work):

(a)∫ 4

0

12 + E · F dx

(b)∫ ∞

0

x2 − E − Fx2 + E + F

dx

(c)∫ π

B+4

0

1x sin((G + 2)x)

dx


Your completed assignment must be submitted together with your solutions to quiz Q5, insideyour answer sheet for that quiz. No other method of submission is acceptable.







The assignment questions Some of the parameters in these problems are based on the digitsof your 9-digit McGill student number, according to the following table:

Parameter name: A B D E F G H J KYour student number:

Before starting to solve the problems below, determine the values of each of these integerconstants; then substitute them into the descriptions of the problems before you begin yoursolution. It is not enough to give the correct answer; in fact, the final answer alone could beworth 0 marks. You should submit a full solution, similar to solutions in Stewart’s textbook orthe Student Solutions Manual, which are where you should look if you have doubts about theamount of detail required in a solution. Not all problems may be graded.

1. For the point with polar coordinates(H,− π

K2 + 5

),

(a) find four other pairs of polar coordinates, two with r ≤ 0 and two with r > 0:

(b) Find the cartesian coordinates, assuming that the positive x-axis is along the polaraxis, the origin is at the pole, and the positive y-axis is obtained by turning the polaraxis through a positive angle of π

2 .

2. For the point whose cartesian coordinates are (F2+1, F2−2), determine polar coordinates(r, θ) with the following properties:

(a) r > 0, 0 ≤ θ < 2π

(b) r < 0, 0 ≤ θ < 2π

(c) r > 0, 5π2 ≤ θ < 9π

2

3. For the following curve given in polar coordinates, determine, showing all your compu-tations, the slope of the tangent at the point with θ =

π

4:

r = A + cos(B + 2)θ .


4. Showing all your work, find the area contained between the outer loop and the inner loopof the curve

r = 1 + 2 sin θ .

Explain carefully how you have established the limits for your definite integral.

5. The curve C is given by the parametric equations

x = 1 + (C + 2)t2

y = t − (E2 − 2)t3

Showing all your work, determine the value ofd2ydx2 (t).

6. Giving an explanation, determine whether or not the following sequences converge.Where a sequence converges, find its limit.

(a) an = ln n − ln(n + A + 1)

(b) an =ln n

ln((A + 2)n)

(c) an =√

n2 + A2 −√

n2 + A2 + 2

D.8 MATH 141 2006 01D.8.1 Solution to Written Assignment W1

Release Date: Friday, 27 January, 2006The Assignment was posted to the class via WeBCT on January 2nd, 2006

Instructions to Students

1. Your solution to this assignment should be brought to your regular tutorial, during theweek 16-20 January, and folded inside your solution sheet to the quiz which will bewritten at that time.

2. Your TA may make special arrangements for submission at other times, but, no solutionsare ever accepted after the end of the week.

3. Your solution must use the data on your own WeBWorK assignment, as described be-low.


The Assignment Question Problem 7 of WeBWorK assignment R1 requires that you eval-uate a definite integral of the form

∫ C

D(Ex2 − Ax + B) dx,

where A, B,C,D, E are various combinations of constants, individualized for each student.Your written assignment is to evaluate a simplified version of this integral as the limit of asequence of Riemann sums, using left endpoints63. You may simplify your problem only inthe following way: you may take the lower limit of the integral (here called D) to be 0; thisis purely to make the algebra a little easier. You should model your proof on that given in [7,Example 2(b), pp. 383-385]. You are expected to provide a full solution for your version of theproblem — it is not enough to supply the correct answer, and you must solve the version ofthe problem on your own WeBWorK assignment R1. In your solution you will need to use thewell known formulæ for the sums of the 1st and 2nd powers of the natural numbers from 1 to n;these formulæ are in your textbook [7, Formulæ ##4, 5, p. 383], and you will not be expectedto prove them here.64 (This is a time-consuming exercise; the purpose of the assignment is toensure that you will have correctly solved a problem of this type during the term.)

The Solution (This solution has been composed with variables, so that it will produce validnumerical solutions for all students. The particular solution for your version of the problemshould look much simpler, because, in place of all variables except n, there will be specificintegers.)

The length of each of n subintervals is

∆x =C − D

n.

Thus xi = D + i∆x = D + i(C − D

n

). Since we are using left endpoints we apply [7, top

formula, p. 381]∫ C

D

(Ex2 − Ax + B

)dx

= limn→∞

n∑

i=1

(Ex2

i−1 − Axi−1 + B)· ∆x

= limn→∞

n∑

i=1

(E

(D2 +

2D(C − D)n

· (i − 1) +(C − D)2

n2 · (i − 1)2)

63Note that the solution given in the textbook uses right endpoints.64Of course, these formulæ may be proved by induction, and students who took MATH 140 2005 09 should

know how to write up such proofs if they had to.


−A(D +

C − Dn· (i − 1)

)+ B

)· ∆x

= limn→∞

(ED2 − AD + B)n∑

i=1

1 +(2DE − A)(C − D)

n

n∑

i=1

(i − 1) +E(C − D)2

n2

n∑

i=1

(i − 1)2

· ∆x

Nown∑

i=1

1 = sum of n 1’s

= nn∑

i=1

(i − 1) =

n−1∑

j=1

j

=(n − 1)n

2by [7, Formula 4, p. 383]

n∑

i=1

(i − 1)2 =

n−1∑

j=1

j2

=(n − 1)n(2n − 1)

6by [7, Formula 5, p. 383].

Substituting the values of these sums yields∫ C

D

(Ex2 − Ax + B

)dx

= limn→∞

C − Dn·((ED2 − AD + B)n +

(2DE − A)(C − D)n

· (n − 1)n2

+E(C − D)2

n2 · (n − 1)n(2n − 1)6

)

= limn→∞

((ED2 − AD + B)(C − D) +

(2DE − A)(C − D)2

2

(1 − 1

n

)

+E(C − D)3

6·(1 − 1

n

) (2 − 1

n

))

= (ED2 − AD + B)(C − D) +(2DE − A)(C − D)2

2· 1 +

E(C − D)3

6· 1 · 2

= (ED2 − AD + B)(C − D) +(2DE − A)(C − D)2

2+

E(C − D)3

3

=E

(C3 − D3

)

3−

A(C2 − D2

)

2+ B(C − D)

It was suggested that you simplify your problem by taking D = 0; in that case the value wouldhave worked out to be

EC3

3− AC2

2+ BC .


The Grading Scheme The assignment was to be graded OUT OF A TOTAL OF 10 MARKS.(In the version of this solution circulated to TA’s, there followed some technical details about the grad-ing.)

D.8.2 Solution to Written Assignment W2

Release Date: Mounted on the Web on February 8th, 2006; corrected on March 16th, 2006Solutions were to be submitted inside answer sheets to Quiz Q2, at tutorials during the period

January 30th – February 2nd, 2006.

The Problem. In [7, Example 6, p. 441] of your textbook, the author solves the followingproblem: “Find the area enclosed by the line y = x − 1 and the parabola y2 = 2x + 6,” byintegrating with respect to y. The textbook then states the following: “We could have foundthe area in Example 6 by integrating with respect to x instead of y, but the calculation ismuch more involved. It would have meant splitting the region in two, and computing theareas labelled A1 and A2 in Figure 14. The method we used in Example 6 is much easier.” Yourassignment is to solve the problem by integrating with respect to x, and you may find the figurein the textbook helpful. The computation is not really very hard, but it does involve workingwith

√2. You should not approximate

√2; just write it that way and work with it carefully,

and the square roots will cancel by the time you finish your solution.You know the numerical answer to this question; the purpose of the assignment is to nudge

you into solving problems in two ways, so that you can verify your work; and to show you thatthere really is nothing to fear, even if you do happen to choose the “wrong” way to approach aproblem.

Your TA will be alert to the possibility that students might be copying their solutions fromothers. Please write up your own solution, so that your TA will not have to waste everyone’stime by sending exact copies to the disciplinary officer of the Faculty. You need to know howto solve problems of this type, since you could be expected to demonstrate that ability at afuture quiz or on the examination.

The Solution. The integration must be carried out separately for two subregions. This isbecause the area to the left of the line x = −1 is bounded above and below by the parabola;while the area to the right of x = −1 is bounded above by the parabola, and below by the liney = x − 1. The method you know for finding the area between two curves requires that theequations of the curves be written as the graphs of functions of x. But the parabola y2 = 2x + 6is not the graph of a function, since it crosses some vertical lines twice. However, we canfactorize the equation in the form

(y −

√2(x + 3)

) (y +

√2(x + 3)

)= 0 :


the parabola is the graph of two functions — the upper arm of the parabola is the graph ofy =

√2(x + 3), while the lower arm is the graph of y = −√2(x + 3). To the left of x = −1

the region A1 is bounded above by the graph of y =√

2(x + 3), and below by the graph ofy = −√2(x + 3); thus the height of the vertical element of area is the difference between thesetwo functions, i.e.,

√2(x + 3) − (−√2(x + 3)) = 2

√2(x + 3), and this will be the integrand for

the integral; the area to the right of x = −1 is bounded above by the parabola y =√

2(x + 3),and below by the line y = x−1, so the element of area for the integral will be

√2(x + 3)−(x−1).

The total area is thus∫ −1

−32√

2(x + 3) dx +

∫ 5

−1

( √2(x + 3) − (x − 1)

)dx

=

[2√

2 · 23· (x + 3)

32

]−1

−3+

[√2 · 2

3· (x + 3)

32 −

(x2

2− x

)]5

−1

=4√

23

[2

32 − 0

]+

2√

23· 8 3

2 −(252− 5

) −2√

23· 2 3

2 −(12

+ 1)

=163

+

[563− 6

]=

163

+383

= 18.

The Grading Scheme. The assignment was to be graded out of a maximum of 15 marks.(In the version of this document prepared for TA’s there were additional details on the markingscheme.)

D.8.3 Solutions to Written Assignment W3

Release Date: March 25th, 2006Solutions were to be submitted inside answer sheets to Quiz Q3, at tutorials during the period

13-16 February, 2006

The Problems

1. Use two integrations by parts to evaluate the integral∫

(sin x) · (sinh x) dx. You will

need, at the last step, to solve an equation. The solution should resemble the solution inyour textbook to [7, Example 4, p. 478], where the author evaluates

∫ex sin x dx.

2. Write your student number (9 digits) in the following chart.

A1 A2 A3 A4 A5 A6 A7 A8 A9

Student # :



Some of the digits are to be used in solving the following problem.

(a) You are to evaluate the following integral by integration by parts:∫ (

A7x2 + A8x1 + A9

)eA1 x dx

Your answer should be simplified as much as possible.

(b) Then you are to differentiate the function you have obtained, to verify that it isindeed an antiderivative of the given integrand.

Remember the rules: McGill’s Code of Student Conduct and Disciplinary Procedures appears inthe Handbook on Student Rights and Responsibilities (PDF English version - French version). Article15(a) of the Code, which is devoted to plagiarism, reads as follows:

No student shall, with intent to deceive, represent the work of another person as his orher own in any academic writing, essay, thesis, research report, project or assignmentsubmitted in a course or program of study or represent as his or her own an entire essay orwork of another, whether the material so represented constitutes a part or the entirety ofthe work submitted.

Please don’t cause yourself embarrassment and waste everyone’s time: it isn’t worth it, and you reallyneed to learn how to solve these kinds of problems yourself.

The Solutions

1. Let u = sin x, v′ = sin x. Then u′ = cos x, v = cosh x (or cosh x plus any real constant— I have taken the constant to be 0, as all choices of constant here will lead to the samesolution.)

∫(sin x) · (sinh x) dx = (cos x) · sinh x −

∫(cosh x) · (cos x) dx .

To evaluate∫

(cosh x) · (cos x) dx I will set U = cos x, V ′ = cosh x, so U′ = − sin x,

V = sinh x.∫

(cosh x) · (cos x) dx = (cos x) · sinh x +

∫(sinh x) · sin x dx .

Combining these results yields∫

(sin x) · (sinh x) dx = (sin x) · (cosh x) − (cos x) · (sinh x) −∫

(sin x) · (sinh x) dx


which we may solve by moving the two integral terms to the same side of the equation:

2∫

(sin x) · (sinh x) dx = (sin x) · (cosh x) − (cos x) · (sinh x) + C

or ∫(sin x) · (sinh x) dx =

(sin x) · (cosh x) − (cos x) · (sinh x)2

+ C′

2. (a) In the first integration by parts I set

u = A7x2 + A8x1 + A9

dv = eA1 xdx⇒ du = (2A7x + A8)dx,

v =1A1

eA1 x .

Hence∫ (

A7x2 + A8x1 + A9

)eA1 x dx

=(A7x2 + A8x + A9

)· 1

A1eA1 x − 1

A1

∫(2A7x + A8) · eA1 x dx

=

(A7

A1x2 +

A8

A1x1 +

A9

A1

)· eA1 x −

∫ (2A7

A1· x +

A8

A1

)· eA1 x dx

To evaluate the subtracted integral we must apply Integration by Parts a secondtime:

U =2A7

A1· x +

A8

A1

dV = eA1 xdx

⇒ dU =2A7

A1dx

V =1A1

eA1 x , so∫ (

2A7

A1· x +

A8

A1

)· eA1 x dx =

(2A7

A1· x +

A8

A1

)· 1

A1eA1 x −

∫2A7

A21

eA1 x dx

=

(2A7

A21

· x +A8

A21

)· eA1 x − 2A7

A31

eA1 x

=

(2A7

A21

· x +A8

A21

− 2A7

A31

)· eA1 x


Combining our results yields∫ (

A7x2 + A8x1 + A9

)eA1 x dx

=

(A7

A1x2 +

(A8

A1− 2A7

A21

)x1 +

(A9

A1− A8

A21

+2A7

A31

))· eA1 x + C

(b) Remember that you were to differentiate the preceding product of polynomial andexponential by the Product Rule to show that, indeed, its derivative is the givenintegrand.


Release Date: March 25th, 2006Solutions were to be submitted at inside answer sheets to Quiz Q4, at tutorials during the

week 06 – 09 March, 2006

The Problems This assignment is based on your WeBWorK assignment R5. You are askedto write out complete solutions to the following modifications of your versions of problems onthat assignment. Note that, in some cases, the question asks for more than was asked on theassignment. You are not permitted to use a Table of Integrals.

1. Problem 10:

(a) Evaluate the following integral with the specific values of the constants given inyour own WeBWorK assignment:

∫Ex2 + Ex + F

x3 + Gx2 + Hx + Jdx .

HINT: −C is a root of the denominator.

(b) No marks will be given unless you verify your integration by differentiating youranswer.

2. Problem 12:

(a) Evaluate the following integral with the specific values of the constants given inyour own WeBWorK assignment:

∫Bx + C

(x2 + A2)2 dx .


(b) No marks will be given unless you verify your integration by differentiating youranswer.

3. Problem 16: Determine whether the following integral (with the constants given in yourown WeBWorK assignment) is divergent or convergent. If it is convergent, evaluate it.If not, prove that fact. ∫ A

−∞

1x2 + 1

dx .

In each case your solution should begin by your writing out the full problem with your data,so your TA does not have to look up your data on the WeBWorK system.

While these problems were generated by WeBWorK, they now constitute a conventionalmathematics assignment, and it does not suffice to make unsubstantiated statements. You areexpected to prove everything you state. Thus, for example, in Problem 10, you have to showhow you use the fact that a certain number is stated to be a root of the denominator.




The Solutions

1. Under a change of variable of the form u =xA

, the integral reduces to one of the form∫Ku + L(u2 + 1

)2 du. The termKu

u2 + 1du simplifies under a substitution v = u2 to a multiple

of∫

dv(v + 1)2 . The term

∫L

(u2 + 1

)2 du simplifies under a substitution θ = arctan u,

eventually proving to be a multiple of θ +tan θ

2(tan2 θ + 1), etc.

2. An antiderivative is arctan x. This is evaluated between an upper limit of some constantA, and a lower limit we may call B. We need to observe that lim

B→−∞arctan B = −π2 .



Release Date: Solutions were to be submitted inside answer sheets to Quiz Q5, at tutorialsduring the week 20 – 23 March, 2006

The Problems This assignment will be graded out of a maximum of 20 MARKS.

Write your student number (9 digits) in the following chart.

A1 A2 A3 A4 A5 A6 A7 A8 A9

Student # :Some of the digits are to be used in solving the following problems.

1. Consider the curve in the plane defined parametrically by

x(t) = t2 + 1y(t) = A7t2 + A8t + A9

(a) [3 MARKS] Showing all your work, determine the slope of the tangent to the curveat the point with parameter value t = 1.

(b) [6 MARKS] Showing all your work, determine the value ofd2ydx2 at the point with

general parameter value t (t , 0). For this problem you must not substitute in anyformula from your class notes or any textbook; you are expected to determine the2nd derivative by differentiation, for example in the manner similar to that done inyour textbook, Example 1, page 661.

(c) [1 MARK] Determine the range of values of t — if there are any such values —where the curve is concave upward.

2. This problem is based on Problem 12 on your WeBWorK assignment R8. For the givenarc of the given curve,

(a) [8 MARKS] determine the area of the surface of revolution of that arc about thex-axis;

(b) [2 MARKS] set up an integral for the area of the surface of revolution of that arcabout the y-axis, but, in this case only, do not evaluate the integral.

In each case you are expected to show all your work.





The Solutions

1. (a)

x(t) = t2 + 1y(t) = A7t2 + A8t + A9

dxdt

= 2t

dydt

= 2A7t + A8

dydx

=

dydtdxdt

=2A7t + A8

2tdydx

∣∣∣∣∣t=1

= A7 +A8

2

(b)

d2ydx2 =

ddx

(2A7t + A8

2t

)

=ddt

(2A7t + A8

2t

)· dt

dx

=ddt

(2A7t + A8

2t

)· 1

dxdt

=ddt

(2A7t + A8

2t

)· 1

2t

=(2A7)(2t) − (2A7t + A8)(2)

(2t)3

= − A8

4t3

(c) If the student’s A8 is 0, the curve is flat. Otherwise it is concave upwards preciselywhen t < 0.


2. CORRECTED ON 20 MARCH 2006. The following solution is valid onlywhen K, the upper limit of integration, is sufficiently small. The correct so-lution is much more complicated, as we need to ensure that the respectivefactors cos θ + θ sin θ and sin θ − θ cos θ remain positive. I have corrected myoriginal version by inserting absolute signs. However, the evaluation of theseintegrals is complicated when we attempt to break the integral up into partsbased on the signs of these factors. This is certainly more difficult than wasintended from students in this course. Please exercise good judgment in grad-ing the two parts of this problem, and do not expect those students for whomthe upper limit K is large to obtain the correct area.

x(θ) = a(cos θ + θ sin θ)y(θ) = a(sin θ − θ cos θ)

⇒ x′ = aθ cos θy′ = aθ sin θ

⇒√

(x′)2 + (y′)2 = |aθ|(a) about the x-axis, PROVIDED K IS SUFFICIENTLY SMALL,

Area =

∫ K

02πa| sin θ − θ cos θ| · |θ| dθ = 2πa

(∫ K

0θ sin θ dθ −

∫ K

0θ2 cos θ dθ

)

for positive K sufficiently small. We must integrate by parts.∫θ sin θ dθ = −θ cos θ +

∫cos θ dθ

= −θ cos θ + sin θ + C∫θ2 cos θ dθ = θ2 sin θ − 2

∫θ sin θ dθ

= θ2 sin θ + 2θ cos θ − 2 sin θ + C′∫(θ sin θ − θ2 cos θ) dθ = −3θ cos θ + 3 sin θ − θ2 sin θ + C′′

Hence the area of the surface of revolution is −3K cos K + 3 sin K − K2 sin K.(b) about the y-axis, PROVIDED K IS SUFFICIENTLY SMALL,

Area =

∫ K

02πa| cos θ + θ sin θ| · |θ| dθ = 2πa

(∫ K

0θ cos θ dθ +

∫ K

0θ2 sin θ dθ

).

D.9 MATH 141 2007 01The use of written assignments was discontinued in 2007.


E Quizzes from Previous Years

E.1 MATH 141 2007 01E.1.1 Draft Solutions to Quiz Q1

Distribution Date: Mounted on the Web on 4 February, 2007corrected 19 January, 2010 — subject to further corrections

There were four different types of quizzes, for the days when the tutorials are scheduled. Eachtype of quiz was generated in multiple varieties for each of the tutorial sections. The order ofthe problems in the varieties was also randomly assigned. All of the quizzes had a heading thatincluded the instructions

• Time = 30 minutes

• No calculators!

• Show all your work: marks are not given for answers alone.

• Enclose this question sheet in your folded answer sheet.

In the following I will either provide a generic solution for all varieties, or a solution to onetypical variety.

Monday version

1. [5 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivativeof the function

g(x) =

∫ a

xb tan(t) dt ,

(where a and b are constants). Then use Part 2 of the Fundamental Theorem to evaluateg(x), by first verifying carefully that ln | sec x| is an antiderivative of tan x.

Solution:

(a) Part 1 of the Fundamental Theorem gives the derivative of a definite integral as afunction of its upper index of integration. Here the variable is the lower index ofintegration.

ddx

∫ a

xb tan(t) dt

=ddx

(−

∫ x

ab tan(t) dt

)


= − ddx

∫ x

ab tan(t) dt

= −b tan x .

Some students may quote a variant of Part 1 which gives the derivative of a definiteintegral with respect to the lower index, and this should be accepted if work hasbeen shown.

(b) Students were expected to first find the derivative of ln | sec x|. Since this is a com-position of 2 functions, the Chain Rule will be needed. Let u = sec x. Then

ddx

ln | sec x| =ddu

ln |u| · ddx

secx

=1u· sec x tan x

=1

sec x· sec x tan x

= tan x .

Henceg(x) = b ln | sec t|ax = b ln

∣∣∣∣∣sec asec x

∣∣∣∣∣ = b ln∣∣∣∣∣cos xcos a

∣∣∣∣∣ .

2. [5 MARKS] Find an antiderivative of the integrand of the integralb∫

a

(k + `y + my2

)dy,

(where a, b, k, `,m are constants), and then use the Fundamental Theorem of Calculus toevaluate the integral. You are not expected to simplify your numerical answer.

Solution:

(a) One antiderivative of ky0 + `y1 + my2 is

k · y1

1+ ` · y2

2+ m · y3

3.

(b) Henceb∫

a

(k + `y + my2

)dy

=

[k · y1

1+ ` · y2

2+ m · y3

3

]b

a

=

(k · b1

1+ ` · b2

2+ m · b3

3

)−

(k · a1

1+ ` · a2

2+ m · a3

3

).



g(x) =

∫ √x

ab

cos tt

dt ,

where a, b are constants.

Solution:

(a) Denote the upper index of the integral by u(x) =√

x.

(b) Then

ddx

g(x) =ddx

∫ √x

ab

cos tt

dt

=ddx

∫ u(x)

ab

cos tt

dt

=d

du

∫ u(x)

ab

cos tt

dt · du(x)dx

= bcos u

u· du(x)

dx

= bcos u

u· 1

2√

x

= bcos√

x√x· 1

2√

x

= bcos√

x2x

4. [10 MARKS] Showing all your work, determine all values of x where the curve y =x∫

0

11 + at + bt2 dt is concave upward, where a, b are constants. (Each version of this

quiz contained specific values for the constants a, b.)

Solution:

(a) By Part 1 of the Fundamental Theorem,

y′(x) =1

1 + ax + bx2 .

(b) Differentiating a second time yields

y′′(x) =ddx

(1

1 + ax + bx2

)


= − 1(1 + ax + bx2)2

· ddx

(1 + ax + bx2

)

= − a + 2bx(1 + ax + bx2)2 .

(c) The curve is concave upward where y′′ > 0:

− a + 2bx(1 + ax + bx2)2 > 0 ⇔ −(a + 2bx) > 0

since the denominator is a square, hence positive⇔ 2bx < −a

⇔

x < − a2b when b > 0

x > − a2b when b < 0

never concave upward when b = 0, a > 0always concave upward when b = 0, a < 0

Tuesday version

1. [5 MARKS] If

c∫

a

f (x) dx = k and

c∫

b

f (x) dx = `, find

b∫

a

f (x) dx. Show your work.

Solution:

(a)c∫

a

f (x) dx =

b∫

a

f (x) dx +

c∫

b

f (x) dx .

(b) Hencec∫

b

f (x) dx =

c∫

a

f (x) dx −b∫

a

f (x) dx .

(c)= k − ` .

2. [5 MARKS] Find an antiderivative of the integrand of the integral∫ a

0

√x dx, and then

use the Fundamental Theorem of Calculus to evaluate the integral. You are not expectedto simplify your numerical answer, but no marks will be given unless all your work isclearly shown.



(a) One antiderivative of x12 is

112 + 1

· x 12 +1 =

23

x32 .

(b) ∫ a

0

√x dx =

[23

x32

]a

0=

23

(a

32 − 0

)=

23

a32 .

3. [10 MARKS] Use Part 1 of the Fundamental Theorem of Calculus to find the derivative

of the function

x2∫

a

b√

1 + tc dt.

Solution:

(a) Let the upper index of the integral be denoted by u = x2. Then

(b)

ddx

x2∫

a

b√

1 + tc dt =ddx

u∫

a

b√

1 + tc dt

=ddu

u∫

a

b√

1 + tc dt · dudx

= b√

1 + uc · dudx

= b√

1 + uc · 2x= b

√1 + x2c · 2x

4. [10 MARKS] If F(x) =

∫ x

1f (t) dt, where f (t) =

∫ t2

1

a + ub

udu and a, b are constants,

find F′′(2).

Solution:

(a) Applying Part 1 of the Fundamental Theorem yields

F′(x) = f (x) =

∫ x2

1

a + ub

udu .


(b) A second application of Part 1 of the Fundamental Theorem yields

F′′(x) = f ′(x) =ddx

∫ x2

1

a + ub

udu .

(c) Denote the upper index of the last integral by v = x2.

(d)

ddx

∫ x2

1

a + ub

udu =

ddx

∫ v

1

a + ub

udu

=ddv

∫ v

1

a + ub

udu · dv

dx

=a + vb

v· dv

dx

=a + vb

v· 2x

=a + x2b

x2 · 2x

=2(a + x2b

)

x.

Wednesday version

1. [5 MARKS] Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integralbπ∫

aπ

cos θ dθ, where a, b are given integers. No marks will be given unless all your work

is clearly shown. Your answer should be simplified as much as possible.

Solution:

(a) One antiderivative of cos θ is sin θ.

(b)bπ∫

aπ

cos θ dθ = [sin θ]bπaπ = sin(bπ) − sin(aπ) .

(c) Students were expected to observe that the value of the sine at the given multiplesof π is 0, so the value of the definite integral is 0.


2. [5 MARKS] Express limn→∞

n∑

i=1

axi sin xi ∆x as a definite integral on the interval [b, c],

which has been subdivided into n equal subintervals.

Solution: ∫ c

bax sin x dx .


g(x) =

∫ bx

ax

t2 + ct2 − c

dt ,

where a, b, c are positive integers.

Solution:

(a) The Fundamental Theorem gives the derivative of a definite integral with respect tothe upper limit of integration, when the lower limit is constant. The given integralmust be expressed in terms of such specialized definite integrals.

g(x) =

∫ bx

ax

t2 + ct2 − c

dt =

∫ 0

ax

t2 + ct2 − c

dt +

∫ bx

0

t2 + ct2 − c

dt

= −∫ ax

0

t2 + ct2 − c

dt +

∫ bx

0

t2 + ct2 − c

dt

(b) For the summand

bx∫

0

t2 + ct2 − c

dt, let u = bx. Then

ddx

bx∫

0

t2 + ct2 − c

dt =ddx

u∫

0

t2 + ct2 − c

dt

=d

du

u∫

0

t2 + ct2 − c

dt · dudx

=u2 + cu2 − c

· dudx

=u2 + cu2 − c

· b

=(bx)2 + c(bx)2 − c

· b


(c) For the summand∫ ax

0

t2 + ct2 − c

dt, let u = ax. Then, analogously to the preceding

step,ddx

∫ ax

0

t2 + ct2 − c

dt =(ax)2 + c(ax)2 − c

· a .

(d)

g′(x) =(bx)2 + c(bx)2 − c

· b − (ax)2 + c(ax)2 − c

· a .

4. [10 MARKS] Find a function f (x) such that

k +

∫ x

a

f (t)t2 dt = `

√x (109)

for x > 0 and for some real number a; k and ` are constants given in the question,different constants to different students. (HINT: Differentiate the given equation.)

Solution: A more proper wording of the problem would have been “Find a function f (x)and a real number a such that...”.

(a) Assume that equation (109) holds. Then differentiation of both sides with respectto x yields

0 +f (x)x2 =

12· ` · 1√

x.

(b) We solve the preceding equation to obtain that f (x) = `2 x

32 .

(c) Substitution into equation (109) yields

k +`

2

∫ x

at−

12 dt = `

√x .

We know how to integrate powers of t:

k +`

cot2 · √t

]x

a= `√

x .

(d) The preceding reduces to k = `√

a, which may be solved to obtain

a =

(k`

)2

.


Thursday version

1. [5 MARKS] Evaluate the integral

πb∫

πa

sin t dt.

Solution:

(a) An antiderivative of sin t is − cos t.

(b)πb∫

πa

sin t dt = [− cos t]πbπa

(c) [1 MARK] Your answer should be simplified as much as possible.

2. [5 MARKS] Evaluate the Riemann sum for f (z) = a−x2, (0 ≤ x ≤ 2) with 4 subintervals,taking the sample points to be the right endpoints. It is not necessary to simplify the finalnumerical answer.

Solution:

(a) The interval 0 ≤ x ≤ 2 is divided by 3 points into 4 subintervals of length ∆x =24 = 1

2 .

(b) The point x∗i selected in the ith interval will always be the right end-point, i.e.,xi = i

(12

)(i = 1, 2, 3, 4).

(c) The Riemann sum is4∑

i=1

f (x∗i )∆x =12

4∑

i=1

(a − i2

4

).


of the function

bx∫

cos x

cos (tc) dt, where a, b, c are real numbers.

Solution:


bx∫

cos x

cos (tc) dt =

0∫

cos x

cos (tc) dt +

bx∫

0

cos (tc) dt


= −cos x∫

0

cos (tc) dt +

bx∫

0

cos (tc) dt

(b) For the summand

bx∫

0

cos (tc) dt, let u = bx. Then

ddx

bx∫

0

cos (tc) dt =ddx

u∫

0

cos (tc) dt

=d

du

u∫

0

cos (tc) dt · dudx

= cos (uc) · b= cos ((bx)c) · b

(c) For the summand

cos x∫

0

cos (tc) dt, let v = cos x.

ddx

cos x∫

0

cos (tc) dt =ddx

v∫

0

cos (tc) dt

=ddv

v∫

0

cos (tc) dt · dvdx

= cos (vc) · (− sin x)= cos (cosc x) · (− sin x)

(d)bx∫

cos x

cos (tc) dt = − cos (cosc x) · (− sin x) + cos ((bx)c) · b .

4. [10 MARKS] Let f (x) =

0 if x < 0x if 0 ≤ x ≤ a2a − x if a < x < 2a0 if x > 2a

and g(x) =

∫ x

0f (t) dt, where a is

a positive constant. Showing all your work, find a formula for the value of g(x) whena < x < 2a.


