information for students in math 141 2004 01

McGILL UNIVERSITY

FACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATH 141 2004 01

CALCULUS II

Information for Students(Winter Term, 2003/2004)

Pages 1 - 18 of these notes may be considered theCourse Outline for this course.

W. G. Brown and N. Sancho

April 8, 2004

Information for Students in MATH 141 2004 01

Contents

1 General Information 11.1 Instructors and Times . . . . . 11.2 Calendar Description . . . . . . 2

1.2.1 Calendar Description . . 21.2.2 Late transfer from MATH

151 . . . . . . . . . . . . 21.3 Tutorials; Tutors’ Coordinates . 21.4 Evaluation of Your Progress . . 3

1.4.1 Your final grade . . . . 31.4.2 WeBWorK . . . . . . 31.4.3 Regular WeBWorK As-

signments . . . . . . . . 51.4.4 Written Assignments . . 61.4.5 Quizzes at the Tutorials;

Submission of Written As-signments at the Quizzes 6

1.4.6 Final Examination . . . 81.4.7 Supplemental Assessments 81.4.8 Machine Scoring . . . . 81.4.9 Plagiarism and Fraud . 81.4.10 Corrections to grades . 9

1.5 Published Materials . . . . . . 91.5.1 Required Text-Book . . 91.5.2 Optional Reference Books 101.5.3 Recommended Video Ma-

terials . . . . . . . . . . 111.5.4 Other Calculus Textbooks 131.5.5 Website . . . . . . . . . 13

1.6 Syllabus . . . . . . . . . . . . . 141.7 Preparation and Workload . . . 15

1.7.1 Prerequisites. . . . . . . 151.7.2 Calculators . . . . . . . 151.7.3 Self-Supervision . . . . . 151.7.4 Escape Routes . . . . . 161.7.5 Terminology . . . . . . 16

1.8 High and Low Technology andMATH 141 . . . . . . . . . . . 171.8.1 Keep your e-mail address

up to date . . . . . . . . 17

1.8.2 Use of Calculators andComputer Algebra Sys-tems . . . . . . . . . . . 17

1.8.3 Use of the Internet . . . 181.9 Special Review Sessions . . . . 18

A Information Specifically for Studentsin Lecture Section 001 1001A.1 Timetable for Lecture Section 001

of MATH 141 2004 01 . . . . . 1001

B Information Specifically for Studentsin Lecture Section 002 2001B.1 Timetable for Lecture Section 002

of MATH 141 2004 01 . . . . . 2001

C Supplementary Notes for Lecturesin Lecture Section 002 2003C.1 Supplementary Notes for the Lec-

ture of January 5th, 2004 . . . 2003C.1.1 §5.1 Areas and Distances 2004C.1.2 §5.2 The Definite Integral 2005

C.2 Supplementary Notes for the Lec-ture of January 7th, 2004 . . . 2010C.2.1 §5.3 The Fundamental The-

orem of Calculus . . . . 2010C.3 Supplementary Notes for the Lec-

ture of January 12th, 2004 . . . 2014C.3.1 §5.4 Indefinite Integrals

and the Net Change The-orem . . . . . . . . . . . 2015

C.3.2 §5.5 The Substitution Rule2019C.4 Supplementary Notes for the Lec-

ture of January 14th, 2004 . . . 2021C.4.1 §5.5 The Substitution Rule

(continued) . . . . . . . 2021C.4.2 §5.6 The Logarithm De-

fined as an Integral . . . 2025C.5 Supplementary Notes for the Lec-

ture of January 19th, 2004 . . . 2027C.5.1 §6.1 Areas between Curves2027C.5.2 §6.2 Volumes . . . . . . 2030


C.6 Supplementary Notes for the Lec-ture of January 21st, 2004 . . . 2033C.6.1 §6.2 Volumes (continued) 2033C.6.2 §6.3 Volumes by Cylin-

drical Shells . . . . . . . 2033C.7 Supplementary Notes for the Lec-

ture of January 26th, 2004 . . . 2037C.7.1 §6.4 Work . . . . . . . . 2038C.7.2 §6.5 Average value of a

function . . . . . . . . . 2038C.8 Supplementary Notes for the Lec-

ture of January 28th, 2004 . . . 2043C.8.1 §7.1 Integration by Parts 2043

C.9 Supplementary Notes for the Lec-ture of February 2nd, 2004 . . . 2050C.9.1 §7.2 Trigonometric Inte-

grals . . . . . . . . . . . 2050C.9.2 §7.3 Trigonometric Sub-

stitution . . . . . . . . . 2054C.10 Supplementary Notes for the Lec-

ture of February 4th, 2004 . . . 2059C.10.1 §7.4 Integration of Ra-

tional Functions by Par-tial Fractions . . . . . . 2061

C.11 Supplementary Notes for the Lec-ture of February 9th, 2004 . . . 2066C.11.1 §7.4 Integration of Ra-

tional Functions by Par-tial Fractions (continued) 2066

C.11.2 §7.5 Strategy for Integra-tion . . . . . . . . . . . 2070

C.12 Supplementary Notes for the Lec-ture of February 11th, 2004 . . 2073C.12.1 §7.5 Strategy for Integra-

tion (continued) . . . . 2073C.12.2 §7.6 Integration Using Ta-

bles and Computer Al-gebra Systems . . . . . 2075

C.12.3 §7.7 Approximate Inte-gration . . . . . . . . . 2075

C.12.4 §7.8 Improper Integrals 2076

C.13 Supplementary Notes for the Lec-ture of February 16th, 2004 . . 2082C.13.1 §8.1 Arc Length . . . . 2082C.13.2 §8.2 Area of a Surface of

Revolution . . . . . . . 2086C.14 Supplementary Notes for the Lec-

ture of February 18th, 2004 . . 2089C.14.1 §8.3 Applications to Physics

and Engineering . . . . 2089C.14.2 §8.4 Applications to Eco-

nomics and Biology . . 2089C.14.3 §8.5 Probability . . . . . 2089C.14.4 §10.1 Curves Defined by

Parametric Equations . 2089C.14.5 §10.2 Calculus with Para-

metric Curves . . . . . . 2092C.15 Supplementary Notes for the Lec-

ture of March 1st, 2004 . . . . 2096C.15.1 §10.2 Calculus with Para-

metric Curves (continued)2096C.15.2 §10.3 Polar Coordinates 2099

C.16 Supplementary Notes for the Lec-ture of March 3rd, 2004 . . . . 2103C.16.1 §10.3 Polar Coordinates

(continued) . . . . . . . 2103C.16.2 §10.4 Areas and Lengths

in Polar Coordinates . . 2105C.17 Supplementary Notes for the Lec-

ture of March 8th, 2004 . . . . 2109C.17.1 §10.4 Areas and lengths

in polar coordinates (con-tinued) . . . . . . . . . 2109

C.17.2 §10.5 Conic Sections . . 2115C.18 Supplementary Notes for the Lec-

ture of March 10th, 2004 . . . . 2116C.18.1 §11.1 Sequences . . . . . 2116

C.19 Supplementary Notes for the Lec-ture of March 15th, 2004 . . . . 2120C.19.1 §11.2 Series . . . . . . . 2120

C.20 Supplementary Notes for the Lec-ture of March 17th, 2004 . . . . 2125


C.20.1 §11.3 The Integral Testand Estimates of Sums . 2125

C.21 Supplementary Notes for the Lec-ture of March 22nd, 2004 . . . 2129C.21.1 §11.4 The Comparison Tests2129

C.22 Supplementary Notes for the Lec-ture of March 24th, 2004 . . . . 2133C.22.1 §11.5 Alternating Series 2133

C.23 Supplementary Notes for the Lec-ture of March 29th, 2004 . . . . 2135C.23.1 §11.6 Absolute Conver-

gence and the Ratio andRoot Tests . . . . . . . 2135

C.24 Supplementary Notes for the Lec-ture of March 31st, 2004 . . . . 2139C.24.1 §11.7 Strategy for Test-

ing Series . . . . . . . . 2139C.25 Supplementary Notes for the Lec-

ture of April 5th, 2004 . . . . . 2140

D Problem Assignments from Previ-ous Years 3001D.1 1998/1999 . . . . . . . . . . . . 3001

D.1.1 Assignment 1 . . . . . . 3001D.1.2 Assignment 2 . . . . . . 3001D.1.3 Assignment 3 . . . . . . 3002D.1.4 Assignment 4 . . . . . . 3002D.1.5 Assignment 5 . . . . . . 3002

D.2 1999/2000 . . . . . . . . . . . . 3003D.2.1 Assignment 1 . . . . . . 3003D.2.2 Assignment 2 . . . . . . 3004D.2.3 Assignment 3 . . . . . . 3006D.2.4 Assignment 4 . . . . . . 3007D.2.5 Assignment 5 . . . . . . 3009D.2.6 Assignment 6 . . . . . . 3010

D.3 2000/2001 . . . . . . . . . . . . 3012D.4 2001/2002 . . . . . . . . . . . . 3012D.5 MATH 141 2003 01 . . . . . . . 3013

E Final Examinations from PreviousYears 3014E.1 Final Examination in Mathemat-

ics 189-121B (1996/1997) . . . 3014

E.2 Final Examination in Mathemat-ics 189-141B (1997/1998) . . . 3015

E.3 Supplemental/Deferred Exami-nation in Mathematics 189-141B(1997/1998) . . . . . . . . . . . 3017









E.12 Final Examination in MATH 1412003 01 . . . . . . . . . . . . . 3032

E.13 Supplemental/Deferred Exami-nation in MATH 141 2003 01 . 3034

F WeBWorK 4001F.1 Frequently Asked Questions (FAQ)4001

F.1.1 Where is WeBWorK? 4001F.1.2 Do I need a password to

use WeBWorK? . . . . 4001F.1.3 Do I have to pay an ad-

ditional fee to use WeB-WorK? . . . . . . . . . 4002

F.1.4 When will assignments beavailable on WeBWorK?4002


F.1.5 Do WeBWorK assign-ments cover the full rangeof problems that I shouldbe able to solve in thiscourse? . . . . . . . . . 4002

F.1.6 May I assume that thedistribution of topics onquizzes and final exami-nations will parallel thedistribution of topics inthe WeBWorK assign-ments? . . . . . . . . . . 4002

F.1.7 WeBWorK provides fordifferent kinds of “Dis-play Mode”. Which shouldI use? . . . . . . . . . . 4002

F.1.8 WeBWorK provides forprinting assignments in“Portable Document Format”(.pdf) or “PostScript” (.ps)form. Which should I use?4003

F.1.9 What is the relation be-tween WeBWorK andWebCT? . . . . . . . . . 4003

F.1.10 Which browser should Iuse for WeBWorK? . . 4003

F.1.11 What do I have to do onWeBWorK? . . . . . . 4004

F.1.12 How can I learn how touse WeBWorK? . . . . 4004

F.1.13 Where should I go if Ihave difficulties with WeB-WorK ? . . . . . . . . . 4004

F.1.14 Can the WeBWorK sys-tem ever break down ordegrade? . . . . . . . . . 4005

F.1.15 How many attempts mayI make to solve a partic-ular problem on WeB-WorK? . . . . . . . . . 4005

F.1.16 Will all Regular WeB-WorK assignments havethe same length? the samevalue? . . . . . . . . . . 4005

F.1.17 Is WeBWorK a good in-dicator of examination per-formance? . . . . . . . . 4006

G Contents of the DVD disks forLarson/Hostetler/Edwards 5001

H References 5001H.1 Stewart Calculus Series . . . . 5001H.2 Other Calculus Textbooks . . . 5002

H.2.1 R. A. Adams . . . . . . 5002H.2.2 Larson, Hostetler, et al. 5002H.2.3 Edwards and Penney . . 5002H.2.4 Others, not “Early Tran-

scendentals” . . . . . . . 5003H.3 Other References . . . . . . . . 5003

Information for Students in MATH 141 2004 01: 1

1 General Information

Distribution Date: This version was mounted on the Web on 6 January, 2004(all information is subject to change)

Pages 1 - 18 of these notes may be considered the Course Outline for this course.

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

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

or on WebCT, at

http://www.mcgill.ca/webct or http://webct.mcgill.ca

1.1 Instructors and Times

INSTRUCTOR: Prof. N. Sancho Prof. W. G. Brown(Course Coordinator)

LECTURE SECTION: 1 2CRN: 856 857

OFFICE: BURN 1130 BURN 1224OFFICE HOURS: MW 09:45 - 11:15 W 13:20 - 14:15

(subject to or by appointment F 10:00 - 11:00change) or by appointment

TELEPHONE: 398–3823 398–3836E-MAIL: SANCHO@ BROWN@

MATH.MCGILL.CA MATH.MCGILL.CACLASSROOM: ADAMS AUD ADAMS AUD

CLASS HOURS: MWF 8:35–9:25 h. MW 16:35–17:55 h.

Table 1: Instructors and Times

Please note that the University considers Tuesday, April 13th, 2003, to be “a Wednes-day for timetable purposes”; this will be the date of the last lectures in both sections ofMATH 141 2004 01.


1.2 Calendar Description

1.2.1 Calendar Description

MATH 1411 CALCULUS II. (4 credits; 3 hours lecture; 2 hours tutorial) (Not opento students who have taken MATH 121 or CEGEP objective 00UP or equivalent; notopen to students who have taken or are taking MATH 130 or MATH 131, except bypermission of the Department of Mathematics and Statistics. Prerequisites: MATH 139or MATH 140 or MATH 150. Each Tutorial section is enrolment limited.) The definiteintegral. Techniques of integration. Applications. Introduction to sequences and series.

1.2.2 Late transfer from MATH 151

Some students from MATH 151 may be permitted to transfer into MATH 141 after theend of the Change of Course Period. If you are in this category, please send an e-mailmessage to Professor Brown as soon as your transfer has been approved.2

1.3 Tutorials; Tutors’ Coordinates

Every student must be registered in one lecture session and one tutorial for this course.Tutorials begin in the week of January 12th, 2004. The last tutorial in all Tutorialsections will be in the week beginning Monday, April 5th, 2004; Table 2 below givestimes, locations, and the tutor’s name for each of the tutorials; Table 3 gives the tutors’coordinates. The information in these tables is subject to change. We try topublicize changes but sometimes we are not informed in advance.3

Tutors The tutors in MATH 141 2004 01 are graduate students in Mathematics andStatistics. Like you, they are students, albeit at the graduate level; they have deadlinesand commitments and personal lives, and the time they have available for MATH 141is limited and controlled by a collective agreement (union contract). Please respect theimportant functions that our tutors provide, and do not ask them to perform servicesthey are not expected to perform:

• Tutors are not expected, nor authorized, to administer a special quiz or a quiz thathas already been administered to others outside of the normal quiz times in theirtutorials.

1The 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.

2This is to ensure that your WeBWorK account is opened, and that your date of entry to the courseis recorded.

3The current room for your tutorial should always be available by clicking on “Class Schedule” onMINERVA FOR STUDENTS, http://www.mcgill.ca/minerva-students/.


• Tutors are not expected, nor authorized, to accept a Written Assignment after theweek when it was due.

• Tutors in MATH 141 2004 01 are not permitted to offer paid, private tuition tostudents in any tutorial section of this course.

If you miss a quiz or assignment for a valid reason (medical or otherwise), please com-municate with Professor Brown, providing a copy of the medical or other supportingdocuments.

1.4 Evaluation of Your Progress

1.4.1 Your final grade

(See Table 5, p. 10) Your grade in this course will be a letter grade, based on a percentagegrade computed from the following components:

1. Assignments submitted over the Web: Ten “Regular” WeBWorK homework as-signments (cf. §1.4.3) — counting together for 10% The Regular Assignments willbe numbered #R1, . . . #R10.

2. Materials graded by your Tutor:

• Five Written Assignments — counting together for 5%. The Written Assign-ments will be numbered #W1, . . . , #W5.

• Five Quizzes given at the tutorials — counting together for 15%. The Quizzeswhich count will be numbered #Q1, . . . , #Q5.

3. The final examination — counting for 70%.

Where a student’s performance on the final examination is superior to her performanceon the tutorial quizzes, the final examination grade will replace the quiz grades in thecalculations. It is not planned to permit the examination grade to replace the grades onWeBWorK assignments or on written assignments.

1.4.2 WeBWorK

We will be using the WeBWorK system, developed at the University of Rochester —which is designed to expose you to a large number of drill problems, and where plagiarismis discouraged. WeBWorK is accessible only over the Web. Details on how to sign onto WeBWorK are contained in Appendix F to these notes, page 4001.

WeBWorK assignments carry a due date and time; only answers submitted by thisdate and time will count.

UPDATED TO April 8, 2004


# CRN Day Begins Ends Room Tutor

T003 858 Tue 08:05 09:55 BURN 1214 S. NashaatT004 859 Tue 12:05 01:55 BURN 1214 M. Al-KhaleelT005 860 Tue 14:05 15:55 BURN 1214 P.-J. BergeronT006 861 Tue 16:05 17:55 BURN 1214 J. CaoT007 862 Tue 16:05 17:55 BURN 1B39 S. ShahabiT008 863 Thurs 14:35 16:25 BURN 920 Y. HanT009 864 Thurs 16:05 17:55 BURN 1214 P. PoulinT010 865 Thurs 16:05 17:55 BURN 1B24 R. KolhatkarT011 866 Mon 13:35 15:25 BURN 1214 M. Fortin-BoisvertT012 867 Mon 14:35 16:25 ARTS W-20 X. LiuT013 868 Mon 14:35 16:25 ADAMS 5 C. FoleyT014 869 Wed 13:35 15:25 BURN 1214 C. FortinT015 870 Wed 14:35 16:25 BURN 1B23 M. HosseiniT016 871 Wed 14:35 16:25 ARTS W-20 G. PaekT017 3583 Mon 13:35 15:25 ENGMD 280 M. SohrabiT018 3584 Wed 13:35 15:25 ENGMD 280 M. Sababheh

Table 2: Schedule and Locations of Tutorials, as of April 8, 2004 (subject to change)

Tutor E-mail address Office Office HoursDay Begins Ends Day Begins Ends

Al-Khaleel, M. [email protected] 1019 Th 14:00 17:00

Bergeron, P.-J. [email protected] 1030 M 09:30 12:30

Cao, J. [email protected] 1218 T 15:00 16:00 T 18:00 20:00

Foley, C. [email protected] 1134 M 09:30 11:30 T 09:00 10:00

Fortin, C. [email protected] 1032 Th 13:00 16:00

Fortin-Boisvert, M. [email protected] 1008 M 12:30 13:30 M 15:30 17:30

Han, Y. [email protected] 1117 T 11:00 12:30 Th 11:00 12:30

Hosseini, M. [email protected] P+P Th 08:00 11:00

Kolhatkar, R. [email protected] 1007 T 08:30 11:30

Liu, X. [email protected] 1020 M 11:30 14:30

Nashaat, S. [email protected] 1115 W 13:00 16:00

Paek, G. [email protected] 1017 W 09:30 11:00 F 09:30 11:00

Poulin, Ph., [email protected] 1032 T 12:00 15:00

Sababheh, M. [email protected] 1021 W 10:30 13:30

Shahabi, S. [email protected] 805 T 15:00 16:00 F 16:00 18:00

Sohrabi, M. [email protected] 805 T 14:30 15:30 F 13:30 15:30

Table 3: Tutors’ Coordinates, as of April 8, 2004


Some of the conditions described below may appear to be strict. They are designedto reduce the chance that you fail the course, not to impose unreasonably bureaucraticrules.