Solution:

(a) The interval where we seek a formula is the third interval into which the domainhas been broken. For x in this interval the integral can be decomposed into

∫ x

0f (t) dt =

∫ a

0f (t) dt +

∫ x

af (t) dt .

The portion of the definition of f for x < 0 is of no interest in this problem, sincewe are not finding area under that portion of the curve; the same applies to theportion of the definition for x > 2a.

(b)∫ a

0f (t) dt =

∫ a

0t dt

=

[t2

2

]t=a

t=0=

a2

2.

(c)∫ x

af (t) dt =

∫ x

a(2a − t) dt

=

[2at − t2

2

]t=x

t=a

=

(2ax − x2

2

)−

(2a2 − a2

2

).

(d)

g(x) =a2

2+

(2ax − x2

2

)−

(2a2 − a2

2

)= 2ax − x2

2− a2 .

E.1.2 Draft Solutions to Quiz Q2

Distribution Date: Posted on the Web on 28 February, 2007Caveat lector! There could be misprints or errors in these draft solutions.




• No calculators!




Monday version

1. [5 MARKS] Evaluate the integral∫ (

xb + a +1

x2 + 1

)dx, (where a and b are given

positive integers).

Solution:

(a)∫

xb dx =xb+1

b + 1+ C1,

(b)∫

a dx = ax + C2

(c)∫

1x2 + 1

dx = arctan x + C3

(d)∫ (

xb + a +1

x2 + 1

)dx =

xb+1

b + 1+ ax + arctan x + C.

2. [5 MARKS] Use a substitution to evaluate the indefinite integral∫

t2 cos(a − t3

)dt,

(where a is a given real number).

Solution:

(a) Try the substitution u = t3.

(b) du = 3t2 dt ⇒ t2 dt = 13 du.

(c)∫

t2 cos(a − t3

)dt =

∫13

cos(a − u) du

= −13

sin(a − u) + C

= −13

sin(a − t3) + C.

(Some students may wish to employ a second substitution v = a− u. Alternatively,a better substitution for the problem would have been to take u = a − t3.)


3. [10 MARKS] Find the area of the region bounded by the parabola y = x2, the tangentline to this parabola at (a, a2), and the x-axis, (where a is a given real number).

Solution: This area can be computed by integrating either with respect to y or withrespect to x.

Integrating with respect to y: (a) Since y′ = 2x, the tangent line through (a, a2) hasequation

y − a2 = 2a(x − a)⇔ y = 2ax − a2 .

(b) To integrate with respect to y we need to express the equations of the parabolaand the line in the form

x = function of y .

The branch of the parabola to the right of the y-axis is x =√

y. The line has

equation x =y

2a+

a2

.

(c) The area of the horizontal element of area at height y is(y + a2

2a− √y

)∆y.

(d) The area is the value of the integral∫ a2

0

(y + a2

2a− √y

)dy .

(e) Integration yields[

y2

4a+

ay2− 2

3y

32

]a2

0=

(14

+12− 2

3

)a3 =

112

a3 .

Integrating with respect to x: (a) As above, the tangent line is y = 2ax − a2. Itsintercept with the x-axis is at x =

a2

.

(b) The area of the vertical element of area at horizontal position x ≤ a2

is(x2 − 0

)∆x.

(c) The area of the vertical element of area at horizontal position x ≥ a2 is

(x2 − (2ax − a2)

)dx =

(x − a)2 ∆x.(d) The area of the region is the sum

∫ a2

0x2 dx +

∫ a

a2

(x − a)2 dx .

(e) Integration yields [x3

3

] a2

0+

[−(x − a)3

3

]a

a2

=a3

12.


4. [10 MARKS] Find the volume of the solid obtained by rotating the region bounded bythe given curves about the line y = 1: y = n

√x, y = x, where n is a given positive integer.

Solution: A favoured method of solution was not prescribed.

Using the method of “washers”: (a) Find the intersections of the curves bounding theregion. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1).

(b) Find the inner and outer dimensions of the washer. Since the axis of revolutionis a horizontal line, the element of area being rotated is vertical. For arbitraryx the lower point on the element is (x, x); the upper point is (x, n

√x). The

distances of these points from the axis are, respectively 1 − x and 1 − n√

x.(c) The volume of the “washer” is, therefore,

π(−(1 − x)2 + (1 − n√x)2

)∆x .

(d) Correctly evaluate the integral:

π

∫ 1

0

(−(1 − x)2 + (1 − n√x)2

)dx

= π

∫ 1

0

(−2x + x2 + 2x

1n − x

2n)

dx

= π

[−x2 +

x2

3+

2nn + 1

xn+1

n − nn + 2

xn+2

n

]1

0

= π

(−1 +

13

+2n

n + 1− n

n + 2

)=

(n − 1)(n + 4)π3(n + 1)(n + 2)

Using the method of cylindrical shells: (a) Find the intersections of the curves bound-ing the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1).

(b) Find the inner and outer dimensions of the washer. Since the axis of revolutionis a horizontal line, the element of area being rotated is also horizontal. Forarbitrary y the left endpoint on the element is (yn, y); the right endpoint is (y, y).The length of the element is, therefore, y−yn; the distances of the element fromthe axis of symmetry is 1 − y.

(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 − y) · (y − yn) · ∆y .


2π∫ 1

0(1 − y)(y − yn) dy


= 2π∫ 1

0

(−yn + yn+1 + y − y2

)dy

= 2π[− 1

n + 1yn+1 +

1n + 2

yn+2 +12

y2 − 13

y3]1

0

= 2π(− 1

n + 1+

1n + 2

+12− 1

3

)− 0

= 2π(16− 1

(n + 1)(n + 2)

)=

(n − 1)(n + 4)π3(n + 1)(n + 2)

Tuesday version

1. [5 MARKS] Evaluate the integral∫

(a − t)(b + t2) dt.

Solution:

(a) Expand the product in the integrand:∫

(a − t)(b + t2) dt =

∫ (ab − bt + at2 − t3

)dt .

(b) Integrate term by term:∫ (

ab − bt + at2 − t3)

dt = ab · t − b2· t2 +

a3· t3 − 1

4· t4 + C .

2. [5 MARKS] Using a substitution, evaluate the indefinite integral∫

cosn x sin x dx, where

n is a fixed, positive integer.

Solution:

(a) Use new variable u, where du = − sin x dx; one solution is u = cos x.

(b)∫

cosn x sin x dx = −∫

un du

= − un+1

n + 1+ C

= − 1n + 1

cosn+1 x + C


3. [10 MARKS] Find the area of the region bounded by the parabola x = y2, the tangentline to this parabola at (a2, a), and the y-axis, where a is a fixed, positive real number.

Solution: The solution is analogous (under the exchange x ↔ y) to that given for Prob-lem 3 of the Monday quiz.

4. [10 MARKS] Find the volume of the solid obtained by rotating the region bounded byy = xn and x = yn about the line x = −1, where n is a given positive integer.

Solution:

Case I: n is even

Using the method of “washers”: (a) Find the intersections of the curves bound-ing the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1).

(b) Find the inner and outer dimensions of the washer. Since the axis of revo-lution is a vertical line, the element of area being rotated is horizontal. Forarbitrary y the farther endpoint on the element is (y

1n , y); the nearer end-

point is (yn, y). The distances of these points from the axis are, respectively1 + n√

y and 1 + yn.(c) The volume of the “washer” is, therefore,

π(−(1 + yn)2 + (1 + n

√y)2

)∆y .


π

∫ 1

0

(−(1 + y)2 + (1 + n

√y)2

)dy

= π

∫ 1

0

(2y

1n + y

2n − 2yn − y2n

)dy

= π

[2n

n + 1y

n+1n +

nn + 2

yn+2

n − 2n + 1

yn+1 − 12n + 1

y2n+1]1

0

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)

Using the method of cylindrical shells: (a) Find the intersections of the curvesbounding the region. Solving the 2 equations yields the points (x, y) =

(0, 0), (1, 1).(b) Since the axis of revolution is a vertical line, the element of area being

rotated is also vertical. For arbitrary x the top endpoint on the element is(x, x

1n ); the lower endpoint is (x, xn). The length of the element is, there-

fore, x1n − xn; the distance of the element from the axis of symmetry is

1 + x.


(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 + x) ·(x

1n − xn

).


2π∫ 1

0(1 + x)(x

1n − xn) dx

= 2π∫ 1

0

(x

1n − xn + x

n+1n − xn+1

)dx

= 2π(

nn + 1

− 1n + 1

+n

2n + 1− 1

n + 2

)

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)

Case II: n is odd

Using the method of “washers”: (a) Find the intersections of the curves bound-ing the region. Solving the 2 equations yields the points (x, y) = (0, 0), (±1,±1).Here there is an issue of interpretation. The textbook usually permits theword region to apply to one that may have more than one component;some authors would not wish to apply the term in such a situation. I willfollow the textbook, and permit a region here to have two components.

(b) Find the inner and outer dimensions of the washer. Since the axis of rev-olution is a vertical line, the element of area being rotated is horizontal.But there are two kinds of elements, depending on whether y is positiveor negative. For arbitrary, positive y the farther endpoint on the element is(y

1n , y); the nearer endpoint is (yn, y). The distances of these points from

the axis are, respectively 1 + n√

y and 1 + yn. For arbitrary, negative y thenearer endpoint on the element is (y

1n , y); the farther endpoint is (yn, y).

The distances of these points from the axis are, respectively 1 + n√

y and1 + yn (both of which are less than 1).

(c) The volume of the “washer” is, therefore,

π∣∣∣−(1 + yn)2 + (1 + n

√y)2

∣∣∣ ∆y .


π

∫ 1

−1

∣∣∣−(1 + y)2 + (1 + n√

y)2∣∣∣ dy

= π

∫ 1

0

(2y

1n + y

2n − 2yn − y2n

)dy


+π

∫ 0

−1

(−2y

1n − y

2n + 2yn + y2n

)dy

= π

[2n

n + 1y

n+1n +

nn + 2

yn+2

n − 2n + 1

yn+1 − 12n + 1

y2n+1]1

0

+π

[− 2n

n + 1y

n+1n − n

n + 2y

n+2n +

2n + 1

yn+1 +1

2n + 1y2n+1

]0

−1

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)

+2(n − 1)(n2 + 3n + 1)π(n + 1)(n + 2)(2n + 1)

=4(n − 1)π

n + 1.


(0, 0), (±1,±1).(b) Since the axis of revolution is a vertical line, the element of area being

rotated is also vertical. For arbitrary, positive x the top endpoint on the el-ement is (x, x

1n ); the lower endpoint is (x, xn); for arbitrary, negative x the

bottom endpoint on the element is (x, x1n ); the upper endpoint is (x, xn).

The length of the element is, therefore,∣∣∣∣x 1

n − xn∣∣∣∣; the distance of the ele-

ment from the axis of rotation is 1 + x.(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 + x) ·∣∣∣∣x 1

n − xn∣∣∣∣ .


2π∫ 1

−1(1 + x)

∣∣∣∣x 1n − xn

∣∣∣∣ dx

= 2π∫ 1

−1

∣∣∣∣x 1n − xn + x

n+1n − xn+1

∣∣∣∣ dx

= 2π(

nn + 1

− 1n + 1

+n

2n + 1− 1

n + 2

)

+2π(

nn + 1

− 1n + 1

− n2n + 1

+1

n + 2

)

=4(n − 1)π

n + 1

Wednesday version


1. [5 MARKS] Showing all your work, evaluate the indefinite integral∫

sin 2tcos t

dt.

Solution:

(a) Apply a “double angle” formula:∫

sin 2tcos t

dt =

∫2 sin t cos t

cos tdt = 2

∫sin t dt .

(b) Complete the integration:

2∫

sin t dt = −2 cos t + C .

2. Using a substitution, evaluate the indefinite integral∫

ex

ex + adx, where a is a non-zero

real number.

Solution:

(a) [2 MARKS] Try the substitution u = ex + a, so du = ex dx.

(b) ∫ex

ex + adx =

∫duu

= ln |u| + C = ln |ex + a| + C .

(If the constant a is positive, then the absolute signs are not required.)

3. [10 MARKS] Find the number b such that the line y = b divides the region boundedby the curves y = ax2 and y = k into two regions with equal area, where a, k are givenpositive constants.

Solution:

(a) Determine the range of values for integration by finding the intersections of the

bounding curves: solving the equations yields the points

∓√

ka, k

.

(b) Determine the portion of the full area which is below the line y = b. We begin byrepeating the calculation of the preceding part: the corner points have coordinates∓

√ba, b

. The area is

∫ √ ba

−√

ba

(b − ax2) dx = 2[bx − ax3

3

]√ ba

0=

43

b

√ba.


(c) As a special case of the foregoing, or by a separate calculation, we can conclude

that the area of the entire region is43

k

√ka

.

(d) The condition of the problem is that

43

b

√ba

=12· 4

3k

√ka

which is equivalent to 4b3 = k3, and implies that the line should be placed whereb = 2−

23 k.

4. [10 MARKS] Use the method of cylindrical shells to find the volume generated by ro-tating the region bounded by the given curves about the specified axis.

y =√

x − 1 , y = 0 , x = a

about y = b, where a, b are fixed positive constants, and b ≥ √a − 1 .

Solution:

(a) Solve equations to determine the limits of integration. Solving x = a with y =√x − 1 yields the single point of intersection

(a,√

a − 1).

(b) The horizontal element of area at height y which generates the cylindrical shell hasleft endpoint (1 + y2, y) and right endpoint (a, y), so its length is a − (1 + y2).

(c) The distance of the horizontal element of area which generates the shell from theaxis of rotation is b − y.

(d) Set up the integral for the volume by cylindrical shells:

2π∫ √

a−1

0(b − y)

(a − (1 + y2)

)dy .

(e) Evaluate the integral

2π∫ √

a−1

0(b − y)

(a − (1 + y2)

)dy

= 2π∫ √

a−1

0

(b(a − 1) − (a − 1)y − by2 + y3

)dy

= 2π[b(a − 1)y − a − 1

2y2 − b

3y3 +

14

y4]√a−1

0

= 2π(b(a − 1)

32 − a − 1

2(a − 1) − b

3(a − 1)

32 +

14

(a − 1)2)

= 2π(a − 1)32

(23· b − 1

4

√a − 1

).



Thursday version

1. [5 MARKS] Showing all your work, evaluate the integral by making a substitution:∫

b(1 + ax)3 dx ,

where a, b are non-zero constants.

Solution:

(a) A substitution which suggests itself is u = 1 + ax, implying that du = a dx, sodx = 1

a du.

(b) ∫b

(1 + ax)3 dx =ba

∫duu3 = − b

2au−2 + C = − b

2a(1 + ax)2 + C .

2. [5 MARKS] Evaluate the indefinite integral∫

seca x tan x dx, where a is a constant,

positive integer.

Solution:

(a) Try the substitution given by du = sec x·tan x dx, of which one solution is u = sec x.

(b) ∫seca x tan x dx =

∫ua−1 du =

ua

a+ C =

seca xa

+ C .

Some students may have integrated by sight.

3. [10 MARKS] Find the number b such that the line divides the region bounded by thecurves x = ay2 and x = k into two regions with equal area.

Solution: The solution is analogous (under the exchange x ↔ y) to that given for Prob-lem 3 of the Wednesday quiz.

4. [10 MARKS] The region bounded by the given curves is rotated about the axis x = −1.Find the volume of the resulting solid by any method:

y = 5, y = x2 − ax + b

Solution: Because there are constraints on the constants, I will work just one variant,with a = 3, b = 7.

Using the method of cylindrical shells: (a) To find the extremes of integration, wesolve the equations y = 5 and y = x2 − 3x + 7, obtaining (x, y) = (1, 5), (2, 5).


(b) The height of a vertical element of area which generates a cylindrical shell is,at horizontal position x, 5 − (x2 − 3x + 7) = −x2 + 3x − 2.

(c) The distance of that vertical element of area from the axis of revolution is 1+x.(d) The volume is given by the integral

2π∫ 2

1(1 + x)(−x2 + 3x − 2) dx

(e) Evaluating the integral:

2π∫ 2

1(1 + x)(−x2 + 3x − 2) dx

= 2π∫ 2

1

(−x3 + 2x2 + x − 2

)dx

= 2π[−1

4x4 +

23

x3 +12

x2 − 2x]2

1

= 2π(−4 +

163

+ 2 − 4 +14− 2

3− 1

2+ 2

)

=5π6

Using the method of “washers” (a) To find the lowest point on the parabola, we solvex2 − 3x + 7 ≥ 0. This can be done by completing the square, or by using the

calculus to find the local minimum. We find it to be(32,

194

).

(b) The horizontal element generating the “washer” at height y extends betweenthe solutions in x to the equation y = x2 − 3x + 7; these are

x =3 ± √

4y − 192

.

(c) The volume of the “washer” at height y is, therefore,

π

1 +

3 +√

4y − 192

2

−1 +

3 − √4y − 192

2 ∆y

= 5π√

4y − 19 ∆y

(d) The volume is given by the integral 5π∫ 5

194

√4y − 19 dy.

(e) Evaluation of the integral:

5π∫ 5

194

√4y − 19 dy =

[5π · 2

3· 1

4(4y − 19)

32

]5

194

=5π6.



Distribution Date: Posted on the Web on 21 March, 2007Caveat lector! There could be misprints or errors in these draft solutions.



• No calculators!




Monday version


t3eat dt, where a is a non-zero constant.

Solution: This problem requires several consecutive applications of integration by partsto reduce the exponent of the power of t to 0:

u = t3, dv = eatdt ⇒ du = 3t2 dt, v =1a

eat

⇒∫

t3eat dt =1a· t3eat − 3

a

∫t2eat dt

U = t2, dV = eatdt ⇒ dU = 2t dt,V =1a

eat

⇒∫

t3eat dt =

(1a· t3 − 3

a2 t2)

eat +6a2

∫teat dt

u = t, dv = eatdt ⇒ du = dt, v =1a

eat

⇒∫

t3eat dt =

(1a· t3 − 3

a2 · t2 +6a3 t

)eat − 6

a3

∫eat dt

=

(1a· t3 − 3

a2 · t2 +6a3 t − 6

a4

)eat + C


the correctness of which integration may be verified by differentiation of the product onthe right.

2. [8 MARKS] Showing all your work, find a reduction formula for the indefinite integral∫cosn ax dx, where a is a non-zero constant, and n is an integer not less than 2.

Solution:

(a) Introduce a symbol for the general indefinite integral sought:

In =

∫cosn ax dx

(b) Integration by parts:

u = cosn−1 ax ⇒ du = −a(n − 1) cosn−2 ax · sin ax dx

dv = cos ax dx ⇒ v =1a· sin ax

In =1a· cosn−1 ax · sin ax + (n − 1)

∫cosn−2 ax · sin2 ax dx

=1a· cosn−1 ax · sin ax + (n − 1)

∫cosn−2 ax ·

(1 − cos2 ax

)dx

=1a· cosn−1 ax · sin ax + (n − 1)In−2 − (n − 1)In .

(c) Solve the last equation for In

In =1

an· sin ax · cosn−1 ax +

n − 1n

In−2 + C .

3. [4 MARKS] Showing all your work, evaluate the integral∫

1

x2√

x2 − a2dx , where a is

a non-zero constant.

Solution: The following solution uses a trigonometric substitution; it is also possible tosolve this problem using a hyperbolic substitution.

As is customary, I will proceed mechanically, taking square roots where necessary with-out much attention to the sign choices; and then verify at the end by differentiation thatthis process has produced a valid indefinite integral. This procedure can be made rigor-ous by defining the new variable in terms of an inverse trigonometric function of x. Thisis a useful exercise, but becomes extremely complicated in this case, because we wouldhave to work with either the inverse cosine or the inverse secant, and the textbook we


are using chooses different domains for these two functions. So I will avoid the nicetiesand proceed as described.

I propose to use a substitution which provides that x = a sec θ. Then

dx = a sec θ · tan θ · dθ ,and

∫1

x2√

x2 − a2dx =

1a2

∫cos θ dθ

=1a2 sin θ + C

=1a2 tan θ · cos θ + C

=sec2 θ − 1a2 sec θ

+ C

=

√x2 − a2

a2x+ C ,

the correctness of which integration may be verified by differentiation of the quotient onthe right.

4. [8 MARKS] Showing all your work, evaluate the integral∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx,

where a, b are distinct constants, c is a non-zero constant, and the limits of integrationk, ` are also prescribed. (The integrand was not presented to students in factored form.)

Solution:

(a) Since the degree of the numerator is not less than the degree of the denominator,begin by dividing the denominator into the numerator, obtaining a quotient and aremainder:

∫x(x − a)(x − b) + c

(x − a)(x − b)dx =

∫x +

c(x − a)(x − b)

dx .

(b) Expand the fraction into partial fractions. Assume that

c(x − a)(x − b)

=A

x − a+

Bx − b

,

take to a common denominator, and equate the resulting polynomials:

c = A · (x − b) + B · (x − a) .


(c) Now either equate coefficients of like powers of x, or, equivalently, give x succes-sive values x = a and x = b:

c = A(a − b)c = B(b − a) ⇒ A =

ca − b

= −B .

(d) The integration reduces to∫

x(x − a)(x − b) + c(x − a)(x − b)

dx =

∫ (x +

ca − b

· 1x − a

− ca − b

· 1x − b

)dx .

(e) Complete the integration:∫

x(x − a)(x − b) + c(x − a)(x − b)

dx =x2

2+

ca − b

(ln |x − a| − ln |x − b|) + C

=x2

2+

ca − b

ln∣∣∣∣∣x − ax − b

∣∣∣∣∣ + C ,

the correctness of which integration may be verified by differentiation of the func-tion on the right.

(f) Provided it is convergent the given definite integral can now be evaluated:

∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx =

[x2

2+

ca − b

ln∣∣∣∣∣x − ax − b

∣∣∣∣∣]`

k

=`2 − k2

2+

ca − b

(ln

∣∣∣∣∣` − a` − b

∣∣∣∣∣ − ln∣∣∣∣∣k − ak − b

∣∣∣∣∣)

=`2 − k2

2+

ca − b

(ln

∣∣∣∣∣(` − a)(k − b)(` − b)(k − a)

∣∣∣∣∣)

(g) All of the preceding is based on the integral being convergent. In some of theversions the integral was divergent. This was because at least one of the roots ofthe polynomial which is the denominator of the integrand was contained in theinterval of integration. In such a case the integral can be seen to diverge.


1√3∫

−1

earctan y

1 + y2 dy . I have stated the problem with just one

pair of possible limits for the integral; the variations of the problem included severalpossible limits in each case, for each of which students should have been familiar withthe arctangent.


Solution: For simplicity, I work a specific instance of this problem. Use the substitution

u = arctan y, so du =dy

1 + y2 . Then an antiderivative can be obtained as follows

∫earctan y

1 + y2 dy =

∫eu du = eu + C = earctan y + C

so the definite integral is equal to

[earctan y

] 1√3

−1= e

π6 − e−

π4 .

Alternatively, the substitution may be executed in the definite integral, replacing thelower limit of −1 by arctan(−1) = −π4 , and the upper limit of 1√

3by arctan 1√

3= π

6 .

Tuesday version


x2 cos ax dx, where a is a

non-zero constant.

Solution: Two applications of integration by parts will be used to reduce the exponentof the power of x to 0.

(a)

u = x2 ⇒ du = 2x dx

dv = cos ax dx ⇒ v =1a

sin ax∫

x2 cos ax dx =x2

a· sin ax − 2

a

∫x · sin ax dx

(b)

U = x ⇒ dU = dx

dV = sin ax dx ⇒ V = −1a· cos ax

∫x2 cos ax dx =

x2

a· sin ax − 2

a

(− x

a· cos ax +

1a

∫cos ax dx

)

=x2

a· sin ax +

2a2 x · cos ax − 2

a2

∫cos ax dx

=x2

a· sin ax +

2a2 x · cos ax − 2

a3 sin ax + C


The integration can be checked by differentiation of the alleged antiderivative.

2. [9 MARKS] Showing all your work, find a reduction formula for the integral∫

xneax dx,

where a is a non-zero constant.

Solution:

(a) Introduce a symbol for the general indefinite integral sought:

In =

∫xneax dx .


u = xn ⇒ du = nxn−1 dx

dv = eax dx ⇒ v =1a· eax

In =1a· eax − n

a

∫xn−1eax dx

=1a· eax − n

aIn−1.

which is the desired reduction formula.

3. [4 MARKS] Showing your work, evaluate the integral∫

sin3 ax · cos2 ax dx, where a is

a non-zero constant.

Solution: This integral is easily evaluated by a substitution giving du = constant ×sin ax dx. So a convenient substitution is u = cos ax, which yields du = −a sin ax dx.

∫sin3 ax · cos2 ax dx =

∫sin2 ax · u2 · −du

a

=

∫ (1 − cos2 ax

)· u2 · −du

a

=

∫ (1 − u2

)· u2 · −du

a

=1a

∫ (u4 − u2

)du

=1a

(u5

5− u3

3

)+ C

=1a

(cos5 ax

5− cos3 ax

3

)+ C ,


which integration may be checked by differentiation. (Of course, there are other, equiv-alent ways of expressing this indefinite integral.)


x3

√x2 + a2

dx, where a is a

given non-zero constant.

Solution: To simplify the surd in the denominator one may use either a trigonometricor a hyperbolic substitution. For students in this course a trigonometric substitution isusually a better choice. To arrange that x = a tan u, we use a substitution

u = arctanxa, (110)

and dx = a sec2 u du. We may assume that a > 0. The interval of validity for substitution(110) is −π2 < x < π

2 , in which the secant is positive.∫

x3

√x2 + a2

dx =

∫a3 tan3 u|a sec u| · a sec2 u du

= a3∫

tan2 u · sec u tan u du

= a3∫ (

sec2 u − 1)· d

dusec u du

= a3(sec3 u

3− sec u

)+ C

effectively by substitution U = sec u

=13

((a tan u)2 − 2a2

) √(a tan u)2 + a2 + C

=13

(x2 − 2a2

) √x2 + a2 + C

which integration may be verified by differentiation.

5. [9 MARKS] (To simplify the exposition of the solution, I work a specific example here.)

Showing all your work, evaluate the integral∫

x + 21(x + 9)(x − 5)

dx.

Solution: Since the degree of the numerator is less than that of the denominator, we candispense with the first step of dividing denominator into numerator.

(a) We need to expand the integrand into a sum of partial fractions; fortunately thefactorization of the denominator has been given. Assuming there are constantsA, B such that

x + 21(x + 9)(x − 5)

=A

x + 9+

Bx − 5


and transforming all fractions to have a common denominator, we find that

x + 21 = A(x − 5) + B(x + 9) .

(b) The values of A, B may be obtained either by comparing coefficients of like powersof x, or by assigning to x successively the “convenient” values 5, -9: we obtain that

26 = 14B ⇒ B =137

12 = −14A ⇒ A = −67.

(c) We may now complete the integration:∫

x + 21(x + 9)(x − 5)

dx =17

∫ (− 6

x + 9+

13x − 5

)dx

= −17

(−6 ln(x + 9) + 13 ln(x − 5)) + C

which can also be expressed in other, equivalent ways. This integration may beverified by differentiation.

Wednesday version

1. [4 MARKS] Showing your work, evaluate

a∫

0

(x2 +1)e−x dx , where a is a given constant.

Solution: I will integrate by parts twice, in order to reduce the degree of the polynomialfactor of the integrand.

(a) First integration by parts:

u = x2 + 1 ⇒ du = 2x dxdv = e−x dx ⇒ v = −e−x

a∫

0

(x2 + 1)e−x dx =[(x2 + 1)

(−e−x)]a

0+

∫ a

02x · e−x dx .

(b) Second integration by parts:

U = 2x ⇒ dU = 2 dxdV = e−x dx ⇒ V = −e−x

a∫

0

(x2 + 1)e−x dx =[(x2 + 1 + 2x)

(−e−x)]a

0+

∫ a

02 · e−x dx .


(c) Completion of the integration:a∫

0

(x2 + 1)e−x dx =[(x2 + 2 + 2x)

(−e−x)]a

0

= −(a2 + 2a + 3)e−a + 3.

2. [9 MARKS] Showing all your work, find a reduction formula for the integral

π2∫

0

sinn ax dx,

where a is a given integer.

Solution: Assume n is an integer greater than 1.

(a) Introduce a symbol for the definite integral sought:

In =

∫ π2

0sinn ax dx .


dv = sin ax dx⇒ v = −cos axa

dx

u = sinn−1 ax⇒ du = a(n − 1) sinn−2 ax · cos ax dx

In =

[−1

a· sinn−1 ax · cos ax

] π2

0+ (n − 1)

∫ π2

0sinn−2 ax · cos2 ax dx

(c) Decomposition of integral:∫ π

2

0sinn−2 ax · cos2 ax dx =

∫ π2

0sinn−2 ax ·

(1 − sin2 ax

)dx

=

∫ π2

0sinn−2 ax dx −

∫ π2

0sinn ax dx

= In−2 − In

(d) Solution of equation to obtain reduction formula

In =

[−1


] π2

0+ (n − 1) (In−2 − In)

⇒ nIn =

[−1


] π2

0+ (n − 1)In−2

⇒ In =1n

(0 − 0) +n − 1

nIn−2


Because of the choice of limits and the fact that a is an integer, the “net change” is0. Thus we obtain a very simple relationship, which can be solved. Students werenot asked to complete this part of the solution. For example, it is possible to proveby induction that, if n = 2m, an even, positive integer, then

I2m =2m − 1

2m· 2m − 3

2m − 2· . . . · 3

2I0

=(2m)!

4mm!m!· π

2.

(This is Exercise 44, page 481 in the textbook.)


x2

(a2 − x2) 3

2

dx, where a is a

non-zero constant.

Solution: Without limiting generality we take a > 0. A trigonometric substitution cansimplify this integral. One such substitution would have x = a sin u; more precisely,u = arcsin x

a , defined for −π2 ≤ x ≤ π2 , in which interval the cosine and secant are

positive. Then dx = a du√1−u2

.

∫x2

(a2 − x2) 3

2

=

∫a2 sin2 ua3 cos3 u

· a cos u du

=

∫ (tan2 u

)du =

∫ (sec2 u − 1

)du

= tan u − u + C

=sin ucos u

− u + C

=sin u√

1 − sin2 u− u + C

=

xa√

1 − x2

a2

− arcsinxa

+ C

=x√

a2 − x2− arcsin

xa

+ C

which may be verified by differentiation.

4. [9 MARKS] Showing all your work, evaluate the integral∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx,

where a, b, k, ` are distinct constants such that the integrand is defined throughout the


given interval, and c is a non-zero constant. (The integrand was not presented to studentsin factored form.)

Solution:


∫x(x − a)(x − b) + c

(x − a)(x − b)dx =

∫x +

c(x − a)(x − b)

dx .


c(x − a)(x − b)

=A

x − a+

Bx − b

,


c = A · (x − b) + B · (x − a) .


c = A(a − b)c = B(b − a) ⇒ A =

ca − b

= −B .


x(x − a)(x − b) + c(x − a)(x − b)

dx =

∫ (x +

ca − b

· 1x − a

− ca − b

· cx − b

)dx .


x(x − a)(x − b) + c(x − a)(x − b)

dx =x2

2+

ca − b

(ln(x − a) − ln(x − b)) + C

=x2

2+

ca − b

lnx − ax − b

+ C ,



(f) Evaluate the definite integral∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx =

[x2

2+

ca − b

(ln |x − a| − ln |x − b|)]`

k

=

[x2

2+

ca − b

lnx − ax − b

]`

k

=`2 − k2

2+

ca − b

ln∣∣∣∣∣(` − a)(k − b)(` − b)(k − a)

∣∣∣∣∣

5. [4 MARKS] Showing all work, evaluate the integral∫

sin3 ax dx, where a is a positive

integer.