1.4.3 Regular WeBWorK Assignments

There will be ten “Regular” WeBWorK assignments. These will be paired — eacheven numbered assignment (##R2, R4, R6, R8, R10) will contain the same types ofproblems as on the preceding odd numbered assignment (##R1, R3, R5, R7, R9). Youwill have an unlimited number of tries on the problems on the odd numbered assignments;but there will be restricted numbers of tries on the even numbered assignments. Theintention is that you should use each odd numbered assignment to thoroughly learn howto solve the problems, and then attempt the following even numbered assignment. Thedata on the even numbered assignment will probably be different — different numbersand/or functions, but the same concepts.4 It is expected that the due date for Regularassignments will be on specified Mondays, at midnight. As mentioned in the WeBWorKFAQ (cf. Appendix §F) if you leave your WeBWorK assignment until the hours closeto the due time on the due date, you should not be surprised if the system is slow torespond. This is not a malfunction, but is simply a reflection of the fact that otherstudents have also been procrastinating! To benefit from the speed that the system candeliver under normal conditions, do not delay your WeBWorK until the last possibleday! If a systems failure interferes with the due date of an assignment, arrangements maybe made to change that date, and an e-mail message may be broadcast to all users (to thee-mail addresses on record), or a note posted in the course announcements on WebCT;but slowness in the system just before the due time will not normally be considered asystems failure.5

BONUS assignment It is hoped that there will be one additional WeBWorK as-signment, which will be optional . The grade on this assignment would replace the gradeon any other WeBWorK assignment with a lower grade, or replace a missed assign-ment. This assignment will have limits on the numbers of attempts. Further details willbe announced in the lectures or on the WebCT or WeBWorK sites.

4The 10 assignments will count equally in your WeBWorK grade, provided you have received a gradeof at least 75% on one Precalculus Assignment; if you have not received that grade on any PrecalculusAssignment by the end of the term, your WeBWorK grade will be 0.

5Should 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 beconsidered a systems failure.


1.4.4 Written Assignments

Written assignments will be posted at about the same time as regular WeBWorKassignments ##R2, R4, R6, R8, R10. These assignments will contain one or moreproblems for which you will be expected to write full solutions, modelled on similartypes of solution in worked examples in the textbook, or solved problems in the StudentSolutions Manual. You will be assigned an individualized version of the problems. Youshould prepare these solutions and carry them to the Tutorial Quiz the following week;you will be expected to hand in your solutions with your quiz paper, and they will begraded by your tutor and returned to you with that graded quiz paper. Please do notattempt any other methods for submitting your written assignments, to minimize therisk of loss.

Written Where to Find How to Submit Date ofAssignment Your Version of the Questions Your Solutions Quiz

#W1 on WeBWorK site with Quiz #Q1 Jan. 19-22#W2 on WeBWorK site with Quiz #Q2 Feb. 02-05#W3 on WeBWorK site with Quiz #Q3 Feb. 16-19#W4 on WeBWorK site with Quiz #Q4 Mar. 08-11#W5 on WeBWorK site with Quiz #Q5 Mar. 22-25

Table 4: Timetable for Written Assignments

1.4.5 Quizzes at the Tutorials; Submission of Written Assignments at theQuizzes

1. There will be 5 short quizzes, numbered #Q1, #Q2, #Q3, #Q4, #Q5, adminis-tered at the tutorials. These quizzes will be graded, and returned (together withthe written assignment that will be handed in with the quiz paper). The primarypurpose of a quiz is to diagnose possible gaps in your understanding, not to drillon examination skills.

No provision is being made for students who miss a quiz. The grading formula per-mits the quiz component of the final grade to be replaced by the final examinationgrade, if this is to a student’s advantage.

2. The quizzes will be based on current topics in the syllabus of the course, most6 ofwhich topics will have been discussed in the lectures before the quiz; the quizzesare not based directly on WeBWorK assignments. To prepare for a quiz you

6but possibly not all


should be working exercises in the textbook based on the material currently underdiscussion at the lectures, and you should have attempted any open WeBWorKassignments. But, unlike the WeBWorK assignments — where the emphasis ison correct answers alone — students may be expected to provide full solutions tosome or all problems on quizzes.7

The quiz time available in MATH 141 is not sufficient to examine on all topics;consequently the quizzes will examine on only a sampling of the topics. Studentsshould not assume that topics not examined are in any subsidiary parts of thesyllabus. Note also that the last quiz will be administered before the full syllabushas been discussed in lectures and/or tutorials, so a portion of the syllabus willnot be examined in quizzes or assignments.

3. You are expected to write the quiz in the tutorial section in which you are registered.Should the classroom become filled while some registered students still cannot beseated, the tutor may insist that any students who are not registered in that tutorialto leave the room.8

4. Submission of Written Assignments at Quizzes Individualized written as-signments ##W1, W2, W3, W4, W5 will be mounted on the WeBWorK siteabout a week before they are to be handed in at a tutorial quiz; you should down-load or copy each written assignment from the web site, write out your solutionsat home, bring the completed assignment with you to the tutorial, and hand it inwith your quiz paper — no other submission method is acceptable. The assignmentmust be ready for submission when you arrive, as you will not be provided time inthe tutorial room to complete it. You will enclose the written assignment in yourfolded answer paper.

5. Your tutors will normally bring graded quizzes and graded assignments submittedwith them to the tutorial to be returned to you. University regulations do notpermit us to leave unclaimed materials bearing names and student numbers inunsupervised locations; you may be able to recover an unclaimed quiz from thetutor who graded it, during her/his regular office hours.

7In Math 141 the general rule for quizzes is that full solutions are expected to all problems, unlessyou receive explicit instructions to the contrary: ALWAYS SHOW YOUR WORK! The solutions in theStudent Solution Manual [3] to the textbook can serve as a guide to what should be included in a “full”solution.

8Anyone who is not registered, and who does not leave the room when so requested may forfeit theright to write quizzes at any tutorial, and could be subject to disciplinary action through their Faculty.We ask you to respect the prior right of students who have registered for each tutorial.


1.4.6 Final Examination

A 3-hour-long final examination will be scheduled during the regular examination periodfor the winter term (April 15th, 2004 through April 30th, 2004). You are advised not tomake any travel arrangements that would prevent you from being present on campus atany time during this period.

1.4.7 Supplemental Assessments

Supplemental Examination. There will be a supplemental examination in this course.(For information about Supplemental Examinations, see the McGill Calendar, [27, §8.1,p. 250-251; or §8.1, p. 52-53].)

There is No Additional Work Option. “Will students with marks of D, F, or Jhave the option of doing additional work to upgrade their mark?” No. (“AdditionalWork” refers to an option available in certain Arts and Science courses, but not availablein this course.)

1.4.8 Machine Scoring

“Will the final examination be machine scored?” It is possible that the final examinationwill contain some questions that will be machine scored. Not more than 20% of themarks available on the examination — possibly less — will derive from problems thatare machine scored. It is likely that some parts of the final examination will consist ofquestions where only you need not show your work, but where you should not expectpart marks for an answer that is not perfectly correct.

1.4.9 Plagiarism and Fraud

Plagiarism While students are not discouraged from discussing methods for solvingWeBWorK assignment problems with their colleagues, all the work that you submit— whether through WeBWorK or written assignments, or on tutorial quizzes or thefinal examination must be your own. It is a violation of University regulations to permitothers to solve your WeBWorK problems, or to extend such assistance to others; youcould be asked to sign a statement attesting to the originality of your work.

The Handbook on Student Rights and Responsibilities states in ¶15(a)9 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,project or assignment submitted in a course or program of study or represent

9http://upload.mcgill.ca/secretariat/G1547E.pdf


as his or her own an entire essay or work of another, whether the material sorepresented constitutes a part or the entirety of the work submitted.”

You are also referred to the following URL:

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

The preceding paragraph was prepared before all faculty were advised that the Senateof the University requires the following message in all course outlines:

“McGill University values academic integrity. Therefore all students must under-stand the meaning and consequences of cheating, plagiarism and other academicoffences under the Code of Student Conduct and Disciplinary Procedures. (Seehttp://www.mcgill.ca/integrity for more information).

“L’universite McGill attache une haute importance a l’honnetete academique. Ilincombe par consequent a tous les etudiants de comprendre ce que l’on entendpar tricherie, plagiat et autres infractions academiques, ainsi que les consequencesque peuvent avoir de telles actions, selon le Code de conduite de l’etudiant et desprocedures disciplinaires. (Pour de plus amples renseignements, veuillez consulterle site http://www.mcgill.ca/integrity).”

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

1.4.10 Corrections to grades

Posted Grades Grades will be eventually posted on WebCT. If you believe a gradehas been recorded incorrectly, you must advise your tutor not later than 4 weeks afterthe grade has been posted, and not later than the day before of the final examinationwhichever of these dates is earlier. It is hoped that grades will be posted within 2 weeksof the due date. You will have to present the graded quiz or assignment to supportyour claim, which must be submitted to the tutor that graded the quiz or assignment.If he/she believes there has been an error, the tutor will advise Professor Brown. Newcorrections to the WebCT posting will appear the next time grades are uploaded toWebCT.

1.5 Published Materials

1.5.1 Required Text-Book

The textbook for the course is J. Stewart, SINGLE VARIABLE CALCULUS: EarlyTranscendentals, Fifth Edition, Brooks/Cole (2003), ISBN 0-534-39330-6, [1].


Item # Due Date Details

RegularWeBWorK

R1 26 Jan 04 No limit to number of attempts at WeBWorK R1problems

Assignments(cf. §1.4.3)

R2 02 Feb 04 Same scope as R1, but with limited numbers of at-tempts

R3 09 Feb 04R4 16 Feb 04 Same scope as R3, but with limited numbers of at-

tempts

10%...

R9 29 Mar 04R10 05 Apr 04 Same scope as R9, but with limited numbers of at-

temptsBONUS 13 Apr 04 see page 5, limited numbers of attempts

Written W1 with Q1 Download; complete at home; hand in with Q1Assignments W2 with Q2 Download; complete at home; hand in with Q2(cf. §1.4.4) W3 with Q3 Download; complete at home; hand in with Q3

W4 with Q4 Download; complete at home; hand in with Q45% W5 with Q5 Download; complete at home; hand in with Q5Quizzes(cf. §1.4.5) Q1 19–22 Jan 04

15% or 0%...

Q5 22–25 Mar 04Final Exam 15–30 Apr 04 Date of exam to be announced by Faculty70% or 85%SupplementalExam

25–26 August04

Only for students who do not obtain standing at thefinal. Supplemental exams count in your average liketaking the course again; exam counts for 100%.

Table 5: Summary of Course Requirements, as of April 8, 2004 (all dates are subject tochange)

This book is the first half of J. Stewart, CALCULUS: Early Transcendentals,Fifth Edition, Brooks/Cole (2003), ISBN 0-534-39321-7, [2]; this edition covers thematerial for Calculus III (MATH 222) as well, but is not the text-book for that courseat the present time.

1.5.2 Optional Reference Books

It is recommended that students make use of the student solution manual:

• D. Anderson, J. A. Cole, D. Drucker, STUDENT SOLUTIONS MANUAL


FOR STEWART’S SINGLE VARIABLE CALCULUS: Early Transcen-dentals, Fifth Edition, Brooks/Cole (2003), ISBN 0-534-39333-0, [3]. Thisbook is also sold “bundled” with either version of the text book; the bundles arenumbered ISBN 0-534-42976-9 [4] and ISBN 0-534-10307-3 [5].

The publishers of the textbook and Solutions Manual also produce

• a “Study Guide”, designed to provide additional help for students who believethey require it: R. St. Andre, STUDY GUIDE FOR STEWART’S SIN-GLE VARIABLE CALCULUS: Early Transcendentals, Fifth Edition,Brooks/Cole (2003), ISBN 0-534-39331-4, [6]. (The “Study Guide” resembles theStudent Solution Manual in appearance: 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 [26].

1.5.3 Recommended Video Materials

Use of the following materials is recommended, but is not mandatory10.

Larson/Hostetler/Edwards DVD Disks A set of video DVD disks produced foranother calculus book, [18] Calculus Instructional DVD Program, for use with (interalia) Larson/Hostetler/Edwards, Calculus of a Single Variable: Early TranscendentalFunctions, Third Edition [19] is produced by the Houghton Mifflin Company. A copyhas been requested to be placed on reserve in the Schulich Library, and the set of 4 disksis for sale in the bookstore among the reference materials for this course. In Appendix§G of these notes there are charts that indicate the contents of these disks that pertainto MATH 140.

Videotapes for Stewart’s Calculus The publisher of Stewart’s Calculus has pro-duced a series of videotapes, [7] Video Outline for Stewart’s Calculus (Early Transcen-dentals), Fifth Edition. These will initially be available for reserve loan at the SchulichLibrary. There may not be VCR viewing equipment in the library; the intention is thatinterested students borrow a tape for viewing on their own equipment at home.

10No one will check whether you have used any of these aids; a student can obtain a perfect gradein the course without ever consulting any of them. No audio-visual or calculator aid can replace thesystematic use of paper and pencil as you work your way through problems. But the intelligent use ofsome of these aids can deepen your understanding of the subject.


Tools for Enriching Calculus This is a CD-ROM included with new copies of Stew-art’s Calculus. From [9, Introduction]:“Tools for Enriching Calculus (TEC) enhances a topicthat is covered in the textbook by providing both broader and deeper coverage of those aspectsfor which technology is particularly useful. The basic format of most modules is a point-and-click laboratory environment in which you can easily visualize functions and their derivatives,experiment with suggested examples and exercises, explore your own choices of examples, andperhaps even test some of your own conjectures. You need to be a well-prepared and activeplayer to reap the benefits from these approaches.

“First, you need to read the textbook materials carefully to gain an understanding of theessential ideas. Next, you need to read the introductory material for each TEC module, whichexplains the basic mathematical approaches and describes how to use the module. Each modulehas several examples which will familiarize you with its basic features. When you have finishedreviewing this material, and have some paper and pencil in hand, you are ready to get the mostbenefit from using the module. You can improve your understanding of the topic by exploringmathematical questions that you find puzzling, and checking your ideas for solutions using themodule. Try to work through some of the exercises in the module to gauge your understandingof the topic. Be willing to use pencil and paper to first guess what the answer might be beforeseeing an electronic graph...

“Another important TEC feature is the homework11 hints. Hints have been created forseveral selected exercises in each section of your textbook to help you understand some keypoints in finding solutions for these exercises. Similar to a good instructor or teaching assistant,these hints ask you questions that will allow you to make progress toward a solution withoutgiving you the actual answer. You need to actively pursue each hint with pencil and paper andfill in many of the computations and details. If you can complete the solution after readingonly one or two hints, you can feel proud of your achievements. If you still have questions aftercompleting all of the hints for a problem, your work should help you to better understand thesolution presented in the Student Solutions Manual.”

Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcen-dentals, 5th Edition [8] This CD-ROM is included with new copies of the textbook.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. Somestudents may find the animations of the examples helpful, although the examples are allworked in the book. You might wish to try some of the quiz questions using paper andpencil, and then check your answers with those given on the CD. It is not recommendedthat you attempt to enter your answers digitally, as this is a time-consuming process,and uses a different input method from your WeBWorK assignments, which serve thesame purpose.

11Note that homework refers to odd-numbered problems in the exercises in [1], not to problems onWeBWorK .


1.5.4 Other Calculus Textbooks

While students may wish to consult other textbooks, instructors and teaching assistantsin Math 141 will normally refer only to the prescribed edition of the prescribed textbookfor the course. Other books can be very useful, but the onus is on you to ensure thatyour book covers the syllabus to at least the required depth; where there are differencesof terminology, you are expected to be familiar with the terminology of the textbook.12

In your previous calculus course(s) you may have learned methods of solving problemsthat appear to differ from those you find in the current textbook. Your instructors willbe pleased to discuss any such methods with you personally, to ascertain whether theyare appropriate to the present course. In particular, any methods that depend uponthe use of a calculator, or the plotting of multiple points, or the tabulation of functionvalues, or the inference of a trend from a graph should be regarded with scepticism.

1.5.5 Website

These notes, and other materials distributed to students in this course, will be accessibleat 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,which many users have on their computers. This free software may be downloaded fromthe following URL:

http://www.adobe.com/prodindex/acrobat/readstep.html 13

The questions on some old examinations will also be available as an appendix to thesenotes on the Web.14 It is expected that most computers in campus labs should have thenecessary software to read the posted materials.

Where revisions are made to distributed printed materials — for example these in-formation sheets — 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 WebCT URL:

http://webct.mcgill.ca

Your grades on assignments and quizzes will be posted on WebCT within about 2 weeksafter they become available. Some other features of WebCT15 have not yet been imple-mented.

12There should be multiple copies of the textbook on reserve in the Schulich library.13At the time of this writing the current version appears to be 6.0.14There is no reason to expect the distribution of problems on quizzes or in assignments and exami-

nations from previous years be related to the frequencies of any types of problems on the examinationthat you will be writing at the end of the term.

15cf. Appendix F to these notes , p. 4001


1.6 Syllabus

In the following list section numbers refer to the text-book [1]. The syllabus will includeall of Chapters 5, 6, 7, 8, 10, 11 with omissions, as listed below.16

Chapter 5: Integrals. §§5.1 – 5.6. The derivation17, in §5.6, of properties of the log-arithm and integral is not examination material. The Midpoint Rule, defined in§5.2, and appearing from time to time subsequently, is not examination material.

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

Chapter 7: Techniques of Integration. §§7.1 – 7.5; §7.8. (§7.6, intended for use inconjunction with integral tables and/or computer algebra systems, is not examina-tion material, but students are to try to solve the problems manually; §7.7 is notexamination material.)

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

Chapter 9: Differential Equations. (No part of this chapter is examination mate-rial; however, 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, 10.7 are not examination material.)

Chapter 11. Infinite Sequences and Series. §§11.1 – 11.7. (§§11.8–11.12 are notexamination material; however, students are urged to peruse these sections.)

Please do not ask the tutors to provide information as to which textbook sections shouldbe emphasized. Unless you are informed otherwise by the instructors in the lecturesections or published notes — printed, or mounted on the Web — you should assumethat all materials listed are included in the syllabus. You are not expected to be able toreproduce proofs of the theorems in the textbook.

16If a textbook section is listed below, you should assume that all material in that section is examina-tion material even if the instructor has not discussed every topic in his lectures; however, the instructorsmay give you information 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 aWeBWorK assignment or a quiz, or if it has not appeared on any of the old examinationsin the course! Some topics to not lend themselves to this type of testing; others may have beenomitted simply because of lack of space, or oversight. By the same token, you need not expect everytopic in the course to be examined on the final examination.

17But N.B., students are expected to know the properties themselves, which were discussed in MATH139 and MATH 140.


1.7 Preparation and Workload

1.7.1 Prerequisites.

It is your responsibility as a student to verify that you have the necessary prerequisite.It would be foolish18 to attempt to take the course without it.

Students who obtained only a grade of C in MATH 140 would be advised to make aspecial effort to reinforce their foundations in differential calculus; if weakness in MATH140 or MATH 139 was a consequence of poor preparation for that course, it is not toolate to strengthen those foundations as well.

In the past students with a grade of D in MATH 140 or MATH 139 have beenpermitted to take MATH 141 at their own risk, provided they have also registered forMATH 140 or have applied to write the supplemental examination in the course in whichthey obtained the D. Students with only a D, or students whose examination in MATH140 has been deferred, who contemplate registering for MATH 141 should recognize thatit places a large number of credits at risk, and could have a substantial negative effect ontheir GPA if they have overestimated their ability; it is safer to obtain “standing” (gradeC or better) in MATH 139 or MATH 140 before registering for MATH 141. Studentswith a grade of F in MATH 139 or MATH 140 are not authorized to registerin MATH 141 2004 01, even if they plan to write a supplemental examinationin MATH 140 or are taking MATH 140 concurrently.

1.7.2 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 beadvised to make particular efforts to avoid the use of a calculator in solving problemsin this course, in order to develop a minimal facility in manual calculation. This meansthat you are urged to do all arithmetic by hand.

1.7.3 Self-Supervision

This is not a high-school course, and McGill is not a high school. The monitoring ofyour progress before the final examination is largely your own responsibility. While thetutors and instructors are available to help you, they cannot do so unless and until youidentify the need for help. WeBWorK and quizzes are designed to assist you in doingthis.

Time Demands of your Other Courses. Be sure to budget enough time to attendlectures and tutorials, for private study, and for the solution of many problems. Don’t be

18and contrary to McGill regulations


tempted to divert calculus study time to courses which offer instant gratification. Whilethe significance of the tutorial quizzes in the computation of your grade is minimal, theseare important learning experiences, and can assist you in gauging your progress in thecourse. This is not a course that can be crammed for: you must work steadily throughthe term if you wish to develop the facilities needed for a strong performance on the finalexamination.

Working Problems on Your Own. An effective way to master the calculus isthrough working large numbers of problems from the textbook. Your textbook wasselected partly because of the availability of an excellent Student Solutions Manual [3];this manual has brief but complete solutions to most of the odd-numbered exercises inthe textbook. The skills you acquire in solving textbook problems could have much moreinfluence on your final grade than either WeBWorK or the quizzes.

The real uses of WeBWorK and the quizzes. Students often misunderstand thetrue significance of WeBWorK assignments and the quizzes. While both contributeto your grade, they help you estimate the quality of your progress in the course. Takeproper remedial action if you are obtaining low grades on quizzes19, or if you requiremany attempts before being able to solve a problem on WeBWorK. However, whileboth WeBWorK and the quizzes have a role to play in learning the calculus, neitheris as important as reading your textbook, working problems yourself, and attending andlistening at lectures and tutorials.

1.7.4 Escape Routes

At any time, even after the last date for dropping the course, students who are experi-encing medical or personal difficulties should not hesitate to consult their advisors or theStudent Affairs office of their faculty. Don’t allow yourself to be overwhelmed by suchproblems; the University has resource persons who may be able to help you.

1.7.5 Terminology

Do not be surprised if your instructors and tutors use different terminology from whatyou have heard in your previous calculus course, particularly if that course was at a highschool. 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 as negative x. To a mathematician the term negative refers to real numbers whichare not squares, i.e. which are less than 0, and −x can be positive if x itself is negative.

19The worst action is to miss the quizzes, and thereby block out an unwelcome message.


However, mathematicians will sometimes refer to the operation of changing a signas the replacement of x by “its negative”; this is not entirely consistent with the usualpractice, but is an “abuse of language” that has crept into the professional jargon.

Inverse trigonometric functions A formula like sin−1 x will be read as the inversesine of x — never as sine to the minus 1 or sine to the negative 1 . However, if we writesinn x, where n is a positive integer, it will always mean (sin x)n. These conventionsapply to any of the functions sin, cos, tan, cot, sec, csc; they also apply to the hyperbolicfunctions, which we have met in [1, §3.9]: sinh, cosh, tanh, coth.... We will usually notwrite exponents on general functions, so a formula like f 2(x) does not have an obviousmeaning, and we will avoid writing it when f is other than a trigonometric or hyperbolicfunction.

Logarithms These days mathematicians rarely use logarithms to the base 10. If youwere taught to interpret log x as being the logarithm to base 10, you should now forgetthat — although it could be the labelling convention of your calculator. Most often,if your instructor speaks of a logarithm, and writes log x, he will be referring to thebase e, i.e. to loge; that is, he is referring to the function that calculus books call ln.When a logarithm to some other base is intended, it will either be denoted by an explicitsubscript, as log2, or some comment will be made at the beginning of the discussion, as“all logarithms in this discussion are to the base 2”. Your instructors try to think likemathematicians even when lecturing to their classes, and so we use the language andterminology we use when talking to each other.

1.8 High and Low Technology and MATH 141

1.8.1 Keep your e-mail address up to date

Both WebCT and WeBWorK contain an e-mail address where we may assume you canbe reached. If you prefer to use another e-mail address, the most convenient way is toforward your mail from your student mailbox, leaving the recorded addresses in thesetwo systems unchanged. You can enter or change a forwarding e-mail address by goingto http://webmail.mcgill.ca, and logging in to your student mailbox at po-box.mcgill.ca.

1.8.2 Use of Calculators and Computer Algebra Systems

Insofar as course content is concerned, we will emphasize the lowest possible use oftechnology: we will avoid the use of calculators and computers. You are urged to do allcalculations manually, and to avoid the use of computer algebra systems, until you havecompleted MATH 141. You should not use a calculator or computer in the solution of


WeBWorK problems, as it prevents you from developing skill for detecting errors inmanual calculations — a skill that you will need for the quizzes and final examination.

1.8.3 Use of the Internet

However, insofar as course delivery is concerned, students are expected to be able toaccess materials through the Internet, whenever required. Here are some of the uses thatare expected:

• To access and submit Regular WeBWorK (§1.4.3) assignments

• To access the individualized problems for a student’s Written Assignments

• To access WebCT, where course grades and announcements will be mounted

• To access the web page for the course — also available through WebCT — wherethese and other notes will be available in .pdf form; the site also contains notesand examinations from previous years.

1.9 Special Review Sessions

All tutorials hold their last sessions during the week 05-08 April, 2003. The followingreview sessions are planned at the end of the term:

(subject to revision)Date Instructor Location Starts Ends


Information for Students in Lecture Section 1 of MATH 141 2004 01 1001

A Information Specifically for Students in Lecture

Section 001

A.1 Timetable for Lecture Section 001 of MATH 141 2004 01

Distribution Date: 0th version: Monday, January 5th, 2004(Subject to correction and change.)

Section numbers refer to the text-book. 20

MONDAY WEDNESDAY FRIDAY

JANUARY05 §5.1 07 §5.2 09 §5.3

Tutorials begin week of January 12th, 200412 §5.4 14 §5.5 16 §5.6

Course changes must be completed on MINERVA by Jan. 18, 200419 §6.1 Q1 W1 21 §6.2, §6.3 Q1 W1 23 §6.2,§6.3 Q1 W1

Deadline for withdrawal with fee refund = Jan. 25, 200426 §6.2,§6.3,§6.5 R1 28 §7.1 30 §7.1

FEBRUARYVerification Period: February 2–6, 2004

02 §7.2 Q2 W2 R2 04 §7.3 Q2 W2 06 §7.4 Q2 W2

Deadline for withdrawal (with W) from course via MINERVA = Feb. 15, 200409 §7.4 R3 11 §7.5 13 §7.816 §8.1 Q3 W3 R4 18 §8.2 Q3 W3 20 X Q3 W3

Study Break: February 23–27, 2004No lectures, no regular office hours, no regular tutorials!

23 NO LECTURE 25 NO LECTURE 27 NO LECTURE

(Page 1002 of the timetable will not be circulated; however, a version is available in theonline version of these notes.)

20

Notation:Rn = Regular WeBWorK Assignment #Rn due at midnight on Monday this weekR© = Read Only

Qn Wn = Quiz #Qn planned for the tutorials this weekHand in Written Assignment #Wn with your Quiz paper

X = reserved for eXpansion or review


MONDAY WEDNESDAY FRIDAYMARCH

01 §§10.1 R5 03 §10.2 05 §10.308 §10.4 Q4 W4 R6 10 §10.4 Q4 W4 12 X, §11.1 Q4 W4

15 §11.1 R7 17 §11.1 19 §11.222 §11.2 Q5 W5 R8 24 §11.2, §11.3 Q5 W5 26 §11.3 Q5 W5

29 §11.4 R9 31 §11.5, §11.6

APRIL02 §11.5, §11.6

05 §11.7 R10 07 X 09 NO LECTURE12 NO LECTURE 13 (TUESDAY) X

RBONUS


B Information Specifically for Students in Lecture

Section 002

B.1 Timetable for Lecture Section 002 of MATH 141 2004 01

Distribution Date: 0th version: Monday, January 5th, 2004(Subject to correction and change.)

Section numbers refer to the text-book.21

MONDAY WEDNESDAY

JANUARY05 §5.1, §5.2 07 §5.3

Tutorials begin week of January 12th, 200412 §5.4, §5.5 14 §5.5, §5.6

Course changes must be completed on MINERVA by Jan. 18, 200419 §6.1, §6.2 Q1 W1 21 §6.3 Q1 W1

Deadline for withdrawal with fee refund = Jan. 25, 200426 §6.5 R1 28 §7.1

FEBRUARYVerification Period: February 2–6, 2004

02 §7.2, §7.3 Q2 W2 R2 04 §7.4 Q2 W2

09 §7.4, §7.5 R3 11 §7.8, XDeadline for withdrawal (with W) from course via MINERVA = Feb. 15, 2004

16 §8.1, §8.2 Q3 W3 R4 18 §8.2, §10.1,§10.2 Q3 W3

Study Break: February23–27, 2004No lectures, no regular office hours, no regular tutorials!

23 NO LECTURE 25 NO LECTURE

(Page 2002 of the timetable will not be circulated; however, a version is available in theonline version of these notes.)

21

Notation:Rn = Regular WeBWorK Assignment #Rn due at midnight on Monday this weekR© = Read Only

Qn Wn = Quiz #Qn planned for the tutorials this weekHand in Written Assignment #Wn with your Quiz paper

X = reserved for eXpansion or review


MONDAY WEDNESDAYMARCH

01 §10.3 R5 03 §§10.3, 10.408 §10.4 Q4 W4 R6 10 §11.1 Q4 W4

15 §11.1, §11.2 R7 17 §11.322 §11.4 Q5 W5 R8 24 §11.5 Q5 W5

29 §11.6 R9 31 §11.7

APRIL05 X R10 07 X12 NO LECTURE 13 (TUESDAY) X RBONUS

Supplementary Notes for Students in Lecture Section 002 2003

C Supplementary Notes for Lectures in Lecture Sec-

tion 002

C.1 Supplementary Notes for the Lecture of January 5th, 2004

Release Date: Tuesday, January 6th, 2004,subject to further revision

Lecture Content A timetable posted on the Web will show you approximately whatis to be discussed at each lecture. It is suggested that you look through the material inadvance. If you have time to try some of the exercises, and find some that cause youdifficulty, you are welcomed to bring them to Professor Brown’s attention; perhaps hewill be able to work some of these examples into the lecture.

What goes on the chalkboard? — Should I Take Notes? Your instructor be-lieves strongly that students should not spend the lecture hour feverishly copying notesfor fear of missing some essential topic; in this course most of what you need to know iscontained in the textbook. You should take notes, but you should be trying to think atthe 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. Evenwhen the material on the board is equivalent to something in your textbook, the act ofwriting may help you remember it.

Graphs Our emphasis here is on qualitative and quantitative properties of the graphsof functions, but not on the production of extremely precise graphs. You can expect to seethe instructor draw on the chalkboard sketches that are extremely crude approximationsof functions, sometimes even caricatures of the true graph. Mathematicians do not baseproofs on sketches of graphs — the role of a sketch is usually only to assist the readerto visualize the verbal or symbolic reasoning which accompanies it. Sometimes a graphis used help one discover a phenomenon, but the result would not be acceptable to amathematician unless it could be proved in a non-graphical way.


Supplementary material on the Web Sometimes it may happen that the discussionof a topic or an exercise evolves during the lecture into one which requires more detailthan is practical to write on the chalkboard. In such cases you may be referred atthe following lecture to supplementary material that will be placed on the Web. Suchevolutions are not planned, and cannot be announced in advance.

These Supplementary Notes The following notes were prepared by your instructorfor his lectures. The section and paragraph headings follow the order of topics in thetextbook. While some of the comments or explanations may be helpful in understandingthe book, the notes are not required reading for examination purposes.

Textbook Chapter 5. INTEGRALS.

C.1.1 §5.1 Areas and Distances

The Area Problem The textbook discusses the approximation of the area under thegraph y = f(x) between x = a and x = b — more precisely, the area between thegraph, the x-axis, and the lines x = a and x = b, as the limit of a sum of areas ofnarrow rectangles. The approximation is first motivated with functions where the areacan be bounded above and below by such sums, which together converge to the samevalue as their number approaches ∞ and their width approaches 0. You should becomecomfortable using the “sigma notation”, where, for integers n1 and n2 with n1 ≤ n2, wewrite

n2∑i=n1

f(i)

to mean the sum f(n1) + f(n1 + 1) + f(n1 + 2) + . . . + f(n2) . The author introduces awell known formula for the sum of the squares of the first n “natural numbers” (i.e., thesquares of the integers 1, 2, ..., n), a topic to which he will return in the next section.

The Distance Problem Here the textbook considers what appears to be a differenttype of problem, and shows that the solution is the same type of sum met in the Areaproblem above. In this case the problem is to determine the distance travelled by aparticle moving so that its velocity at time t is a given function f(t). It will be seen thatthe distance can again be interpreted as a limit of a sum — the same sum that wouldbe seen if we attempted to determine the area under the graph y = f(t).

5.1 Exercises


C.1.2 §5.2 The Definite Integral

The formal definition of the integral involves a number of technical difficulties which weshall not consider in this course. You should read the definition the textbook gives of theintegral [1, p. 380], but you are not going to be asked to work with it in full generality;in fact the definition given in the textbook is more simplified than the definition that isnormally used for the Riemann Integral. We would need to appeal to this definition ifwe wished to formally prove all the properties that the author is going to ascribe to theintegral; but we shall not attempt to provide such proofs.

The usual definition of the integral would permit the widths of the subintervals, heredenoted by ∆x, to be different, and would then require that the largest of them shouldapproach zero. This technicality is needed for general functions, but will not be discussedfurther in this course.

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

• definite integral of f from a to b

• integral sign

• integrand

• limits of integration

• lower limit, upper limit

• Riemann sum

Evaluating Integrals Among the integrals discussed in this subsection are severalthat require the following formulæ for sums of powers of the natural numbers:

n∑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

to which we could add the following trivial result:

n∑i=1

i0 = n . (1)


These formulae are proved in an appendix, but you are not expected to have read thoseproofs. (The proofs given are “by Mathematical Induction”.22) The author discussessome properties of the “sigma” notation; these could be called the linearity propertiesof the operator Σ, and are all special cases of the following:

k+∑

i=k

(rai + sbi) = r

k+∑

i=k

ai + s

k+∑

i=k

bi .

We may have more to say about the sigma notation after we discuss [1, §5.5], where weshall encounter 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 beperformed 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+∑J=k

aJ

The Midpoint Rule The “Midpoint Rule” is an approximation formula for definiteintegrals. Use of an approximation formula entails a willingness to accept an error inthe calculation. Mathematicians normally expect to see an estimate of how good or howbad an approximation can be before recommending their use. The justification of theMidpoint Rule is contained in [1, §7.7], a section that is to be omitted from the syllabus.For that reason you are asked to omit this subsection: you will not be expected to beable to use the Midpoint Rule.