Solution:∫

sin3 ax dx =

∫ (1 − cos2 ax

)· sin ax dx

under substitution u = cos ax, where du = −a sin ax dx

=

∫(1 − u2)(−1

a) du

= −1a

(u − u3

3

)+ C

= −cos axa

+cos3 ax

3a+ C .

Thursday version

1. [4 MARKS] Showing all your work, evaluate the integral∫ b

ae√

t dt.

Solution: Begin with a substitution√

t = u, so t = u2, dt = 2u du. When t = a, u =√

a,etc.: ∫

e√

t dt =

∫2ueu du .

Now apply integration by parts:

U = u ⇒ dU = dudV = 2eu du ⇒ V = 2eu∫

2ueu du = u · 2eu −∫

2eu du

= u · 2eu − 2eu + C = 2(u − 1)eu + C= 2(

√t − 1)e

√t + C


The definite integral given is then equal to[2(√

t − 1)e√

t]b

a.

2. [9 MARKS] Showing all your work, find a reduction formula for the integral∫

(ln(ax + 1))n dx,

where a is a given, positive constant.

Solution: This problem is a slight generalization of Exercise 45, p. 481 in the textbook,an odd-numbered problem for which there is a solution in the Student Solutions Manual,and also hints on one of the CD-Roms supplied with the textbook.

(a) Introduce a symbol for the definite integral sought:

In = (ln(ax + 1))n dx .

(b) Change the variable (a step which is helpful, but not necessary)

u = ax + 1 ⇒ du = a dx

In =1a

∫(ln u)n du

(c) Integration by parts:

U = (ln u)n ⇒ dU = n(ln u)n−1 · 1u

dV = 1 du ⇒ V = u

In = u(ln u)n − n∫

(ln u)n−1 du

= (ax + 1)(ln(ax + 1))n − na∫

(ln(ax + 1))n−1 dx

= (ax + 1)(ln(ax + 1))n − na · In−1


cos4 at dt, where a is a

given non-zero constant.

Solution: We have to apply the following double angle identity twice:

cos 2θ = 2 cos2 θ − 1 .

∫cos4 at dt =

∫ (1 + cos 2at

2

)2

dt

=14

(∫cos2 2at dt + 2

∫cos 2at dt +

∫1 dt

)


=14

(∫1 + cos 4at

2dt + 2

∫cos 2at dt +

∫1 dt

)

=18· 1

4asin 4at +

12· 1

2asin 2at +

38

t + C

=1

32asin 4at +

14a

sin 2at +38

t + C

which may be verified by differentiation. (Of course, the integral may be expressed inother ways under transformation by trigonometric identities.)

4. [4 MARKS] Showing your work, evaluate the integral∫

dx

x√

x2 + a, where a is a given

positive integer, not a perfect square.

Solution: The surd in the denominator may be simplified by either a trigonometric or ahyperbolic substitution. For students in this course the trigonometric substitutions areusually easier.

x =√

a · tan θ ⇒ dx =√

a sec2 θ dθ∫dx

x√

x2 + a=

∫sec2 θ dθ

tan θ · √a · | sec θ| .

The actual substitution is given by θ = arctan x√a , valid for −π2 < x < π

2 . In that intervalthe secant function is positive, so the absolute signs may be dropped. The integral isequal to

1√a

∫csc θ dθ =

1√a

ln | csc θ − cot θ| + C

=1√a

ln∣∣∣∣∣sec θ − 1

tan θ

∣∣∣∣∣ + C

=1√a

ln

∣∣∣∣∣∣∣

√tan2 θ − 1

tan θ

∣∣∣∣∣∣∣ + C

=1√a

ln

∣∣∣∣∣∣∣

√x2 + a − √a

x

∣∣∣∣∣∣∣ + C

which can be verified by differentiation.

5. [9 MARKS] Showing all your work, evaluate the integral

`∫

k

x(x − a)(x − b) + c(x − a)(x − b)

dx,

where a, b, k, ` are distinct constants such that a, b are not contained in the interval whoseend-points are k, `, and c is a non-zero constant.

Solution: I will first determine an indefinite integral.



∫x(x − a)(x − b) + c

(x − a)(x − b)dx =

∫x +

c(x − a)(x − b)

dx .


c(x − a)(x − b)

=A

x − a+

Bx − b

,


c = A · (x − b) + B · (x − a) .


c = A(a − b)c = B(b − a) ⇒ A =

ca − b

= −B .


x(x − a)(x − b) + c(x − a)(x − b)

dx =

∫ (x +

ca − b

· 1x − a

− ca − b

· cx − b

)dx .


x(x − a)(x − b) + c(x − a)(x − b)

dx =x2

2+

ca − b

(ln |x − a| − ln |x − b|) + C

=x2

2+

ca − b

ln∣∣∣∣∣x − ax − b

∣∣∣∣∣ + C ,


(f) ∫ `

k

x(x − a)(x − b) + c(x − a)(x − b)

dx =

[x2

2+

ca − b

ln∣∣∣∣∣x − ax − b

∣∣∣∣∣]`

k



Distribution Date: Posted on the Web on 06 April, 2007; corrected on 09 April, 2007.Caveat lector! There could be misprints or errors in these draft solutions.

There were four different types of quizzes, for the days when the tutorials are scheduled. Eachtype of quiz was generated in multiple varieties for each of the tutorial sections. The order ofthe problems in the varieties was also randomly assigned. Each version of the quiz was gradedout of a maximum of 30 marks, but 2 of the versions had 5 problems and 2 had 4 problems.All of the quizzes had a heading that included the instructions


• No calculators!




Monday version

1. [6 MARKS] Showing all of your work, find the length of the following curve for theinterval 0 < a ≤ u ≤ b :

y = ln(eu + 1eu − 1

).

Solution:

(a)

dydx

=eu − 1eu + 1

(eu(eu − 1) − (eu + 1)eu

(eu − 1)2

)

= − 2eu

e2u − 1

1 +

(dydx

)2

=

(e2u + 1e2u − 1

)2

= coth2 u

(b)

arc length =

∫ b

a

√1 +

(dydx

)2

du =

∫ b

a| coth u| du


(c) Successful completion of the integration:∫ b

a| coth u| du =

∫ b

acoth u du

since a < b[ln sinh u]b

a

= lnsinh bsinh a

= ln(eb − e−b

ea − e−a

)

0.0

2.5

−1.0

0.5

1.51.00.5

3.0

2.0

−0.5

1.5

1.0

0.0

−1.5

Figure 27: The limacon r = 1 + 2 sin θ

2. [10 MARKS] (see Figure 27 on page 5071) The graph of the following curve is given.


Showing detailed work, find the area that is enclosed between the inner and the outerloops: r = a(1 + 2 sin θ), where a is a positive constant.

Solution:

(a) Determination of the limits of integration: we need first to locate where the curvecrosses itself. Since its formula is in terms of sin θ, the curve is periodic with period(at most) 2π. As θ ranges over the values from 0 to 2π, the values r(θ) range overuniquely determined values. How then can the curve cross itself? This can happeneither

i. at points (r(θ1), θ1) and (r(θ1 + π), θ1 + π) where r(θ1 + π) = −r(θ1); orii. at the pole, where r = 0 for two distinct values of θ.

The first possibility would, for the present curve, require that

1 + 2 sin(θ1 + π) = − (1 + 2 sin θ1)

which is equivalent to

2 + 2 sin θ1 + 2 sin(θ1 + π) = 0

which is equivalent to 2 = 0, a contradiction. Thus the present curve can crossitself only at the pole. That occurs where 1 + 2 sin θ = 0, i.e., where sin θ = − 1

2 .The values of θ satisfying this equation are 2nπ− π

6 and (2n+1)π+ π6 , where n is any

integer. The outer loop of this limacon is traced out, for example, for −π6 ≤ θ ≤ 7π6 .

The inner loop is traced for 7π6 ≤ θ ≤ 11π

6 .

(b) The area of the region bounded by the larger, outer loop is

a2∫ 7π

6

− π6

12· (1 + 2 sin θ)2 dθ

= a2∫ 7π

6

− π6

12·(1 + 4 sin θ + 4 sin2 θ

)dθ

= a2∫ 7π

6

− π6

12· (1 + 4 sin θ + 2 − 2 cos 2θ) dθ

=a2

2

∫ 7π6

− π6(3 + 4 sin θ − 2 cos 2θ) dθ

=a2

2[3θ − 4 cos θ − sin 2θ]

7π6− π6

=a2

2

(7π2− 4 cos

7π6− sin

7π3

)− a2

2

(−π

2− 4 cos

−π6− sin

−π3

)


= a2

2π +3√

32

.

(c) The area of the inner loop is

a2∫ 11π

6

7π6

12· (1 + 2 sin θ)2 dθ

=a2

2[3θ − 4 cos θ − sin 2θ]

11π6

7π6

=a2

2

(11π

2− 4 cos

11π6− sin

11π3

)− a2

2

(7π2− 4 cos

7π6− sin

7π3

)

= a2

π − 3√

32

.

(d) The area of the region between the loops is the excess of the area inside the outerloop over the area inside the inner loop, i.e.,

a2

2π +

3√

32

−π − 3

√3

2

= a2(π + 3

√3) .

Note that a cleaner way of solving this problem would have been to first integratefrom 0 to 2π, which would give the area between the loops plus twice the areainside the smaller loop; and then to subtract twice the area inside the smaller loop.This method is better because the first integral is very easy to evaluate, since theperiodic terms contribute nothing.

This curve is discussed in Exercise 10.4.21, on page 683 of the textbook, and is solvedin the Student Solutions Manual and also on one of the CD-Roms which accompany thetextbook.

3. [4 MARKS] Showing full details of your work, find the exact length of the curve x =

et + e−t, y = a − 2t, 0 ≤ t ≤ b, where a, b are constants.

Solution:(dxdt

)2

+

(dydt

)2

=(et − e−t)2

+ 4

=(et + e−t)2

arc length =

∫ b

0

(et + e−t) dt

=[et − e−t]b

0 = eb − e−b = 2 sinh b .


4. [4 MARKS] Find the value of the limit for the sequence. If it diverges, prove that fact:arctan

(3n

3n + 1

).

Solution: As n → ∞, 3n3n+1 = 1 − 1

3n+1 → 1. Since the arctangent function is continuousat the point 1, the limit of the sequence is the arctangent of 1, i.e., π

4 .

5. [6 MARKS] The given curve is rotated about the y-axis. Find the area of the resultingsurface: x = 1

2√

2

(y2 − ln y

), (1 ≤ y ≤ a), where a is a real constant greater than 1.

Solution:

(a)

dxdy

=1

2√

2

(2y − 1

y

)

1 +

(dxdy

)2

= 1 +18

(4y2 +

1y2 − 4

)

=18

(4y2 +

1y2 + 4

)

=

(1

2√

2

(2y +

1y

))2

(b)

surface area =

∫ a

12πx

√1 +

(dxdy

)2

dy

=π

4

∫ a

1

(y2 − ln y

) (2y +

1y

)dy

=π

4

∫ a

1

(2y3 + y − 2y ln y − ln y

y

)dy

=π

4

[y4

2+

y2

2− 1

2(ln y)2

]a

1− π

2

∫ a

1y ln y dy

(c) One way to integrate∫

y ln y dy is by parts, with u = ln y, v′ = y: u′ = 1y , v =

y2

2 ,

∫y ln y dy =

y2

2· ln y −

∫y2

dy

=y2

4(2 ln y − 1) + C .


Another way is to use the substitution w = y2, so dw = 2y dy, ln w = 2 ln y:∫y ln y dy =

∫ln w

2· dw

2=

14

∫ln w dw =

14

w (ln w − 1) + C etc.

Thus the surface area isπ

4

[23

y3 + ln y − 12

(ln y)2 − y2

4(2 ln y − 1)

]a

1

=π

4

(a4

2+

a2

2− 1

2(ln a)2 − a2

4(2 ln a − 1) − 1

)

Tuesday version

1. [4 MARKS] Showing detailed work, find the arc length function for the curve y = ax32

with starting point P0(1, a), where a is a positive constant. That is, find a function f (x)whose value is the distance along the curve from the starting point to the point withabscissa x.

Solution:

(a)dydx

= a · 32· √x

1 +

(dydx

)2

= 1 +9a2x

4

(b)

f (x) =

∫ x

1

√1 +

9a2t4

dt

=

23· 4

9a2

(1 +

9a2t4

) 32

x

1

=

(4 + 9a2x

) 32 −

(4 + 9a2

) 32

27a2 .

2. [6 MARKS] Showing detailed work, find the area of the surface obtained by rotating thefollowing curve about the x-axis:

y =x2

4− ln x

2(a ≤ x ≤ b)

where a, b are two positive real constants, a < b.

Solution:


(a)

dydx

=12

(x − 1

x

)

1 +

(dydx

)2

= 1 +14

(x2 +

1x2 − 2

)

=14

(x2 +

1x2 + 2

)

=

(12

(x +

1x

))2

(b)

surface area =

∫ b

a2πy

√1 +

(dydx

)2

dx

= π

∫ b

a

(x2

4− ln x

2

) (x +

1x

)dx

= π

∫ b

a

(x3

4+

x4− x ln x

2− ln x

2x

)dx

= π

[116

x4 +x2

8− 1

4(ln x)2

]b

a− π

∫ b

ax ln x dx

(c) One way to integrate∫

x ln x dx is by parts, with u = ln x, v′ = x: u′ = 1x , v = x2

2 ,

∫x ln x dx =

x2

2· ln x −

∫x2

dx

=x2

4(2 ln x − 1) + C .

Another way is to use the substitution w = x2, so dw = 2x dx, ln w = 2 ln x:∫

x ln x dx =

∫ln w

2· dw

2=

14

∫ln w dw =

14

w (ln w − 1) + C etc.

Thus the surface area is

π

[x4

16+

x3

8− 1

4(ln x)2 − x2

4(2 ln x − 1)

]b

a


3. [4 MARKS] Find the exact length of the curve given by

x = et cos t y = et sin t (0 ≤ t ≤ a)

where a is a positive constant.

Solution:

(a)

dxdt

= et(cos t − sin t)

dydt

= et(sin t + cos t)(dxdt

)2

+

(dydt

)2

= e2t(2 cos2 t + 2 sin2 t) = 2e2t

(b) The arc length is ∫ a

0

√2et dt =

√2(ea − 1) .

4. [10 MARKS] Working only with polar coordinates, find the area of the region that liesinside the first curve and outside the second curve: r = b sin θ, r = a, where a and b arepositive constants.

Solution:

(a) Both of these curves are circles; we need to determine the coordinates of the pointsof intersection. Solving the equations yields

r = a sin θ =ab.

One point of intersection will be (r, θ) =(a, arcsin a

b

). Another point of intersection

will be (r, θ) =(a, π − arcsin a

b

)— remember that the values of the arcsine function

are in the interval[−π2 , π2

]. It appears from a drawing that we have all the points of

intersection. If we solve the equation r = −a with r = b sin θ we obtain preciselythe same points, albeit with different coordinates. If we attempt to replace theequation r = b sin θ with that obtained under the identification (r, θ) → (−r, θ + π)there is no change. This algebraic investigation discloses all possible points ofintersection except the pole, which must be checked separately. But the pole cannotlie on r = a, since the pole has first coordinate 0 always. Thus we have, indeed,found all the points of intersection.


(b) The area bounded by the arcs can be considered to consist of the disk r = b sin θdiminished by a sector of the circle r = a for arcsin a

b ≤ θ ≤ π − arcsin ab , together

with two small segments of the disk r = b sin θ bounded by the rays θ = arcsin ab

and θ = π − arcsin ab . What I have provided is one prescription for computing the

area. An easier way would be to take the integral

12

∫ π−arcsin ab

arcsin ab

((b sin θ)2 − a2

)dθ

=

∫ π2

arcsin ab

((b sin θ)2 − a2

)dθ

by symmetry around the line θ = π2

=

∫ π2

arcsin ba

(b2

(1 − cos 2θ

2

)− a2

)dθ

=

∫ π2

arcsin ba

((b2

2− a2

)− b2

2cos 2θ

)dθ

=

[(b2

2− a2

)θ − b2

4sin 2θ

] π2

arcsin ba

=b2 − 2a2

2

(π

2− arcsin

ab

)+

a2

√b2 − a2

5. [6 MARKS] Determine whether the series is convergent or divergent. If it is convergent,

find its sum. Otherwise prove that it is divergent:∞∑

n=1

b

n(n + a)

, where a, b are positive

integers.

Solution:

(a) Expand the general term into partial fractions: there exist constants A, B such that

bn(n + a)

=An

+B

n + a.

To determine the coefficients A, B we can proceed in several ways. If we take thefractions to a common denominator n(n + a), we obtain the polynomial identity

b = A · (N + A) + b · a .In this identity, if we set the variable n equal to −a, we obtain that b = B(−a), soB = − b

a ; and, setting n = 0, we obtain A = ba , hence

bn(n + a)

=ba

(1n− 1

n + a

).


(b) For sufficiently large N the Nth partial sum is equal to

N∑

n=1

(b

n(n + a)

)

=ba

(11

+12

+ . . . +1

a − 1− 1

N + 1− 1

N + 2− . . . − 1

N + a

)

→ ba

(11

+12

+ . . . +1

a − 1− 0 − 0 − . . . − 0

)

as N → ∞. Hence the series converges to the sum

ba

(11

+12

+ . . . +1

a − 1

).

Convergence could be proved in other ways, thereby earning the student part marks. Forexample, using the Comparison or Limit Comparison Tests, or the Integral Test.

Wednesday version

1. [10 MARKS] Showing detailed work, find all points different from the origin on thefollowing curve where the tangent is horizontal; a is a positive constant:

x = a(cos θ − cos2 θ), y = a(sin θ − sin θ cos θ) .

Solution:

(a)

dxdθ

= a (− sin θ − 2 cos θ(− sin θ))

= a(sin θ)(2 cos θ − 1)dydθ

= a(− cos θ + 2 cos2 θ − 2 sin2 θ

)

= a(− cos θ + 2 cos2 θ − 1

)

= a(2 cos θ + 1)(cos θ − 1)

Actually, you weren’t expected to find dxdt .

(b) There will be a horizontal tangent at the point with parameter value t if dydx = 0,

i.e., ifdydθdxdθ

= 0, implying that dydθ must be 0, provided dx

dθ , 0 at the same value of t.


(This last requirement is subtle, and you weren’t expected to actually check it. It isbecause of this restriction that I explicitly excluded the origin from consideration.)The equations we have to solve are

cos θ = 1

andcos θ = −1

2.

They are not being solved simultaneously: we are looking for points t that satisfyat least one — meaning, here, either — of the equations.

(c) The first of these equations is satisfied when θ is an even integer multiple of π. Butthis parameter value corresponds always to the origin, which was excluded fromconsideration.

(d) The second is satisfied when θ is of the form

(2n + 1)π ± 13π .

(e) The functions defining this curve are all periodic with period 2π. Thus we canstudy the curve completely by examining its behavior for parameters θ chosen overan interval of length 2π, e.g. 0 ≤ θ < 2π. There are precisely 3 points here wheredydθ = 0:

(x(0), y(0)) = (0, 0)(x(2π3

), y

(2π3

))=

−a4,

3√

3a4

(x(4π3

), y

(4π3

))=

−a4,−3√

3a4

.

Of these, you were specifically instructed to exclude the first, which is the origin.

2. [5 MARKS] Showing detailed work determine the total length of the portion of thefollowing curve which is in the first quadrant: x = a cos3 θ, y = a sin3 θ, where a is apositive constant.

Solution:

dxdθ

= −3a cos2 θ · sin θ

dydθ

= 3a sin2 θ · cos θ


⇒(dxdθ

)2

+

(dydθ

)2

= 9a2 sin2 θ · cos2 θ ·(cos2 θ + sin2 θ

)

= 9a2 sin2 θ cos2 θ .

The curve is in the first quadrant when both coordinates are positive; as each of these isa cube of a sine or cosine, this means that the portion of the curve in the first quadrant isthat given by 0 ≤ θ ≤ π

2 . The length of the arc is

∫ π2

0

√9a2 sin2 θ cos2 θ dθ =

∫ π2

03a sin θ · cos θ dθ = 3a

[sin2 θ

2

] π2

0=

3a2.

3. [10 MARKS] Find the area of the region that lies inside both of the following curvesr = a + 2 sin θ, r = a − 1, where a is a suitable positive constant.

Solution:

(a) Determination of the limits of integration: we need first to locate where the curvescross. We begin by solving the two given equations, and find that

sin θ = −12⇒ θ =

7π6

or11π

6

or any angle obtained from these by adding an integer multiple of 2π. This yieldsthe points (

a − 1,7π6

),

(a − 1,

11π6

).

Students weren’t expected to pursue this question further. Strictly speaking, theyshould then have solved r = a + 2 sin θ, r = −(a− 1), which would have yielded nopoints; then solved r = −a + 2 sin θ, r = a − 1, which would again yield no points;then r = −a + 2 sin θ, r = −(a − 1), which would have yielded the 2 points alreadyfound.

(b) This problem could then be approached in several ways. To find the area “directly”would require finding the sum of the integrals

12

∫ 7π6

−pi6

(a − 1)2 dθ

and12

∫ 11π6

7pi6

(a + 2 sin θ)2 dθ .


The first of these is just 2/3 of the area of a disk of radius a − 1, i.e., 2π(a−1)2

3 . Thesecond is

12

∫ 11π6

7pi6

(a + 2 sin θ)2 dθ

=12

∫ 11π6

7pi6

(a2 + 4a sin θ + 2(1 − cos 2θ)

)dθ

=12

[a2θ − 4a cos θ + 2θ − sin 2θ

] 11π6

7π6

=1π3· a2 − 2

√3a +

2π3.

(Another way to solve this would be to find the area of the disk of radius a − 1 andsubtract from it the portion that is cut off.These integrals could have been slightly more efficiently computed by taking onlythe area up to the y-axis and doubling it.)Hence the net area of the region inside both of the curves is

5π9

a2 −(4π3

+ 2√

3)

+4π3.

4. [5 MARKS] Showing detailed work, determine whether the following series is conver-gent or divergent. If it is convergent, find its sum. Otherwise explain why it diverges:

∞∑

n=1

an + bn

(ab)n ,

where a, b are integers greater than 1.

Solution: The nth term is

an =

(1b

)n

+

(1a

)n

.

The nth partial sum is, therefore, the sum of the partial sums of two geometric series;both of the geometric series are convergent, since the common ratios are less than 1in magnitude. Since the two separate partial sums approach a limit, the sum of thesesequences approaches as its limit the sum of the limits of the two sequences. In the caseof the series whose nth term is

(1b

)n, the limit of the partial sums is

1b

1 − 1b

=1

b − 1;


similarly, the second series sums to 1a−1 . Hence the given series sums to the sum of these

limits, i.e.1

b − 1+

1a − 1

=a + b − 2

(a − 1)(b − 1).

Thursday version

1. [10 MARKS] The curve x = a(1 − 2 cos2 t), y = (tan t)(1 − 2 cos2 t), where a is agiven positive integer, crosses itself at some point (x0, y0). Showing all your work, findthe point of crossing, and the equations of both tangents at the point. (In determiningthe point of crossing you are expected to investigate the parametric functions: it is notsufficient to simply plot a finite number of points on the curve.)

Solution:

(a) Since the functions are all periodic with period π, it suffices to take an interval ofthis length for t, and that will reveal all aspects of the behavior of this curve. (Moreprecisely, the tangent function has period π, and, while the cosine function hasperiod 2π, its square has period π.) So, without limiting generality, let’s consider−π2 ≤ t ≤ π

2 : we have to exclude both end points of this interval, since the tangentfunction is not defined at either of them.Suppose that the curve crosses itself at the points with parameter values t = t1 andt = t2; without limiting generality, we can assume that these parameter values havebeen so labelled that t1 < t2. Since the x-coordinates will need to be the same,

a(1 − 2 cos2 t1) = a(1 − 2 cos2 t2) (111)

socos t1 = ± cos t2 . (112)

Since the y-coordinates must also coincide, we also have

(tan t1)(1 − 2 cos2 t1) = (tan t2)(1 − 2 cos2 t2) (113)

which implies that either

cos2 t1 = cos2 t2 =12, (114)

ortan t1 = tan t2 . (115)

In the interval we have chosen for t, the cosines are always positive; the only solu-tion to (114) is t1 = −π4 , t2 = +π

4 . In that same interval for t there will be no solu-tions to (115), since the tangent function is increasing there. Thus the only possible


crossing points are t = ±π4 , and the point of crossing is the origin, (x, y) = (0, 0).While it isn’t required in the solution, note that as t → ±π2 , x → 1: this curve isasymptotic to the vertical line x = 1.

(b) Tangent at the point with parameter value t = −π4 :

dxdt

= 4a(cos t)(sin t) = 2a sin 2t = −2a

y = (tan t)(1 − 2 cos2 t) = tan t − sin 2tdydt

= sec2 t − 2 cos 2t = 2 − 0

dydx

=

dydtdxdt

= − 22a

= −1a,

and the tangent has equation y = − xa .

(c) Tangent at the point with parameter value t = π4 :

dxdt

= 4a(cos t)(sin t) = 2a sin 2t = 2a

dydt

= sec2 t − 2 cos 2t = 2 − 0

dydx

=

dydtdxdt

=22a

=1a,

and the tangent has equation y = xa .

2. [5 MARKS] Showing detailed work, find the surface area generated by rotating thefollowing curve about the y-axis.

x = at2 , y = bt3 , 0 ≤ t ≤ 5.

Solution:

dxdt

= 2at

dydt

= 3bt2

(dxdt

)2

+

(dydt

)2

= 4a2t2 + 9b2t4

area about y-axis = 2π∫ 5

0at2√

4a2t2 + 9b2t4 dt

= 2πa∫ 5

0t3√

4a2 + 9b2t2 dt


Under the substitution u = 4a2 + 9b2t2,

du = 18b2t dt

2πa∫ 5

0t3√

4a2 + 9b2t2 dt =aπ

81b4

∫ 4a2+225b2

4a2(u

32 − 4a2u

12 ) du

=πa

81b4

[25

u52 − 8a2

3u

32

]4a2+225b2

4a2

=2πa

(81)(15)b4

((4a2 + 225b2)

32 (2a2 + 675b2) − 16a5

)

= ...

3. [10 MARKS] There is a region in the first quadrant that is bounded by arcs of both ofthe following curves. Showing your work in detail, find the area of the region:

r2 = a sin 2θ r2 = a cos 2θ .

Solution:

(a) The given curves are expressed only in terms of sine and cosine of 2θ. The givenfunctions are periodic with period π. When, for either of these curves, we permitθ to range over an interval of length π, we will trace out the entire curves. Theintersections of the curves in the first quadrant will be where sin 2θ = cos 2θ ispositive: thus the only point we have found by this algebraic solution of the twoequations is at θ = π

8 .However there are other ways in which curves can intersect, since points haveinfinitely many different sets of polar coordinates. If we transform either of thegiven equations under the substitution (r, θ) → (−r, θ + π), we find that there isno change in the equation. Thus we haven’t missed any points because of theconvention that permits the first coordinate to be negative.But there is another situation that leads to multiple sets of coordinates; that is at thepole, where the second — angular — coordinate is totally arbitrary; the pole canlie on a curve simply because of the fact that its distance coordinate r = 0, with noreference to θ. To determine whether the pole lies on a curve we must investigatewhether the equation is satisfied by r = 0 with any value of θ. We find the curver2 = a sin 2θ does contain the pole: when r = 0 the equation is satisfied by any θsuch that sin 2θ = 0; so two solutions are θ = 0 and θ = π

2 . Similarly, the pole lieson the curve r2 = a cos 2θ with θ = π

4 . I have given only the coordinates in the firstquadrant. To summarize: there are 2 intersection points in the first quadrant:

(r, θ) =

(a

12 2−

14 ,π

4

)


where the point lies on both of the curves with the same pair of coordinates; andthe pole, which lies on the two curves with different sets of coordinates.

(b) We can find the area by joining the point( √

a4√2, π8

)to the pole and calculating the

sum of two integrals:

πa2

∫ π4

0sin 2θ dθ +

πa2

∫ π2

π4

cos 2θ, dθ

=πa4

[− cos 2θ]π40 +

πa4

[sin 2θ]π2π4

=πa4

+πa4

=πa2.

4. [5 MARKS] Showing detailed work, express the number below as a ratio of integers.

0.ab = 0.abababab...

where a, b are any two digits. You are expected to simplify your answer as much aspossible.

Solution: The repeating decimal is the sum of an infinite series(10a + b

100

)+

(10a + b

100

)1

100+

(10a + b

100

)1

1002 +

(10a + b

100

)1

1003 + . . .

= limN→∞

N∑

n=0

(10a + b

100

)1

100n

=

(10a + b

100

)lim

N→∞

N∑

n=0

(1

100

)n

=

(10a + b

100

)lim

N→∞

1 −(

1100

)N+1

1 − 1100

=

(10a + b

100

)1

1 − 1100

=

(10a + b

100

)10099

=10a + b

99


Release Date: Mounted on the Web on Friday, 01 February, 2008 (but subject to correction)




• No calculators!




Monday version

1. [5 MARKS] If

c∫

a

f (x) dx = k and

c∫

b

f (x) dx = `, find

b∫

a

f (x) dx. Show your work.

Solution:

(a)c∫

a

f (x) dx =

b∫

a

f (x) dx +

c∫

b

f (x) dx .

(b) Hencec∫

b

f (x) dx =

c∫

a

f (x) dx −b∫

a

f (x) dx .

(c)= k − ` .

2. [5 MARKS] Find an antiderivative of the integrand of the integral∫ a

0

√x dx, and then

use the Fundamental Theorem of Calculus to evaluate the integral. You are not expectedto simplify your numerical answer, but no marks will be given unless all your work isclearly shown.


(a) One antiderivative of x12 is

112 + 1

· x 12 +1 =

23

x32 .

(b) ∫ a

0

√x dx =

[23

x32

]a

0=

23

(a

32 − 0

)=

23

a32 .

3. [10 MARKS] Showing all your work, differentiate the function g(x) =

x4∫

tan x

1√2 + t2

dt.

Solution:

(a) First split the interval of integration into 2 parts at a convenient place:

g(x) =

0∫

tan x

1√2 + t2

dt +

x4∫

0

1√2 + t2

dt .

(b) Then reverse the limits in the first summand and change its sign, so that the variablelimit is the upper one:

g(x) = −tan x∫

0

1√2 + t2

dt +

x4∫

0

1√2 + t2

dt .

(c) Denote the upper limit of the first integral by u = tan x. Then

ddx

tan x∫

0

1√2 + t2

dt =d

du

u∫

0

1√2 + t2

dt · dudx

=1√

2 + u2· sec2 x

=2 sec x · tan x√

2 + tan2 x=

2 sec x · tan x√1 + sec2 x

.