Properties of the Definite Integral The textbook lists many properties of the Defi-nite Integral, proving some of them. Some of these properties can be derived from others,or can be combined into a more general formula. So, for example we can prove that

∫ b

a

(rf(x) + sg(x)) dx = r

∫ b

a

f(x) dx + s

∫ b

a

g(x) dx (2)

∫ c

a

f(x) dx =

∫ b

a

f(x) dx +

∫ c

b

f(x) dx (3)

for any real numbers r, s, a, b, c, and these two equations are equivalent to the propertiesthat the textbook numb3ers 2, 3, 4, 5, as well as to the two equations given on [1, p.387]. Following this list of properties — which you should know and be able to use —

22Students in Math 140 this past fall were introduced to Mathematical Induction.


the textbook lists some “comparison properties”, of which the property [1, 7, p. 389]

f(x) ≥ g(x) for a ≤ x ≤ b ⇒∫ b

a

f(x) dx ≥∫ b

a

g(x) dx

implies property [1, 6., p. 389], as well as the following

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

a

f(x) dx ≤ M(b− a) . (4)

5.2 Exercises

[1, Exercise 17, 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: When ∆x = 5−1n

,

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, butwill only be able to approximate.)

[1, Exercise 22, p. 391] “Use the form of the definition of the integral...to evaluate

the integral∫ 4

1(x2 + 2x− 5) dx.”

Solution: We have ∆x = 4−1n

, xi = 1+i∆x = 1+ 3in, and xi−1 = 1+(i−1)∆x = 1+

3(i−1)n

. If we choose the sample points to be the right end-points of the subintervals,

∫ 4

1

(x2 + 2x− 5) dx

= limn→∞

n∑i=1

(x2i + 2xi − 5)∆x

= limn→∞

n∑i=1

((1 +

3i

n

)2

+ 2

(1 +

3i

n

)− 5

)3

n

= limn→∞

n∑i=1

((1 + 2 · 3i

n+

9i2

n2

)+ 2

(1 +

3i

n

)− 5

)3

n

= limn→∞

(n∑

i=1

1 +6

n

n∑i=1

i +9

n2

n∑i=1

i2 + 2n∑

i=1

1 +6

n

n∑i=1

i− 5n∑

i=1

1

)3

n


= limn→∞

(n∑

i=1

1 +6

n

n∑i=1

i +9

n2

n∑i=1

i2 + 2n∑

i=1

1 +6

n

n∑i=1

i− 5n∑

i=1

1

)3

n

= limn→∞

((1 + 2− 5)

n∑i=1

1 +6 + 6

n

n∑i=1

i +9

n2

n∑i=1

i2

)3

n

= limn→∞

(−2n +

12

n· n(n + 1)

2+

9

n2· n(n + 1)(2n + 1)

6

)3

n

= limn→∞

(−2n + 6(n + 1) +

(3n +

9

2+

3

2n

))3

n= 21

If we choose the sample points to be the left end-points, we will again obtain alimit of 21.

[1, Exercise 40, p. 393] “Evaluate the integral by interpreting it in terms of areas:10∫0

|x− 5| dx.”

Solution: The portion of the integral from 0 to 5 is a right-angled triangle whosehypotenuse is on the line y = −(x− 5), with base of length 5, and height 5, so itsarea is 52

2= 25

2. The portion of the integral from 5 to 10 is the mirror the area of

the mirror image of the triangle described above, this time with hypotenuse alongthe line y = x − 5; its area is the same as the previous one, so the value of theintegral is 25.

[1, Exercise 42, p. 392] “Evaluate1∫1

x2 cos x dx.”

Solution: If in (3)

∫ c

a

f(x) dx =

∫ b

a

f(x) dx +

∫ c

b

f(x) dx

we set c = a and b = a, we obtain

∫ a

a

f(x) dx =

∫ a

a

f(x) dx +

∫ a

a

f(x) dx

from which it follows that ∫ a

a

f(x) dx = 0 .

The present integral is of this type — the limits are equal.


This result could also have been inferred from (4), by taking b = a in (4), i.e. in

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

a

f(x) dx ≤ M(b− a) :

0 = m · 0 ≤∫ a

a

f(x) df ≤ M · 0 = 0 ,

implying that the integral is equal to 0.

[1, Exercise 58, p. 392] “Use (4) to estimate the value of the integral∫ 2

0(x3 − 3x +

3) dx.”

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

1 ≤ x3 − 3x + 3 ≤ 5 .

The length of the interval is 2 − 0 = 2. The value of the integral is, thereforebounded between 2 and 10. (Eventually we shall be able to evaluate this integralexactly, and shall be able to show that its value is

24

4− 3

2· 22 + 3 · 2 = 4− 6 + 6 = 4 .

)

Discovery Project: Area Functions


C.2 Supplementary Notes for the Lecture of January 7th, 2004

Release Date: Thursday, January 8th, 2004, subject to further revision

C.2.1 §5.3 The Fundamental Theorem of Calculus

In a number of areas of mathematics there are theorems that have acquired the name“The Fundamental Theorem of...”. The present section is devoted to a theorem, alsoknown as “The Fundamental Theorem of (the) Integral Calculus”, one part of whichrelates the value of a definite integral to antiderivatives of its integrand, and provides amethod for evaluating such integrals without the need for computing limits of compli-cated sums.

Differentiation and Integration as Inverse Processes

Theorem C.1 (The Fundamental Theorem of Calculus) Suppose that f is con-tinuous on [a, b].

1.d

dx

(∫ x

a

f(t) dt

)= f(x).

2. F ′ = f ⇒∫ b

a

f(x) dx = F (b)− F (a) .

5.3 Exercises

[1, Exercise 10, p. 402] “Use Part 1 of the Fundamental Theorem of Calculus to find

the derivative of the function g(u) =

∫ u

3

1

x + x2dx.”

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

x + x2where x = u, i.e.

1

u + u2.

(Eventually we will see that g(u) = ln u+1u− ln 4

3; students may thus verify that the

Fundamental Theorem is giving us the correct derivative.)

[1, Exercise 12, p. 402] “Use Part 1 of the Fundamental Theorem of Calculus to find

the derivative of the function g(u) =

∫ 10

x

tan θ dθ.”

Solution: First observe that g(u) = − ∫ x

10tan θ dθ; (see the first observation in

the problem solved next). 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, whenθ = x, i.e., − tan x.

(Eventually we will see that g(u) = ln cos x − ln cos(10); students may thus againverify that the Fundamental Theorem is giving us the correct derivative.)

[1, Exercise 50, p. 403] “Find the derivative of the function y =

∫ 5x

cos x

cos(u2) du.”

Solution: Two observations are necessary:

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

• The Fundamental Theorem is concerned with an integral whose upper limitis the independent variable under consideration; should we wish to permitthe upper limit to vary in a more complicated way, we will need to apply theChain 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]into two parts at a “convenient” point. It is not even necessary that the pointwe choose be inside the interval, since the property we are applying, (3), doesnot require that fact. I will choose the constant 0 to be the point where thesplitting occurs:

∫ 5x

cos x

cos(u2) du =

∫ 0

cos x

cos(u2) du +

∫ 5x

0

cos(u2) du

= −∫ cos x

0

cos(u2) du +

∫ 5x

0

cos(u2) du

Hence

d

dx

∫ 5x

cos x

cos(u2) du = − d

dx

∫ cos x

0

cos(u2) du +d

dx

∫ 5x

0

cos(u2) du (5)

2. Each of these derivatives will be computed using the Chain Rule, with theintermediate variable being the function appearing as the variable upper limit.In the first case, if we take the intermediate variable to be, say z = cos x, wehave

d

dx

∫ cos x

0

cos(u2) du =d

dz

∫ z

0

cos(u2) du · dz

dx


= cos(z2) · dz

dx= cos(cos2 x) · (− sin x)

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

d

dx

∫ 5x

0

cos(u2) du = − d

dw

∫ w

0

cos(u2) du · dw

dx

= cos(w2) · dw

dx= cos(25x2) · 5

Combining the two results gives

d

dx

∫ 5x

cos x

cos(u2) du = − d

dx

∫ cos x

0

cos(u2) du +d

dx

∫ 5x

0

cos(u2) du

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

[1, Exercise 26, p. 402] “Use Part 2 of the Fundamental Theorem of Calculus to eval-

uate 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]. Thismeans that we cannot even speak of the integral at this time. (Later we willgeneralize our definition of integral to permit us to consider certain “improper”integrals where there is an infinite discontinuity. But even that generalization willnot apply to this problem, although it is premature to consider it here. Look atthis problem again when we study [1, §7.8].) The theorem is not applicable becausethe integral is not defined.

[1, Exercise 54, p. 403] “Find the interval on which the curve

y =

∫ x

0

1

1 + t + t2dt

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 square, so it is al-

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

x < −1

2: this is where the graph is concave upward.


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

y =2√3

(arctan

2x + 1√3

− π

6

).

Students may differentiate to check that we have found the correct second deriva-tive.)


C.3 Supplementary Notes for the Lecture of January 12th,2004

Release Date: Monday, January 12th, 2004, subject to further revision

“The Distance Problem” In the lecture of January 5th, 2004, I discussed the useof the definite integral to solve problems of areas under curves, but did not discuss thesecond motivational application mentioned by the textbook [1, §5.1, pp. 376-378]: to findthe distance traversed by a moving particle. If a particle is known to be moving along thex-axis at a velocity of v(t) = x′(t) cm/s, how much distance is traversed between timet = a and time t = b? If by distance we mean displacement or “signed distance”, wheremovement to the right counts positively, and to the left negatively, then the distance is

x(b)− x(a) =

∫ b

a

d

dtx(t) dt .

Note that, with this definition, a particle that moves around and then returns to thesame point will have travelled a distance of 0. When it is intended to consider all motionas non-negative — the way the odometer of an automobile measures distance, then wewould want to integrate not the velocity, v(t) = x′(t), but the speed , |v(t)| = |x′(t)|, andthe distance travelled would be

∫ b

a

|v′(t)| dt =

∫ b

a

∣∣∣∣d

dtx(t)

∣∣∣∣ dt .

But in practice the word distance is often used with both meanings, so care is required.

Comparison Properties of the Definite Integral Another topic for which therewas no time at the lecture of Monday, January 5th, 2004, was the relationships thathold between integrals of different integrands over the same integral [a, b] of integration(a ≤ b) [1, §5.2, pp. 389-390]. If g(x) ≤ f(x) throughout the interval, then the areas willhave the same relationship:

∫ b

a

g(x) dx ≤∫ b

a

f(x) dx

(provided a ≤ b). In the special case where g(x) = 0 for all x, we obtain the propertythat

0 =

∫ b

a

0 dx ≤∫ b

a

f(x) dx


so the integral of a non-negative integrand from a to b, where a ≤ b, is always non-negative. And, if we can bound the function f on two sides by constant functions, i.e.,where

m ≤ f(x) ≤ M ,

then we have

m(b− a) = mx]ba =

∫ b

a

mdx ≤∫ b

a

f(x) dx ≤∫ b

a

M dx = Mx]ba = M(b− a) .

This provides a way of estimating the value of an integral, and can be used where anintegral is difficult for us to evaluate directly.

C.3.1 §5.4 Indefinite Integrals and the Net Change Theorem

Indefinite Integrals The traditional symbol for the family of all antiderivatives of a

function f(x) is

∫ b

a

f(x) dx, which is called the indefinite integral of f(x). The convention

is that

∫f(x) dx represents not simply one antiderivative, but all antiderivatives over

some clearly indicated domain. Since two antiderivatives differ by a constant, we usuallywrite statements in the form

∫f(x) dx = F (x) + C

where C is a constant of integration, intended to range over all real numbers. Theparticular real number that applies in a particular situation has to be determined fromadditional information that is usually available in the problem at hand. Much of thiscourse will be concerned with methods for finding indefinite integrals. While the findingof the indefinite integral may be a difficult problem, the verification that a function Fthat is claimed to be an antiderivative of f is not, since all that needs to be done is todifferentiate F and to check whether the derivative is f .

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 (2) of the definite integral, we can also prove that

∫(rf(x) + sg(x)) dx = r

∫f(x) dx + s

∫g(x) dx (6)


The “Net Change” Theorem This is simply the author’s name for the second partof the Fundamental Theorem. It is not a term in standard usage, and we will not beusing it in this course.23

As a beginning in the development of general techniques for determining indefiniteintegrals, we can reformulate results that we developed for derivatives. For example,since we know that

d

dxsin 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.From the result that

d

dxxn dx = n · xn−1 when n 6= 0

we can conclude that ∫xn−1 dx =

1

nxn when n 6= 0

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

∫xm dx =

1

m + 1xm+1 when m 6= −1 .

Applications

5.4 Exercises

[1, 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 recog-nizable as the derivative of arctan x; the two terms at the beginning are multiples

23Note, however, that while we have been using the 2nd part of the theorem to express the value ofa definite integral in terms of the “net change” in the antiderivative, we can also apply the result inthe opposite way: that the net change can be found by evaluating the integral. We will see an examplebelow in the solution of [1, Exercise 58, p. 412].


of powers of x, and we have observed earlier how to integrate them. Thus∫ (

x2 + 1 +1

x2 + 1

)dx =

∫ (x2 + 1

)dx +

∫1

x2 + 1dx

=1

3x3 +

1

1x + arctan x + C

where the letter C represents the constant of integration.

[1, Exercise 11, p. 411] “Find the general indefinite integral of

∫ (2−√x

)2dx.”

Solution: One might be tempted, at first, to consider the possibility that theantiderivative of the given 2nd power is a multiple of the 3rd power of 2 − √

x;unfortunately that temptation will have to be resisted, as the resulting functionswill not have the correct derivative. The simplest approach is to expand the square:since (2−√x)2 = 4− 4x

12 + x,

∫ (2−√x

)2dx =

∫ (4− 4

√x + x

)dx

= 4x− 4 · 2

3x

32 +

1

2x2 + C

[1, Exercise 30, p. 411] “Evaluate the integral

∫ 9

1

3x− 2√x

dx.”

Solution: Simplify the integrand by dividing the denominator into the two sum-mands of the 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 .


∫ π3

0

sin θ + sin θ · tan2 θ

sec2 θdθ .”


Solution: When the integrand involves trigonometric functions, one may have toapply a familiar identity to simplify the integration. There is often more than oneway 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

0

sin θ dθ

= [− cos θ]π30 = − cos

π

3+ cos 0 = −1

2+ 1 =

1

2.

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

[1, Exercise 54, p. 412] The velocity function is v(t) = t2−2t−8 for a particle movingalong a line. Find (a) the displacement and (b) the distance travelled by the particleduring the time interval 1 ≤ t ≤ 6.

Solution:

(a)

displacement =

∫ 6

1

(t2 − 2t− 8) dt

=

[1

3t3 − t2 − 8t

]6

1

= (72− 36− 48)−(

1

3− 1− 8

)= −10

3

(b)

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 intervalof integration) and again at t = 4, which is inside the interval of integration.We will break the integral up into two parts at x = 4, and then we can expresseach of the parts 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

= −[1

3t3 − t2 − 8t

]4

1

+−[1

3t3 − t2 − 8t

]6

4

= −[−18] +

[44

3

]=

98

3

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

[1, Exercise 58, p. 412] “Water flows from the bottom of a storage tank at a rate ofr(t) = 200− 4t litres/min, where 0 ≤ t ≤ 50. Find the amount of water that flowsfrom the tank during the first 10 minutes.”

Solution: Let’s denote the amount of water in the tank by W (t). We are told that

d

dtW (t) = r(t) = 200− 4t (0 ≤ t ≤ 50).

Then

W (10)−W (0) =

∫ 10

0

dW

dtdt

=

∫ 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.

C.3.2 §5.5 The Substitution Rule

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

Theorem C.2 Let u = g(x) and f(x) be respectively differentiable and continuous ona given interval; 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 wetry to 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 whoseintegrand is one what we are able to integrate. In practice one works with the differentialsdx and du in a “mechanical” way that can be justified by the theorem.

In the first problems in the list of exercises the author suggests a substitution whichwill be helpful; eventually students are expected to find an appropriate substitution ontheir own — often there are several possible choices.

Example C.3 Earlier we considered [1, Exercise 11, p. 411], which was concerned with

the indefinite integral of

∫ (2−√x

)2dx. 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

)10000dx?

Solution: While we could expand the integrand in this case too, the result would have10,001 terms, each of which would have to be integrated. The result we would obtainwould not be very useful. Consider the following alternative approach, that could alsohave been used when the exponent was 2. Define u = 2−√x, so that du = − 1

2√

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

√x

10002− 4

(10001)(10002)

)+ C



Release Date: Wednesday, January 14th, 2004, subject to further revision

C.4.1 §5.5 The Substitution Rule (continued)

As we saw in the last lecture, the Substitution Rule is simply a restatement of the ChainRule. Let F and g be differentiable. Then

F ′(g(x)) · g′(x) =d

dxF (g(x)) .

Integrating with respect to x, we have∫

F ′(g(x)) · g′(x) dx =

∫d

dxF (g(x)) dx = F (g(x)).

If u = g(x), and if we define f = F ′, then∫

f(g(x)) · g′(x) dx =

∫d

dxF (g(x)) dx = F (u) =

∫f(u) du.

In practice this “substitution” is often applied by “mechanically”, substituting functionsand differentials, and such operations can be shown to be fully justifiable. The generalidea in looking for substitutions is to try to reduce the complication of the originalindefinite integral. This is a subjective term, and different users may find a variety ofdistinct substitutions which they will find helpful in evaluating an indefinite integral.

Example C.4 [1, Example 6, p. 417] “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 x

cos xthe author shows that it can be simplified by

treating cos x as the new variable. That is, by defining u = cos x, which implies thatdu = − sin x dx, we can evaluate the integral as follows:

∫tan x dx =

∫sin x dx

cos x

= −∫

d(cos x)

cos x

which has the form−∫

du

u, which we should recognize as d(ln |u|). Thus one substitution

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.


Definite Integrals The Substitution Rule may be applied to a definite integral

∫ b

a

f(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 antideriva-tive; or

• A variant of the Substitution Rule can be formulated specifically for definite inte-grals. Using the same functions as described earlier, it is

∫ b

a

f(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 = aand x = b.

These two methods are equivalent — use whichever you are more comfortable with.

Symmetry The author takes this opportunity to prove a result about the integrals ofeven and odd functions over an interval that is symmetric around 0. He shows that thedefinite integral of an even function f over the interval −a ≤ x ≤ +a will be twice theintegral over the interval 0 ≤ x ≤ a; and the integral of an odd function over the sameinterval will be 0, because the integral to the left of 0 will cancel the integral to the rightof 0. The proofs are simple applications of the Substitution Rule.

5.5 Exercises


∫x(4 + x2)10 dx by making the substi-

tution u = 4 + x2.”

Solution: First we should observe that we could solve this problem without anysubstitution at all. We could expand the 10th power of the binomial into a polyno-mial of degree 20, multiply each of the terms by an additional x, and then integratethe resulting sum of 11 terms one by one. The resulting polynomial would be thecorrect solution. But what would you do it the exponent were not 10 by 10000? The


method of substitution works in all cases. We set u = g(x) = 4 + x2, f(u) = u10,

sodu

dx= g′(x) = 2x. Then

∫x(4 + x2)10 dx =

∫(4 + x2)10 · 1

2

d

dx(4 + x2) dx

=1

2

∫d

dx

((4 + x2)11

)dx

=1

22(4 + x2)11 + C .

In practice we often operate mechanically with differentials; from u = 4 + x2 wehave du = 2x dx, so

∫x(4 + x2)10 dx =

∫xu10dx =

∫u10 · x dx

=

∫u10 · 1

2du =

1

22u10 + C .

But it would be bad form to stop here, since our answer has been expressed interms of u, not the original variable x; so we continue

=1

22(4 + x2)11 + C .

[1, Exercise 22, p. 421] “Evaluate the indefinite integral

∫tan−1 x

1 + x2dx.”

Solution: In looking for a substitution, the intention is to try to simplify theintegrand. In the present integrand, one might be expected to consider the arct-angent factor the most complicated, so I will try to simplify by the substitution

u = tan−1 x. This implies that du =1

1 + x2dx, which is also present in the integral.

The integral becomes∫

tan−1 x

1 + x2dx =

∫u du =

1

2u2 =

(tan−1 x)2

2+ C .


∫x

1 + x4dx.”

Solution: Examining the integral we see that the powers of x present are x1 in thenumerator, and x4 in the denominator. A first idea might be to try u = x4. Thatwould yield ∫

x

1 + x4dx =

1

4

∫1√

u(1 + u)du


which looks more complicated than before. However, we could then undertake asecond substitution, v =

√u, which would correspond to an original substitution

of v = x2; this would produce∫

x

1 + x4dx =

1

2

∫1

1 + v2dv

=1

2arctan v + C =

1

2arctan

(x2

)+ C

An experienced student would have been able to see the substitution v = x2 im-mediately.


∫sin x

1 + cos2 xdx.”

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

becomes

∫ −du

1 + u2du = − arctan u + C = − arctan(cos x) + C.

[1, Exercise 55, p. 421] “Evaluate the definite integral∫ π

6

−π6tan3 θ dθ.”

Solution: Eventually you will be able to integrate the cube of the tangent; but, atthis point, you may not be able to do that. However, you can observe that theintegrand is an odd function of θ. (Why?) Hence the integral from −π

6to 0 will be

equal in magnitude but opposite in sign to the integral from 0 to π6; so the given

integral is 0.

[1, Exercise 63, p. 421] “Evaluate the definite integral∫ 2

1x√

x− 1 dx.”

Solution: This integral can be simplified by defining u = x− 1, so du = dx. Then∫

x√

x− 1 dx =

∫(u + 1)

√u du

=

∫ (u

32 + u

12

)dx

=2

5u

52 +

2

3u

32 + C

=2

5(x− 1)

52 +

2

3(x− 1)

32 + C

Taking the differences of the values of this antiderivative (i.e. with any specificvalue for C, e.g., with C = 0) at the limits yields

[2

5(x− 1)

52 +

2

3(x− 1)

32

]2

1


=

(2

5(2− 1)

52 +

2

3(2− 1)

32

)−

(2

5(1− 1)

52 +

2

3(1− 1)

32

)

=

(2

5+

2

3

)− (0 + 0) =

16

15

Alternatively, the Definite Integral version of the Substitution Rule could havebeen used, to obtain

[2

5u

52 +

2

3u

32

]2−1

1−1

=

(2

51

52 +

2

31

32

)−

(2

50

52 +

2

30

32

)

etc.

Symmetry The author reviews the definitions of even and odd functions, and considerstheir integrals over a finite interval centred at the origin. He observes that such an integralof an odd function will be zero, while such an integral of an even function will be equalto twice the area from 0 to the right endpoint. These results are contained in the presentsection of the textbook because the proofs are based on an appropriate substitution.

C.4.2 §5.6 The Logarithm Defined as an Integral

The Natural Logarithm Previously we had “defined” the logarithm to be the inverseof the exponential function, whose definition was, at best, intuitive. Now we are goingto present a more rigorous definition of both functions, beginning with the logarithm.You should be comfortable with all of the results printed in red boxes in the textbook.There will not be time to prove them in the lectures, and you need not try to memorizethe proofs. This theory is located here because it is based on substitutions.

We define

ln x =

∫ x

1

1

tdt

i.e. it is the area under the right branch of the hyperbola y = 1t, between t = 1 and

t = x. From the first part of the Fundamental Theorem we immediately obtain the factthat d

dxln x = 1

x. We can also prove the following, using basic properties of the integral:

Theorem C.5 1. ln(xy) = ln x + ln y

2. ln 1 = 0


3. ln 1x

= − ln x

4. ddx

ln |x| = 1x

5. limx→∞

ln x = ∞

6. limx→0+

ln x = −∞

At this point we can define the exponential function as the inverse of the logarithm. Webegin by calling the function exp(x); only after we prove that it behaves the way weexpect a power to behave, do we revert to the familiar notation.

Students are not expected to be able to prove the results in this section, althoughthey are not difficult.

The Natural Exponential Function

General Logarithmic Functions

The Number e Expressed as a Limit In class it was shown that, for any constanta,

limy→∞

(1 +

a

y

)y

= ea ,

or, equivalently (by the substitution x = 1y), that

limx→0+

(1 + ax)1x = ea .

In fact, the result is true for a 2-sided limit:

limx→0

(1 + ax)1x = ea .

But the primary definition of e is that it is the real number x for which∫ x

11tdt = 1.

Since we know that ddx

ln x = 1x, the logarithm function is increasing. We can show that

limx→∞

ln x = +∞. Being differentiable, it is continuous. Hence there is a point where its

value is 1; that point, which we can prove to be unique, is defined to be e.

5.6 Exercises

Review

Problems Plus




Textbook Chapter 6. APPLICATIONS OF INTEGRATION.

C.5.1 §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. if the graph of f is not lower than the graph of g for that interval, then the areabounded by the graphs 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, we make several observations:

• When g is the constant 0 function, 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 canbe described as above. Sometimes there is more than one way to decompose theregion, but all decompositions should yield the same area.

• An analogous formula can be applied if we consider graphs of the form x = f(y),x = g(y). In that case we say that we are integrating with respect to y or integratingalong the y-axis .

• Often we wish to find the area between two graphs, determined by points wherethey intersect. In these cases the vertical sides of the region may have zero length.

• In solving problems it is useful to make a sketch showing an “element of area” —a thin rectangle whose width is shown as dx or ∆x in the case of integration withrespect to x, and analogously for the case of integration with respect to y. It isdifficult for me to include such sketches in these notes, but they will be shown onthe chalkboard. The sketch is not a formal part of the solution, but you are likelyto find it helps you formulate your solution.

• Where two graphs cross in several places, so that the area between them is inseveral pieces, be sure you understand what you intend if you write an integralthat extends over several pieces: if the curves change positions, with the upper onebecoming the lower, then the signs of the areas will change, and some of the areas


will cancel. If this is not what you intend, you must either write the integrandwith absolute signs, as |f(x) − g(x)|, or, equivalently, find the area of each of thepieces, and add their magnitudes so as to prevent cancellation.

6.1 Exercises

[1, Exercise 16, p. 442] “To find the area of the region bounded 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 equationsshows that the intersections occur when x = ±2, y = 4, i.e. in the points (±2, 4).But we are asked to find the area from x = −3 to x = 3. Thus this region hasthree 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 +

16

3

)− (−24 + 18)

]+

[(16− 16

3

)−

(−16 +

16

3

)]

−[(24− 18)−

(16− 16

3

)]

=92

3


This attack is naive in that we have not taken advantage of symmetry. Since theintegrand is an even function, and the interval of integration is symmetric aboutx = 0, we can find the integral from 0 to 3 and double it: that means evaluating 2integrals instead of 3.

The areas could also be found by integrating with respect to y, where we wouldhave to sum three pieces (or two if we make use of symmetry). For example, themiddle region could be described as being double the following region: boundedby the y-axis, the curve x =

√y, and the curve x =

√8− y. This approach

is cumbersome, as the region has to be broken up further into two pieces: onebounded by the y-axis, x =

√y, and y = 4, and the other bounded by the y-axis,

y = 4, and x =√

y − 8.

[1, Exercise 46, p. 443] 1. “Find the number a such that the line x = a bisects thearea under the curve y = 1

x2 , 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

1

x2dx =

∫ 4

a

1

x2dx

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 hasequation y = 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 316

. As this is less than halfthe total, we know that b > 1

16. This fact is important, as it tells us that we

can represent the upper half of 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 —now viewed as being “under” x = 1√

yand “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 =3

8


⇔ (√

b− 1)2 =3

8

⇔√

b = 1±√

3

8

⇔ b =11

8±

√3

2

One of the values we obtain is greater than 1, which contradicts our assump-tion that 0 ≤ b ≤ 1. The other gives the bisecting line

y =11

8−

√3

2≈ 0.150255129.

The value we rejected has a geometric significance: it is the height of a hori-zontal line at which the area bounded by the curve y = 1

x2 , the line x = 1 andthat line is equal to 3

8.

C.5.2 §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 invarious ways. In this course you will see volumes expressed either as sums of thin sliceswith planar sides (called laminæ), and — for solids with rotational symmetry (calledsolids of revolution) — as sums of thin shells with cylindrical sides. Some problems willlend themselves only to one type of dissection; where a problem can be approached inmore than one way, it is instructive to try it in several ways, in order to verify youranswer and also to gain experience in choosing the method that is more efficient fordifferent types of problems. In this section we will be considering dissections into thinslices; where the slice has the shape of a disk with a concentric disk cut from its centre,the author uses the term washer 24.

6.2 Exercises

[1, Exercise 2, p. 452] “Find the volume of the solid obtained by rotating the regionbounded by the curves y = ex, y = 0, x = 0, x = 1 about the x-axis.” We will firstsolve the problem as stated, and then consider some related problems obtained bychanging some of the data.

24Students whose native language is not English sometimes cannot understand why the author willuse the name of a household appliance here; this is another meaning of the English word washer , whichrefers to a thin disk with a hole in the middle


Solution: (Remember to come back to this problem when we study [1, §6.3], tosolve it using the methods of that section.)

We will decompose the solid into laminæ that are thin disks. The “washers” willbe obtained by rotating about the x-axis the element that we would have used forthe area if we had found the area by integrating with respect to x. For the diskwhose faces are centred at points (x, 0) and (x + ∆x, 0), the volume obtained byhanging the element from its upper left corner on the curve is πy2∆x = πe2x∆x.By the same reasoning that led us to express areas as definite integrals, we have

Volume =

∫ 1

0

πe2xdx

=π

2e2x]10 =

π

2· (e2 − 1)

Now let us change the problem, asking that the solid be rotated about the y-axis.Here the description of the cross sections will depend on the height of the crosssection. We will 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), thecross sections are disks of radius 1, and the volume of that part of the solid is

∫ 1

0

π12dy = πy]10 = π.

For y ≥ 1 the cross sections are “washers”, with outer radius 1 and inner radius x;to express this in terms of y we need to rewrite the equation of the curve y = ex

in an equivalent way that expresses x in terms of y, as x = ln y. The washers atheight y have volume

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. Youcan make a substitution like x = ln y, under which the integral transforms toπ

∫ 1

0(1− x2)ex dx, and you know how to integrate part of this integral, as

π

∫ 1

0

1ex dx = πex]10 = 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 bydifferentiation). With that we know that

Volume = π[ex − (x2 − 2x + 2)ex]10 = π[(−x2 + 2x− 1)ex]10 = π .

As another variant on this problem, consider the following: “Find the volume ofthe solid obtained by rotating the region bounded by the curves y = ex, y = 0,x = 0, x = 1 about the line y = −3.” The analysis is similar to what we didoriginally, except that the laminæ now have a hole of radius 3 in the middle, andthe outer radius is 3 units larger. This leads to an integral

Volume =

∫ 1

0

π((ex + 3)2 − 32

)dx =

∫ 1

0

π(e2x + 6ex

)dx

= π

[1

2e2x + 6ex

]1

0

= π

[(1

2e2 + 6e

)−

(1

2+ 6

)]=

π

2

(e2 + 12e− 13

)


C.6 Supplementary Notes for the Lecture of January 21st, 2004

Release Date: Wednesday, January 21st, 2004, subject to further revision

C.6.1 §6.2 Volumes (continued)

Example C.6 [1, 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 ofS is the parabolic region

(x, y)|x2 ≤ y ≤ 1 ;

cross-sections perpendicular to the y-axis are equilateral triangles.”Solution: The cross-section of the base at level y has ends with coordinates (±√y, y),so the length 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

=

∫ 1

0

√3y dy =

[√3

2y2

]1

0

=

√3

2(12 − 02) =

√3

2

C.6.2 §6.3 Volumes by Cylindrical Shells

Read your textbook to see the rationale for representing the volume of a cylindrical shellas the product of the circumference of a circle at the base, the height of the cylinder,and the thickness of the cylinder, which, in the limit, is represented by the differentialin the definite integral. If you choose to approach these problems by substituting informulæ, you are urged to remember how to generalize to situations where the axis ofcircular symmetry is parallel to but different from one of the coordinate axes. I do notrecommend memorizing formulae.

6.3 Exercises

[1, Exercise 14, p. 459] “Use the method of cylindrical shells to find the volume ofthe solid obtained by rotating the region bounded by the given curves about thex-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 coor-dinate axes.

1. First we follow the instructions, using the method of cylindrical shells. Theelements of area that generate the shells will be narrow horizontal rectangles;


at height y the rectangle has width (4 − (y − 1)2) − (3 − y), obtained byexpressing the equations in the form x =function of y. The circumference ofthe base of the generated cylinder 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

=27π

2.

2. Now let us compute the volume using washers. The equation of the line maybe rewritten as y = 3 − x; but the parabola now splits into 2 curves —the upper branch has equation y = 1 +

√4− x, and the lower has equation

y = 1 − √4− x. The description of the element of area that generates thewasher will change at x = 3. To the left of x = 3 the element of area athorizontal 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 lengthof the element is not enough, as we need to know the distance from the axisabout which it is revolving. The outer radius of the washer is 1+

√4− x, and

the inner radius is 3− x, so the area of the annulus is the difference betweenthe 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, andouter radius 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

1

4√

u(−1) du

using substitution u = 4− x, du = −x, x = 4− u

=27π

2as before.

[1, Exercise 32, p. 459] “(The) integral represents the volume of a solid. Describe thesolid.”

Solution: The same integral could easily represent more than one solid. One in-terpretation is the following: The solid is a solid of revolution around the hor-izontal line x = 2π, generated by revolving the region bounded by the y-axis)and the graphs y = cos x and y = sin x up to the point where they intersect,(x, y) =

(π4, π

4

). But the region could be deformed vertically without changing the

volume. For example, the region could be taken to be bounded by the lines y = 0,x = 0, y = cos x− sin x, on the positive sign of the y-axis, from x = 0 to x = π

4.

[1, Exercise 20, p. 459] “Use the method of cylindrical shells to find the volume gen-erated by rotating the region bounded by the given curves about the specified axis.Sketch the region 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 horizontaldimension 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 revolutionis 2π(y − (−1)) (since the radius is the distance between a point (x, y) andthe point (x,−1) below it), so the volume is

∫ 1

0

(√y − y2

)2π(y + 1) dy

= 2π

∫ 1

0

(y

32 − y3 +

√y − y2

)dy

= 2π

[2

5y

52 − 1

4y4 +

2

3y

32 − 1

3y3

]1

0

=29

30π .

2. It wasn’t asked for in the problem, but let’s find the same volume by themethod of washers. 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 section will be π times the difference of the squares of theradii, i.e.,

π((√

x + 1)2 − (x2 + 1)2)

.

The integral from x = 0 to x = 1 is again equal to 2930

π.




Some tutorial sections are closed to unregistered students Students who arenot registered in the following tutorial section(s) must not write quizzes there, as a gradeof 0 will be recorded for unregistered students:

Tutorial TA Location Time CRNT008 Ms. Y. Han BURN920 Thursdays 14:35-16:25 863T009 Mr. Ph. Poulin BURN1214 Thursdays 16:05-17:55 864T011 Ms. M. Fortin-Boisvert BURN1224 Mondays 13:35-15:25 866

This list will probably change.

I received full marks for the problem; doesn’t that mean that my solution wasperfect? If your quiz or written assignment is graded carefully, full marks are usuallyassociated with a solution that is essentially correct. Sometimes there are reasons whyyour work will be graded unusually generously, and there could be some aspects thatcould still be improved substantially. If you have any doubts about the correctness ofa solution, bring it to your instructors or a TA and ask them to check it. (Even if thegrading of a question was incorrectly high, no one is planning to lower your grade.)

It can happen occasionally — very rarely — that a WeBWorK problem will assignfull marks for a solution that your instructors may find to be wanting.25 Here again, ifyou have any doubts, ask a TA or an instructor: no one is planning to lower your grade.