(d) Denote the upper limit of the second integral by v = x4. Then

ddx

x4∫

0

1√2 + t2

dt =ddv

v∫

0

1√2 + t2

dt · dvdx


=1√

2 + v2· 4x3

=4x3

√2 + x8

.

(e) Henceddx

g(x) = −2 sec x · tan x√2 + tan2 x

+4x3

√2 + x8

.

4. [10 MARKS] If F(x) =

∫ x

1f (t) dt, where f (t) =

∫ t2

1

a + ub

udu and a, b are constants,

find F′′(2).

Solution:

(a) Applying Part 1 of the Fundamental Theorem yields

F′(x) = f (x) =

∫ x2

1

a + ub

udu .

(b) A second application of Part 1 of the Fundamental Theorem yields

F′′(x) = f ′(x) =ddx

∫ x2

1

a + ub

udu .

(c) Denote the upper index of the last integral by v = x2.

(d)

ddx

∫ x2

1

a + ub

udu =

ddx

∫ v

1

a + ub

udu

=ddv

∫ v

1

a + ub

udu · dv

dx

=a + vb

v· dv

dx

=a + vb

v· 2x

=a + x2b

x2 · 2x

=2(a + x2b

)

x.


Tuesday version


g(x) =

∫ a

xb tan(t) dt ,

(where a and b are constants). Then use Part 2 of the Fundamental Theorem to evaluateg(x), by first verifying carefully that ln | sec x| is an antiderivative of tan x.

Solution:

(a) Part 1 of the Fundamental Theorem gives the derivative of a definite integral as afunction of its upper index of integration. Here the variable is the lower index ofintegration.

ddx

∫ a

xb tan(t) dt

=ddx

(−

∫ x

ab tan(t) dt

)

= − ddx

∫ x

ab tan(t) dt

= −b tan x .

Some students may quote a variant of Part 1 which gives the derivative of a definiteintegral with respect to the lower index, and this should be accepted if work hasbeen shown.

(b) Students were expected to first find the derivative of ln | sec x|. Since this is a com-position of 2 functions, the Chain Rule will be needed. Let u = sec x. Then

ddx

ln | sec x| =ddu

ln |u| · ddx

secx

=1u· sec x tan x

=1

sec x· sec x tan x

= tan x .

Henceg(x) = b ln | sec t|ax = b ln

∣∣∣∣∣sec asec x

∣∣∣∣∣ = b ln∣∣∣∣∣cos xcos a

∣∣∣∣∣ .


2. [5 MARKS] Evaluate the limit by first recognizing the sum as a Riemann sum for afunction defined on [0, 1]:

limn→∞

1n

√

7n

+

√14n

+

√21n

+ . . . +

√7nn

.

Solution:

(a) We are told to take the interval of integration to be [0, 1]; when this is divided into

n equal parts, each has length ∆x =1n

. Such a factor has been explicitly written inthe sum.

(b) The typical summand is — aside from the common factor 1n — of the form

√7in

.

Since the distance of the left end-point of the ith subinterval from 0 is i∆x = in , we

may interpret

√7in

=√

7x .

(c) Thus the limit must be equal to∫ 1

0

√7x dx =

√7 · 2

3· x 3

2

]1

0=

23

√7.


g(x) =

∫ √x

ab

cos tt

dt ,

where a, b are constants.

Solution:

(a) Denote the upper index of the integral by u(x) =√

x.

(b) Then

ddx

g(x) =ddx

∫ √x

ab

cos tt

dt

=ddx

∫ u(x)

ab

cos tt

dt

=d

du

∫ u(x)

ab

cos tt

dt · du(x)dx

= bcos u

u· du(x)

dx


= bcos u

u· 1

2√

x

= bcos√

x√x· 1

2√

x

= bcos√

x2x

4. [10 MARKS] Showing all your work, determine all values of x where the curve y =x∫

0

11 + at + bt2 dt is concave downward, where a, b are constants.

Solution:

(a) By Part 1 of the Fundamental Theorem,

y′(x) =1

1 + ax + bx2 .

(b) Differentiating a second time yields

y′′(x) =ddx

(1

1 + ax + bx2

)

= − 1(1 + ax + bx2)2

· ddx

(1 + ax + bx2

)

= − a + 2bx(1 + ax + bx2)2 .

(c) The curve is concave downward where y′′ < 0:

− a + 2bx(1 + ax + bx2)2 > 0 ⇔ −(a + 2bx) > 0

since the denominator is a square, hence positive⇔ 2bx < −a

⇔

x < − a2b when b > 0

x > − a2b when b < 0

never concave upward when b = 0

Wednesday version



πb∫

πa

sin t dt.

Solution:

(a) An antiderivative of sin t is − cos t.

(b)πb∫

πa

sin t dt = [− cos t]πbπa

(c) Your answer should be simplified as much as possible.

2. [5 MARKS] Evaluate the following limit by first recognizing the sum as a Riemann

sum for a function defined on [0, 1]: limn→∞

n∑

i=1

i8

n9 . A full solution is required — it is not

sufficient to write only the value of the limit.

Solution:

(a) We are told to take the interval of integration to be [0, 1]; when this is divided into

n equal parts, each has length ∆x =1n

.

(b) The typical summand is — aside from the common factor1n

— of the form( in

)8

.

Since the distance of the left end-point of the ith subinterval from 0 is i∆x =in

, we

may interpret( in

)8

= x8 .

(c) Thus the limit must be equal to

∫ 1

0x8 dx =

19

x9]1

0=

19.


of the function

bx∫

cos x

cos (tc) dt, where a, b, c are real numbers.

Solution:



bx∫

cos x

cos (tc) dt =

0∫

cos x

cos (tc) dt +

bx∫

0

cos (tc) dt

= −cos x∫

0

cos (tc) dt +

bx∫

0

cos (tc) dt

(b) For the summand

bx∫

0

cos (tc) dt, let u = bx. Then

ddx

bx∫

0

cos (tc) dt =ddx

u∫

0

cos (tc) dt

=d

du

u∫

0

cos (tc) dt · dudx

= cos (uc) · b= cos ((bx)c) · b

(c) For the summand

cos x∫

0

cos (tc) dt, let v = cos x.

ddx

cos x∫

0

cos (tc) dt =ddx

v∫

0

cos (tc) dt

=ddv

v∫

0

cos (tc) dt · dvdx

= cos (vc) · (− sin x)= cos (cosc x) · (− sin x)

(d)bx∫

cos x

cos (tc) dt = − cos (cosc x) · (− sin x) + cos ((bx)c) · b .


4. [10 MARKS] Let f (x) =

0 if x < 0x if 0 ≤ x ≤ a2a − x if a < x < 2a0 if x > 2a

and g(x) =

∫ x

0f (t) dt, where a is

a positive constant. Showing all your work, find a formula for the value of g(x) whena < x < 2a.

Solution:

(a) The interval where we seek a formula is the third interval into which the domainhas been broken. For x in this interval the integral can be decomposed into

∫ x

0f (t) dt =

∫ a

0f (t) dt +

∫ x

af (t) dt .

The portion of the definition of f for x < 0 is of no interest in this problem, sincewe are not finding area under that portion of the curve; the same applies to theportion of the definition for x > 2a.

(b)∫ a

0f (t) dt =

∫ a

0t dt

=

[t2

2

]t=a

t=0=

a2

2.

(c)∫ x

af (t) dt =

∫ x

a(2a − t) dt

=

[2at − t2

2

]t=x

t=a

=

(2ax − x2

2

)−

(2a2 − a2

2

).

(d)

g(x) =a2

2+

(2ax − x2

2

)−

(2a2 − a2

2

)= 2ax − x2

2− a2 .

Thursday version


1. [5 MARKS] Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integralbπ∫

aπ

cos θ dθ, where a, b are given integers. No marks will be given unless all your work

is clearly shown. Your answer should be simplified as much as possible.

Solution:

(a) One antiderivative of cos θ is sin θ.

(b)bπ∫

aπ

cos θ dθ = [sin θ]bπaπ = sin(bπ) − sin(aπ) .

(c) Students were expected to observe that the value of the sine at the given multiplesof π is 0, so the value of the definite integral is 0.

2. [5 MARKS] Express limn→∞

n∑

i=1

axi sin xi ∆x as a definite integral on the interval [b, c],

which has been subdivided into n equal subintervals.

Solution: ∫ c

bax sin x dx .


g(x) =

∫ bx

ax

t2 + ct2 − c

dt ,

where a, b, c are positive integers.

Solution:


g(x) =

∫ bx

ax

t2 + ct2 − c

dt =

∫ 0

ax

t2 + ct2 − c

dt +

∫ bx

0

t2 + ct2 − c

dt

= −∫ ax

0

t2 + ct2 − c

dt +

∫ bx

0

t2 + ct2 − c

dt


(b) For the summand

bx∫

0

t2 + ct2 − c

dt, let u = bx. Then

ddx

bx∫

0

t2 + ct2 − c

dt =ddx

u∫

0

t2 + ct2 − c

dt

=d

du

u∫

0

t2 + ct2 − c

dt · dudx

=u2 + cu2 − c

· dudx

=u2 + cu2 − c

· b

=(bx)2 + c(bx)2 − c

· b

(c) For the summand∫ ax

0

t2 + ct2 − c

dt, let u = ax. Then, analogously to the preceding

step,ddx

∫ ax

0

t2 + ct2 − c

dt =(ax)2 + c(ax)2 − c

· a .

(d)

g′(x) =(bx)2 + c(bx)2 − c

· b − (ax)2 + c(ax)2 − c

· a .

4. [10 MARKS] Find the derivative of the function f (x) =

x3∫

√x

√t cos t dt.

Solution:

(a) First split the interval of integration into 2 parts at a convenient place:

f (x) =

0∫

√x

√t cos t dt +

x3∫

0

√t cos t dt


(b) Then reverse the limits in the first summand and change its sign, so that the variablelimit is the upper one:

f (x) = −√

x∫

0

√t cos t dt +

x3∫

0

√t cos t dt .

(c) Denote the upper limit of the first integral by u =√

x. Then

ddx

√x∫

0

√t cos t dt =

ddx

u∫

0

√t cos t dt

=d

du

u∫

0

√t cos t dt · du

dx

=√

u cos u · dudx

=

√√x cos

√x · 1

2√

x

=cos√

x

2x14

.

(d) Denote the upper limit of the second integral by v = x3. Then

ddx

x3∫

0

√t cos t dt =

ddx

v∫

0

√t cos t dt

=ddv

v∫

0

√t cos t dt · dv

dx

=√

v cos v · dvdx

=√

x3 cos(x3

)· 3x2

= 3x72 cos

(x3

)

(e) Henceddx

f (x) = −cos√

x

2x14

+ 3x72 cos

(x3

).



Distribution Date: Mounted on the Web on Monday, March 03rd, 2008Caveat lector! There could be misprints or errors in these draft solutions.



• No calculators!




Monday version

1. [10 MARKS] Showing all your work, find the volume of the solid obtained by rotatingabout the line y = 1 the region bounded by the curves y = n

√x and y = x, where n is a

given positive integer.

Solution: A favoured method of solution was not prescribed.

Using the method of “washers”: (a) The solution I am giving is for the case where nis even.

(b) Find the intersections of the curves bounding the region. Solving the 2 equa-tions yields the points (x, y) = (0, 0), (1, 1).

(c) It’s not clear from the wording of the problem whether it was intended, in thecase of odd n, to permit the second intersection point (x, y) = (−1,−1); thedecision was left to the individual TA’s. The remainder of this solution coversthe case of even n; for odd n this solution does not consider the solid generatedby rotating the region with vertices (x, y) = (−1,−1), (0, 0).

(d) Find the inner and outer dimensions of the washer. Since the axis of revolutionis a horizontal line, the element of area being rotated is vertical. For arbitraryx the lower point on the element is (x, x); the upper point is (x, n

√x). The

distances of these points from the axis are, respectively 1 − x and 1 − n√

x.


(e) The volume of the “washer” is, therefore,

π(−(1 − x)2 + (1 − n√x)2

)∆x .

(f) Correctly evaluate the integral:

π

∫ 1

0

(−(1 − x)2 + (1 − n√x)2

)dx

= π

∫ 1

0

(−2x + x2 + 2x

1n − x

2n)

dx

= π

[−x2 +

x2

3+

2nn + 1

xn+1

n − nn + 2

xn+2

n

]1

0

= π

(−1 +

13

+2n

n + 1− n

n + 2

)=

(n − 1)(n + 4)π3(n + 1)(n + 2)

Using the method of cylindrical shells: (a) The solution I am giving is for the casewhere n is even.

(b) Find the intersections of the curves bounding the region. Solving the 2 equa-tions yields the points (x, y) = (0, 0), (1, 1).

(c) It’s not clear from the wording of the problem whether it was intended, in thecase of odd n, to permit the second intersection point (x, y) = (−1,−1); thedecision was left to the individual TA’s. The remainder of this solution coversthe case of even n; for odd n this solution does not consider the solid generatedby rotating the region with vertices (x, y) = (−1,−1), (0, 0).

(d) Find the inner and outer dimensions of the washer. Since the axis of revolutionis a horizontal line, the element of area being rotated is also horizontal. Forarbitrary y the left endpoint on the element is (yn, y); the right endpoint is (y, y).The length of the element is, therefore, y−yn; the distances of the element fromthe axis of symmetry is 1 − y.

(e) The volume of the cylindrical shell element of volume is, therefore,

2π(1 − y) · (y − yn) · ∆y .

(f) Correctly evaluate the integral:

2π∫ 1

0(1 − y)(y − yn) dy

= 2π∫ 1

0

(−yn + yn+1 + y − y2

)dy


= 2π[− 1

n + 1yn+1 +

1n + 2

yn+2 +12

y2 − 13

y3]1

0

= 2π(− 1

n + 1+

1n + 2

+12− 1

3

)− 0

= 2π(16− 1

(n + 1)(n + 2)

)=

(n − 1)(n + 4)π3(n + 1)(n + 2)


(a − t)(b + t2) dt.

Solution:

(a) Expand the product in the integrand:∫

(a − t)(b + t2) dt =

∫ (ab − bt + at2 − t3

)dt .

(b) Integrate term by term:∫ (

ab − bt + at2 − t3)

dt = ab · t − b2· t2 +

a3· t3 − 1

4· t4 + C .

3. [10 MARKS] Showing all your work, determine a number b such that the line x = bdivides into two regions of equal area the region bounded by the curves x = ay2 andx = k.

Solution: The solution is analogous (under the exchange x ↔ y) to that given for Prob-lem 1 of the Tuesday quiz.

4. [5 MARKS] Showing all your work, use a substitution to evaluate the indefinite integral∫ex

ex + adx, where a is a non-zero real number.

Solution:

(a) Try the substitution u = ex + a, so du = ex dx.

(b) ∫ex

ex + adx =

∫duu

= ln |u| + C = ln |ex + a| + C .

(If the constant a is positive, then the absolute signs are not required.)


Tuesday version

1. [10 MARKS] Showing all your work, find a number b such that the line y = b dividesthe region bounded by the curves y = ax2 and y = k into two regions with equal area,where a, k are given positive constants.

Solution:

(a) Determine the range of values for integration by finding the intersections of the

bounding curves: solving the equations yields the points

∓√

ka, k

.

(b) Determine the portion of the full area which is below the line y = b. We begin byrepeating the calculation of the preceding part: the corner points have coordinates∓

√ba, b

. The area is

∫ √ ba

−√

ba

(b − ax2) dx = 2[bx − ax3

3

]√ ba

0=

43

b

√ba.

(c) As a special case of the foregoing, or by a separate calculation, we can conclude

that the area of the entire region is43

k

√ka

.

(d) The condition of the problem is that

43

b

√ba

=12· 4

3k

√ka

which is equivalent to 4b3 = k3, and implies that the line should be placed whereb = 2−

23 k.

2. [10 MARKS] The region bounded by the curves y = 5 and y = x2 − ax + b is rotatedabout the axis x = −1. Showing all your work, find the volume of the resulting solid.

Solution: Because there are constraints on the constants, I will work just one variant,with a = 3, b = 7.

Using the method of cylindrical shells: (a) To find the extremes of integration, wesolve the equations y = 5 and y = x2 − 3x + 7, obtaining (x, y) = (1, 5), (2, 5).

(b) The height of a vertical element of area which generates a cylindrical shell is,at horizontal position x, 5 − (x2 − 3x + 7) = −x2 + 3x − 2.

(c) The distance of that vertical element of area from the axis of revolution is 1+x.


(d) The volume is given by the integral

2π∫ 2

1(1 + x)(−x2 + 3x − 2) dx

(e) Evaluating the integral:

2π∫ 2

1(1 + x)(−x2 + 3x − 2) dx

= 2π∫ 2

1

(−x3 + 2x2 + x − 2

)dx

= 2π[−1

4x4 +

23

x3 +12

x2 − 2x]2

1

= 2π(−4 +

163

+ 2 − 4 +14− 2

3− 1

2+ 2

)

=5π6

Using the method of “washers” (a) To find the lowest point on the parabola, we solvex2 − 3x + 7 ≥ 0. This can be done by completing the square, or by using the

calculus to find the local minimum. We find it to be(32,

194

).

(b) The horizontal element generating the “washer” at height y extends betweenthe solutions in x to the equation y = x2 − 3x + 7; these are

x =3 ± √

4y − 192

.

(c) The volume of the “washer” at height y is, therefore,

π

1 +

3 +√

4y − 192

2

−1 +

3 − √4y − 192

2 ∆y

= 5π√

4y − 19 ∆y

(d) The volume is given by the integral 5π∫ 5

194

√4y − 19 dy.

(e) Evaluation of the integral:

5π∫ 5

194

√4y − 19 dy =

[5π · 2

3· 1

4(4y − 19)

32

]5

194

=5π6.


3. [5 MARKS] Showing all your work, use a substitution to evaluate the indefinite integral∫cosn x sin x dx, where n is a fixed, positive integer.

Solution:

(a) Use new variable u, where du = − sin x dx; one solution is u = cos x.

(b)∫

cosn x sin x dx = −∫

un du

= − un+1

n + 1+ C

= − 1n + 1

cosn+1 x + C

4. [5 MARKS] Showing all your work, evaluate the integral∫ (

xb + a +1

x2 + 1

)dx, (where

a and b are given positive integers).

Solution:

(a)∫

xb dx =xb+1

b + 1+ C1,

(b)∫

a dx = ax + C2

(c)∫

1x2 + 1

dx = arctan x + C3

(d)∫ (

xb + a +1

x2 + 1

)dx =

xb+1

b + 1+ ax + arctan x + C.

Wednesday version

1. [10 MARKS] Showing all your work, find the volume of the solid obtained by rotatingabout the line x = −1 the region bounded by y = xn and x = yn, where n is a givenpositive integer.

Solution:

Case I: n is even

Using the method of “washers”: (a) Find the intersections of the curves bound-ing the region. Solving the 2 equations yields the points (x, y) = (0, 0), (1, 1).


(b) Find the inner and outer dimensions of the washer. Since the axis of revo-lution is a vertical line, the element of area being rotated is horizontal. Forarbitrary y the farther endpoint on the element is (y

1n , y); the nearer end-

point is (yn, y). The distances of these points from the axis are, respectively1 + n√

y and 1 + yn.(c) The volume of the “washer” is, therefore,

π(−(1 + yn)2 + (1 + n

√y)2

)∆y .


π

∫ 1

0

(−(1 + y)2 + (1 + n

√y)2

)dy

= π

∫ 1

0

(2y

1n + y

2n − 2yn − y2n

)dy

= π

[2n

n + 1y

n+1n +

nn + 2

yn+2

n − 2n + 1

yn+1 − 12n + 1

y2n+1]1

0

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)


(0, 0), (1, 1).(b) Since the axis of revolution is a vertical line, the element of area being

rotated is also vertical. For arbitrary x the top endpoint on the element is(x, x

1n ); the lower endpoint is (x, xn). The length of the element is, there-

fore, x1n − xn; the distance of the element from the axis of symmetry is

1 + x.(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 + x) ·(x

1n − xn

).


2π∫ 1

0(1 + x)(x

1n − xn) dx

= 2π∫ 1

0

(x

1n − xn + x

n+1n − xn+1

)dx

= 2π(

nn + 1

− 1n + 1

+n

2n + 1− 1

n + 2

)

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)


Case II: n is oddUsing the method of “washers”: (a) Find the intersections of the curves bound-

ing the region. Solving the 2 equations yields the points (x, y) = (0, 0), (±1,±1).Here there is an issue of interpretation. The textbook usually permits theword region to apply to one that may have more than one component;some authors would not wish to apply the term in such a situation. I willfollow the textbook, and permit a region here to have two components.

(b) Find the inner and outer dimensions of the washer. Since the axis of rev-olution is a vertical line, the element of area being rotated is horizontal.But there are two kinds of elements, depending on whether y is positiveor negative. For arbitrary, positive y the farther endpoint on the element is(y

1n , y); the nearer endpoint is (yn, y). The distances of these points from

the axis are, respectively 1 + n√

y and 1 + yn. For arbitrary, negative y thenearer endpoint on the element is (y

1n , y); the farther endpoint is (yn, y).

The distances of these points from the axis are, respectively 1 + n√

y and1 + yn (both of which are less than 1).

(c) The volume of the “washer” is, therefore,

π∣∣∣−(1 + yn)2 + (1 + n

√y)2

∣∣∣ ∆y .


π

∫ 1

−1

∣∣∣−(1 + y)2 + (1 + n√

y)2∣∣∣ dy

= π

∫ 1

0

(2y

1n + y

2n − 2yn − y2n

)dy

+π

∫ 0

−1

(−2y

1n − y

2n + 2yn + y2n

)dy

= π

[2n

n + 1y

n+1n +

nn + 2

yn+2

n − 2n + 1

yn+1 − 12n + 1

y2n+1]1

0

+π

[− 2n

n + 1y

n+1n − n

n + 2y

n+2n +

2n + 1

yn+1 +1

2n + 1y2n+1

]0

−1

=2(n − 1)(3n2 + 7n + 3)π(n + 1)(n + 2)(2n + 1)

+2(n − 1)(n2 + 3n + 1)π(n + 1)(n + 2)(2n + 1)

=4(n − 1)π

n + 1.


(0, 0), (±1,±1).


(b) Since the axis of revolution is a vertical line, the element of area beingrotated is also vertical. For arbitrary, positive x the top endpoint on the el-ement is (x, x

1n ); the lower endpoint is (x, xn); for arbitrary, negative x the

bottom endpoint on the element is (x, x1n ); the upper endpoint is (x, xn).

The length of the element is, therefore,∣∣∣∣x 1

n − xn∣∣∣∣; the distance of the ele-

ment from the axis of rotation is 1 + x.(c) The volume of the cylindrical shell element of volume is, therefore,

2π(1 + x) ·∣∣∣∣x 1

n − xn∣∣∣∣ .


2π∫ 1

−1(1 + x)

∣∣∣∣x 1n − xn

∣∣∣∣ dx

= 2π∫ 1

−1

∣∣∣∣x 1n − xn + x

n+1n − xn+1

∣∣∣∣ dx

= 2π(

nn + 1

− 1n + 1

+n

2n + 1− 1

n + 2

)

+2π(

nn + 1

− 1n + 1

− n2n + 1

+1

n + 2

)

=4(n − 1)π

n + 1


sin 2tcos t

dt.

Solution:

(a) Apply a “double angle” formula:∫

sin 2tcos t

dt =

∫2 sin t cos t

cos tdt = 2

∫sin t dt .

(b) Complete the integration:

2∫

sin t dt = −2 cos t + C .

3. [10 MARKS] Showing all your work, find the area of the region bounded by the parabolay = x2, the tangent line to this parabola at (a, a2), and the x-axis, (where a is a given realnumber).

Solution: This area can be computed by integrating either with respect to y or withrespect to x.


Integrating with respect to y: (a) Since y′ = 2x, the tangent line through (a, a2) hasequation

y − a2 = 2a(x − a)⇔ y = 2ax − a2 .

(b) To integrate with respect to y we need to express the equations of the parabolaand the line in the form

x = function of y .

The branch of the parabola to the right of the y-axis is x =√

y. The line has

equation x =y

2a+

a2

.

(c) The area of the horizontal element of area at height y is(y + a2

2a− √y

)∆y.

(d) The area is the value of the integral∫ a2

0

(y + a2

2a− √y

)dy .

(e) Integration yields[

y2

4a+

ay2− 2

3y

32

]a2

0=

(14

+12− 2

3

)a3 =

112

a3 .

Integrating with respect to x: (a) As above, the tangent line is y = 2ax − a2. Itsintercept with the x-axis is at x =

a2

.

(b) The area of the vertical element of area at horizontal position x ≤ a2

is(x2 − 0

)∆x.

(c) The area of the vertical element of area at horizontal position x ≥ a2 is

(x2 − (2ax − a2)

)dx =

(x − a)2 ∆x.(d) The area of the region is the sum

∫ a2

0x2 dx +

∫ a

a2

(x − a)2 dx .

(e) Integration yields [x3

3

] a2

0+

[−(x − a)3

3

]a

a2

=a3

12.


seca x tan x dx,

where a is a constant, positive integer.

Solution:


(a) Try the substitution given by du = sec x·tan x dx, of which one solution is u = sec x.

(b) ∫seca x tan x dx =

∫ua−1 du =

ua

a+ C =

seca xa

+ C .

Some students may have integrated by sight.

Thursday version

1. [10 MARKS] Showing all your work, use the method of cylindrical shells to find thevolume generated by rotating about the axis x = b the region bounded by the curvesy =√

x − 1, y = 0, x = a, where a, b are fixed real constants. b ≥ √a − 1 .

Solution:

(a) Solve equations to determine the limits of integration. Solving x = a with y =√x − 1 yields the single point of intersection

(a,√

a − 1).

(b) The horizontal element of area at height y which generates the cylindrical shell hasleft endpoint (1 + y2, y) and right endpoint (a, y), so its length is a − (1 + y2).

(c) The distance of the horizontal element of area which generates the shell from theaxis of rotation is b − y.

(d) Set up the integral for the volume by cylindrical shells:

∫ √a−1

0(b − y)

(a − (1 + y2)

)dy .

(e) Evaluate the integral

∫ √a−1

0(b − y)

(a − (1 + y2)

)dy

=

∫ √a−1

0

(b(a − 1) − (a − 1)y − by2 + y3

)dy

=

[b(a − 1)y − a − 1

2y2 − b

3y3 +

14

y4]√a−1

0

= b(a − 1)32 − a − 1

2(a − 1) − b

3(a − 1)

32 +

14

(a − 1)2

= (a − 1)32

(23· b − 1

4

√a − 1

).


2. [5 MARKS] Showing all your work, use a substitution to evaluate the integral∫

b(1+ax)3 dx ,

where a, b are non-zero constants.

Solution:

(a) A substitution which suggests itself is u = 1 + ax, implying that du = a dx, sodx = 1

a du.

(b) ∫b

(1 + ax)3 dx =ba

∫duu3 = − b

2au−2 + C = − b

2a(1 + ax)2 + C .

3. [5 MARKS] Showing all your work, use a substitution to evaluate the indefinite integral∫t2 cos

(a − t3

)dt, (where a is a given real number).

Solution:

(a) Try the substitution u = t3.

(b) du = 3t2 dt ⇒ t2 dt = 13 du.

(c)∫

t2 cos(a − t3

)dt =

∫13

cos(a − u) du

= −13

sin(a − u) + C

= −13

sin(a − t3) + C.

(Some students may wish to employ a second substitution v = a− u. Alternatively,a better substitution for the problem would have been to take u = a − t3.)

4. [10 MARKS] Showing all your work, determine the area of the region bounded by theparabola x = y2, the tangent line to this parabola at (a2, a), and the y-axis, where a is afixed, positive real number.

Solution: The solution is analogous (under the exchange x ↔ y) to that given for Prob-lem 3 of the Wednesday quiz.


Release Date: Mounted on the Web on 05 April, 2008These draft solutions could contain errors, and they must be subject to correction. Caveat

lector!




• No calculators!




Monday version


x5 ln(20x) dx.

Solution:

(a) I will integrate by parts, setting u = ln(20x), dv = x5 dx. Then du =dxx

, v =x6

6.

(b)∫

x5 ln(20x) dx = (ln(20x)) ·(

x6

6

)− 1

6

∫x5 dx

=x6 ln(20x)

6− x6

36+ C .


1 − cot2 xcsc2 x

dx.

Solution:∫

1 − cot2 xcsc2 x

dx =

∫sin2 x − cos2 xsin2 x · csc2 x

dx

=

∫ (sin2 x − cos2 x

)dx

=

∫(− cos 2x) dx

= −12

sin 2x + C .


3. [10 MARKS] Use a substitution to transform this integral into the integral of a rationalfunction; then integrate, and express your answer in terms of x:

∫1

e−3x − e−x dx

Solution:

(a) I would try the substitution u = e−x, so du = −e−x dx = −u dx.

(b) Then∫

1e−3x − e−x dx =

∫du

u2(1 − u2)du

=

∫ 1u2 +

12

1 + u+

12

1 − u

du

= −1u

+12

ln |1 + u| − 12

ln |1 − u| + C

= −1u− 1

2ln

∣∣∣∣∣1 − u1 + u

∣∣∣∣∣ + C

= −ex − ln

√∣∣∣∣∣1 − e−x

1 + e−x

∣∣∣∣∣ + C

There are other, equivalent ways in which this last class of antiderivatives can be ex-pressed. (For example, if we define K = eC, so C = ln K, we can bring the logarithmterms together.)


x2

(4 − x2) 3

2

dx.

Solution:

(a) Use a trigonometric substitution, e.g., x = 2 cos θ, i.e., θ = arccos x2 . Then dx =

−2 sin θ dθ.

(b)∫

x2

(4 − x2) 3

2

dx = −∫

cos2 θ sin θsin3 θ

dθ

= −∫

cot2 θ dθ

= −∫ (

csc2 θ − 1)

dθ


= cot θ + θ + C

=cos θsin θ

+ θ + C

=cos θ√

1 − cos2 θ+ arccos

x2

+ C

=

x2√

1 − x2

4

+ arccosx2

+ C

=x√

4 − x2+ arccos

x2

+ C .

Tuesday version


e5x cos(2x) dx .

Solution:

(a) Use integration by parts. In this case the factors e5x and cos 2x are rendered neither“more complicated” nor “simpler” under either integration or differentiation. Twoapplications of integration by parts, with appropriate choices of functions, willyield an equation that can be solved for the value of the indefinite integral. For thefirst application take, for example, u = e5x and dv = cos 2x. Then du = 5e5x dx,and v = 1

2 sin 2x.

(b) ∫e5x cos(2x) dx = e5x · 1

2sin 2x − 5

2

∫e5x sin 2x dx . (116)

(c) Now we apply integration by parts to evaluate∫

e5x sin 2x dx, taking U = e5x anddV = sin 2x. Then dU = 5e5x dx, and V = − 1

2 cos 2x.

(d) ∫e5x sin(2x) dx = e5x ·

(−1

2cos 2x

)+

52

∫e5x cos 2x dx . (117)

(e) Combining equations (116), (117) yields∫

e5x cos(2x) dx = e5x · 12

sin 2x − 52

(−e5x · 1

2· cos 2x +

52

∫e5x sin 2x dx

)

= e5x ·(12· sin 2x +

54

cos 2x)− 25

4

∫e5x cos(2x) dx .


(f) Solving the last equation yields∫

e5x cos(2x) dx =1

29e5x · (2 sin 2x + 5 cos 2x) + C .

2. [10 MARKS] Showing all your work, use a substitution to change this integral into theintegral of a rational function; then integrate and express your solution in terms of x:

∫1 + 7ex

1 − ex dx .

Solution:

(a) Let u = ex, so du = ex dx.

(b) Then∫

1 + 7ex

1 − ex dx =

∫1 + 7u1 − u

· duu

and we proceed to expand the integrand into

a sum of partial fractions.

(c) Assuming that1 + 7u

u(1 − u)=

Au

+B

1 − u, and multiplying both sides by u(1 − u), we

obtain the identity in u 1 + 7u = A(1 − u) + Bu. Setting u = 0 and u = 1 yields theequations 1 = A and 8 = B. We can now continue integration.

(d)∫

1 + 7u1 − u

du =

∫ (1u

+8

1 − u

)du

= ln |u| − 8 ln |1 − u| + C= ln ex − 8 ln |1 − ex| + C= x − 8 ln |1 − ex| + C

which solution can be checked by differentiation.


dx(x2 + 8x + 17

)2 dx.

Solution:

(a) Completing the square of the polynomial in the denominator, we obtain

x2 + 8x + 17 =

(x +

82

)2

+

17 −(82

)2 = (x + 4)2 + 1 .

Accordingly, we can simplify the integral by taking u = x + 4, du = dx.


(b) The preceding substitution is not sufficient, however. All we obtain is∫

dx(x2 + 8x + 17

)2 dx =

∫du

(u2 + 1

)2 .

We can simplify this further by taking u = tan θ, i.e., by taking θ = arctan u, so

dθ =du

1 + u2 .

(c)∫

du(1 + u2)2 =

∫sec2 θ

sec4 θdθ

=

∫cos2 θ dθ

=

∫1 + cos 2θ

2dθ =

θ

2+

sin 2θ4

+ C

=12

arctan u +u

2(1 + u2)+ C

=12

(arctan(x + 4) +

x + 4x2 + 8x + 17

)+ C

which can be verified by differentiation. You should always verify this type ofintegration by differentiation, in order to locate silly algebra mistakes (or worse).

Wednesday version


x cos(18x) dx.

Solution:

(a) This can be solved using integration by parts. Define u = x, dv = cos(18x) dx, so

du = dx and v =sin(18x)

18.

(b)∫

x cos(18x) dx = x · sin(18x)18

−∫

sin(18x)18

=x sin(18x)

18+

cos(18x)182 + C ,

which can be verified by differentiation.



1 − sin xcos x

dx.

Solution:∫

1 − sin xcos x

dx =

∫(sec x − tan x) dx

= ln | sec x + tan x| + ln | cos x| + C= ln |(sec x + tan x) · (cos x)| + C= ln |1 + sin x| + C = ln(1 + sin x) + C .

Note that the absolute signs are not needed, since 1 + sin x cannot be negative.


π3∫

π6

ln tan x(sin x) · (cos x)

dx.

Solution:

(a) In view of the complicated nature of the integrand, one would be advised to seeka substitution that could render it more amenable. But the integrand involves bothsines and cosines. However, note that

(sin x)(cos x) = (tan x)(cos2 x) =tan xsec2 x

=tan x

tan2 x + 1.

Taking u = tan x, we have du = sec2 x dx. When x =π

3, u =

√3; when x =

π

6,

u =1√3

.

(b)

π3∫

π6

ln tan x(sin x) · (cos x)

dx

=

√3∫

1√3

ln uu

u2+1

duu2 + 1

=

√3∫

1√3

ln uu

du


(c) Now the integral looks as though it could be simplified by a substitution v = ln u,

so dv =duu

. When u =√

3, v = ln 32 ; when u = 1√

3, v = − ln 3

2 .

(d)√

3∫

1√3

ln uu

du =

∫ ln 22

− ln 32

v dv

=

[v2

2

] ln 32

− ln 32

= 0.


et√

49 − e2t dt.

Solution:

(a) Clearly a substitution of the form u = et is indicated, in order to simplify theintegrand. We find that du = et dt.

(b) We obtain∫

et√

49 − e2t dt =

∫ √49 − u2 du. Now a trigonometric substitution

is indicated. Take u = 7 sin θ — more precisely, θ = arcsin u7 (the inverse cosine

could also have been used), so du = 7 cos θ dθ:∫ √

49 − u2 du = 49∫

cos2 θ dθ.

(c)

49∫

cos2 θ dθ =492

∫(1 + cos 2θ) dθ

=492

(θ + sin θ · cos θ) + C

=492

arcsinu7

+u2

√49 − u2 + C

=492

arcsinet

7+

et√

49 − e2t

2+ C .

Thursday version


sin3 9x dx.


Solution: The integrand is an odd power of the sine function. I will substitute u = cos 9x,so du = −9 sin 9x dx.

∫sin3 9x dx =

∫ (1 − u2

) (19

)du

= −u9

+u3

27+ C

= −cos x9

+cos3 x

27+ C .


π2∫

π4

cot2 x dx. Your answer

should be simplified as much as possible; the instructors are aware that you do not havethe use of a calculator.

Solution:

(a) Recall that cot2 x = csc2 x − 1, and thatddx

cot x = − csc2 x.

(b)

π2∫

π4

cot2 x dx =

π2∫

π4

(csc2 x − 1

)dx

= [− cot x − x]π4π2.

(c) = −π2

+ 1 +π

4= 1 − π

4.

3. [10 MARKS] Showing all your work, evaluate the integral∫ √

9 − x2

xdx.

Solution:

(a) To simplify the square root, substitute x = 3 cos θ, i.e., θ = arccos x3 . Then dx =

−3 sin θ dθ.

(b)

∫ √9 − x2

xdx = −3

∫sin2 θ

cos θdθ


= −3∫

1 − cos2 θ

cos θdθ

= −3∫

(sec θ − cos θ) dθ

= −3 ln | sec θ + tan θ| + 3 sin θ + C

(c)

−3 ln | sec θ + tan θ| + 3 sin θ + C = −3 ln

∣∣∣∣∣∣∣∣∣

1 +

√1 − x2

9x3

∣∣∣∣∣∣∣∣∣+ 3

√1 − x2

9+ C

= −3 ln

∣∣∣∣∣∣∣3 +√

9 − x2

x

∣∣∣∣∣∣∣ +√

9 − x2 + C

4. [10 MARKS] Showing all your work, determine whether the following integral is con-vergent or divergent. Evaluate it if it is convergent; in such a case you are expected tosimplify your answer as much as is consistent with not having the use of a calculator:

∫ 1

06(ln 7x√

x

)dx

Solution:

(a) Let’s look first at the associated indefinite integral,∫

ln 7x√x

dx. The integrand

can be expressed as a product, in which one of the factors “simplifies” upon dif-ferentiation, while the other does not become significantly “more difficult” uponintegration. So we will integrate by parts, taking u = ln 7x, and dv = x−

12 dx. Then

du = 1x , and we may take v = 2

√x.

(b)∫

ln 7x√x

dx = (ln 7x)(2√

x) − 2∫

dx√x

= (ln 7x)(2√

x) − 4√

x + C .

(c) The integrand is not defined at x = 0 in the given interval of integration. By thedefinition of an improper integral, we have

∫ 1

0

ln 7x√x

dx = lima→0+

∫ 1

a

ln 7x√x

dx


= lima→0+

[(ln 7x)(2

√x) − 4

√x]1

a

= lima→0+

[(2 ln 7 − 4) − √a · (2 ln(7a) − 4)

]

= 2 ln 7 − 4 − lima→0+

(ln 7a)(2√

a)

(d) The limit can be expressed as that of a ratio where numerator and denominator bothbecome infinite. Thus l’Hospital’s Rule may be used:

lima→0+

(ln 7a)(2√

a) = 2 lima→0+

ln 7a

a−12

= 2 lima→0+

1a

− 12a−

32

= 2 lima→0+

(−2√

a) = 0.

Thus the original integral is convergent, and its value is 6(2 ln 7 − 4).

Problems not used

1. [10 MARKS] Make a substitution to express the integrand as a rational function, andthen evaluate the integral. ∫ 4

1

√x

x − 16dx .

Solution:

(a) Start by substituting u =√

x, so x = u2, dx = 2u du. When x = 1, u = 1; whenx = 4, u = 2.

(b)∫ 4

1

√x

x − 16dx =

∫ 2

1

2u2

u2 − 36du

=

∫ 2

1

(2 +

72u2 − 36

)du

=

∫ 2

1

(2 +

6u − 6

− 6u + 6

)du

=[u2 + 6 ln |u − 6| − 6 ln |u + 6|

]2

1

= 3 + 6 ln7

10.



Release Date: Mounted on the Web on Wednesday, April 9th, 2008

These are draft solutions that were prepared when the quizzes were being designed. It was in-tended that Teaching Assistants would consult these draft solutions when they graded their stu-dents’ quizzes, and would report any errors or omissions. As the Teaching Assistants may be-lieve that they are inhibited from communicating with the instructor who manages this course,it is not clear that the solutions have been thoroughly checked. The solutions are being releasedwith the cautionary warning, Caveat lector! — Let the reader beware! Use them at your ownrisk.

There were four different types of quizzes, for the days when the tutorials are scheduled.Each type of quiz was generated in multiple varieties for each of the tutorial sections. Theorder of the problems in the varieties was also randomly assigned. All of the quizzes had aheading that included the instructions.


• No calculators!




Monday version

1. [10 MARKS] Showing all your work, find the length of the curve y =x5

30+

12x3 (1 ≤

x ≤ 2). Simplify your answer as much as possible; the instructors are aware that you donot have the use of a calculator.

Solution:

dydx

=x4

6− 3

2x4

⇒√

1 +

(dydx

)2

=

√1 +

x8

36+

94x8 −

12

=

√x8

36+

94x8 +

12

=

√(x4

6+

32x4

)2

=

∣∣∣∣∣∣x4

6+

32x4

∣∣∣∣∣∣ .


The absolute signs may be dropped, since the given square root is a sum of positivemultiples of even powers, and must be non-negative. The length is

∫ 2

1

(x4

6+

32x4

)dx =

[x5

30− 1

2x3

]2

1

=353240

.

2. [10 MARKS] Showing all your work, find the area of the surface obtained by rotating

x =12

(y2 + 2

) 32 (7 ≤ y ≤ 10) about the x-axis. Simplify your answer as much as

possible; the instructors are aware that you do not have the use of a calculator.

Solution:dxdy

=13· 3

2·(y2 + 2

) 12 · 2y = y

√y2 + 2 .

Thus √1 +

(dxdy

)2

=√

1 + y2(2 + y2) = y2 + 2 .

The area of the surface is∫ 10

7

(1 + y2

)· (2πy) dy = 2π

[y2

2+

y4

4

]10

7

=7701π

2.

3. [10 MARKS] Showing all your work, find the area enclosed by the curve (in polarcoordinates) r = 7 + 2 sin 6θ.

Solution: The curve surrounds the pole, and is periodic with period 2π. The area may beexpressed as an integral over an interval of length 2π; for example, as

12

∫ 2π

0(7 + 2 sin 6θ)2 dθ =

12

∫ 2π

0(49 + 28 sin 6θ + 4 sin2 6θ) dθ

=12

∫ 2π

0(49 + 28 sin 6θ + 2(1 − cos 12θ)) dθ

=12

[49θ − 28

6cos 6θ + 2θ − 2

12sin 12θ

]2π

0

=12

(51)(2π) = 51π .


4. [10 MARKS] Showing all your work, find the exact length of the curve x = 6 + 3t2,y = 2 + 2t3 (0 ≤ t ≤ 1).

Solution:

dxdt

= 6t

dydt

= 6t2

dydx

=

dydtdxdt

= t

⇒√

1 +

(dydx

)2

=√

1 + t2

Arc Length =

∫ 1

0

√(dxdt

)2

+

(dydt

)2

=

∫ 1

0

√36t2 + 36t4 dt

= 6∫ 1

0t√

1 + t2 dt

= 6 · 12· 2

3

[(1 + t2

) 32]1

0= 2

[(1 + t2

) 32]1

0

= 2(2

32 − 1

)= 4√

2 − 2 .

5. [10 MARKS] Showing all your work, sum a series in order to express the followingnumber as a ratio of integers: 0.35 = 0.35353535 . . ..

Solution:

0.35 =35

100

(1 +

1100

+1

1002 +1

1003 + . . .

)

=

35100

1 − 1100

=3599

.

Tuesday version

1. [10 MARKS] Showing all your work, find the length of the curve x =13√

y · (y − 3)(49 ≤ y ≤ 64). Simplify your answer as much as possible; the instructors are aware thatyou do not have the use of a calculator.


Solution:

dxdy

=13· 3

2√

y − 33· 1

2y−

12 =

12

(√y − 1√

y

)

⇒√

1 +

(dxdy

)2

=

√1 +

14

(y +

1y− 2

)

=

√14

(y +

1y

+ 2)

=

√14

(√y +

1√y

)2

=12

∣∣∣∣∣∣√

y +1√y

∣∣∣∣∣∣ .

The absolute signs may be dropped, since the square root is non-negative. The length is

12

∫ 64

49

(√y +

1√y

)dy =

12

[23· y 3

2 + 2√

y]64

49=

[13· y 3

2 +√

y]64

49

=1723

.

2. [10 MARKS] Showing all your work, find the area of the surface obtained by rotatingthe curve x = 6 + 2y2 (0 ≤ y ≤ 3) about the x-axis.

Solution:

dxdy

= 4y

⇒√

1 +

(dxdy

)2

=√

1 + 16y2

⇒ Area =

∫ 3

0

√1 + 16y2 · 2πy dy

=π

24

[(1 + 16y2)

32]3

0

=π

24

((145)

32 − 1

).

3. [15 MARKS] Showing all your work, find the area of the region that lies inside the curver = 15 cos θ, and outside the curve r = 5 + 5 cos θ.


Solution: The first curve is a circle; the second is a cardioid whose axis of symmetryis the initial ray. If we solve the equations we find that the curves intersect at θ =

± arccos 12 = ±π3 . They also intersect at the pole, which appears on the circle when

θ = π2 , etc., and on the cardioid when θ = π, etc. The region whose area we seek lies

between the two curves when −π3 ≤ θ ≤ π3 and r is positive. Integration shows the area

to be

12

∫ π3

− π3

((15 cos θ)2 − (5 + 5 cos θ)2

)dθ

=252

∫ π3

− π3

(8 cos2 θ − 2 cos θ − 1

)dθ

=252

∫ π3

− π3(4(1 + cos 2θ) − 2 cos θ − 1) dθ

=252

∫ π3

− π3(3 + 4 cos 2θ − 2 cos θ) dθ

= 25 [3θ + 2 sin 2θ − 2 sin θ]π30

= 25(π +√

3 −√

3)

= 25π .

4. [15 MARKS] Showing all your work, find the length of the loop of the curve x = 18t −6t3, y = 18t2.

Solution:

dxdt

= 18 − 18t2

dydt

= 36t√(

dxdt

)2

+

(dydt

)2

=

√182 (

1 − t2)2+ 362t2 = 18

(1 + t2

).

We must determine where the curve crosses itself. The student was expected to show thatshe knew how to find this crossing point systematically, not just by guessing or exam-ining a rough graph. The crossing point can be found by solving for distinct parametervalues t1 , t2 the equations x = 18t1 − 8t3

1 = 18t2 − 8t32, y = 18t2

1 = 18t2. Collectingterms and factorizing yields the system of equations

(t1 − t2)(18 − 6

(t21 + t1t2 + t2

2

))= 0 ,

18 (t1 − t2) (t1 + t2) = 0 .


Since we are looking for a solution where t1 , t2, we may divide by t1− t2, which cannotequal 0, and obtain the system

3 −(t21 + t1t2 + t2

2

)= 0 ,

t1 + t2 = 0 .

From the second equation we see that t2 = −t1, and then the first equation yields 3 = t21,

so the solutions are t1 = −t2 = ±√3; we may take the loop as beginning with parametervalue −√3 and ending with parameter value +

√3. The length of the arc will be

∫ +√

3

−√318

(1 + t2

)dt

= 2∫ √

3

018

(1 + t2

)dt

since the integrand is even and the interval is symmetric around 0

= 36[t +

13

t3]√3

0= 72

√3 .

Wednesday version

1. [10 MARKS] Showing all your work, find the length of the curve y = ln sec x. Simplifyyour answer as much as possible.

Solution:

dydx

=1

sec x· (sec x tan x) = tan x

⇒√

1 +

(dydx

)2

=√

1 + tan2 x = | sec x| .

In the following integral I will drop the absolute signs because the secant is positive overthe entire interval of integration; the length is

∫ π4

− π4| sec x| dx =

∫ π4

− π4sec x dx

= [ln | sec x + tan x|] π4− π4= ln(

√2 + 1) − ln(

√2 − 1)

= ln

√2 + 1√2 − 1

= ln(√

2 + 1)2

2 − 1

= ln((√

2 + 1)2)

= 2 ln(√

2 + 1) .


It would be sufficient for a student to obtain the different of logarithms above. Thesubsequent steps simplify the argument, and would be useful if the user did not have theuse of a calculator.

2. [10 MARKS] The curve y = 3√

x (1 ≤ y ≤ 3) is rotated about the y-axis. Showing allyour work, find the area of the resulting surface.

Solution: The data are given partly in terms of x and partly in terms of y, so some care isneeded. Since the limits are given in terms of y, I will integrate with respect to y; it willbe convenient to rewrite the equation of the curve as x = y3.

dxdy

= 3y2

⇒√

1 +

(dxdy

)2

=√

1 + 9y4

Area =

∫ 3

1

√1 + 9y4 · 2πy3 dy

=

[19· 1

4· 2π · 2

3

(1 + 9y4

) 32

]3

1

=π

27

(730

32 − 10

32).

3. [10 MARKS] Showing all your work, find the area of the region enclosed by the innerloop of the curve r = 9 + 18 sin θ.

Solution: The function 9 + 18 sin θ is periodic with period 2π, so the entire curve istraced out as θ passes through an interval of that length. If, for example, we consider theinterval 0 ≤ θ ≤ 2π, we find that the curve passes through the pole only at θ = 7π

6 andat θ = 11π

6 . Between these values the smaller loop is traced out; the larger loop is tracedout, for example, for −π6 ≤ θ ≤ 7π

6 . We can find the area of the small loop by integratingbetween the appropriate limits; the area is

12

∫ 11π6

7π6

(9 + 18 sin θ)2 dθ

=812

∫ 11π6

7π6

(1 + 4 sin θ + 4 sin2 θ) dθ

=812

∫ 11π6

7π6

(1 + 4 sin θ + 2(1 − cos 2θ)) dθ

=812

∫ 11π6

7π6

(3 + 4 sin θ − 2 cos 2θ) dθ


=812

[3θ − 4 cos θ − sin 2θ]11π

67π6

=812

(11π

2− 4 cos

11π6− sin

11π3

)− 81

2

(7π2− 4 cos

7π6− sin

7π3

)

=812

11π2− 2√

3 − −√

32

− 812

7π2

+ 2√

3 −√

32

=812

(2π − 3√

3) = 81π − 243√

32

.

4. [10 MARKS] Showing all your work, find the area of the surface obtained by rotatingthe curve x = 3t − t3, y = 3t2 (0 ≤ t ≤ 4) about the x-axis.

Solution:

dxdt

= 3 − 3t2

dydt

= 6t√(

dxdt

)2

+

(dydt

)2

=

√32 (

1 − t2)2+ 62t2 = 3

(1 + t2

).

The area of the surface of revolution about the x-axis will be∫ 4

03(1 + t2

)· 2π

(3t2

)dt = 18π

∫ 4

0

(t2 + t4

)dt

= 18π[13

t3 +15

t5]4

0=

6 × 64 × 535

=20352

5.

5. [10 MARKS] Showing all your work, determine the value of c, if it is known that∞∑

n=2

(5+

c)−n = 2.

Solution: We are told that the geometric series converges; this implies that its commonratio is less than 1 in magnitude, i.e., that |5 + c| < 1, which implies that −1 < 5 + c <1, equivalently, that −6 < c < −4. The sum of the geometric series on the left is

1(5+c)2 · 1

1− 15+c

= 1(5+c)(4+c) . Equating this to 2, we obtain c2 + 9c + 18 = 0, implying that

(c + 3)(c + 6) = 0, so c = −3,−6. Of these two values, −6 lies outside of the admissibleinterval, and would yield a divergent series. Thus c can be equal only to −3.


Thursday version

1. [10 MARKS] Showing all your work, find the exact length of the polar curve r = 7e4θ

(0 ≤ θ ≤ 2π).

Solution:

drdθ

= 28e4θ

√r2 +

(drdθ

)2

=

√(72 + 282) e8θ

= 7√

17e4θ .

The length of the arc is then

7√

17∫ 2π

0e4θ dθ =

7√

174

e4θ

2π

0

=7√

174

(e8π − 1

).

2. [10 MARKS] The curve y = 4 − x2 (1 ≤ x ≤ 3) is rotated about the y-axis. Showing allyour work, find the area of the resulting surface.

Solution:

dydx

= −2x⇒√

1 +

(dydx

)2

=√

1 + 4x2

Area =

∫ 3

1

√1 + 4x2 · 2πx dx

= 2π · 14· 1

2· 2

3·[(

1 + 4x2) 3

2]3

1

=π

6

(37

32 − 5

32).

3. [10 MARKS] Showing all your work, find the area of the region enclosed by the outerloop of the curve r = 9 + 18 sin θ: this region will include the entire inner loop.

Solution: The function 9 + 18 sin θ is periodic with period 2π, so the entire curve istraced out as θ passes through an interval of that length. If, for example, we consider theinterval 0 ≤ θ ≤ 2π, we find that the curve passes through the pole only at θ = 7π

6 andat θ = 11π

6 . Between these values the smaller loop is traced out; the larger loop is traced


out, for example, for −π6 ≤ θ ≤ 7π6 . We can find the area of the large loop by integrating

between those limits, and the area will include the entire smaller loop. The area is

12

∫ 11π6

7π6

(9 + 18 sin θ)2 dθ

=812

∫ 7π6

−π6

(1 + 4 sin θ + 4 sin2 θ) dθ

=812

∫ 7π6

−π6

(1 + 4 sin θ + 2(1 − cos 2θ)) dθ

=812

∫ 7π6

−π6

(3 + 4 sin θ − 2 cos 2θ) dθ

=812

[3θ − 4 cos θ − sin 2θ]7π6− π6

=812

(7π2− 4 cos

7π6− sin

7π3

)− 81

2

(−π

2− 4 cos

π

6+ sin

π

3

)

=812

7π2

+ 2√

3 −√

32

− 812

−π2 − 2√

3 +

√3

2

= 162π +243√

32

.

4. [10 MARKS] Showing all your work, sum a series in order to express the followingnumber as a ratio of integers: 4.645 = 4.645454545 . . ..

Solution:

4.645 = 4.6 +110

45100

(1 +

1100

+1

1002 +1

1003 + . . .

)

= 4.6 +

451000

1 − 1100

=4610

+1

10· 45

99=

511110

.

5. [10 MARKS] Showing all your work, find an equation for the tangent to the curve x =

cos θ + sin 7θ, y = sin θ + cos 2θ (−∞ < θ < +∞) at the point corresponding to θ = 0.

Solution: The slope of the tangent is

dxdθ

= − sin θ + 7 cos 7θ


dydθ

= cos θ − 2 sin θ

dydx

=

dydθdxdθ

=cos θ − 2 sin θ− sin θ + 7 cos 7θ

=17

when θ = 0 .

The line through (x(0), y(0)) = (1, 1) with slope 17 has equation y − 1 =

17

(x − 1), i.e.,x − 7y = −6.

Problems prepared but not used

1. The curve x =√

c2 − y2 (0 ≤ y ≤ c2 ) is rotated about the y-axis. (c is a fixed real number.)

Showing all your work, find the area of the resulting surface.

Solution:

dxdy

=12· 1√

c2 − y2· (−2y) = − y√

c2 − y2√

1 +

(dxdy

)2

=

√1 +

y2

c2 − y2 =|c|√

c2 − y2

Area =

∫ c2

0

|c|√c2 − y2

· 2π√

c2 − y2 dy

= 2π|c|∫ c

2

0dy = πc2 .

2. Showing all your work, find the slope of the tangent line to the curve with equation in

polar coordinates r =12θ

, at the point corresponding to θ = π.

Solution:

drdθ

= −12θ2

dydx

=sin θ · dr

dθ + r · cos θ

cos θ · drdθ − r · sin θ

=− sin θ + θ cos θ− cos θ − θ sin θ

.

At the point θ = π this ratio is equal to −π.


3. Showing all your work, find the area of one of the regions bounded by the line θ = π2 and

the closed curve r = 8 + 6 sin θ.

Solution: (The actual wording of the problem referred to a figure which it is not conve-nient to include in these notes.) The region can be interpreted as being swept out by aradius vector from the pole moving between −π2 and +π

2 . The area is this

12

∫ + π2

− π2(8 + 6 sin θ)2 dθ =

∫ + π2

− π2

(32 + 48 sin θ + 18 sin2 θ

)dθ

=

∫ + π2

− π2(32 + 48 sin θ + 9(1 − cos 2θ)) dθ

=

∫ + π2

− π2(41 + 48 sin θ − 9 cos 2θ) dθ

=

[41θ − 48 cos θ − 9

2sin 2θ

]+ π2

− π2= 41π .

4. Showing all your work, find the area enclosed by the curve (in polar coordinates) r =

9 + cos 2θ.

Solution: The curve surrounds the pole, and is periodic with period 2π. The area may beexpressed as an integral over an interval of length 2π; for example, as

12

∫ 2π

0(9 + cos 2θ)2 dθ =

12

∫ 2π

0(81 + 18 cos 2θ + cos2 2θ) dθ

=12

∫ 2π

0(81 + 18 cos 2θ +

1 + cos 4θ2

) dθ

=12

[81θ + 9 sin 2θ +

12θ +

18

sin 4θ]2π

0

=1634· 2π =

163π2

.

5. Showing all your work, find the exact length of the polar curve r = 4θ2 (0 ≤ θ ≤ 2π).

Solution:

drdθ

= 8θ√

r2 +

(drdθ

)2

=√

16θ4 + 64θ2

= 4|θ|√

4 + θ2 .


Over the interval in question θ is positive, and the absolute signs may be dropped. Thelength of the arc is

∫ 2π

04θ√

4 + θs dθ =43

[(4 + θ2)

32]2π

0

=323

((1 + π2)

32 − 1

).

6. Showing all your work, find equations of the tangents to the curve x = 3t2 +4, y = 2t3 +3that pass through the point (7, 5).

Solution: We might, in error think that we need first to determine the parameter valueassociated with the given point. We would then solve the system of equations

3t2 + 4 = 72t3 + 3 = 5

to obtain t = +1. This would be an error. It happens that the given curve passes throughthe point (7, 5), but that is fortuitous: we want the tangents to pass through the point, notthe curve! And we can’t find the points of contact of the tangents directly. So let’s firstdetermine the general tangent to the curve, at the point with parameter value t.

dxdt

= 6t

dydt

= 6t2

dydx

=

dydtdxdt

=6t2

6t= t ,

so the slope of the tangent at the point (3t2 + 4, 2t3 + 3) on the curve is t; the equation ofthat tangent is y −

(2t3 + 3

)= t

(x −

(3t2 + 4

)), or

y = tx − t3 − 4t + 3 . (118)

We now impose the condition that this line pass through the point (x, y) = (7, 5), i.e., thatits equation be satisfied by (x, y) = (7, 5), obtaining t3 − 3t + 2 = 0, whose left memberfactorizes to (t − 1)2(t + 2) = 0, so the points of contact of the tangents are t = 1 andt = −2. The equations of the tangents through the given point are found by giving theparameter t these two values in equation (118):

y = x − 2 and y = −2x + 19 .


7. Showing all your work, use methods of polar coordinates to find the length of the polarcurve r = 15 sin θ

(0 ≤ θ ≤ 4π

15

).

Solution:

drdθ

= 15 cos θ√

r2 +

(drdθ

)2

= 15 .

The length of the arc is, therefore

∫ 4π15

015 dθ = 4π .


Instructions to students

1. Show all your work. Marks may not be given for answers not supported by a full solu-tion. For future reference, the form of your solutions should be similar to those shownin the textbook or Student Solutions Manual for similar problems.




Monday Versions

1. [10 MARKS] Use the Fundamental Theorem of Calculus and the chain rule to find thederivative of the function

f (x) =

∫ √x

√3

sin(t)t5 dt

Solution:


• (This step may not be shown explicitly, but it underlies the successful implemen-tation of the Chain Rule.) Introduction of an intermediate variable: If the new

variable/function is called u = u(x) =√

x, then f ′(x) =ddu

u∫

√3

sin(t)t5 dt · du

dx

• application of the Fundamental Theorem

ddu

u∫

√3

sin(t)t5 dt =

sin(u)u5 .

• completionsin(√

x)√

x5 · 12

x−12 =

sin√

x2x3


(a)∫ 0

−π/38 sec2(x) dx

Solution: [5 MARKS TOTAL]• state one antiderivative, e.g., 8 tan x• indicate that the value of the integral is the net change, 8 tan x]0

− π3• compute the final answer. Students should know the trigonometric functions

of simple submultiples of π.(b)

∫ 1

0

√x(5x2 + 4x − 5) dx

Solution: [5 MARKS TOTAL]• state one antiderivative, here the obvious method is to express as a sum of

fractional powers and to integrate each separately:∫ (

5x52 + 4x

32 − 5x

12)

dx = 5 · 27

x72 + 4 · 2

5x

52 − 5 · 2

3x

32 + C

• indicate that the value of the integral is the net change,[5 · 2

7x

72 + 4 · 2

5x

52 − 5 · 2

3x

32

]1

0

• compute the final answer correctly = 107 + 8

5 − 103


Tuesday Versions


f (x) =

∫ ex

5

√7 + ln6 t

tdt

Solution:

• (This step may not be shown explicitly, but it underlies the successful implemen-tation of the Chain Rule.) Introduction of an intermediate variable: If the newvariable/function is called u = u(x) = ex, then

f ′(x) =d

du

∫ u

5

√7 + ln6 t

tdt · du

dx


ddu

∫ u

5

√7 + ln6 t

tdt =

√7 + ln6 u

u.

• completion √7 + (ln(ex))6

ex · ex =

√7 + x6

ex · ex =√

7 + x6 .

While you may use some judgment about how much simplification you expect, I don’tbelieve it would not be appropriate to accept a composition like ln(ex) not simplified.


(a)∫ 0

−π/6[8 sec(x) tan(x) + 7 cos(x)] dx

Solution: [5 MARKS TOTAL]

• state one antiderivative, e.g., 8 sec x + 7 sin x• indicate that the value of the integral is the net change in the antiderivative,

e.g., [8 sec x + 7 sin x]0− π6


• compute the final answer correctly.