For model solutions to the types of problems you meet in this course you are referredto the Student Solutions Manual [3].

Submission of written assignments W2, W3, ... If you submit your assignmentin any way other than at the quiz in your registered tutorial, you expose yourself to therisk that it will be lost. In particular, assignments submitted through the Math/StatReception Desk (BURN 1005) must show the tutor’s name clearly; it is not enough toindicate the time or location of your tutorial, and it is certainly not enough to show onlythe course number or your lecturer’s name. Even with this information there is a risk ofloss, and you are urged to submit your assignment at the proper time and in the properway. Assignments submitted after your tutorial session are an inconvenience to your TA,whose expects to have all the assignments when she/he begins grading. No assignmentcan be accepted after 4 p.m. on the Friday of the week when it is due.

25This can derive from the way in which WeBWorK evaluates functions, which is based on a randomsampling procedure — two functions may be equal at the points sampled, and still be different elsewhere.


C.7.1 §6.4 Work

This section has been omitted from the syllabus because it involves physical conceptsthat some students from outside of the Faculties of Science and Engineering might notbe prepared for. If you are a Science or Engineering student, you are urged to perusethe section and try the problems. Your instructors and TA’s will be happy to help youwith any difficulties.

C.7.2 §6.5 Average value of a function

In this section the textbook defines what is meant by the term average of a continuousfunction or a function “pieced” together from continuous functions over an interval a ≤x ≤ b. The definition is a generalization of the definition familiar to you of the averageof a finite number of 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 ofunit width, situated so that its width is placed on the x-axis with vertices (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− 1

∫ b

a

f(x) dx ,

and so our definition is consistent with the earlier definition when the function is definedat a finite number of points x1, . . . , xn: we are simply extending the definition of thatfunction by extending the value at any integer point to the entire interval of unit lengthto the left of the point.

The Mean Value Theorem for Integrals If f is continuous on [a, b], the MeanValue Theorem may be applied to the function g(x) =

∫ x

af(t) dt, which we know from


the Fundamental Theorem to be differentiable. It asserts the existence of a point c in(a, b) such that g(b)−g(a)

b−a= g′(c), i.e., such that∫ b

a

f(t) dt = g(b)− g(a) = f(c) · (b− a) .

[1, Exercise 23, p. 467].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 theorem is not constructive: it proves the existence without telling you how to findthe point. (A constructive proof of a theorem proves existence by providing a way offinding the point; the proof we have given for the present theorem is not constructive.)

Example C.7 [1, Exercise 13, p. 467] The textbook asks you to prove that, if f is any

continuous function for which

∫ 3

1

f(x) dx = 8, 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 that1 ≤ x ≤ 3 and

1

3− 1

∫ 3

1

f(x) dx = f(c)

and the left side of this equation is exactly 82

= 4.

Average velocity Suppose that the position of a particle moving along the x-axis isf(t) at time t. In [1, §3.3, p. 199] the textbook has defined the Average Velocity of theparticle over a time 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 wasalso a generalization of the traditional meaning recalled above for the average of a finiteset of numbers. 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, theAverage Velocity 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 “Average Velocity” was a two-word name for a concept, and you would not bejustified in treating the first word as a modifier of the second; now you may interpret thatas a conventional use of language, where average is an adjective. When mathematiciansname concepts we try to make the nomenclature intuitive, and consistent with earlierusage.


6.5 Exercises

[1, Exercise 4, p. 467] Find the average value of the function g(x) = x2√

1 + x3 onthe interval [0, 2].

Solution:

average =1

2− 0

∫ 2

0

x2√

1 + x3 dx

We apply the substitution u = x3, so du = 3x2 dx and

average =1

2− 0

∫ 23

0

√1 + u · 1

3du

=1

6· 2

3(1 + u)

32

]8

0

=1

9

[9

32 − 1

32

]

=26

9

(In the lectures I used the substitution u =√

1 + x3.) Had the problem asked usto “sketch a rectangle whose area is the same as the area under the graph of g,”we could take a rectangle based on the interval 0 ≤ x ≤ 2, with height 26

9.

Example C.8 (taken from a problem book for students in Russian technical universi-ties) [29, Problem 1647, p. 126] Suppose that a trough has a parabolic cross-section withequation of the form y = Kx2, and measures 1 metre across the top and is 1.5 metresdeep. Determine the average depth.Solution: From the data we know the cross section passes through the point

(12, 3

2

), so

32

= K(

32

)2, and K = 6. The depth of the trough at position x is 3

2− 6x2, so

average depth =

∫ 12

− 12

(3

2− 6x2

)dx

= 2

∫ 12

0

(3

2− 6x2

)dx

by symmetry, since the integrand is an even function

= 2

[3x

2− 2x3

] 12

0

= 1

so the average depth is 1 metre, i.e., two-thirds of the way to the bottom of the trough.(Try to prove that this property is completely general, and independent of the horizontaldimension of the parabolic trough.)


Applied Project: Where to Sit at the Movies

Review

Problems Plus

6 Review

[1, Exercise 32, p. 469] “Let R1 be the region bounded by y = x2, y = 0, and x = b,where b > 0. Let R2 be the region bounded by y = x2, x = 0, and y = b2, to theright of the line x = 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 rotatedabout the x-axis and the y-axis?

3. “Is there a value of b such that R1 and R2 sweep out the same volume whenrotated about the x-axis?

4. “Is there a value of b such that R1 and R2 sweep out the same volume whenrotated about the y-axis?”

Solution: Note that the statement of the problem in the textbook is ambiguous, asthe description of R2 applies to one region to the right of the y-axis and anotherto the left; for that reason I have added the italicized words. Now the two regionscombine to form the rectangle (x, y)| 0 ≤ x ≤ b, 0 ≤ y ≤ b2.

1. We are asked to investigate solutions of the equation

∫ b

0

x2 dx =

∫ b

0

(b2 − x2) dx ⇔ b3

3=

2b3

3,

which has no positive solution. (We could have evaluated either or both ofthese areas by integration along the y-axis, writing the equation of the rightbranch of the parabola 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 aboutthe y-axis using the same elements of area, using cylinders. Equating the twoareas I obtain the equation

∫ b

0

π(x2

)2dx =

∫ b

0

2πx · 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 tobe πb5

5. If we compute the volume obtained by rotating R2 about the x-axis,

we find it to be∫ b

0

π((

b2)2 − (

x2)2

)dx = π

[b4x− x5

5

]b

0

=4

5b5 using washers, and

∫ b2

0

2πy · √y dy = 2π · 2

5y

52

]b2

0=

4π

5b5 using cylindrical shells.

Equating the volume obtained by rotating R1 about the same axis to this, weobtain πb5

5= 4πb5

5, which has no positive solution.

4. The volume obtained by rotating R1 about the y-axis has been determinedabove to be π

2b4. The volume obtained by rotating R2 about the y-axis is

∫ b2

0

π(√

y)2 dy = π

∫ b2

0

y dy = π

[y2

2

]b2

0

=π

2b4

using washers, and

∫ b

0

2πx(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 allvalues of the parameter b.

Note: We have not developed in the lectures the familiar formulæ for the volumes ofcones, spheres, and pyramids. Students are expect to try such problems on their own; orto read applicable solutions in the textbook. For example, the volume of a right circularcone of height h and base radius r is found in the solution to [1, Exercise 47, p. 453],which is solved in the Student Solutions Manual and also on one of the CR-ROM’s thataccompanies the textbook; the volume of the sphere is a special case of [1, Exercise 49,p. 453], which has solutions in the same places.



Release Date: Wednesday, January 28th, 2004,subject to further revision

Textbook Chapter 7. TECHNIQUES OF INTEGRATION.

C.8.1 §7.1 Integration by Parts

Earlier we developed the Substitution Rule for evaluating integrals, from the Chain Rulefor differentiation. Now we will develop another procedure for evaluation of integrals,called Integration by Parts , based on the Product Rule for differentiation. As with theSubstitution Rule, 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 limitsin definite integrals.

Starting from the Chain Rule,

d

dx[f(x) · g(x)] =

d

dxf(x) · g(x) + f(x) · d

dxg(x) ,

we integrate all 3 members with respect to x:∫

d

dx[f(x) · g(x)] dx =

∫ [d

dxf(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 termsaround gives the Rule of Integration by Parts:

f(x) · g(x) =

∫ [d

dxf(x) · g(x)

]dx +

∫ [f(x) · d

dxg(x)

]dx

⇔∫ [d

dxf(x) · g(x)

]dx = f(x) · g(x) +

∫ [f(x) · d

dxg(x)

]dx

∫ [f(x) · d

dxg(x)

]dx = f(x) · g(x) +

∫ [d

dxf(x) · g(x)

]dx

Or, compactly,

∫u dv = uv −

∫v du

These equations are always true for differentiable functions, but we shall be applyingthem when they tend to replace a difficult integral by one that is “easier” to evaluate.Usually the applications will be such that the integrand admits a “natural” factorizationinto two factors, where either one of them becomes “simpler” under differentiation, orone becomes “simpler” under integration.


Example C.9 Sometimes the factorization of the integral is less than obvious. Considerthe problem of integrating ln x (cf. [1, Example 2, p. 477]). The “factorization” we chooseis

u = ln x, dv = dx . (7)

See the textbook for the full solution. This is a derivation you should remember, as weoften need an antiderivative of a logarithm.

Example C.10 Similar to the preceding example is the integration of arctan x (cf. [1,Example 5, p. 479]). Here again the function does not admit a well defined factorization,but we can try substitution (7) and obtain

∫arctan x dx = x · arctan x−

∫x

1 + x2dx .

Now apply a substitution to the remaining integral, either

v = x2 ⇒ dv = 2x dx or

w = 1 + x2 ⇒ dw = 2x dx

Hence ∫arctan x dx = x · arctan x−

∫1

wdw

= 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 logarithmfunction is evidently non-negative.

There are several “standard” situations where we need to use integration by parts.Suppose we need to integrate a function of one of the following forms, where P (x) issome 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 “oflower degree”), while the integral of the other factor is “not more complicated”. Repeatedapplications cause the polynomial to disappear, leaving only an integral involving thesecond factor. Here again one should remember the derivation, but not memorize theformulæ, since they can be easily reconstructed, and there are too many variations tomemorize.

The rule of integration by parts need not be used in isolation: it may be necessaryto precede or follow its use by substitutions, and several applications of integration byparts could be needed to complete the solution to a problem.


Two applications of Integration by Parts? We saw in connection with substitu-tions that we might need to use the procedure more than once. Can that occur withIntegration by Parts? It can always occur, but, if not done carefully, the second iterationwill reverse the action of the first, and return us to the integral that we began with.Where the purpose of using Integration by Parts is to truly simplify the integrand, thenit is unlikely you will make this tactical error. But consider the first example in the nextparagraph.

Solving an equation to evaluate an indefinite integral Sometimes the applicationof integration by parts does not appear to make any progress, but a second or moreapplications may eventually produce a constraint on the integral, which enables one toevaluate it. This idea will be extended later.

Example C.11 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 partsagain: U = ex, dV = sin x, dU = ex dx, V = − cos 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?Fortunately not. The same integral appears on both sides of the equation, but withdifferent coefficients. 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 (8)

which implies that ∫ex cos x dx =

1

2(ex sin x + ex cos x) + C (9)

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 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.

2. Why did the constant of integration appear in equations (8), (9) but not in thepreceding 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 understandsthat each side is a set of functions which differ by a constant; no more generality isachieved if we add the notational comment that one may add a constant to one side.But, when one side would consist of a single function — here it is ex sin x+ ex cos x— the inclusion of a constant changes the meaning from one specific function toall functions obtained from it by adding any real number.

3. When we integrate the dv term, we appear to be selecting a specific antiderivative.Is this restrictive? Should we be including a constant of integration? Try toconvince yourself that he selection of a specific antiderivative is not at all restrictive;that is, if you do include a constant of integration, the changes to the formula willcancel each other out. For that reason you should always choose the simplestantiderivative of v that is convenient.

Reduction Formulæ I mentioned above that integration by parts should be able toassist in the integration of the product of a polynomial and a sine, cosine, exponential, orhyperbolic function. But what happens if we have to integrate other products of thesefunctions? In some cases the differentiations and integrations do not produce majorchanges of simplicity, but they can lead to information that enables us to determinethe integral, in the way in which the preceding example was solved. This method isparticularly important when we wish to obtain an algorithm for evaluating certain generalclasses of integrals. Sometime we will need to use the methods of the preceding paragraphin this connection, and sometimes not.


Example C.12 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 . (10)

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.13 [1, Exercise 48, p. 481] Let n be an integer greater than 1. Find aprocedure — i.e. a reduction formula —that can be used to evaluate secn x dx.Solution:

∫secn x dx =

∫secn−2 x · (sec2 dx

)

Applying integration by parts with u = secn−2 x, dv = sec2 x dx, we have 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 thesame integral 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− 2

n− 1

∫secn−2 x dx . (11)

Since we know the integrals of the 0th and 2nd powers of the secant we can now findthe integral of any even positive power. For the odd positive powers we may reduce theproblem to the integration of sec x, which has not been achieved yet.

7.1 Exercises


∫arcsin x dx.”

Solution: Since you probably don’t know an antiderivative of the inverse sine, butdo know its derivative, we will 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 thegiven integral into another; but if you are unable to evaluate the new integral, youhaven’t fully solved the problem, and sometimes you may have made it even moredifficult 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 +1

2· 2√w + C

= x arcsin x +√

1− x2 + C

Other substitutions that could have been used are w = x2, x =√

1− x2.

[1, Exercise 36, p. 480] Follow a substitution by integration by parts to integrate∫x5ex2

dx.

Solution: The obvious substitution is u = x2, so du = 2x dx. Then

∫x5ex2

dx =1

2

∫u2eu du

to which we apply integration by parts with


U = u2, dV = eu du, dU = 2u du, V = eu

=1

2

(u2eu − 2

∫u du

)

=1

2u2eu −

∫ueu du

to which we apply integration by parts with

u = u, dv = eu du, du = du, v = eu

=1

2u2eu −

(ueu −

∫eu du

)

=1

2u2eu − (ueu − eu) + C

=

(1

2u2 − u + 1

)eu + C

=

(1

2x4 − x2 + 1

)ex2

+ C

Your solution is incomplete unless you express the integral in terms of the originalvariable.

[1, Exercise 30, p. 480] Several methods were discussed for evaluating

∫ 1

0

r3

√4 + r2

dr.

One solution using integration by parts is based on

u = r2, dv =r√

4 + r2dr ⇒ du = 2r dr, v =

√4 + r2 .

Some students might have had difficulty observing the integral of dv, but a substi-tution could make that phase easier.

For a solution that does not use integration by parts, try the substitution u = 4+r2.Then du = 2r dr.

∫ 1

0

r3

√4 + r2

dr =

∫ √5

2

u− 4√u

du

=

∫ √5

1

(√u− 4u−

12

)


C.9 Supplementary Notes for the Lecture of February 2nd,2004

Release Date: Monday, February 2nd, 2004,subject to further revision

∫sin2 x dx and

∫cos2 x dx. One way of evaluating these integrals is to use one of the

double angle formulæ from trigonometry:

sin 2θ = 2 sin θ · cos θ (12)

cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1− 2 sin2 θ (13)

which follow from the formulæ for the sine and cosine of a sum. The formulæ involvingcos 2θ may be rewritten as

sin2 θ =1

2(1− cos 2θ) (14)

cos2 θ =1

2(1 + cos 2θ) (15)

From these we obtain∫

sin2 x dx =1

2

∫(1− cos 2x) dx

=1

2

(x− 1

2sin 2x

)+ C

∫cos2 x dx =

1

2

∫(1 + cos 2x) dx

=1

2

(x +

1

2sin 2x

)+ C

Another way to evaluate these integrals is through integration by parts. See [1, Exercises41(a), 42(a,b) p. 481] which describe how to use the reduction formula in [1, Example 6,pp. 479-480] and an analogue for cosines for these purposes.

C.9.1 §7.2 Trigonometric Integrals

This section is concerned with integrating functions that can be expressed simply in termsof trigonometric functions. The techniques rely on heavy use of familiar trigonometricidentities. 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 isalso considered. Strategies are developed for products of these types. However, studentsmay well be able to use techniques already seen to integrate certain integrals of thesetypes in ways other than those suggested here. To reiterate: YOU MAY BE ABLE TOINTEGRATE CERTAIN FUNCTIONS OF THE TYPES LISTED BY USING OTHERMETHODS. The objective in all of these procedures is to “simplify” the integration;where the function is a product of trigonometric functions, this “simplification” is usuallymeasured by a reduction in the total degree of the product, i.e., in the total number oftrigonometric factors. Since the total is finite, repeated applications of such procedureswill eventually result in successful integration.

Strategy for evaluating

∫sinm x · cosn x dx

0. This is a recursive procedure: if the first 2 steps do not lead to a substitutionproducing an integral that may be evaluated immediately, the last 2 steps will leadto a simplification in 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 thecosine factors into a function of sines. Then apply the substitution u = sin x withdu = cos x dx.

2. If m is odd, proceed analogously to the preceding: use the same identity to convertall but one of the sine factors into a function of cosines. Then apply the substitutionu = cos x, with du = − sin x dx.

3. If both m and n are odd, use the identities cos2 x = 2 cos2 x − 1 and sin2 x =1−2 sin2 x to express the integrand as a sum of products of sines and cosines of 2x.The degrees of these terms will be less than the degree of the integrand we startedwith, so that we have simplified the problem, and can repeat the procedure untilwe have completed the integration.

4. When both m and n are odd, other identities may also be used to simplify theintegrand; for example, we can use sin 2x = 2 sin x · cos x combined with the twodouble angle formulæ mentioned immediately above.

Strategy for evaluating

∫tanm x · secn x dx


0. When n is even or m is odd, this procedure leads immediately to an integralthat can be evaluated. When n is odd and m is even — the remaining case — theprocedure can lead to various solutions, one of which is the application of reductionformulæ.