(8 + 0) −(8 · 2√

3− 7 · 1

2

)=

92− 16√

3

(b)∫ 4

1

−3x−1 + 5x + 3√x

dx

Solution: [5 MARKS TOTAL]

• state one antiderivative, here the obvious method is to express as a sum offractional powers and to integrate each separately:

∫ (−3x−1 + 5x + 3√x

)dx = −3 ·

(−2

1

)x−

12 + 5 · 2

3x

32 + 3 · 2

1x

12 + C

• indicate that the value of the integral is the net change,

[−3 ·

(−2

1

)x−

12 + 5 · 2

3x

32 + 3 · 2

1x

12

]4

1

• compute the final answer correctly

=

(62

+103· 8 + 6 · 2

)−

(6 +

103

+ 6)

=973

Wednesday Versions


f (x) =

∫ ln x

5et√

1 + t2 dt

Solution:

• (This step may not be shown explicitly, but it underlies the successful implemen-tation of the Chain Rule.) Introduction of an intermediate variable: If the newvariable/function is called u = u(x) = ln x, then

f ′(x) =d

du

∫ u

5et√

1 + t2 dt · dudx



ddu

∫ u

5et√

1 + t2 dt = eu√

1 + u2 .

• completion

eln x√

1 + (ln x)2 · 1x

=√

1 + (ln x)2

It is essential that eln x be simplified to x for full marks in this part.

2. [10 MARKS] Compute∫ π

4

− π6f (x) dx, where

f (x) =

4 sin x if x ≤ 0

5 sec x tan x if 0 < x < π2

Solution:

• decompose the interval into subintervals matching the intervals where the 2 partsof the definition apply:

∫ π4

− π6f (x) dx =

∫ 0

− π6f (x) dx +

∫ π4

0f (x) dx

• matching the different functions to the appropriate subintervals:∫ 0

− π6f (x) dx +

∫ π4

0f (x) dx =

∫ 0

− π64 sin x dx +

∫ π4

05 sec x tan x dx

• shift the constants outside of the integration:∫ 0

− π64 sin x dx +

∫ π4

05 sec x tan x dx = 4

∫ 0

− π6sin x dx + 5

∫ π4

0sec x tan x dx

• find antiderivatives for both of the 2 integrands, e.g., − cos x and sec x

• indicate that the value of each integral is the net change,

4[− cos x]0− π6 + 5[sec x]

π40

• correctly complete the computations

4(− cos 0 + cos

π

6

)+5

(sec

π

4− sec 0

)= 4

−1 +

√3

2

+5(√

2−1) = 2√

3+5√

2−9


Thursday Versions


f (x) =

∫ 7x

3x

√8 + 9t2

tdt

Solution:

• The integral must be split into two, at a convenient place, each integral with onefixed and one variable limit; note that the point where the integral is split CANNOTBE 0, since the integrand is undefined there:

f (x) =

∫ 7x

3x

√8 + 9t2

tdt =

∫ 1

3x

√8 + 9t2

tdt +

∫ 7x

1

√8 + 9t2

tdt

• one integral must be reversed so that the dependence on x is in the upper limit:

f (x) = −∫ 3x

1

√8 + 9t2

tdt +

∫ 7x

1

√8 + 9t2

tdt

• differentiate each of the integrals separately, using the Fundamental Theorem, andmultiply by the factor of the form du

dx from the Chain Rule (see problems on earlierversions)

ddx

f (x) = −√

8 + 9(3x)2

3x· d(3x)

dx+

√8 + 9(7x)2

7x· d(7x)

dx

• completion

−√

8 + 9(3x)2

3x· 3 +

√8 + 9(7x)2

7x· 7 =

−√

8 + 81x2 +√

8 + 9(49)x2

x

2. [10 MARKS] Compute∫ √

3

0f (x) dx, where

f (x) =

3x if 0 ≤ x ≤ 16

1+x2 if x > 1

Solution:


• for decomposing the interval into subintervals matching the intervals where the 2parts of the definition apply:

∫ √3

0f (x) dx =

∫ 1

0f (x) dx +

∫ √3

1f (x) dx

• for matching the different functions to the appropriate subintervals:

∫ √3

0f (x) dx =

∫ 1

03x dx +

∫ √3

1

61 + x2 dx

• for shifting constants outside of the integration:

∫ 1

03x dx +

∫ √3

1

61 + x2 dx =

32

∫ 1

02x dx + 6

∫ √3

1

11 + x2 dx

• for finding antiderivatives for both of the 2 integrands, e.g., x2 and arctan x

• for indicating that the value of each integral is the net change,

32

[x2]10 + 6[arctan x]

√3

1

• for correctly completing the computations

32

[x2]10+6[arctan x]

√3

1 =32

(1−0)+6(arctan

√3 − arctan 1

)=

32

+6(π

3− π

4

)=

32

+π

2




2. In your folded answer sheet you must enclose this question sheet: it will be returnedwith your graded paper. WITHOUT THIS SHEET YOUR QUIZ WILL BE WORTH 0.All submissions should carry your name and student number.




Monday Versions


(a)∫

sec2 x1 + 6 tan x

dx

(b)∫ 9/2

1/2

e√

2x

√2x

dx

Solution:

(a) [4 MARKS] for this indefinite integral

• [1 MARK] for stating the substitution• [1 MARK] for rewriting the indefinite integral in terms of the new variable• [1 MARK] for finding an antiderivative in terms of the new variable• [1 MARK] for restating the antiderivative in terms of the original variable

(b) [6 MARKS] for this definite integral

• [1 MARK] for stating the substitution• [3 MARKS] for transforming the definite integral, including the upper and

lower limit• [1 MARK] for finding an antiderivative• [1 MARK] for the final answer

(Some students may, instead, find an antiderivative [4 MARKS] and then find thenet change [2 MARKS].)

2. [10 MARKS] Compute the volume of the solid of revolution about the x-axis obtainedby revolving the region bounded by the x-axis, the lines x =

π

6and x =

π

3, and the curve

y =√

4 cos x.

Solution: It was intended that students solve this problem using the “Method of Wash-ers”. A solution using Cylindrical Shells would certainly be acceptable, but would bemore difficult, as students do not yet know how to integrate arccos y, and may not havemastered integration by parts. If they complete part of such a solution, allocate the markssimilarly to the scheme for Washers.

• [4 MARKS] for the integrand


• [2 MARKS] for the limits of integration

• [2 MARKS] for finding an antiderivative

• [2 MARKS] for completing the integration.

3. [10 MARKS] Let R be the region in the xy-plane bounded by the x-axis, the lines x = π4

and x = π2 , and the curve y = 9 sin x. Compute the volume of the solid of revolution

obtained by revolving the region R about the y-axis. Hint: Use the method of cylindricalshells.

Solution:

• [3 MARKS] for determining the integrand correctly

• [2 MARKS] for determining the limits of integration correctly

• [4 MARKS] for applying integration by parts and correctly determining the fullantiderivative

• [1 MARK] for apparently completing the integration correctly

Tuesday Versions


(a)∫

sec x tan x−7 − 8 sec x

dx

(b)

∫ e4

e

cos(9 ln x)x

dx

Solution:

(a) [4 MARKS] Same scheme as for Problem 1(a) on Monday Versions.

(b) [6 MARKS] Same scheme as for Problem 1(b) on Monday Versions.

2. [10 MARKS] Compute the volume of the solid of revolution about the x-axis obtainedby revolving the region bounded by the lines y = 1, x = −π

4, x =

π

4, and the curve

y = 3 sec x.

Solution: Same scheme as for Problem 2 on Monday Versions.


3. [10 MARKS] Let R be the region in the xy-plane bounded by the x-axis, the lines x = 1and x = 5, and the curve y = ln(5x). Compute the volume of the solid of revolutionobtained by revolving the region R about the y-axis. Hint: Use the method of cylindricalshells.

Solution:

• [3 MARKS] for determining the integrand correctly

• [2 MARKS] for determining the limits of integration correctly

• [4 MARKS] for applying integration by parts and correctly determining the fullantiderivative

• [1 MARK] for apparently completing the integration correctly

Wednesday Versions


(a)∫

9x + 2√9x2 + 4x

dx

(b)∫ 1

0

e3x

e6x + 1dx

Solution:



2. [10 MARKS] Compute the volume of the solid of revolution about the y-axis obtainedby revolving the region bounded by the y-axis, the lines y = ln 3 and y = ln 4, and thecurve y = ln

(x2

).

Solution: Note that this is a solid of revolution about the y-axis. To use the Methodof Washers, which is intended, students will have to rewrite the equation of the curvein the form x = 2ey. A correct solution using the Method of Cylindrical Shells wouldcertainly be acceptable. Follow the same grading scheme as shown above for Question2 of Monday Versions.


3. [10 MARKS] Find the average value of the function

x√

8 + x

on the interval [−8,−4].

Solution:

• [2 MARKS] for setting up the integral correctly, with correct integrand and limitsof integration

• [7 MARKS] for the evaluation of this integral — more than one method is feasible:

Integration by Parts: – [2 MARKS] for a correct selection of u and dv– [2 MARKS] for determining du and v– [2 MARKS] for applying integration by parts– [1 MARK] for the integration of

∫v du.

Substitution: – [2 MARKS] for selection of an appropriate substitution u =

u(x)– [2 MARKS] for transforming the integrand correctly into terms of u– [2 MARKS] for correctly changing the limits of integration into terms of

the new variable– [1 MARKS] for correctly evaluating the new definite integral

• [1 MARK] for dividing the weighted integral by the length of the interval, andobtaining the final answer.

Thursday Versions


(a)∫

x3

√x2 + 4

dx

(b)∫ 1

0

x + 2x2 + 1

dx

Solution:




2. [10 MARKS] Compute the volume of the solid of revolution about the y-axis obtained

by revolving the region bounded by the lines x = 1 and y = 2 and the curve y =3x

.

Solution: Note that this is a solid of revolution about the y-axis; moreover, it has a holein the middle. It was intended that students solve this problem using the “Method ofWashers”. However, the Method of Cylindrical Shells would be acceptable, and is notdifficult.

• [4 MARKS] for the integrand

• [2 MARKS] for the limits of integration; note that the data given are partly in termsof x-coordinates and partly in terms of y-.

• [2 MARKS] for finding an antiderivative

• [2 MARKS] for completing the integration.

3. [10 MARKS] Find the average value of the function

x√3 + x

on the interval [−2, 1].

Solution:

• [2 MARKS] for setting up the integral correctly, with correct integrand and limitsof integration

• [7 MARKS] for the evaluation of this integral — more than one method is feasible:

Integration by Parts: – [2 MARKS] for a correct selection of u and dv– [2 MARKS] for determining du and v– [2 MARKS] for applying integration by parts– [1 MARK] for the integration of

∫v du.

Substitution: – [2 MARKS] for selection of an appropriate substitution u =

u(x)– [2 MARKS] for transforming the integrand correctly into terms of u– [2 MARKS] for correctly changing the limits of integration into terms of

the new variable– [1 MARKS] for correctly evaluating the new definite integral

• [1 MARK] for dividing the weighted integral by the length of the interval, andobtaining the final answer.








Monday Versions

1. [10 MARKS]

(a) Use integration by parts to compute the integral x3/5 ln x dx.

(b) Make a substitution and then use integration by parts to compute the integral∫

x3ex2dx .

Solution:

(a) [4 MARKS]

• [2 MARKS] for a correct choice of u and dv and correctly determining du andv

• [2 MARKS] for correctly implementing the selection of u and v and complet-ing the integration correctly

(b) [6 MARKS]

• [1 MARK] for correctly implementing an appropriate substitution• [2 MARKS] for a correct choice of u and dv and correctly determining du and

v• [2 MARKS] for correctly implementing the selection of u and v and complet-

ing the integration in terms of the new variable• [1 MARK] for expressing the final, correct answer in terms of the original

variable


2. [10 MARKS] Use a trigonometric substitution to compute∫

1

(√

25 − x2)3dx . Verify

your answer by differentiating it!

Solution:

• [2 MARKS] for selecting a correct substitution (either a sine, a cosine will do).(Strictly speaking, the substitution should be expressed first in terms of an inversesine or inverse cosine, but it is common practice not to make that step explicit, soone can’t expect students to be better than the textbooks.)

• [2 MARKS] for implementing the substitution correctly and writing the integral interms of the the square of the secant or cosecant.

• [2 MARKS] for correctly integrating in terms of the new variable

• [2 MARKS] for transforming the integral into terms of the original variable x.

• [2 MARKS] for correctly differentiating the antiderivative and thereby obtainingthe original integrand

item [10 MARKS] Find the arc length of the parameterized curve x(t) = e2t +e−2t, y(t) =

4t − 2 . for t between 0 and12

.

Solution: Grading instructions:

• [1 MARKS] for an integral of the correct form

• [4 MARKS] for correctly computing the derivatives of x and y

• [3 MARKS] for correctly finding an antiderivative

• [2 MARKS] for correctly completing the evaluation of the integral.

Tuesday Versions

1. [10 MARKS]

(a) Use integration by parts to compute the integral∫

x sec2(5x) dx

(b) Make a substitution and then use integration by parts to compute the integral∫

x3 cos(x2) dx


Solution:

(a) [4 MARKS]

• [2 MARKS] for a correct choice of u and dv and correctly determining du andv

• [2 MARKS] for correctly implementing the selection of u and v and complet-ing the integration correctly

(b) [6 MARKS]

• [1 MARK] for correctly implementing an appropriate substitution• [2 MARKS] for a correct choice of u and dv and correctly determining du and

v• [2 MARKS] for correctly implementing the selection of u and v and complet-

ing the integration in terms of the new variable• [1 MARK] for expressing the final, correct answer in terms of the original

variable

2. [10 MARKS] Use a trigonometric substitution to compute∫

x2

√4 − x2

dx . Verify your

answer by differentiating it!

Solution:

• [2 MARKS] for selecting a correct substitution (either a sine, a cosine will do).(Strictly speaking, the substitution should be expressed first in terms of an inversesine or inverse cosine, but it is common practice not to make that step explicit, soone can’t expect students to be better than the textbooks.)

• [2 MARKS] for implementing the substitution correctly and writing the integral interms of the the square of the secant or cosecant.

• [2 MARKS] for correctly integrating in terms of the new variable

• [2 MARKS] for transforming the integral into terms of the original variable x. Thisantiderivative does not include any term with plus/or/minus: there are no ambigu-ities of signs! If a student shows an ambiguity, this means he has not properlycompleted the differentiation of the next part, since only one of the signs will yieldthe correct derivative.

• [2 MARKS] for correctly differentiating the antiderivative and thereby obtainingthe original integrand

3. [10 MARKS] Find the arc length of the parameterized curve x(t) = −3 + e2t cos t, y(t) =

5 + e2t sin t for t between 0 and12

.







Wednesday Versions

1. [10 MARKS] Compute the integral∫ e9π

1cos(ln(x)) dx. Hint: A solution by integration

by parts could begin from the observation that cos(ln(x)) = 1 · cos(ln(x)). You could alsoapply integration by parts after making a substitution.

Solution: There appear to be several ways of attacking this problem, but the attacks willrequire 2 applications of integration by parts, followed by the solving of an equation.

Applying Integration by Parts immediately: • [1 MARK] for a correct selectionof u and dv for the first integration by parts

• [1 MARKS] for correctly determining du and v• [2 MARKS] for correctly applying integration by parts and expressing the

given integral as the value of uv] minus a second integral, which will thenrequire a second application of integration by parts

• [1 MARK] for a correct selection of U and dV for the second integration byparts

• [1 MARKS] for correctly determining dU and V• [2 MARKS] for correctly applying integration by parts and expressing the

original integral as a sum of [uv + UV] minus the same integral• [2 MARKS] for solving the equation for the desired integral and completing

all calculations apparently correctly

Preceding Integration by Parts by a Substitution: • [0 MARKS] for selecting a cor-rect substitution, and implementing that substitution correctly both in the inte-grand and the limits of integration, so that the integral is now written in a formwhere the use of integration by parts is well indicated.

• [1 MARK] for a correct selection of u and dv for the first integration by parts• [1 MARKS] for correctly determining du and v• [2 MARKS] for correctly applying integration by parts and expressing the

given integral as the value of uv] minus a second integral, which will thenrequire a second application of integration by parts


• [1 MARK] for a correct selection of U and dV for the second integration byparts

• [1 MARKS] for correctly determining dU and V• [2 MARKS] for correctly applying integration by parts and expressing the

original integral as a sum of [uv + UV] minus the same integral• [2 MARKS] for solving the equation for the desired integral and completing

all calculations apparently correctly

2. [10 MARKS] Compute∫

13(x + 3)(x2 + 4)

dx .

Solution:

• [2+2 MARKS] for correctly factorizing the denominator and expressing the needto expand the function into a sum of 2 partial fractions, one with a linear denomi-nator, the other having a general numerator of degree 1 and denominator of degree2. Reserve a full 2 MARKS for the numerator of the fraction with the quadraticdenominator.

• [3 MARKS] for determining correctly the 3 undetermined constants

• [3 MARKS] for completing the integration correctly

3. [10 MARKS] Find the surface area of the solid of revolution obtained by revolving thegraph of the parametric curve x(t) = 2t − 3, y(t) = t2 − 3t − 1, 0 ≤ t ≤ 3/2 about they-axis.






Thursday Versions

1. [10 MARKS] Use a substitution and then integration by parts to compute the integral∫ 6

3

e3/x

x3 dx .

Solution:


• [5 MARKS] for selecting an appropriate substitution to simplify the integral, andcorrectly changing the integrand and the limits of integration. Some students mayarrive at the ultimate substitution through several composed substitutions. If theydo not succeed in completing a substitution and thus cannot begin seriously inte-gration by parts, you should grade their work out of a maximum of 5 MARKS.

• [2 MARKS] for a correct choice of u and dv and correctly determining du and v

• [2 MARKS] for correctly implementing the selection of u and v and completingthe integration in terms of the new variable

• [1 MARK] for expressing the final, correct answer in terms of the original variable

2. [10 MARKS] Compute∫

25x(x + 5)2 dx .

Solution:

• [2+2 MARKS] for correctly factorizing the denominator and expressing the needto expand the function into a sum of either 2 or 3 partial fractions:

– One expansion would have one partial fraction with a linear denominator, theother having a general numerator of degree 1 and denominator of degree 2consisting of the 2nd power of a linear function.

– Alternatively the function could be expressed as the sum of 3 partial fractions:the first being associated with the factor of multiplicity 1; the second havingin its denominator the first power of the other linear factor; and the last havingin its denominator the second power of the other linear factor

In both cases there are 3 constants to be determined. Reserve a full 2 MARKS forthe numerator of the fraction with the quadratic denominator, or, alternatively, the2 terms associated with that linear factor of the denominator.

• [3 MARKS] for determining correctly the 3 undetermined constants

• [3 MARKS] for completing the integration correctly

3. [10 MARKS] Find the surface area of the solid of revolution obtained by revolving the

graph of the parametric curve x(t) =23

(t + 4)3/2, y(t) = 2√

t + 4, −4 ≤ t ≤ 0 about they-axis.







F Final Examinations from Previous Years

F.1 Final Examination in Mathematics 189-121B (1996/1997)1. [4 MARKS] Find the derivative of the function F defined by

F(x) =

∫ x4

x2sin√

t dt .

2. [4 MARKS] Evaluate∫ π

−π2

f (x) dx , where

f (x) =

cos x, −π

2 ≤ x ≤ π3

3πx + 1, π

3 < x ≤ π .

3. [7 MARKS] Evaluate∫

x sin3 x2 cos x2 dx .


(x5 + 4−x) dx .

5. [10 MARKS] Calculate the area of the region bounded by the curves x = y2 andx − y = 2 .

6. [10 MARKS] The region bounded by f (x) = 4x − x2 and the x-axis, betweenx = 1 and x = 4 , is rotated about the y-axis. Find the volume of the solid that isgenerated.


x ln x dx .


sin2 x cos5 x dx .

9. [6 MARKS] Determine the partial fraction decomposition of the following ratio of poly-nomials:

x5 + 2x2 − 1

.

10. [4 MARKS] Determine whether or not the following sequence converges as n→ ∞ .If it does, find the limit: (

1 +xn

)3n.


11. [4 MARKS] Determine the following limit, if it exists:

limx→0+

√x√

x + sin√

x.

12. [6 MARKS] Determine whether the series∞∑

k=2

ke−k2converges or diverges.

13. [6 MARKS] Test the following series for

(a) absolute convergence,

(b) conditional convergence.

∞∑

k=10

(−1)k

√k(k + 1)

.

14. [10 MARKS] Find the area of the region that consists of all points that lie within thecircle r = 2 cos θ , but outside the circle r = 1 .

15. [10 MARKS] Determine the length of the curve

r = 5(1 − cos θ) , (0 ≤ θ ≤ 2π) .

F.2 Final Examination in Mathematics 189-141B (1997/1998)1. [10 MARKS]

(a) Sketch the region bounded by the curves

y = x2 and y = 3 + 5x − x2 .

(b) Determine the area of the region.

2. [10 MARKS] The triangular region bounded by the lines

y = x , y =32− x

2, and y = 0

is revolved around the line y = 0. Determine the volume of the solid of revolution whichis generated.


3. [10 MARKS] Find the length of the curve y =x2

2− ln 4√x from x = 1 to

x = 2 .

4. [5 MARKS] Determine, at x = 12 , the value of the function sin−1 x and the slope of its

graph.

5. [5 MARKS] Evaluate limx→2

x3 − 8x4 − 16

.

6. [5 MARKS] Showing all your work, evaluate limx→0+

xx .


x3e−x2dx .


x3 − 1x3 + x

dx .


x3

√1 − x2

dx , where |x| < 1 .

10. [10 MARKS] Find the area of the region that lies within the limacon r = 1 + 2 cos θand outside the circle r = 2 .

11. [5 MARKS] Showing all your work, obtain a second-degree Taylor polynomial for

f (x) =

∫ x

0et(1−t)dt at x = 0 .

12. [5 MARKS] Showing all your work, determine whether the following infinite seriesconverges or diverges. If it converges, find its sum.

∞∑

n=0

3n − 2n

4n

13. [5 MARKS] Showing all your work, determine whether or not the following series con-verges:

∞∑

n=1

21n

n2

14. [5 MARKS] Showing all your work, determine whether the following series converges:

∞∑

n=1

1n · 2n


F.3 Supplemental/Deferred Examination in Mathematics 189-141B (1997/1998)1. [10 MARKS]

(a) Sketch the region bounded by the curves

y =8

x + 2and x + y = 4 .

(b) Determine the area of the region.

2. [10 MARKS] The triangular region bounded by the lines

y = x , y =32− x

2, and y = 0

is revolved around the line y = 0. Determine the volume of the solid of revolution whichis generated.

3. [10 MARKS] Find the area of the surface of revolution generated by revolving the curve

y =12

(ex + e−x) (0 ≤ x ≤ 1)

about the x-axis.

4. [5 MARKS] Determine, at x = 12 , the value of the function cos−1 x and the slope of its

graph.

5. [10 MARKS] Evaluate limx→2

x − 2 cos πxx2 − 4

.

6. [5 MARKS] Evaluate limx→∞

(cos

1x2

)x4

.


e2x

1 + e4x dx .


x2 cos x dx .


x3 − 1x3 + x

dx .

10. [10 MARKS] Evaluate∫ √

a2 − u2 du , where |u| < a.


11. [10 MARKS] Find the area of the region that lies within the limacon r = 1 + 2 cos θand outside the circle r = 2 .

12. [5 MARKS] Showing all your work, obtain a second-degree Taylor polynomial for

f (x) =

∫ x

0es(1−s)ds at x = 0 .

13. [5 MARKS] Showing all your work, determine whether the following infinite seriesconverges or diverges. If it converges, find its sum.

∞∑

n=0

1 + 2n + 3n

5n

14. [5 MARKS] Showing all your work, determine whether or not the following series con-verges.

∞∑

n=1

ln nn

15. [5 MARKS] Showing all your work, determine whether the following series convereges.

∞∑

n=1

n2 + 1en(n + 1)2

F.4 Final Examination in Mathematics 189-141B (1998/1999)1. [8 MARKS] Find the area of the region bounded by the curves y2 = x and (y−1)2 = 5−x.

2. [8 MARKS] Find the volume of the solid of revolution generated by revolving about theline x = 1 the region bounded by the curve (x−1)2 = 5−4y and the line y = 1 .

3. [8 MARKS] Find the volume of the solid generated by revolving about the line x = 0the region bounded by the curves

y = sin xy = −2x = 0

and x = 2π .

4. [8 MARKS] Find the area of the surface obtained by revolving the curve y = x2 (0 ≤x ≤ √2) about the y-axis.


5. Define the function F by F(x) =

x∫

0

et3dt .

(a) [4 MARKS] Showing all your work, explain clearly whether or not the followinginequalities are true.

e < F(e) < ee3+1 .

(b) [4 MARKS] Determine the value ofddx

F(x3) at each of thefollowing points:

i. at x = 0 .ii. at x = 2 .

6. [4 MARKS] Showing all your work, evaluate∫

sin3 πx dx .


x2e−x dx .


x − 1x3 − x2 − 2x

dx .


x3 + x2 + x − 1x2 + 2x + 2

dx .

10. [8 MARKS] Find the area of the region inside the curve r = 3 sin θ and outside the curver = 2 − cos θ.

11. Showing all your work, determine whether each of the following integrals is convergentor divergent:

(a) [4 MARKS]

∞∫

0

sin x dx .


(b) [4 MARKS]

2∫

0

dx1 − x2 .

12. Showing all your work, determine whether each of the following sequences is convergentor divergent.

(a) [4 MARKS]n sin

π

n

(b) [4 MARKS](2 n + 1) e−n

13. Showing all your work, determine whether each of the following infinite series is con-vergent or divergent:

(a) [4 MARKS]∞∑

n=1

14n3 .

(b) [4 MARKS]∞∑

n=1

(1n

+1n2

).

14. Showing all your work, determine whether each of the following series is convergent,divergent, conditionally convergent and/or absolutely convergent.

(a) [4 MARKS]∞∑

n=1

(−1)n n + 2n(n + 1)

.

(b) [4 MARKS]∞∑

n=1

(−1)n cos nn2 .

F.5 Supplemental/Deferred Examination in Mathematics 189-141B (1998/1999)1. [8 MARKS] Find the area of the region bounded by the curves y2 = x and y = 6 − x.

2. [8 MARKS] Find the volume of the solid of revolution generated by revolving about theline x = 0 the region bounded by the curve y = 4 − x2 and the lines x = 0 andy = 0 .

3. [8 MARKS] Find the volume of the solid generated by revolving about the line x = 0the region bounded by the curves

y = sin xy = 2x = 0

and x = 2π .


4. [8 MARKS] Find the area of the surface obtained by revolving the curve y = −x2 (0 ≤x ≤ √2) about the y-axis.

5. Define the function F by F(x) =

x∫

0

sin10 t dt .

(a) [4 MARKS] Showing all your work, explain clearly whether or not the followinginequalities are true.

0 < F(e) < e .

(b) [4 MARKS] Determine the value ofddx

F(x) at each of thefollowing points:

i. at x = 0 .ii. at x =

π

2.


x5e−x2dx .


x3 − x2 + x + 1x2 − 2x + 2

dx .

8. [8 MARKS] Find the area of the region inside the curve r = 6 sin θ and outside the curver = 4 − 2 sin θ.

9. Showing all your work, determine whether each of the following integrals is convergentor divergent:

(a) [4 MARKS]

∞∫

0

cos x dx .

(b) [4 MARKS]

4∫

0

dx4 − x2 .

10. Showing all your work, determine whether each of the following sequences is convergentor divergent.


(a) [4 MARKS]n sin

π

n

(b) [4 MARKS](2 n + 1) e−n

11. Showing all your work, determine whether each of the following infinite series is con-vergent or divergent:

(a) [4 MARKS]∞∑

n=1

14n5 .

(b) [4 MARKS]∞∑

n=1

(1n− 1

n3

).

F.6 Final Examination in Mathematics 189-141B (1999/2000)1. [11 MARKS] Find the area of the region bounded by the curves x = y2 and x =

−y2 + 12y − 16 .

2. [11 MARKS] Let C denote the arc of the curve y = cosh x for −1 ≤ x ≤ 1 . Findthe volume of the solid of revolution generated by revolving about the line x = −2

the region bounded by C and the line y =e2 + 1

2e.

3. (a) [5 MARKS] Showing all your work, evaluate

∫ 92

32

√6t − t2 dt .

(b) [6 MARKS] Showing all your work, evaluate

∫ 3π4

π4

√1 − sin u du .

4. (a) [7 MARKS] Showing all your work, determine a reduction formula which ex-

presses, for any integer n not less than 2, the value of∫

xn sin 2x dx in

terms of∫

xn−2 sin 2x dx.

(b) [4 MARKS] Use your reduction formula to determine the indefinite integral∫

x2 sin 2x dx.



8x2 − 21x + 6(x − 2)2(x + 2)

dx .

6. [11 MARKS] Find the area of the region inside the curver = 1 + cos θ and outside the curve r = 1 − cos θ .

7. [11 MARKS] Determine whether the following integral is convergent or divergent. If itis convergent, find its value. Show all your work.

∫ 3

0

1

(x − 1)45

dx

8. [11 MARKS] Showing all your work, determine whether the following infinite series is

convergent or divergent:∞∑

n=1

n! e−(n − 1)2.

9. Showing all your work, determine whether each of the following series is convergent,divergent, conditionally convergent and/or absolutely convergent.

(a) [6 MARKS]∞∑

n=1

(−1)n(√

n + 2 − √n)

.

(b) [6 MARKS]∞∑

n=1

(−1)n nln

(n2) .

F.7 Supplemental/Deferred Examination in Mathematics 189-141B (1999/2000)1. [11 MARKS] Determine the area of the region bounded by the curves y = x4 and

y = 2 − x2 .

2. [11 MARKS] Determine the volume of the solid generated by rotating the region boundedby the curves y = 2x2 and y2 = 4x around the x-axis.

3. Evaluate the integrals:

(a) [5 MARKS]∫

x7

√1 − x4

dx .

(b) [6 MARKS]∫

x2

√4 − x2

dx .


4. [11 MARKS] Showing all your work, find∫ π/2

0e2x sin 3x dx .

5. [11 MARKS] Determine∫

6x3 − 18x(x2 − 1)(x2 − 4)

dx .

6. [11 MARKS] Find the area of the region inside the curve r = 2+2 sin θ and outsider = 2 .

7. [11 MARKS] Determine whether the following improper integral converges:∫ 1

0

ln xx2 dx .

8. [11 MARKS] Showing all your work, determine whether the following infinite series

converges:∞∑

n=1

1√15n3 + 3

.

9. Showing all your work, determine, for each of the following series, whether it is conver-gent, divergent, conditionally convergent and/or absolutely convergent.

(a) [6 MARKS]∞∑

n=1

(−1)n ln nn

.

(b) [6 MARKS]∞∑

n=1

cos nπn

.

F.8 Final Examination in Mathematics 189-141B (2000/2001)1. Showing all your work, determine, for each of the following infinite series, whether or

not it converges.

(a) [3 MARKS]∞∑

i=1

nn3 + 1

.

(b) [3 MARKS]∞∑

n=1

ln( n3n + 1

).

(c) [6 MARKS]∞∑

n=2

(−1)n(3n + 1)4

5n .

2. [12 MARKS] Determine the volume of the solid of revolution generated by revolvingabout the y-axis the region bounded by the curves

y = e−x2,


y = 0 ,x = 0 ,x = 1 .

3. [12 MARKS] Determine the area of the surface of revolution generated by revolvingabout the x-axis the curve

y = cos x ,(0 ≤ x ≤ π

6

).

[Hint: You may wish to make use of the fact that

2∫

sec3 θ dθ = sec θ tan θ + ln | sec θ + tan θ| + C .]