1. If n is even, the integrand may be written as a function of tan x multiplied bysec2 x. One may then apply the substitution u = tan x to convert the problem tothe integration of a polynomial.

2. If m is odd and n > 0, the integrand may be written as a function of sec x multipliedby tan x sec x. One may then apply the substitution u = sec x to convert theproblem to the integration of a polynomial. (The case of m odd and n = 0 couldbe treated under the preceding case, as 0 is even.)

3. If n is odd and m is even, one method would be to transform the entire integrandinto powers of sec x and then to use the reduction formula [1, Exercise 48, p. 481]to reduce everything to the problem of integrating sec x. Other reductions arepossible: 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 theproblem to the type we shall meet in [1, §7.3]; alternatively, one may express theintegrand as a sum of powers of tan x, and develop a reduction formula for them(cf. [1, endpapers, item 75 p. 9.]).

The integral

∫tan x dx. You are reminded that we have seen that

∫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 this result, although you mayexpect that in many of the problems you consider, where sec x+tan x is positive, omissionof the signs may not produce any visible error.

While it is possible to prove the validity of the preceding equation simply by differ-entiation of the alleged antiderivative, a more direct proof requires some ingenuity.26

26One way to derive this result is to observe that

sec x =1

cos x=

cosx

1− sin2 x


“Other” trigonometric identities By adding and subtracting the expansions ofsin(A±B) and cos(A±B), one may obtain the following identities

sin A · cos B =1

2(sin(A−B) + sin(A + B)) (16)

sin A · sin B =1

2(cos(A−B)− cos(A + B)) (17)

cos A · cos B =1

2(cos(A−B) + cos(A + B)) (18)

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

2

and B byV − U

2, we obtain

sin U + sin V = sinV + U

2· cos

V − U

2(19)

sin U − sin V = sin−V + U

2· cos

V + U

2(20)

cos U + cos V = cosV + U

2· cos

V − U

2(21)

cos U − cos V = sinV + U

2· sin V − U

2(22)

Example C.14 Integrate

∫(sin 50x · cos 12x) dx.

Solution: (One possible solution)

∫(sin 50x · cos 12x) dx =

1

2

∫(sin 38x + sin 62x)) dx

= −1

2· 1

38· cos 38x− 1

2· 1

62· cos 62x + C

7.2 Exercises

=cosx

(1− sin x)(1 + sin x)

=12

(cosx

1− sin x+

cosx

1 + sin x

)

using ideas of partial fractions that we shall be meeting in [1, §7.4]. (See [23, pp. 505-506 ].)


C.9.2 §7.3 Trigonometric Substitution

In this section we describe a type of substitutions which simplify certain commonlymet integrals. We approach these substitutions in the reverse direction from that usedearlier. Whereas earlier we investigated substitutions of the form u = g(x), this time wewill usually formulate our substitutions first in the form x = h(u); that is, we will lookfor a substitution that will simplify the integrand, and then try to implement it. Weusually proceed “mechanically” in these problems; but, in principle, we are postulatingthe existence of an inverse function, and should be checking that there really is aninverse whenever we use the method. I will try to go through some of those steps in thefirst examples we consider, but, in practice, we often become careless and don’t checkeverything unless something indicates a problem. This is unwise; the reason that it“works” is that we usually confine the substitutions to certain well understood pairs offunctions/inverses, where all of the snags have already been worked out.

Example C.15 [1, Exercise 4, p. 494] To evaluate

∫ 2√

3

0

x3

√16− x2

dx.

Solution: A first look at the integrand suggests that the complication comes from theexpression

√16− x2 in the denominator. We could try to simplify by giving this a new

name, u =√

16− x2. That leads to

du = −√

16− x2

xdx ⇒ x dx = −u du

and so the integral becomes∫ 2

4

(16− u2) du =

[−16u +

u3

3

]2

4

=40

3.

But we would like to illustrate the notion of trigonometric substitution here (in aproblem 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 appli-cation of Pythagoras’s Theorem to a right-angled triangle; equivalently, from the identitythat

sin2 θ + cos2 θ = 1 .

To do this, we can first divide out the factor 16, which, when it leaves the square root,will reappear outside as 4:

√16− x2 = 4

√1− x2

16= 4

√1−

(x

4

)2

.

This suggests a substitution

x

4= sin v or, equivalently, x = 4 sin v


that will makex

4into a sine or a cosine — either one will work. When we express the

substitution that way we are really working with the inverse — to conform with ourearlier theory about substitutions we should be expressing the new variable in terms ofthe old; so we should be starting with

v = arcsinx

4.

The range of values of x that interest us is 0 ≤ x ≤ 2√

3, equivalently 0 ≤ x

4≤√

3

2, and

we know that the inverse sine function is defined over this domain. So, beginning withthe substitution above, we obtain

dv =1

1− (x4

)2 ·1

4dx =

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

∫ π3

0

sin3 v dv

which we proceed to evaluate using the methods of the preceding section:

64

∫ π3

0

sin3 v dv = 64

∫ π3

0

sin v(1− cos2 v

)dv

= 64

[− cos v +

1

3cos3 v

]π3

0

=40

3

In practice this method works smoothly, but one must occasionally be careful aboutthe evaluation of the inverse function, remembering our original definitions of the re-stricted interval where the inverse was taken. The method is indicated whenever we seeexpressions like

√1− x2 or, more generally,

√a2 − x2, since we can transform the latter

into the former by division by a positive real number. The inverse cosine could be usedinstead of the inverse sine, and the results would be no more difficult.

Table of trigonometric substitutions I can expand the table of substitutions ofthis type given in the textbook:


Expression Inverse Substitution Substitution Identity√a2 − x2 θ = arcsin x

ax = a sin θ 1− sin2 θ = cos2 θ

(−a ≤ x ≤ +a)(−π

2≤ θ ≤ π

2

)√

a2 − x2 θ = arccos xa

x = a cos θ 1− cos2 θ = sin2 θ(−a ≤ x ≤ +a) (0 ≤ θ ≤ π)√

a2 + x2 θ = arctan xa

x = a tan θ 1 + tan2 θ = sec2 θ(−∞ < x < +∞)

(−π2≤ θ ≤ π

2

)√

x2 − a2 θ = arcsecxa

x = a sec θ sec2 θ − 1 = tan2 θ(−∞ < x ≤ −a or

(π ≤ θ < 3π

2or

a ≤ x < ∞) 0 ≤ θ < π2

)

I have shown both sine and cosine versions of the first substitution, and could similarlyhave produced a cotangent version of the tangent substitution, and a cosecant version ofthe secant substitution; in practice the latter two variants are not used frequently, andoffer no advantages over the substitutions shown.

Hyperbolic substitutions It is possible to achieve the same sorts of simplificationsby using inverse hyperbolic functions. Since we have spent little time in becoming com-fortable with the hyperbolic functions, I will not discuss these substitutions in general,but may apply them in specific cases. For example, consider [1, Exercise 31(b), p. 495].

Example C.16 We have avoided computing the area of a disk until now. It is trivialwith a trigonometric substitution. If the disk is the set of points (x, y) such that x2+y2 ≤R2, then it is the region bounded by the 2 graphs y = ±√R2 − x2. Its area is

∫ R

−R

(√R2 − x2 −

(−√

R2 − x2))

dx =

∫ R

−R

2√

R2 − x2 dx

=

∫ R

0

4√

R2 − x2 dx since the integrand is even.

The substitution u = cos−1 xR

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 = π

2to u = cos−1 1 = 0;

dx = −R sin u du;√

R2 − x2 = R| sin u| = R sin u, since the sine is positive in thisinterval. The integral transforms to

4

∫ R

0

√R2 − x2 dx = 4

∫ 0

π2

(R sin u)(−R sin u) du

= −4R2

∫ 0

π2

sin2 u du


= −4R2

[u

2− sin

2u

4

]0

π2

= −4R2[0− π

4

]= πR2

If we had not used the evenness of the integrand to reduce the original problem tointegrating over the interval 0 ≤ x ≤ R, we would not have been able to use this cosinesubstitution. The reason is that the function cos−1 was defined for 0 ≤ x ≤ π, andcannot be taken to act on numbers in the interval −R ≤ x < 0. This is a detail thatcannot be avoided simply by using only the inverse sine function, since it could alsoencounter problems if the interval of integration was outside of −π

2≤ x ≤ π

2where it is

defined.

7.3 Exercises

[1, Exercise 24, p. 494] Evaluate

∫1√

t2 − 6t + 13dt.

Solution: We will complete the square of the quadratic polynomial in the denom-inator in order that, after a first substitution, this integral will be of a type thatwe 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 + 13dt =

∫du√

u2 + 1.

Now we 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 sec θ > 0

= ln∣∣∣(1 + u2θ

) 12 + u

∣∣∣ + C

= ln

∣∣∣∣∣∣

(1 +

(t− 3

2

)2) 1

2

+

(t− 3

2

)∣∣∣∣∣∣+ C

= ln∣∣∣(t2 − 6t + 13

) 12 + (t− 3)

∣∣∣ + (C − ln 2)

And we could rename the constant with a single letter, e.g. K = C − ln 2.


C.10 Supplementary Notes for the Lecture of February 4th,2004

Release Date: Wednesday, February 4th, 2004, subject to further revision

More examples on trigonometric substitution


∫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− 2

2

)2)

.

Hence∫

x2

√4x− x2

dx =

∫x2

2√

1− (x−2

2

)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 2v

2dv − 8 cos v + 4v + C

= 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 positive; hence

cos v = +√

1− sin2 v =√

1− u2 =1

2

√4x− x2 .

We conclude that∫

x2

√4x− x2

dx = 6 arcsinx− 2

2− x + 6

2·√

4x− x2 + C .



∫ √x2 − 4

x4dx.

Solution: We need a substitution that will convert x2 to 4 times the square of asecant.27 One way to achieve this is to make u have the property that

x = 2 sec u ;

so the actual substitution will be

u = arcsecx

2,

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 we need the absolute signs? Recall that the arcsecant takes its valuesin the two intervals 0 ≤ u < π

2and π ≤ u ≤ 3π

2. In these two intervals the tangent

is always positive, so the absolute signs may be dropped.

∫ √x2 − 4

x4dx =

∫2 tan u

16 sec4 u· 2 sec u · tan u du

=1

4

∫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 theoriginal variable x. Since

sin u = tan u · cos u

=tan u

sec u

=12

√x2 − 4x2

,

1

12sin3 u =

1

12· (x2 − 4)

32

x3

Hence ∫ √x2 − 4

x4dx =

(x2 − 4)32

12x3+ C .

27Alternatively, we could make x2 four times the square of a hyperbolic cosine.


C.10.1 §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 sys-tematic algebraic decomposition procedure, followed by specific methods for the variouscomponents into which we decompose the functions.

Polynomials Recall that a polynomial [1, p. 29] is a function of the form

P (x) = anxn + an−1xn−1 + · · ·+ a1x

1 + a0x0

where a0, . . . , an are real numbers, called the coefficients of the polynomial. Except forthe zero polynomial , which is the constant function 0, all polynomials will have a largestinteger m such that am 6= 0; m is called the degree of the (non-zero) polynomial, andam is called the leading coefficient; a0 is called the constant term. We usually writex0 simply as 1, and x1 simply as x. It can be proved that every polynomial admits adecomposition as a product of factors of the forms either x − a, where a is a constant,or x2 + bx + c, where b and c are constants, and where x2 + bx + c does not factorizefurther, i.e. where b2 − 4c < 0. The procedure we shall develop depends on having sucha factorization. In this course we shall not be concerned with finding the factorizationitself, 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, andby “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 + b2n = (a− b)(a2n−1 − a2n−2b + . . . + ab2n−2 − b2n−1)

Sometimes a factor x− a may divide P (x), so that P (x) = (x− a)Q(x), and then x− amay also divide Q(x), and possibly divide more quotients. We 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 dividea polynomial A(x) into a polynomial B and obtain a quotient polynomial and a remain-der polynomial. The procedure is similar to long division of integers, and you will bereminded very briefly of it at the lecture. (You will see an example of long division ofpolynomials in the bottom left hand corner of [1, p. 496], but the way you learned towrite such a calculation may be slightly different from that used in the textbook.)


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 theroots of B, if there are any.

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 essentialstep; if it is omitted, it may cause the subsequent steps to fail. However, it is not necessaryif the degree of B is greater than the degree of A, since, in that case, the quotient will be0 and the remainder will be A(x).

The goal of partial fraction decompositions In a partial fraction decompositionwe express a ratio of polynomials as a sum of “partial” fractions — fractions that havespecial properties. In these “partial” fractions the denominator polynomials will alwaysbe powers of one polynomial that is irreducible, i.e., it cannot be factored further. It isa theorem of algebra that

Theorem C.17 If a non-zero real polynomial is irreducible then it must be of one ofthe following forms:

1. a non-zero constant

2. a polynomial of degree 1, of the form ax + b, where a 6= 0.

3. a polynomial of degree 2 without real roots, i.e., of the form ax2 + bx + c wherea 6= 0 and b2 < ac.

The numerators of these fractions will always have degree less than the degree of thedenominator. 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).


Some examples to illustrate the general procedure Before describing the generalprocedure we will consider some examples that will make it easier to comprehend.

[1, Exercise 14, p. 504] Evaluate the integral

∫1

(x + a)(x + b)dx.28

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 sub-stitution u = x + a.

∫1

(x + a)2dx = − 1

x + a+ C .

Case a 6= b: We try to decompose the integrand into the sum of fractions whosedenominators are, respectively, x+a, and x+b. The degrees of the numeratorsmust be less than these polynomials of degree 1, so they have to have degree0, i.e., they have to be constants. So suppose that

1

(x + a)(x + b)dx =

α

x + a+

β

x + b.

It can be shown algebraically that such a decomposition is always possible;all that is missing is to know the values of the constants α and β. The lastequation, after multiplication 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 thecoefficients of corresponding powers of x:

degree 0: 1 = α · b + β · adegree 1: 0 = α + β

Solving these equations gives

α =1

b− a

β =1

a− b

28In the lecture I solved the special case a = 4, b = −1, which is

[1, Exercise 10, p. 504] .


2. This equation must be true for all values of x. Give x “convenient”values to obtain equations in α and β, and solve those equations. Two“convenient” values are x = −a and x = −b. They give, respectively, thefollowing 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+

1

x + b

)

=1

a− b(− ln(x + a) + ln(x + b)) + C

=1

a− bln|x + b||x + a| + C

=1

a− bln

∣∣∣∣x + b

x + a

∣∣∣∣ + C

This family of functions could be written in other ways. For example, sinceC · (a− b) = ln eC·(a−b), we could call eC·(a−b) K, and write the family as

1

a− b

(ln

∣∣∣∣x + b

x + a

∣∣∣∣ + ln K

)=

1

a− bln

∣∣∣∣K · x + b

x + a

∣∣∣∣

where K serves as the constant of integration.∫

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 degree of the numerator — presently 3 — be lessthan the degree of the denominator — 2. If we divide the denominator into thenumerator, 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 bex2

2+

1

a− bln

∣∣∣∣x + b

x + a

∣∣∣∣ + C


What would have happened if we had attempted to expand this function into partialfractions? No such decomposition can exist. Using the first method (equatingcoefficients of corresponding powers) we would have obtained 4 equations whichwould overdetermine the constants α and β, and which would be inconsistent —there would be no solution. But, if we used the second method, and didn’t takeenough equations, we might not notice that there was an inconsistency, and thealleged partial fraction would be incorrect.



Release Date: Monday, February 9th, 2004, subject to further revision

C.11.1 §7.4 Integration of Rational Functions by Partial Fractions (contin-ued)

Example C.18 [1, Exercise 28, p. 504] 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 andmultiplicity 2, and x2 + 1, an irreducible quadratic factor of degree 2 and multiplicity 1.The degree of the numerator is less than 4, which is the degree of the denominator, sothere is no need for any long division. One type of partial fraction decomposition is ofthe 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 degreeof the denominator. This is not the most useful type of partial fraction decomposition,but we will carry this step out and then improve on it. Multiplying both sides by(x− 1)2(x2 + 1), we obtain a 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 theterms of degrees 3, 2, 1, 0:

α + γ = 0

β − 2γ + δ = 1

α + 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 + 1

x2 + 1.


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 thesubstitution u = x− 1:

∫u− 1

u2du =

∫ (1

u− 1

u2

)du

= ln |u|+ 1

u+ C

= ln |x− 1|+ 1

x− 1+ C

In practice we anticipate the results of this substitution by refining the partial fractiondecomposition: in place of a summand of the form

an−1xn−1 + an−2x

n−2 + . . . + a0

(x− a)n

we repeatedly divide x− a into the numerator, so that we can express the numerator asa sum of powers of x− a; then we decompose the fraction and divided excess powers ofx− a, to obtain a 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 beaccomplished as a second phase of partial fraction decomposition, or immediately, byassuming a decomposition 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)2dx +

∫1

(x− 1)1dx +

∫ −x

x2 + 1dx +

∫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

Repeated irreducible quadratic factors

Example C.19 [1, Exercise 38, p. 504] To integratex4 + 1

x(x2 + 1)2dx.

Solution: What distinguishes this example from those studied earlier is the multiplicityof the irreducible quadratic factor x2 + 1 in the denominator; this is the first case wehave seen where that multiplicity exceeds 1. we proceed analogously to the last phase ofthe preceding example, always taking the numerator to be the most general polynomialwhose degree is less than the degree of the irreducible factor in the denominator, but nowtaking separate summands for the powers of that irreducible factor. Thus we assume adecomposition of the form

x4 + 1

x(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 = 1

η = 0

2α + β + δ = 0

γ + η = 0

α = 1

implying that

(α, β, γ, δ, η) = (1,−2, 0, 0, 0) .

x4 + 1

x(x2 + 1)2=

1

x+

−2x

(x2 + 1)2 .


∫x4 + 1

x(x2 + 1)2dx =

∫1

xdx +

∫ −2x

(x2 + 1)2 dx

= ln |x|+ 1

x2 + 1+ C

∫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 ,

which, in principle, we know how to integrate.