4. [12 MARKS] Find the area that is inside the circler = 3 cos θ and outside the curve r = 2 − cos θ .


x(x − 1)(x2 + 4)

dx .

6. For the curve given parametrically by x = t3 + t2 + 1 , y = 1 − t2 , determine

(a) [6 MARKS] The equation of the tangent line at the point(x, y) = (1, 0) , written in the form y = mx + b , where m and b are con-stants;

(b) [6 MARKS] the value ofd2ydx2 at the point (x, y) = (1, 0) .

7. (a) [10 MARKS] Use integration by parts to determine the value of∫

ex cos x dx .

(b) [4 MARKS] Evaluate∫ 0

−∞ex cos x dx .

8. [12 MARKS] Find the area of the region bounded by the curves y = x2 − 4 andy = −2x2 + 5x − 2 .


F.9 Supplemental/Deferred Examination in Mathematics 189-141B (2000/2001)1. (a) [6 MARKS] Showing all your work, find F′(1) when

F(t) =

∫ 2t

1

xx3 + x + 7

dx .

(b) [6 MARKS] Showing all your work, evaluate∫ 6

0|x − 2| dx .

2. Showing all of your work, evaluate each of the following integrals:

(a) [4 MARKS]∫

x + 1√9 − x2

dx;

(b) [4 MARKS]∫

12x3 + x

dx;

(c) [4 MARKS]∫

sin2 2x cos2 2x dx;

(d) [4 MARKS]∫

ln x dx

3. [15 MARKS] Showing all your work, find the area of the region bounded below by the

line y =12

, and above by the curve y =1

1 + x2 .

4. [15 MARKS] Showing all your work, find the volume generated by revolving about they-axis the smaller region bounded by the circle x2 + y2 = 25 and the line x = 4 .

5. Showing all your work,

(a) [2 MARKS] sketch the curve r = 1 − sin θ ;

(b) [6 MARKS] find the length of the portion of the curve that lies in the region givenby r ≥ 0 , −π

2≤ θ ≤ π

2;

(c) [5 MARKS] find the coordinates of the points on the curve where the tangent lineis parallel to the line θ = 0 .

6. For each of the following integrals, determine whether it is convergent or divergent; if itis convergent, you are expected to determine its value. Show all your work.

(a) [7 MARKS]∫ 2

−1

1x3 dx ;

(b) [7 MARKS]∫ ∞

1xe−x2

dx .


7. Showing all your work, determine, for each of the following series, whether or not itconverges:

(a) [5 MARKS]∞∑

n=2

1n(ln n)2 ;

(b) [5 MARKS]∞∑

n=1

(−1)n

(n2 − 1n2 + 1

);

(c) [5 MARKS]∞∑

n=1

n + 13n .

F.10 Final Examination in Mathematics 189-141B (2001/2002)1. Showing all your work, evaluate each of the following indefinite integrals:

(a) [3 MARKS]∫

x3

√4 − x2

dx

(b) [3 MARKS]∫

1

y√

ln ydy

(c) [3 MARKS]∫

sec u1 + sec u

· tan u du

(d) [3 MARKS]∫

et

1 + e2t dt

2. Let K denote the curvey = x2 , (0 ≤ x ≤ 1) .

(a) [6 MARKS] Determine the area of the surface of revolution generated by revolvingK about the y-axis.

(b) [6 MARKS] Determine the volume of the solid of revolution formed by revolvingabout the line y = 0 the region bounded by K and the lines x = 1 andy = 0 .

3. Consider the arc C given by r = θ2 (0 ≤ θ ≤ π).

(a) [4 MARKS] Express the length of C as a definite integral. Then evaluate the inte-gral.

(b) [4 MARKS] Determine the area of the region subtended by C at the pole — i.e. ofthe region bounded by the arc C and the line θ = 0.


(c) [4 MARKS] The given curve can be represented in cartesian coordinates paramet-rically as x = θ2 cos θ, y = θ2 sin θ. Determine the slope of the tangent to this curve

at the point (x, y) =

(0,

(π2

)2).


40 − 16x2

(1 − 4x2) (1 + 2x)

dx .

5. [12 MARKS] Showing all your work, determine the area of the region bounded by thecurves y = arctan x and 4y = π x in the first quadrant.

6. (a) [4 MARKS] Showing all your work, determine the value of∫sin3 x cos2 x dx .

(b) [4 MARKS] Showing all your work, determine the value of∫tan4 x dx .

(c) [4 MARKS] Investigate the convergence of the integral

π2∫

0

tan4 x dx .

7. [12 MARKS] Showing all your work, determine the value of

d2

dx2

∫ x

0

∫ e2t

1

√u + 1 du

dt

when x = 0.

8. Showing all your work, determine, for each of the following infinite series, whether it isabsolutely convergent, conditionally convergent, or divergent.

(a) [4 MARKS]∞∑

n=5

(−1)n n2 − 16n2 + 4

.

(b) [4 MARKS]∞∑

n=2

(−1)n

n(ln n)2 .

(c) [4 MARKS]∞∑

n=2

(−1)n(n+1)

2

2n .

(d) [4 MARKS]∞∑

n=0

n + 52n .


F.11 Supplemental/Deferred Examination in Mathematics 189-141B (2001/2002)1. Showing all your work, evaluate each of the following, always simplifying your answer

as much as possible:

(a) [3 MARKS]∫

ex sin x dx

(b) [3 MARKS]∫ 1

2

0

sin−1 y√1 − y2

dy

equivalently,∫ 1

2

0

arcsin y√1 − y2

dy

.

(c) [3 MARKS]∫

(u2 + 2u)e−u du

(d) [3 MARKS]∫

1 + cos tsin t

dt

2. Let K denote the curve

y =√

2x − x2 ,

(0 ≤ x ≤ 1

2

).

(a) [6 MARKS] Showing all your work, use an integral to determine the area of thesurface of revolution generated by revolving K about the x-axis.

(b) [6 MARKS] Determine the volume of the solid of revolution formed by revolvingabout the line y =

√3

2 the region bounded by K and the lines x = 0 andy =

√3

2 .

(You may assume that∫ √

2x − x2 dx =x − 1

2

√2x − x2 +

12

arccos(1 − x) .)

3. A curve C in the plane is given by parametric equations

x = t3 − 3t2

y = t3 − 3t .

(a) [6 MARKS] Showing all your work, determine all points (x, y) on C where thetangent is horizontal.

(b) [6 MARKS] By determining the value ofd2ydx2 as a function of t, determine all

points (x, y) on C at which the ordinate (y-coordinate) is a (local) maximum, andall points at which the ordinate is a (local) minimum.



4x3

(x2 − 9

)(3x + 9)

dx .

5. [12 MARKS] Showing all your work, determine the area of the region bounded by thecurves x − 2y + 7 = 0 and y2 − 6y − x = 0 .

6. (a) [6 MARKS] Showing all your work, evaluate∫

sin4 x cos2 x dx .

(b) [4 MARKS] Showing all your work, evaluate∫

tan5 x dx .

(c) [4 MARKS] Investigate the convergence of the integral

π2∫

π4

tan5 θ dθ .

7. [12 MARKS] Showing all your work, determine the value of

d2

dx2

∫ x

0

(∫ π3

−2t

√4 + sin(−2u) du

)dt

when x = π4 . Your answer should be simplified, if possible.


(a) [4 MARKS]∞∑

n=5

(−1)n 1√n + 1

.

(b) [4 MARKS]∞∑

n=2

(−1)2n

n(ln n)3 .

(c) [4 MARKS]∞∑

n=2

(2n

1 + 5n

)3n

.

(d) [4 MARKS]∞∑

n=1

sin(

1n

)

cos(

1n

) · 1n

.


9. [10 MARKS] Prove or disprove the following statement: The point with polar coordi-nates

r = 2(√

2 − 1)θ = −π + arcsin((

√2 − 1)2)

lies on the intersection of the curves with polar equations

r2 = 4 sin θ,r = 1 + sin θ .

You are expected to justify every statement you make, but you do not need to sketch thecurves.

F.12 Final Examination in MATH 141 2003 011. [10 MARKS] Find the area of the region bounded in the first quadrant by the curves

y = ex, y = e−x, y = e2x−3 .

Simplify your answer as much as possible. (Your instructors are aware that you do nothave the use of a calculator.)

2. Showing all your work, evaluate each of the following indefinite integrals:

(a) [5 MARKS]∫

1x2 + 2x + 17

dx

(b) [5 MARKS]∫

ln(ln x)x

dx

3. [12 MARKS] For each of the following integrals,

(a) [2 MARKS] Explain why the integral is improper.

(b) [10 MARKS] Determine its value, or show that the integral does not converge.

Show all your work.

I1 =

∫ ∞

2

2(x2 − x + 1)(x − 1)(x2 + 1)

dx , I2 =

∫ 1

0

2(x2 − x + 1)(x − 1)(x2 + 1)

dx

4. Let Ω denote the region in the first quadrant bounded by the curves x =√

16 + y2, y = 0,x = 0, and y = 3.


(a) [5 MARKS] Showing all your work, determine the volume of the solid of revolu-tion obtained by rotating Ω about the y-axis.

(b) [7 MARKS] Showing all your work, determine the area of the surface of revolutionobtained by rotating the arc

x =√

16 + y2, (0 ≤ y ≤ 3)

about the y-axis. You may assume that

ddθ

(sec θ tan θ + ln |sec θ + tan θ|) = 2 sec3 θ .

.

5. Consider the arc C given parametrically by

x =

∫ t

0

√4(1 − cos θ)θ2 dθ

y = cos t + t sin t

(−π ≤ t ≤ 2π) .

Showing all your work

(a) [4 MARKS] Find the slope of the tangent to C at the point with parameter value

t =3π2

.

(b) [6 MARKS] Find the length of C.

6. Give, for each of the following statements, a specific example to show that the statementis not a theorem:

(a) [3 MARKS] If an∞n=0 is a sequence such that limn→∞

an = 0, then∞∑

n=0an converges.

(b) [3 MARKS] If the series∞∑

n=0an and

∞∑n=0

bn are both divergent, then∞∑

n=0(an + bn) is

divergent.

(c) [3 MARKS] If a series∞∑

n=0an converges, then

∞∑n=0

a2n converges.


(a) [4 MARKS]∞∑

n=0

(−1)n

4n2 + 1.


(b) [4 MARKS]∞∑

n=2

(n − 1

n

)n2

(c) [4 MARKS]∞∑

n=2

1√n(n + 1)

.

8. [10 MARKS] Showing all your work, determine the area of the part of one “leaf” of the“4-leafed rose” r = 2 cos(2θ) that is inside the circle r = 1.

F.13 Supplemental/Deferred Examination in MATH 141 2003 011. [10 MARKS] Find the area of the region bounded by the curves

y = ex − 1, y = x2 − x, x = 1.

2. Showing all your work, evaluate each of the following:

(a) [5 MARKS]∫

e1x

x2 dx

(b) [5 MARKS]∫ 5

2|x2 − 4x| dx

3. [12 MARKS] For each of the following integrals,

(a) [2 MARKS] Explain precisely whether the integral is improper.

(b) [10 MARKS] Determine its value, simplifying as much as possible; or show thatthe integral does not converge. (The examiners are aware that you do not haveaccess to a calculator.)

Show all your work.

I1 =

∫ 1

0

2x(x − 1)(x2 + 1)

dx , I2 =

∫ 3

2

2x(x − 1)(x2 + 1)

dx

4. Let Ω denote the region bounded by the curves y = sin x, y = 0, x = π2 , x = π.

(a) [6 MARKS] Showing all your work, determine the volume of the solid of revolu-tion obtained by rotating Ω about the y-axis.

(b) [6 MARKS] Showing all your work, determine the volume of the solid of revolu-tion obtained by rotating Ω about the x-axis.


5. Consider the arc C given parametrically by

x = 2t(t2 − 3)

y = 6t(−t)

(−1 ≤ t ≤ 1) .

Showing all your work

(a) [4 MARKS] Find the slope of the tangent to C at the point with parameter valuet = − 1

2 .

(b) [6 MARKS] Find the area of the surface obtained by rotating the curve about thex-axis.

6. Give, for each of the following statements, a specific example to show that the statementis not a theorem:

(a) [3 MARKS] If bn∞n=0 is a sequence such that limn→∞

bn = 1, then∞∑

n=0bn converges.

(b) [3 MARKS] If a series∞∑

n=0b2

n converges, then∞∑

n=0bn converges.

(c) [3 MARKS] If the series∞∑

n=0an and

∞∑n=0

bn are both divergent, then∞∑

n=0(anbn) is di-

vergent.


(a) [4 MARKS]∞∑

n=3

(−1)n

e1n

.

(b) [4 MARKS]∞∑

n=0

(1

(n + 5)(n + 6)

)

(c) [4 MARKS]∞∑

n=2

(−1)n−1 2√n − 1

.

8. [10 MARKS]Showing all your work, find the area of the region that lies inside the curver = 2 − 2 sin θ and outside the curve r = 3 .

F.14 Final Examination in MATH 141 2004 01One of several versions


1. BRIEF SOLUTIONS

[3 MARKS EACH] Give the numeric value of each of the following limits, sums, inte-grals. If it does not exist write “DIVERGENT”.

(a)∞∑

n=1

1 + 2n−1

3n =

ANSWER ONLY

(b)∫ ∞

−∞xex2

dx =

ANSWER ONLY

(c) The limit of the Riemann sum limn→∞

n∑i=1

2n

((3 + 2i

n

)2 − 6(3 + 2i

n

)5)

=

ANSWER ONLY

(d)∫ −∞

∞xe−x2

dx =

ANSWER ONLY

(e)∞∑

n=3

4(2n + 1)(2n + 3)

=

ANSWER ONLY

2. BRIEF SOLUTIONS

[3 MARKS EACH] Simplify your answers as much as possible.


(a) For the point with polar coordinates(3, π7

)give another set of polar coordinates

(r, θ) in which r < 0 and θ > 2.

ANSWER ONLY

(b) Determine the length of the arc of the curve r = θ2 from (0, 0) to (1, 1).

ANSWER ONLY

(c) A curve is given parametrically by x(t) =∫ t

0e−u2

du, y(t) =∫ 4

teu2

du. Find theslope of the tangent to the curve at (x(1), y(1)).

ANSWER ONLY

(d) Give a definite integral whose value is the area of the surface obtained by rotating

the curve x =y3

6+

12y

(12 ≤ y ≤ 1

)about the y-axis. You need not evaluate the

integral.

ANSWER ONLY

(e) On the interval 0 ≤ x ≤ 4 the average value of the function

f (x) =

√1 − x if 0 ≤ x ≤ 1

x − 2 if 1 < x ≤ 4is


ANSWER ONLY

3. BRIEF SOLUTIONS

[3 MARKS EACH] Give the value of each of the following indefinite integrals:

(a)∫

x3x2 + 1

dx =

ANSWER ONLY

(b)∫

ex√

1 + ex dx =

ANSWER ONLY

(c)∫

(sin2 x − 3 cos2 x) dx =

ANSWER ONLY

(d)∫

tan2 3x dx =

ANSWER ONLY


(e)∫

sec3 x tan3 x dx =

ANSWER ONLY


Let R be the finite region bounded by the curves x = y2 and x = 4 − 3y4.

(a) [5 MARKS] Find the area of R.

(b) [5 MARKS] Find the volume of the solid generated by revolving R about the y-axis.


[12 MARKS] Evaluate the definite integral

∫ 12

− 12

4x(2x2 − x − 2

)

(x2 + 1)(x2 − 1)dx .


(a) [4 MARKS] Show that, for any positive integer n,∫

(ln x)2n dx = x(ln x)2n − 2n∫

(ln x)2n−1 dx

(b) [7 MARKS] Evaluate the integral∫ 1

0

y√2y − y2

dy .


[4 MARKS EACH] Determine for each of the following series whether it

• converges absolutely;

• converges conditionally; or

• diverges.


(a)∞∑

n=1

(1 +

1n

)2

e−n

(b)∞∑

n=10

(−1)n√

2n1 + 2

√n

(c)∞∑

n=2

(−1)n ·√

n + 2 − √n − 1n

(d)∞∑

n=0

2π + cos n6n


[6 MARKS] Find the area bounded by one loop of the curve r = cos 3θ.

Another version

1. BRIEF SOLUTIONS


(a)∞∑

n=1

1 + 3n−1

4n =

ANSWER ONLY

(b)∫ ∞

−∞yey2

dy =

ANSWER ONLY


n∑i=1

2n

((4 + 2i

n

)2 − 7(4 + 2i

n

)5)

=

ANSWER ONLY


(d)∫ −∞

∞ye−y2

dy =

ANSWER ONLY

(e)∞∑

n=3

4(2n − 1)(2n + 1)

=

ANSWER ONLY

2. BRIEF SOLUTIONS


(a) On the interval 0 ≤ x ≤ 4 the average value of the function

f (x) =

√1 − x if 0 ≤ x ≤ 1

x − 2 if 1 < x ≤ 4is

ANSWER ONLY

(b) For the point with polar coordinates(3, π5

)give another set of polar coordinates

(r, θ) in which r < 0 and θ > 2.

ANSWER ONLY

(c) Determine the length of the arc of the curve r = θ2 from (0, 0) to (1, 1).

ANSWER ONLY


(d) A curve is given parametrically by x(t) =∫ t

0e−v2

dv, y(t) =∫ 4

tev2

dv. Find theslope of the tangent to the curve at (x(1), y(1)).

ANSWER ONLY

(e) Give a definite integral whose value is the area of the surface obtained by rotating

the curve x =y3

6+

12y

(12 ≤ y ≤ 1

)about the y-axis. You need not evaluate the

integral.

ANSWER ONLY

3. BRIEF SOLUTIONS


(a)∫

sec3 x tan3 x dx =

ANSWER ONLY

(b)∫

x5x2 + 1

dx =

ANSWER ONLY

(c)∫

ex√

1 + ex dx =


ANSWER ONLY

(d)∫

(3 sin2 x − cos2 x) dx =

ANSWER ONLY

(e)∫

tan2 4x dx =

ANSWER ONLY


Let S be the finite region bounded by the curves y = x2 and y = 4 − 3x4.

(a) [5 MARKS] Find the area of S .

(b) [5 MARKS] Find the volume of the solid generated by revolving S about the x-axis.


[12 MARKS] Evaluate the definite integral

∫ 12

− 12

4x(4x2 − 2x − 4

)

(x2 + 1)(x2 − 1)dx .


(a) [4 MARKS] Show that, for any positive integer m,∫

(ln y)2m dy = y(ln y)2m − 2m∫

(ln y)2m−1 dy


(b) [7 MARKS] Evaluate the integral∫ 1

0

x√2x − x2

dx .





• diverges.

(a)∞∑

n=10

(−1)n√

2n1 + 2

√n

(b)∞∑

n=1

(1 +

1n

)2

e−n

(c)∞∑

n=0

(cos n) − 2π4n

(d)∞∑

n=2

(−1)n ·√

n + 2 − √n − 1n


[6 MARKS] Find the area bounded by one loop of the curve r = cos 3θ.

F.15 Supplemental/Deferred Examination in MATH 141 2004 011. BRIEF SOLUTIONS


(a)∞∑

n=1

en − e−n

3n =

ANSWER ONLY


(b)∫ 0

−∞xex dx =

ANSWER ONLY


π

n

n∑

i=1

sin2( iπ

n

)=

ANSWER ONLY

(d)∞∑

n=0

4(2n + 1)(2n + 3)

=

ANSWER ONLY

2. BRIEF SOLUTIONS


(a) For the point with polar coordinates (r, θ) =(− 10π

3 ,−π6)

give another set of polarcoordinates (r1, θ1) in which r1 > 0 and θ1 > 2 .

ANSWER ONLY

(b) Find all points — if there are any — where the curvesr = 1 − cos θ and r = −3

2 intersect.


ANSWER ONLY

(c) Find the exact length of the curve r = e2θ , (0 ≤ θ ≤ π) .

ANSWER ONLY

(d) On the interval ln 12 ≤ x ≤ π the average value of the function

f (x) =

sinh x if ln 1

2 ≤ x ≤ 0sin x if 0 < x ≤ π is

ANSWER ONLY

3. BRIEF SOLUTIONS


(a)∫

14x2 + 1

dx =

ANSWER ONLY


(b)∫

dx√x2 − 25

=

ANSWER ONLY

(c)∫

(cos x + 1)(cos x − 2) dx =

ANSWER ONLY

(d)∫

tan 3x dx =

ANSWER ONLY


Let C be the arc x =13

( √y2 + 2

)3, (−√2 ≤ y ≤ 0) .

(a) [6 MARKS] Find the area of the surface obtained by revolving C about the x-axis.

(b) [6 MARKS] Find the volume of the solid generated by revolving about the y-axisthe region bounded by C, the coordinate axes, and the line y = −√2.



x3 − 8x − 1(x2 − 1)(x + 1)

dx .



Simplify your answers as much as possible.

(a) [6 MARKS] Evaluate the indefinite integral∫ √

1 − 9x2 dx .

(b) [6 MARKS] Evaluate the definite integral∫ −√3

0arctan x dx .





• diverges.

(a)∞∑

n=2

(−1)n ln√

nln(n2)

(b)∞∑

n=1

1 · 3 · 5 · . . . · (2n − 1)3nn!

(c)∞∑

n=1

sin 2n1 + 2n

(d)∞∑

n=1

(−1)n√

n3n − 1


[12 MARKS] Use polar coordinates — no other method will be accepted — to find the

area of the region bounded by the curve r = 2 and the line r =1

cos θ, and containing the

pole.

F.16 Final Examination in MATH 141 2005 011. SHOW ALL YOUR WORK!


0|x − 1| dx .



∫ 5

x

√4 + t2 dt .


∫ x2

2πsec t dt .


x5( 3√

x3 + 1)

dx .

SHOW ALL YOUR WORK!





(a) [4 MARKS]∞∑

n=2

2 − cos nn

(b) [4 MARKS]∞∑

n=0

n(−3)n

4n

(c) [4 MARKS]∞∑

n=2

1n ln n

3. BRIEF SOLUTIONS Express the value of each of the following as a definite integralor a sum, product, or quotient of several definite integrals, but do not evaluate the inte-gral(s). It is not enough to quote a general formula: your integrals must have integrandand limits specific to the given problems:

(a) [6 MARKS] The area of the region bounded by the parabola y = x2, the x-axis, andthe tangent to the parabola at the point (1, 1).



(b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 4 theregion bounded by x = 0 and the curve x =

√sin y (0 ≤ y ≤ π).


(c) [3 MARKS] The area of the surface obtained by revolving about the y-axis thecurve y = ex, 1 ≤ y ≤ 2.


(d) [2 MARKS] The average value of the function2x

(1 + x2)2 over the interval 0 ≤ x ≤2.




2x3 + 3x2 + 3x2 + x − 12

dx .


(a) [9 MARKS] Use integration by parts to prove that, for integersm ≥ 2, ∫

cosm x dx =1m


m

∫cosm−2 x dx



0cos6 x dx.



x = 2 cos t − cos 2ty = 2 sin t − sin 2t .

(a) [8 MARKS] Determine the points where the arc of the curve given by

π

4≤ t ≤ 7π

4

has a vertical tangent.

(b) [4 MARKS] Determine the length of the arc of the curve given by

0 ≤ t ≤ 2π .


(a) [5 MARKS] Determine whether the following integral is convergent; if it is con-vergent, determine its value: ∫ 1

−1

dx√1 − x2


∞∑

n=1

(−1)n n!nn

(c) [3 MARKS] Determine whether the sequence an = ln(n + 1) − ln n is convergent;if it is convergent, carefully determine its limit.


[12 MARKS] Find the area of the region bounded by the curves

r = 4 + 4 sin θr sin θ = 3

which does not contain the pole.


F.17 Supplemental/Deferred Examination in MATH 141 2005 01

Instructions



3. Calculators are not permitted.

4. This examination booklet consists of this cover, Pages 1 through 7 containing questions; andPages 8, 9, and 10, which are blank.


• Most of the questions on this paper require that you SHOW ALL YOUR WORK!Their solutions are to be written in the space provided on the page where the questionis printed. When that space is exhausted, you may write on the facing page. Anysolution may be continued on the last pages, or the back cover of the booklet, but youmust indicate any continuation clearly on the page where the question is printed!







−2|x2 − 1| dx .


∫ 4

xet2 dt .


∫ x2

1

dt1 + t5 .


x sin(x2) dx .

SHOW ALL YOUR WORK!






(a) [4 MARKS]∞∑

n=2

(2n

3n − 1

)n

(b) [4 MARKS]∞∑

n=2

(−1)n

√n − 1

(c) [4 MARKS]∞∑

n=2

((−4)n

3n + 2n

)

3. BRIEF SOLUTIONS Express the value of each of the following as a definite integral– possibly improper — or a sum, product, or quotient of several such integrals, but donot evaluate the integral(s). It is not enough to quote a general formula: your integralsmust have integrand and limits specific to the given problems:

(a) [6 MARKS] The area of the infinite region containing the point(0, 1

2

)bounded by

the curve y = ex, the x-axis, and the tangent to the curve at the point (1, e).


(b) [3 MARKS] The volume generated by rotating the region bounded by the curvesy =√

x − 1, y = 0, x = 5 about the line y = 3.



(c) [3 MARKS] The area of the surface obtained by revolving about the x-axis thecurve x = ln y, 1 ≤ x ≤ 3.




x3 − x2 − 2x − 2x(x2 + x + 1)

dx .


(a) [9 MARKS] Use integration by parts to prove that, for integersm , 1,

∫secm x dx =

1m − 1

secm−2 x · tan x +m − 2m − 1

∫secm−2 x dx


0sec3 x dx.


The curve C has equations x = t3 + 4t, y = 6t2.

(a) [8 MARKS] Determine the points on C where the tangent is parallel to the linewith equations x = −7t, y = 12t − 5.

(b) [4 MARKS] Determine a definite integral whose value is the length of the arc ofC between the points with parameter values t = 1 and t = 2. YOU ARE NOTEXPECTED TO EVALUATE THE INTEGRAL!


[12 MARKS] Find the area of the region between the inner loop and the outer loop ofthe curve r = 1 − 2 cos θ.


F.18 Final Examination in MATH 141 2006 01 (One version)Instructions


2. Do not tear pages from this book; all your writing — even rough work — must be handedin. You may do rough work for this paper anywhere in the booklet.

3. Calculators are not permitted. This is a closed book examination. Regular and transla-tion dictionaries are permitted.

4. This examination booklet consists of this cover, Pages 1 through 8 containing questions;and Pages 9, 10, and 11, which are blank. Your neighbour’s version of this test may bedifferent from yours.


• Most of the questions on this paper require that you SHOW ALL YOUR WORK!Their solutions are to be written in the space provided on the page where the ques-tion is printed. When that space is exhausted, you may write on the facing page.Any solution may be continued on the last pages, or the back cover of the booklet,but you must indicate any continuation clearly on the page where the question isprinted!

• Some of the questions on this paper require only BRIEF SOLUTIONS ; for theseyou are expected to write the correct answer in the box provided; you are not askedto show your work, and you should not expect partial marks for solutions that arenot correct.


You are advised to spend the first few minutes scanning the problems. (Please informthe invigilator if you find that your booklet is defective.)




−1|x| dx .


e3∫

1

dt

t√

1 + ln t.



∫ x2

0et2 dt .

(d) [4 MARKS] Evaluate

limn→∞

1n

(0n

)7

+

(1n

)7

+

(2n

)7

+ . . . +

(n − 1

n

)7 .

SHOW ALL YOUR WORK!





(a) [4 MARKS]∞∑

n=1

1(tanh n)2 + 1

(b) [4 MARKS]∞∑

n=1

n2ne−n2

(c) [4 MARKS]∞∑

n=1

n2 − 85n + 12n(n + 6)2


(a) [3 MARKS] Expressed as integral(s) along the x-axis only, the area of the regionbounded by the parabola y2 = 2x + 6 and the line y = x − 1. An answer involvingintegration along the y-axis will not be accepted.

(b) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 theregion bounded by the curves y = x3 and y = x2. For this question you are to useonly the method of “washers”.


(c) [3 MARKS] The volume of the solid obtained by rotating about the line y = 1 theregion bounded by the curves y = x3 and y = x2. For this question you are to useonly the method of “cylindrical shells”.

(d) [3 MARKS] The length of the curve whose equation is

x2

4+

y2

9= 1 .



x5 + xx4 − 16

dx .



(a) [4 MARKS]∫

cos x · cosh x dx

(b) [5 MARKS]

1∫

−3

√x2 + 2x + 5 dx

(c) [4 MARKS]∫

sin2 x · cos2 x dx



x = x(t) = 10 − 3t2

y = y(t) = t3 − 3t ,

where −∞ < t < +∞.

(a) [8 MARKS] Determine the value ofd2ydx2 at the points where the tangent is horizon-

tal.

(b) [4 MARKS] Determine the area of the surface of revolution about the x-axis of thearc

(x(t), y(t)) : −√

3 ≤ t ≤ 0.




∫ 0

−1

dx

x23

.


∞∑

n=1

(−1)n

n − ln n

(c) [3 MARKS] Give an example of a sequence an with the property that limn→∞

an = 0

but∞∑

n=1

an = +∞. You are expected to give a formula for the general term an of

your sequence.


[12 MARKS] The arcr = 1 − cos θ (0 ≤ θ ≤ π)

divides the area bounded by the curve

r = 1 + sin θ (0 ≤ θ ≤ 2π)

into two parts. Showing all your work, carefully find the area of the part that containsthe point (r, θ) =

(12 ,

π2

).

F.19 Supplemental/Deferred Examination in MATH 141 2006 01

Instructions





4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10, and 11, which are blank.








Simplify your answers as much as possible.

(a) [4 MARKS] Evaluate∫ ee

e2

dtt ln t

.

(b) [4 MARKS] Evaluate∫ + π

2

− π2| cosh x| dx .

(c) [4 MARKS] Evaluate

limn→∞

1n

(4 +

0n

)2

+

(4 +

1n

)2

+

(4 +

2n

)2

+ . . . +

(4 +

n − 1n

)2 .

(d) [3 MARKS] Evaluateddx

∫ −3x

0

√1 + t2 dt when x = 1.

SHOW ALL YOUR WORK!


• [1 MARK] Name the test(s) that you are using.


• [1 MARK] Explain why the test(s) you have chosen is/are applicable to the givenseries.

• [2 MARKS] Use the test(s) to conclude whether or not the series is convergent.

(a) [4 MARKS]∞∑

n=1

(−1)n nn2 + 4

(b) [4 MARKS]∞∑

n=1

ln( n3n + 2

)

(c) [4 MARKS]∞∑

n=1

3n + 65n


(a) [4 MARKS] The length of the curve whose equation is

x = 1 + et , y = t2 , (−3 ≤ t ≤ 3).

(b) [4 MARKS] The volume of the solid obtained by rotating about the x-axis theregion bounded by the curves y = x and y = x2. For this question you are expectedto use only the method of “cylindrical shells”.

(c) [4 MARKS] The volume of the solid obtained by rotating about the line y = 2 theregion bounded by the curves y = x and y = x2. For this question you are expectedto use only the method of “washers”.



x3

(x2 + 4)(x − 2)dx .


Showing all your work, evaluate each of the following. Simplify your answers as muchas possible.


(a) [4 MARKS]∫

8x cos 2x dx

(b) [4 MARKS]

2−√2∫

0

1√4x − x2

dx

(c) [4 MARKS]∫

ex sin x dx



x = x(t) = 3t − t3

y = y(t) = 3t2 ,

where 0 ≤ t ≤ 1.

(a) [6 MARKS] Determine the value ofd2ydx2 at the point with parameter value t =

12

.

(b) [6 MARKS] Determine the area of the surface of revolution of C about the x-axis.



∫ 6

−∞xe

x3 dx .

(b) [4 MARKS] Give an example of a series which is convergent but not absolutelyconvergent. Justify all of your statements.

(c) [4 MARKS] Give an example of 2 divergent sequences an, bn with the propertythat the sequence anbn is convergent. You are expected to give formulas for thegeneral terms an, bn of both of your sequences.


[12 MARKS] The curves r = 2 cos 2θ and r = 2 sin θ define a number of regions in theplane. Let R denote the region containing the point (r, θ) = (1, 0), bounded by arcs ofboth of the curves. Showing all your work, carefully find the area of R.


F.20 Final Examination in MATH 141 2007 01 (One version)Instructions













−1|x|2 dx .


0∫

1

t4dt√t5 + 1

.

(c) [3 MARKS] Determine the value of

1n

(0n

)3

+

(1n

)3

+

(2n

)3

+ . . . +

(n − 1

n

)3 .


(d) [3 MARKS] Suppose it is known that f ′(x) = 4 cosh x for all x. Showing all yourwork, determine the value of f (1)− f (−1), expressed in terms of the values of eitherexponentials or hyperbolic functions.


∫ x2

12

ett dt when x = 1 .

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests for conver-gence or divergence to determine whether the series is absolutely convergent, condition-ally convergent, or divergent. All tests used must be named, and all statements must becarefully justified.

(a) [4 MARKS]∞∑

n=1

(−n − 2)n(n − 2)n

(2n2 + 1)n

(b) [4 MARKS]∞∑

n=1

(−1)n+1 n!n22n

(c) [4 MARKS]∞∑

n=1

(−1)n sin1n


R is defined to be the region enclosed by the curves x + y = 6 and y = x2; C is the arcy = 3x (−1 ≤ x ≤ 2).

(a) [3 MARKS] The region R is rotated about the x-axis. Give an integral or sum ofintegrals whose value is the volume of the resulting solid.


(b) [3 MARKS] The region R is rotated about the line x = 5. Give an integral or sumof integrals whose value is the volume of the resulting solid.



(c) [3 MARKS] Express in terms of integrals — which you need not evaluate — theaverage length that R cuts off from the vertical lines which it meets.




(e) [3 MARKS] Given an integral whose value is the area of the surface generated byrotating C about the line y = −1; you need not evaluate the integral.



[12 MARKS] Evaluate the indefinite integral∫ x

(x2 − 4

)(x − 2) + 4

(x2 − 4

)(x − 2)

dx .




(a) [4 MARKS]∫

e−x · cos x dx

(b) [5 MARKS]

52∫

− 12

x√8 + 2x − x2

dx

(c) [4 MARKS]∫ (

cos2 x +1

cos2 x

)· tan2 x dx



x = x(t) = cos t + t sin ty = y(t) = sin t − t cos t ,

where 0 ≤ t ≤ π2 .

(a) [6 MARKS] Determine as a function of t the value ofd2ydx2 .

(b) [6 MARKS] Determine the area of the surface generated by revolving C about they-axis.



∫ π

π2

sec x dx .


n=3

4n ln n

is convergent.(c) [3 MARKS] Showing all your work, determine whether the following sequence

converges; if it converges, find its limit:

a1 = 1.a2 = 1.23a3 = 1.2345a4 = 1.234545a5 = 1.23454545a6 = 1.2345454545




[10 MARKS] The polar curves

r = 2 + 2 sin θ (0 ≤ θ ≤ 2π)and

r = 6 − 6 sin θ (0 ≤ θ ≤ 2π)

divide the plane into several regions. Showing all your work, carefully find the area ofthe region bounded by these curves which contains the point (r, θ) = (1, 0).

F.21 Supplemental/Deferred Examination in MATH 141 2007 01 (Oneversion)

Instructions



3. The use of calculators is not permitted. This is a closed book examination. Use of regularand translation dictionaries is permitted.

4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10, and 11, which are blank. Your neighbour’s version of this examination may bedifferent from yours.


• Most of the questions on this paper require that you SHOW ALL YOUR WORK!

Their solutions are to be written in the space provided on the page where the questionis printed. When that space is exhausted, you may write on the facing page. Anysolution may be continued on the last pages, or the back cover of the booklet, but youmust indicate any continuation clearly on the page where the question is printed!








− 1√2

1√1 − x2

dx .


t3 cosh(t4)

dt .

(c) [3 MARKS] Determine one antiderivative of x ln x.

(d) [3 MARKS] Evaluate the integral∫ x

−xtet2 dt.


∫ 1

sin x(ln | sec t + tan t|) dt when x =

π

4.

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests to determinewhether the series is convergent or divergent. All tests used must be named, and allstatements must be carefully justified.

(a) [4 MARKS]∞∑

n=3

1(n + 2)(n − 2)

(b) [4 MARKS]∞∑

n=1

∞∑

i=n

3−i

(c) [4 MARKS]∞∑

n=1

(n + 1

n

)n2


R is defined to be the region enclosed by the curves y − x = 9 and y = (x + 3)2; C is thearc x = t, y = e3t (−2 ≤ t ≤ 1).


(a) [3 MARKS] The region R is rotated about the y-axis. Give an integral or sum ofintegrals whose value is the volume of the resulting solid.


(b) [3 MARKS] The region R is rotated about the line y = −3. Give an integral or sumof integrals whose value is the volume of the resulting solid.


(c) [3 MARKS] Let f (t) denote the vertical distance of the point(t, e3t

)from the x-

axis. Express in terms of integrals — which you need not evaluate — the averagevalue of f (t) over the interval −2 ≤ t ≤ 1.




(e) [3 MARKS] Given an integral whose value is the area of the surface generated byrotating C about the line x = 1; you need not evaluate the integral.





∫ x3(x4 − 4x2

)− 16

x4 − 4x2 dx .


(a) [4 MARKS] Showing all your work, evaluate∫

sin2(3x) cos2(3x) dx .


1

x2√

9x2 − 16dx .

(c) [5 MARKS] Assume that

f (x) =

∫ x

0sec100 t dt

is known. Showing all your work, express the value of∫ x

0sec102 t dt in terms of

f (x). (You are not expected to determine f (x) explicitly.)


Consider the closed arc C defined by

x = x(t) = 3t2

y = y(t) = t3 − 3t ,

where −√3 ≤ t ≤ √3.

(a) [3 MARKS] Determine the area bounded by C.

(b) [3 MARKS] Determine the equation of the tangent to C at the point with parameter

value t =12

.

(c) [6 MARKS] Determine the area of the surface generated by revolving C about they-axis.



∫ ∞

−∞

xx2 + 4

dx .

(b) [7 MARKS] Showing all your work, carefully determine whether the series

∞∑

n=3

(−1)n

√ln nn

is conditionally convergent, absolutely convergent, or divergent.


[10 MARKS] Find the area inside the larger loop and outside the smaller loop of thelimacon r = 2 sin θ − 1.

F.22 Final Examination in MATH 141 2008 01 (one version)This examination was written during a labour disruption, when the services of Teaching As-sistants were not available for grading purposes. The following additional instructions weredistributed with the examination.

VERSION nMcGILL UNIVERSITYFACULTY OF SCIENCEFINAL EXAMINATION

IMPORTANT ADDITIONAL INSTRUCTIONS

MATHEMATICS 141 2008 01CALCULUS 2EXAMINER: Professor W. G. Brown DATE: Monday, April 14th, 2008ASSOCIATE EXAMINER: Mr. S. Shahabi TIME: 09:00 – 12:00 hours

A. Part marks will not be awarded for any part of any question worth [4 MARKS] or less.


B. To be awarded part marks on a part of a question whose maximum value is 5 marks ormore, a student’s answer must be deemed to be more than 75% correct.

C. While there are 100 marks available on this examination 80 MARKS CONSTITUTE APERFECT PAPER. You may attempt as many problems as you wish.

All other instructions remain valid. Where a problem requires that all work be shown, thatremains the requirement; where a problem requires only that an answer be written in a boxwithout work being graded, that also remains the requirement.

Students are advised to spend time checking their work; for that purpose you could verifyyour answers by solving problems in more than one way. Remember that indefinite integralscan be checked by differentiation.

W. G. Brown, Examiner.

Instructions






• Most of the questions on this paper require that you SHOW ALL YOUR WORK!Their solutions are to be written in the space provided on the page where the questionis printed; in some of these problems you are instructed to write the answer in a box,but a correct answer alone will not be sufficient unless it is substantiated by your work,clearly displayed outside the box. When space provided for that work is exhausted, youmay write on the facing page. Any solution may be continued on the last pages, or theback cover of the booklet, but you must indicate any continuation clearly on the pagewhere the question is printed!









4 − 6x1 + x2 dx .

ANSWER (SHOW YOUR WORK OUTSIDE THE BOX)


2∫

0

y2√

y3 + 1 dy .



(c) [3 MARKS] Evaluate∫

sin(18 θ) cos(30 θ) dθ .



(a) [3 MARKS] Simplifying your answer as much as possible, evaluateddx

∫ √3

−xearcsin z dz .



(b) [4 MARKS] For the interval 2 ≤ x ≤ 5 write down the Riemann sum for thefunction f (x) = 3 − x, where the sample points are the left end-point of each of nsubintervals of equal length.

ANSWER ONLY

(c) [4 MARKS] Determine the value of the preceding Riemann sum as a function ofn, simplifying your work as much as possible. (NOTE: You are being asked todetermine the value of the sum as a function of n, not the limit as n→ ∞.)




For each of the following series determine whether the series diverges, converges condi-tionally, or converges absolutely. All of your work must be justified; prior to using anytest you are expected to demonstrate that the test is applicable to the problem.

(a) [4 MARKS]∞∑

n=3

(1

n√

ln n

)

(b) [4 MARKS]∞∑

n=1

(−1)n+1

√4n + 5

3n + 10

(c) [4 MARKS]∞∑

n=1

(cot−1

(1

n + 1

)− cot−1

(1n

))

4. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by thecurves 2y = x, y = 2x, and x2 + y2 = 5.

(a) [4 MARKS] The region R is rotated about the line x = −1. Give an integral or sumof integrals whose value is the volume of the resulting solid.


(b) [4 MARKS] Let L(a) denote the length of the portion of line y = a which lies insideR. Express in terms of integrals — which you need not evaluate — the average ofthe positive lengths L(a).



(c) [4 MARKS] Let C1 be the curve x(t) = t, y(t) = cosh t (0 ≤ t ≤ ln 2). Simplifyingyour answer as much as possible, find the length of C1.

ANSWER ONLY



36(x + 4)(x − 2)2 dx .

(b) [4 MARKS] Determine whether

∞∫

3

36(x + 4)(x − 2)2 dx converges.

If it converges, find its value.



(a) [4 MARKS]∫ √

e√

x dx



(b) [5 MARKS]

0∫

− 12

x√3 − 4x − 4x2

dx


(c) [4 MARKS]∫ π

0sin2 t cos4 t dt .




Consider the curve C2 defined by x = x(t) = 1 + e−t , y = y(t) = t + t2 .

(a) [2 MARKS] Determine the coordinates of all points where C2 intersects the x-axis.


(b) [2 MARKS] Determine the coordinates of all points of C2 where the tangent ishorizontal.



(c) [6 MARKS] Determine the area of the finite region bounded by C2 and the x-axis.



(a) [5 MARKS] Showing all your work, determine whether the series∞∑

n=2

√n(√

n + 2 −√

n − 2)

is convergent or divergent.

(b) [5 MARKS] Showing all your work, determine whether the following sequenceconverges; if it converges, find its limit:

a1 = 3.a2 = 3.14a3 = 3.1414a4 = 3.141414a5 = 3.14141414a6 = 3.1414141414




Curves C3 and C4, respectively represented by polar equations

r = 4 + 2 cos θ (0 ≤ θ ≤ 2π) (119)and

r = 4 cos θ + 5 (0 ≤ θ ≤ 2π) , (120)

divide the plane into several regions.

(a) [8 MARKS] Showing all your work, carefully find the area of the one region whichis bounded by C3 and C4 and contains the pole.

(b) [4 MARKS] Find another equation — call it (120*) — that also represents C4, andhas the property that there do not exist coordinates (r, θ) which satisfy equations(119) and (120*) simultaneously. You are expected to show that equations (119)and (120*) have no simultaneous solutions.

F.23 Supplemental/Deferred Examination in MATH 141 2008 01 (oneversion)

Instructions



3. The use of calculators is not permitted. This is a closed book examination. Use of regularand translation dictionaries is permitted.

4. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10, and 11, which are blank. Your neighbour’s version of this examination may bedifferent from yours.










− 1√2

1√1 − x2

dx .


t3 cosh(t4)

dt .

(c) [3 MARKS] Determine one antiderivative of x ln x.

(d) [3 MARKS] Evaluate the integral∫ x

−xtet2 dt.


∫ 1

sin x(ln | sec t + tan t|) dt when x =

π

4.

SHOW ALL YOUR WORK!

2. For each of the following series you are expected to apply one or more tests to determinewhether the series is convergent or divergent. All tests used must be named, and allstatements must be carefully justified.

(a) [4 MARKS]∞∑

n=3

1(n + 2)(n − 2)

(b) [4 MARKS]∞∑

n=1

∞∑

i=n

3−i

(c) [4 MARKS]∞∑

n=1

(n + 1

n

)n2


R is defined to be the region enclosed by the curves y − x = 9 and y = (x + 3)2; C is thearc x = t, y = e3t (−2 ≤ t ≤ 1).


(a) [3 MARKS] The region R is rotated about the y-axis. Give an integral or sum ofintegrals whose value is the volume of the resulting solid.


(b) [3 MARKS] The region R is rotated about the line y = −3. Give an integral or sumof integrals whose value is the volume of the resulting solid.


(c) [3 MARKS] Let f (t) denote the vertical distance of the point(t, e3t

)from the x-

axis. Express in terms of integrals — which you need not evaluate — the averagevalue of f (t) over the interval −2 ≤ t ≤ 1.




(e) [3 MARKS] Given an integral whose value is the area of the surface generated byrotating C about the line x = 1; you need not evaluate the integral.





∫ x3(x4 − 4x2

)− 16

x4 − 4x2 dx .


(a) [4 MARKS] Showing all your work, evaluate∫

sin2(3x) cos2(3x) dx .


1

x2√

9x2 − 16dx .

(c) [5 MARKS] Assume that

f (x) =

∫ x

0sec100 t dt

is known. Showing all your work, express the value of∫ x

0sec102 t dt in terms of

f (x). (You are not expected to determine f (x) explicitly.)


Consider the closed arc C defined by

x = x(t) = 3t2

y = y(t) = t3 − 3t ,

where −√3 ≤ t ≤ √3.

(a) [3 MARKS] Determine the area bounded by C.

(b) [3 MARKS] Determine the equation of the tangent to C at the point with parameter

value t =12

.

(c) [6 MARKS] Determine the area of the surface generated by revolving C about they-axis.




∫ ∞

−∞

xx2 + 4

dx .


n=3

(−1)n

√ln nn

is conditionally convergent, absolutely convergent, or divergent.


[10 MARKS] Find the area inside the larger loop and outside the smaller loop of thelimacon r = 2 sin θ − 1.

F.24 Final Examination in MATH 141 2009 01 (one version)Instructions



3. This examination booklet consists of this cover, Pages 1 through 8 containing questions; andPages 9, 10 and 11, which are blank. A TOTAL OF 75 MARKS ARE AVAILABLE ON THISEXAMINATION.

4. You are expected to simplify all answers wherever possible.

• Most questions on this paper require that you SHOW ALL YOUR WORK!Solutions are to be begun on the page where the question is printed; a correct answeralone will not be sufficient unless substantiated by your work. You may continue yoursolution on the facing page, or on the last pages, or the back cover of the booklet, butyou must indicate any continuation clearly on the page where the question is printed!To be awarded partial marks on a part of a question a student’s answer for that partmust be deemed to be more than 50% correct. Most of these questions will requirethat the answer be written in a box provided on the page where the question is printed;even if you continue your work elsewhere, the answer should be in the box provided.


• Some questions on this paper require only BRIEF SOLUTIONS ; for these you mustwrite the correct answer in the box provided; you are not asked to show your work, andyou should not expect partial marks for solutions that are not correct. Check your work!



t3 cos t2 dt .


(b) [4 MARKS] Simplifying your answer as much as possible, evaluate the derivativeddt

∫ t2

0tanh x2 dx .





(a) [4 MARKS] Evaluate

12∫

1√2


.




2y√y2 − y + 1

dy


SHOW ALL YOUR WORK!

3. For each of the following series determine whether the series diverges, converges condi-tionally, or converges absolutely. All of your work must be justified; prior to using anytest you are expected to demonstrate that the test is applicable to the problem.

(a) [4 MARKS]∞∑

n=1

(−1)n+1(cos n

2

)n

(b) [4 MARKS]∞∑

n=2

(−1)n 1

n√

ln n.

(c) [4 MARKS]∞∑

n=4

(−1)n 1n

ln (3n + 1)

4. [9 MARKS] SHOW ALL YOUR WORK!

(a) [3 MARKS] Evaluate limn→∞

1n·

n∑

r=1

cos2(rπ

n

) . (Hint: This could be a Riemann

sum.)

(b) [3 MARKS] Showing all your work, prove divergence, or find the limit of an =

arctan(−2n) as n→ ∞.



(c) [3 MARKS] Showing all your work, prove divergence, or find the value of∞∑

n=1

∞∑

i=1

15i

n

.



t + 12t2 − t − 1

dt .

(b) [2 MARKS] Determine whether the following improper integral converges or di-verges; if it converges, find its value:

∞∫

4

t + 12t2 − t − 1

dt .


Consider the curve C1 defined by x = 3t2 , y = 2t3 , (t ≥ 0).

(a) [7 MARKS] Showing all your work, determine the area of the surface generatedwhen the arc 0 ≤ t ≤ 1 of C1 is rotated about the y-axis.

(b) [2 MARKS] Showing all your work, determine all points — if any — where thenormal to the curve is parallel to the line x + y = 8.


Curves C3 and C4, are respectively represented by polar equations

r = 3 + 3 cos θ (0 ≤ θ ≤ 2π) (121)and

r = 9 cos θ (0 ≤ θ ≤ π) . (122)

(a) [7 MARKS] Showing all your work, carefully find the area of the region lyinginside both of the curves.

(b) [3 MARKS] Determine the length of the curves which form the boundary of theregion whose area you have found.

8. BRIEF SOLUTIONS R is defined to be the region in the first quadrant enclosed by the

curves y =8x2 , x = y, x = 1.


(a) [3 MARKS] The region R is rotated about the line x = −1 to create a 3-dimensionalsolid, S 1. Give an integral or sum of integrals whose value is the volume of S 1;you are not asked to evaluate the integral(s).


(b) [3 MARKS] The region R is rotated about the x- axis to create a 3-dimensionalsolid, S 2. Give an integral or sum of integrals whose value is the volume of S 2

obtained only by the Method of Cylindrical Shells; you are not asked to evaluatethe integral(s).


(c) [3 MARKS] Calculate the area of R.

ANSWER ONLY


G WeBWorK

G.1 Frequently Asked Questions (FAQ)G.1.1 Where is WeBWorK?

WeBWorK is located on Web servers of the Department of Mathematics and Statistics, and isaccessible at the following URL:

http://msr02.math.mcgill.ca/webwork2/MATH141 WINTER2010/

If you access WeBWorK through WebCT, the link on your page will have been programmedto take you to the correct WeBWorK server automatically.

G.1.2 Do I need a password to use WeBWorK?

You will need a user code and a password.

Your user code. Your user code will be your 9-digit student number.

Your password. The WeBWorK system is administered by the Mathematics and StatisticsDepartment, and is not accessible directly through the myMcGill Portal; your initial passwordwill be different from your MINERVA password, but you could change it to that if you wish.Your initial password will be your 9-digit student ID number. You will be able to change thispassword after you sign on to WeBWorK.65

Your e-mail address. The WeBWorK system requires each user to have an e-mail address.After signing on to WeBWorK, you should verify that the e-mail address shown is the one thatyou prefer. You should endeavour to keep your e-mail address up to date, since the instructorsmay send messages to the entire class through this route.

We suggest that you use either your UEA66 or your po-box address. You may be able toforward your mail from these addresses to another convenient address, (cf. §4.)

G.1.3 Do I have to pay an additional fee to use WeBWorK?

WeBWorK is available to all students registered in the course at no additional charge.

65If you forget your password you will have to send a message to Professor Brown so that the system adminis-trator may be instructed to reset the password at its initial value.

66Uniform E-mail Address



G.1.4 When will assignments be available on WeBWorK?

Each assignment will have a begin date and a due date. The assignment is available to youafter the begin date; solutions will be made available soon after the due date.

G.1.5 Do WeBWorK assignments cover the full range of problems that I should be ableto solve in this course?

The questions on the WeBWorK assignments (A1 through A6) are a sampling of some typesof problem you should be able to solve after successfully completing this course. Some typesof calculus problems do not lend themselves to this kind of treatment, and may not appear onthe WeBWorK assignments. Use of WeBWorK does not replace studying the textbook —including the worked examples, attending lectures and tutorials, and working exercisesfrom the textbook — using the Student Solutions Manual [3] to check your work. Studentsare cautioned not to draw conclusions from the presence, absence, or relative frequencies ofproblems of particular types, or from particular sections of the textbook. Certain sectionsof the textbook remain examination material even though no problems are included in theWeBWorK assignments.

G.1.6 May I assume that the distribution of topics on quizzes and final examinationswill parallel the distribution of topics in the WeBWorK assignments?

No! While the order of topics on WeBWorK assignments should conform to the order of thelectures, there are some topics on the syllabus that will not appear in WeBWorK questions.Use WeBWorK for the areas it covers, and supplement it by working problems from yourtextbook. Also, remember that WeBWorK — which checks answer only — cannot ascertainwhether you are using a correct method for solving problems. But, if you write out a solutionto an odd-numbered textbook problem, you can compare it with the solution in the SolutionsManual; and, if in doubt, you can show your work to a Teaching Assistant at one of the manyoffice hours that they hold through the week.

G.1.7 WeBWorK provides for different kinds of “Display Mode”. Which should I use?

“Display mode” is the mode that you enter when you first view a problem; and, later, whenyou submit your answer. You may wish to experiment with the different formats. The defaultis jsMath mode, which should look similar to the version that you print out (cf. next question).


G.1.8 WeBWorK provides for printing assignments in “Portable Document Format”(.pdf), “PostScript” (.ps) and “TEXSource” forms. Which should I use?

Most newer home computers have already been loaded with the Acrobat Reader for .pdf files; ifthe Reader has not been installed on your computer67, you will find instructions for download-ing this (free) software in §1.6.5 of these notes. Most computers available to you on campusshould be capable of printing in .pdf format.

G.1.9 What is the relation between WeBWorK and WebCT?

There is none. WebCT is the proprietary system of Web Course Tools that has been imple-mented by McGill University. You may access the web page for this course, and WeBWorKthrough your WebCT account68, and WebCT will link you to the appropriate server for WeB-WorK. If you follow this route to WeBWorK, you will still have to log in when you reachthe WeBWorK site. At the present time we will be using WebCT primarily for the posting ofgrades, and as a convenient repository for links to notes and announcements in the course. Weare not planning to use the potential WebCT sites that exist for the tutorial sections: use onlythe site for the lecture section in which you are registered.

G.1.10 What do I have to do on WeBWorK?

After you sign on to WeBWorK, and click on “Begin Problem Sets”, you will see a list of As-signments, each with a due date. Since there is no limit to the number of attempts at problemson P0 or the other “Practice” assignments, you may play with these assignments to learn howto use the WeBWorK software.

You may print out a copy of your assignment by clicking on “Get hard copy”. This isyour version of the assignment, and it will differ from the assignments of other students inthe course. You should spend some time working on the assignment away from the computer.When you are ready to submit your solutions, sign on again, and select the same assignment.This time click on “Do problem set”. You can expect to become more comfortable with thesystem as you attempt several problems; but, in the beginning, there are likely to be situationswhere you cannot understand what the system finds wrong with some of your answers. It isuseful to click on the Preview Answers button to see how the system interprets an answer thatyou have typed in. As the problems may become more difficult, you may have to refer to the“Help” page, and also to the “List of functions” which appears on the page listing the problems.Don’t submit an answer until you are happy with the interpretation that the Preview Answersbutton shows that the system will be taking of your answer.

67At the time these notes were written, the latest version of the Reader was 9.0, but recent, earlier versionsshould also work properly.

68http://mycourses.mcgill.ca


G.1.11 How can I learn how to use WeBWorK?

As soon as your instructor announces that the WeBWorK accounts are ready, sign on and tryassignment P0, which does not count. The system is self-instructive, so we will not burdenyou with a long list of instructions.

You will need to learn how to enter algebraic expressions into WeBWorK as it is coded toread what you type in a way that may different from what you expect. For example, the symbolˆ is used for writing exponents (powers). If you type 2ˆ3, WeBWorK will interpret this as23 = 8. However, if you type 2ˆ3+x, WeBWorKwill interpret it as 23 + x, i.e. as 8 + x; if youwish to write 23+x, you have to type 2ˆ(3+x). You may obtain more information from the Listof Available Functions, available online, or at

http://webwork.maa.org/wiki/Mathematical notation recognized by WeBWorK

G.1.12 Where should I go if I have difficulties with WeBWorK ?

If you have difficulties signing on to WeBWorK, or with the viewing or printing functions onWeBWorK, or with the specific problems on your version of an assignment, you may sendan e-mail distress message directly from WeBWorK by clicking on the Feedback button.You may also report the problem to your instructor and/or your tutor, but the fastest way ofresolving your difficulty is usually the Feedback . Please give as much information as youcan. (All of the instructors and tutors are able to view from within WeBWorK the answersthat you have submitted to questions.)

If your problem is mathematical, and you need help in solving a problem, you shouldconsult one of the tutors at their office hours; you may go to any tutor’s office hours, not onlyto the hours of the tutor of the section in which you are registered.

G.1.13 Can the WeBWorK system ever break down or degrade?

Like all computer systems, WeBWorK can experience technical problems. The systems man-ager is continually monitoring its performance. If you experience a difficulty when online,please click on the Feedback button and report it. If that option is not available to you,please communicate with either instructor by e-mail.

If you leave your WeBWorK assignment until the hours close to the due time on the duedate, you should not be surprised if the system is slow to respond. This is not a malfunction,but is simply a reflection of the fact that other students have also been procrastinating! Tobenefit from the speed that the system can deliver under normal conditions, do not delay yourWeBWorK until the last possible day! If a systems failure interferes with the due date of anassignment, arrangements could be made to change that date, and an e-mail message could


be broadcast to all users (to the e-mail addresses on record), or a message could be posted onWeBCT or the WeBWorK sign-on screen.69

G.1.14 How many attempts may I make to solve a particular problem on WeBWorK?

Practice Assignments P1 — P6 are intended to prepare you for Assignments A1 — A6, andpermit unlimited numbers of attempts; your grades on these “Practice” do not count in yourterm mark. For the problems on assignments A1 — A6 you will normally be permitted about 5tries: read the instructions at the head of the assignment.

G.1.15 Will all WeBWorK assignments have the same length? the same value?

The numbers of problems on the various assignments may not be the same, and the individualproblems may vary in difficulty. Assignments A1 — A6 will count equally in the computationof your grade.

G.1.16 Is WeBWorK a good indicator of examination performance?

A low grade on WeBWorK has often been followed by a low grade on the examination.A high grade on WeBWorK does not necessarily indicate a likely high grade on the exam-

ination.To summarize: WeBWorK alone is not enough to prepare this course; but students who

don’t do WeBWorK appear to have a poor likelihood of success in MATH 141: that is onereason why we have made the WeBWorK assignments compulsory.

69But slowness of the system just before the due time will not normally be considered a systems failure.


H Contents of the DVD disks forLarson/Hostetler/Edwards

These excellent disks were produced to accompany the textbook, Calculus of a Single Vari-able: Early Transcendental Functions, 3rd Edition[28] (called LHE in the charts below). Thecorrespondence shown to sections of [7] are only approximate. (NOTE THAT THIS BOOKDOES NOT FOLLOW STEWART’S CONVENTIONS FOR INVERSE SECANT/COSECANT!)

[All references in this table are to the 5th edition of Stewart, [7].]

DVD LHE Stewart# Section Subject Minutes Section1 P Chapter P: Preparation for Calculus1 P.1 Graphs and Models 451 P.2 Linear Models and Rates of Change 27 A101 P.3 Functions and Their Graphs 48 1.11 P.4 Fitting Models to Data 21 1.21 P.5 Inverse Functions 48 1.61 P.6 Exponential and Logarithmic Functions 30 1.5

DVD LHE Stewart# Section Subject Minutes Section1 1 Chapter 1: Limits and Their Properties1 1.1 A Preview of Calculus 11 2.11 1.2 Finding Limits Graphically and Numerically 25 2.2, 2.41 1.3 Evaluating Limits Analytically 28 2.31 1.4 Continuity and One-Sided Limits 22 2.51 1.5 Infinite Limits 18 2.6

DVD LHE Stewart# Section Subject Minutes Section1 2 Chapter 2: Differentiation1 2.1 The Derivative and the Tangent Line Problem 68 2.11 2.2 Basic Differentiation Rules and Rates of

Change34 2.3

1 2.3 The Product and Quotient Rules and HigherOrder Derivatives

25 3.2, 3.7


DVD LHE Stewart# Section Subject Minutes Section2 2 Chapter 2 (continued): Differentiation2 2.4 The Chain Rule 44 3.52 2.5 Implicit Differentiation 50 3.62 2.6 Derivatives of Inverse Functions 17 3.5, 3.8, 3.92 2.7 Related Rates 34 3.102 2.8 Newton’s Method 26 4.9

DVD LHE Stewart# Section Subject Minutes Section2 3 Chapter 3: Applications of Differentiation2 3.1 Extrema on an Interval 41 4.12 3.2 Rolle’s Theorem and the Mean Value Theo-

rem15 4.2

2 3.3 Increasing and Decreasing Functions and theFirst Derivative Test

19 4.3

2 3.4 Concavity and the Second Derivative Test 24 4.32 3.5 Limits at Infinity 23 2.62 3.6 A Summary of Curve Sketching 43 4.52 3.7 Optimization Problems 37 4.72 3.8 Differentials 51 3.11

DVD LHE Stewart# Section Subject Minutes Section3 4 Chapter 4: Integration3 4.1 Antiderivatives and Indefinite Integration 40 4.10

DVD LHE Stewart# Section Subject Minutes Section4 7 Chapter 7: Integration by Parts Trigonomet-

ric Substitution Partial Fractions L’Hopital’sRule

4 7.7 Indeterminate Forms and L’Hopital’s Rule 22 4.4

(The coverage extends to part of the material for Math 141 as well.)

class notes

Documents

c supplementary notes

numbers of pages

relevant pages

specific pages

students insection

calculus textbooks

lecture style

lecture section001