Irreducible quadratic factors with 1st degree term Where the irreducible quadraticfactor is of the form x2 + bx + c (where b2 − 4c < 0) one can, by a substitution of the

form u = x+b

2followed by a transformation of the form v = constant×u transform the

integral to one involving multiples of1

u2 + 1.

Rationalizing Substitutions In some integrals in which the integrand is not origi-nally a rational function, it can be transformed into a rational integrand by an appro-priate substitution.

7.4 Exercises

[1, 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 substi-tution, u2 = x + 2, dx = 2u du,

∫1

x−√x + 2dx =

∫2u

(u− 2)(u + 1)du

=2

3

∫ (2

u− 2+

1

u + 1

)du


=2

3(2 ln |u− 2|+ ln |u + 1|) + C

=2

3

(ln

((√

x + 2− 2)2)|√x + 2 + 1|

)+ C

C.11.2 §7.5 Strategy for Integration

The textbook gives some a suggested 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 andyour intuition, and are occasionally not appropriate, as sometimes the best way toattach a problem will be obscured by these methods. So be prepared to try again.As for experience, you need to work many problems to acquire it.

Table of integration formulæ Many of these formulæ are just recasts of familiardifferentiation formulæ that you already know. But you should remember the integralsof 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 mostof the functions you will meet in this course it will be possible to integrate them. Ifthat is not the case, it could be that we are asking you indirectly to do something elsethan to integrate. You are not expected to be able to detect which functions cannot beintegrated in terms of elementary functions: this is a difficult problem even for a trainedmathematician. “You may be assured, though, that the integrals in the...exercises areall elementary functions.”

7.5 Exercises


∫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 througha substitution, but that could will render the denominator rather complicated. Ipropose to try integration by parts, and to assign u and v so that the ln x term ispart of u. But how much 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 anideal first step:

u = ln x ⇒ du =dx

x

dv =x√

x2 − 1dx

But what is v? If you don’t see, use the substitution U = x2:

∫x√

x2 − 1dx =

1

2

∫1√

U − 1dU .

You should be able to evaluate this last integral by sight; but, if you can’t, try thesubstitution V = U − 1:

v =1

2

∫1√

U − 1dU =

1

2

∫1√V

dV =√

V + C =√

x2 − 1 + C .

Hence ∫x ln x√x2 − 1

dx = (ln x)√

x2 − 1−∫ √

x2 − 1

xdx .

The problem isn’t solved yet, but at least the logarithm is gone. Now I propose totry a trigonometric substitution to simplify the square root:

w = arcsecx ⇒ x2 − 1 = tan2 w, dx = secw 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 thebest! But I have written down the way I solved it first. One shorter, but lessintuitive way would be to use the substitution u =

√x2 − 1, which implies that

du =x√

x2 − 1dx. Then

∫x ln x√x2 − 1

dx =

∫ln√

u2 + 1 du =1

2

∫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

1

2

∫ln(u2 + 1) du =

1

2u ln(u2 + 1)−

∫u2

u2 + 1du .

The point is that the last integrand is rational, and we know how to integrate allsuch functions!

Another strategy would be to use the same trigonometric inverse substitution im-mediately: 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.This suggests 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 Try to integrate

∫ 2π

0

√1 + sin x dx. A solution will be provided next day.




C.12.1 §7.5 Strategy for Integration (continued)

On the integral∫ √

1 + sin x dx. This is an interesting integral which, at first glance,does not appear to fit into any of the families of integrals we have been studying. How-ever, it can easily be seen that, except for the determination of a sign, the integral is notvery difficult to evaluate. But the sign question is delicate, and even the previous editionof a well known text book overlooked this difficulty. In the Student Solution Manual tothat 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.Unfortunately, the answer that was given in that textbook was correct only within certainintervals.

Solution following the suggestion

∫ √1 + sin x dx =

∫ √1 + sin x ·

√1− sin x√1− sin x

dx =

∫ √1− sin2 x

1− sin xdx

=

∫ √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 to which 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+1

2· 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 xmust be positive; the correct value 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 6= 0

Other ways to find the indefinite integral While we cannot remove the signdifficulties in this problem, we can show that the problem does, in fact, lend itself to amore systematic integration — i.e., the hint given above is not really necessary. One wayto see this is to remember that we have a trigonometric identity that expresses 1 + cos θas a square. But, as the 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 observethat

√1 + sin x =

√sin2 x

2+ cos2

x

2+ 2 sin

x

2· cos

x

2

=∣∣∣cos

x

2+ sin

x

2

∣∣∣ .

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(x

2+

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 an antiderivative of√

1 + sin x? We have found that, if weconfine ourselves to one interval of the form 4n+1

2π < x < 4n+3

2π, any antiderivative

has the form −2√

1− sin x + C; and, if we confine ourselves to one interval of the form4n+3

2π < x < 4n+5

2π, any antiderivative has the form 2

√1− sin x + C. By choosing the

constants to make the function continuous (indeed, differentiable) we can patch suchsubfunctions together to form 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 < −12

π −4√

2 + 2√

1− sin x−12

π ≤ x < 12π 0

√2− 2

√1− sin x

12π ≤ x < 3

2π 0

√2 + 2

√1− sin x

32π ≤ x < 5

2π 4

√2− 2

√1− sin x

52π ≤ x < 7

2π 4

√2 + 2

√1− sin x

72π ≤ x < 9

2π 8

√2− 2

√1− sin x

92π ≤ x < 11

2π 8

√2 + 2

√1− sin x

112π ≤ x < 13

2π 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.12.2 §7.6 Integration Using Tables and Computer Algebra Systems

This section “is not examination material, but students are to try to solve the problemsmanually.”

C.12.3 §7.7 Approximate Integration

This section is not part of the syllabus of this course.


C.12.4 §7.8 Improper Integrals

Piecewise continuous integrands Consider a function f that is continuous every-where in an interval [a, b], including continuity from the right at b and from the left ata. For such functions we have developed the theory of the definite integral, and theFundamental Theorem applies. Now suppose that there is a point c such that a < c < b,where f has a jump discontinuity: lim

x→c−f(x) and lim

x→c+f(x) both exist, but are different.

It is possible to define the integral of f as follows:

∫ b

a

f(x) dx =

∫ c

a

f(x) dx +

∫ b

c

f(x) dx ,

and it can be shown that the familiar properties of the definite integral hold here, withthe exception of the Fundamental Theorem. We can proceed in the usual way withsuch integrals, until we need to actually evaluate them; then we must split them up atthe jump discontinuity before we attempt to apply the Fundamental Theorem. Moregenerally, this definition “works” for functions with many jump discontinuities.2930

Other types of generalizations In this section we wish to consider other types ofgeneralizations related to a condition of the original definition of the definite integral thatfails to be satisfied. We will follow the terminology of the textbook, calling two typesof “impropriety” Type 1 and Type31 2 , but students should be aware that these termsare not in universal use. The general spirit of these definitions, and of the precedinggeneralization to jump discontinuities, is that the familiar properties of integrals provedfor continuous functions should hold for these broader classes wherever they make sense.We shall still have to keep away from any situation that might lead us to attempt to,for example, add +∞ to −∞, and any other undefinable operations. Remember that, inthis theory, you must rely on the definition, and not attempt to write down what “makessense”. The restrictions in some of these definitions are needed to avoid paradoxeselsewhere.

Type 1: Infinite Intervals Our original definition of the definite integral was givenfor a finite interval. If we wish to speak of an integral where either or both limits areinfinite, we need to define what these are to mean. The definition we give is one thatis consistent with the definition for finite intervals, and preserves those properties of

29We will not explore here what limits — if any — exist on the number of discontinuities.30We 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 splitting is not by choice, it is by our definition of the meaning of the generalized symbol.

31Note that the terms Type 1 and Type 2 refer to a type of “impropriety” — not to a type of integral.One improper integral could contain multiple instances of both Types.


the integral that are meaningful when the limits before infinite. We repeat the boxeddefinition on [1, p. 531]:

Definition C.1 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” integralon the left is then said to converge or to be convergent . If the limit does not exist,the improper integral is divergent .

2. An analogous definition holds when the upper limit is −∞, or when the lower limitis either of ±∞. Read the textbook.

3. When both limits of the definite integral are infinite, we define the value to be thesum of two integrals obtained by splitting the domain. It can be shown that itdoesn’t matter where the line is split, the following definition will always give thesame value: ∫ ∞

−∞f(x) dx =

∫ a

−∞f(x) dx +

∫ ∞

a

f(x) dx

4. It is ESSENTIAL to understand that, in the definition of∫∞−∞, TWO limits have

to be determined 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 finite number — and that includes being infinite — thesum is not finite, and the improper integral does not exist.

Type 2: Discontinuous Integrands I have observed above how to cope with afinite, jump discontinuity in the integrand. Here we are interested in other types ofdiscontinuity, in particular, discontinuities where the function has a vertical asymptote.If the discontinuity occurs at the 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” integrais defined 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 basedon splitting 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 twolimits have to be evaluated independently. It is important to understand that in this caseit is not acceptable that the two limits be linked so that one may affect the other.

Example C.20 The improper integral

∫ 2

0

1

x− 1dx does not converge, since neither of

the limits

limt→1−

∫ t

0

1

x− 1dx , lim

t→1+

∫ 2

t

1

x− 1dx

exists. We say that this improper integral diverges.

Without this severe definition we might find that some of the properties we wish theintegral to possess might not be present.

Comparison Test for Improper Integrals The definitions we have sketched concernlimits of the values of certain integral. Where those integrals do not exist, we can justify asimilar theorem to that we saw in connection with finite integrals. Read [1, pp. 536–537].

7.7 Exercises

[1, Example 4, p. 533] “For what values of p is the integral

∫ ∞

1

1

xpdx convergent?”

Solution:

∫ ∞

0

1

xdx = lim

t→∞

∫ t

1

1

xdx

=

limt→∞

[ln x]t1 when p = 1

limt→∞

1

1− p

[x1−p

]t

1when p 6= 1

=

limt→∞

[ln x− 0] = ∞ when p = 1

1

1− plimt→∞

[1

xp−1− 1

]=

1

p− 1when p > 1

1

1− plimt→∞

[x1−p − 1

]= ∞ when p < 1

This result will be needed in Chapter 11 in connection with the “p-series test” [1,p. 725]. In the special case p = 1, it states that the area under the hyperbolay = 1

xfrom x = 1 indefinitely to the right is unbounded. By symmetry, that region


should have the same area as the area between x = 0 and x = 1 under the samecurve, and over the line y = 1. That area is the improper integral

∫ 1

0

1

xdx = lim

t→0+

∫ 1

t

1

xdx

= limt→0+

[ln x]1t

= − limt→0+

ln t = +∞

WeBWorK Assignment R1 The following problem, in a number of other variations,was inadvertently included in a WeBWorK assignment: “Evaluate the definite

integral

∫ e3

1

dx

x√

ln x.” Show that the integral is a convergent, improper integral,

and find its value. (While the problem was premature on R1, a naive solutionwould still produce the correct value.).

Solution: The integrand is not defined at x = 1, since one factor in the denominatoris expressed in 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]3

ln t= 2

√3− 2 lim

t→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 the right at t = 1.

[1, Exercise 52, p. 538] “Use the Comparison Theorem to determine whether the in-

tegral

∫ ∞

1

x√1 + x6

dx is convergent or divergent.”

Solution: . At first glance, this integral suggests a substitution u = x2. While thatwould simplify its form, it would not enable us to integrate it immediately, and itis not necessary, since we can prove the convergence without this step. A simpler

attack is to observe that 1 + x6 > x6, so√

1 + x6 > x3,1√

1 + x6<

1

x3. Hence

we can consider the limit limt→∞

∫ t

1

1

x3dx = lim

t→∞

[− 2

x2

]t

1

= 2 − 2 limt→∞

1

t2= 2. The


convergence of the larger, improper integral implies the convergence of the givenone.

[1, Exercise 75, p. 540] “Show that

∫ ∞

0

x2e−x2

dx =1

2

∫ ∞

0

e−x2

dx .”

Solution: This is an interesting question, because we cannot express an antideriva-tive of e−x2

in terms of elementary functions. So, at first glance, one wonders howit will be possible to work with these integrals. Before doing that, I must provethat the integrals are convergent — otherwise we don’t have any right to includethem as numbers in an equation.

I note that

x ≥ 1

⇒ −x2 ≤ −x (multiplying the inequality by a negative number)

⇒ e−x2 ≤ e−x (exponential function is increasing)

⇒∫ t

1

e−x2

dx ≤∫ t

1

e−x dx =1

e− 1

et

Hence, as t →∞, the integral on the right approaches1

e, i.e., the improper integral

∫ ∞

1

e−xdx =1

e.

By the [1, Comparison Theorem, p. 536], the improper integral

∫ ∞

1

e−x2

dx is also

convergent, and is less than1

e. 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 reasoninggiven above is valid only for x ≥ 1. For example,

−x2 < 0 ⇒ e−x2

< e0 = 1

⇒∫ 1

0

e−x2

dx <

∫ 1

0

1 dx = 1

Hence ∫ ∞

0

e−x2

dx =

∫ 1

0

e−x2

dx +

∫ ∞

1

e−x2

dx < 1 +1

e.


This is not the exact value of the integral; in fact, it can be shown that

∫ ∞

0

e−x2

dx =

√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−x2

dx and u = x, so v = −12e−x2

, and du = dv:

∫ t

0

x2e−x2

dx = −1

2xe−x2

]t

0

+1

2

∫ t

0

e−x2

dx

=t

et2+

1

2

∫ t

0

e−x2

dx.

By l’Hospital’s Rule,

limt→∞

t

et2= lim

t→∞1

2tet2= 0 .

Hence∫ ∞

0

x2e−x2

dx = limt→∞

∫ t

0

x2e−x2

dx

=1

2limt→∞

∫ t

0

e−x2

dx

=1

2

∫ ∞

0

e−x2

dx .

Review

Problems Plus



Release Date: Monday, February 16th, 2004, subject to further revision

Textbook Chapter 8. FURTHER APPLICATIONS OFINTEGRATION.

C.13.1 §8.1 Arc Length

Just as with the earlier concepts of area, volume, and average, we are faced first withadopting a definition that appears to have the properties that we associate with theconcept, and, at the same time, is workable in practice. The length of an arc willbe defined to be the limit — if there is a limit — of the sum of the lengths of anapproximating polygon formed by choosing points closer and closer together on thecurve, and joining them by line segments. Note that we haven’t even defined what wemean in general by a curve, so the definition we give will apply at first only to the graphof a function.

Suppose that we wish to find the length of the arc of the graph of y = f(x) betweenthe points (a, f(a)) and (b, f(b)). We can subdivide the interval [a, b] on the x-axis byintermediate vertices, so that we have a sequence a = x0, x1, x2, . . . , xn = b of points onthe x-axis. If we define

∆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 incrementsappear in these formulæ only as magnitudes (or squares). When we pass to the limit, asthe “mesh” of points selected on the x-axis become closer and closer together, the first ofthese expressions, gives rise in the limit to the following integral representing the lengthof 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 lengthfrom the point (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 they give the same length.32

Evaluation of these integrals can often require an approximation method, as theintegrands tend to be of types for which a function expressible in terms of elementaryfunctions is unavailable. For that reason the problems that one meets in calculus booksare often confined to a 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 expressedas an integral with a variable upper limit, and is signed , so that, in effect, we haveparameterized the curve with a variable — usually denoted by the symbol s — uniquelydenoting the position of a point on a path along the curve. This practice differs fromthat employed when we are evaluate the length of the arc between two points, whereonly the magnitude is of interest. I have written “...on a path along the curve” ratherthan “...on the curve”, because we shall be generalizing these ideas in [1, Chapter 10],to consider curves that are not the graphs of functions; in those generalizations a curvemay cross itself, and the same point could be traversed more than once by a point whosepath we are studying: in that case it is the length of the path that will be given by thearc length function.

Example C.21 Circumference of a circle. What is the circumference of the circlex2 + y2 = R2 (where R > 0).?Solution: Since the equation given is not the graph of a function (because the curvecrosses some vertical lines more than once), let’s find the length of the upper arc fromx = −R to x = +R, and double it. This is given by the function

f(x) =√

R2 − x2 = R

√1−

( x

R

)2

.

y′ = −xR√

1− (xR

)2

32Passage between the two forms can be seen to result from the change of variable given by y = f(x).


1 + (y′)2=

1

1− (xR

)2

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 change of variable u =x

R

= 4R lima→1−

∫ a

0

1√1− u2

du

under the change of variable a =t

R

= 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

= 4R lima→1−

[v]arcsin a0

= 4R arcsin 1 by continuity of arcsin

= 4R · π

2= 2πR .

Of course, we didn’t need to apply this last substitution v = . . . because we know 2

antiderivatives of1√

1− u2:

4R limt→R−

∫ tR

0

1√1− u2

du = 4R limt→R−

[arcsin u]tR0

= 4R limt→R−

[arcsin

t

R− 0

]


= 4R [arcsin 1− 0] by continuity of arcsin

= 4R[π

2− 0

]= 2πR .

8.1 Exercises

[1, 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 descriptionof the underlying curve, here given by an equation, and specifications of the portionof the curve whose length is to be determined, here given by an interval 2 ≤ x ≤ 4or in the alternative notation x ∈ [2, 4]. Only the absolute value of the length is ofinterest, so we need not be careful about which of the end-points 2, 4 is in whichlimit of the integral; alternatively, it is the absolute value of the integral that weseek.

y =x2

2− ln x

4⇒ dy

dx= x− 1

4x

⇒√

1 + (y′)2 =

√1 +

(x2 − 1

2+

1

16x2

)

⇒√

1 + (y′)2 =

√x2 +

1

2+

1

16x2=

∣∣∣∣x +1

4x

∣∣∣∣

Hence the arc length between x = 2 and x = 4 (where the function

∣∣∣∣x +1

4x

∣∣∣∣ is

equal to x +1

4x) is

∫ 4

2

∣∣∣∣x +1

4x

∣∣∣∣ dx =

∫ 4

2

(x +

1

4x

)dx

=

[x2

2+

ln x

4

]4

2

= 8 +2 ln 2

4− 2− ln 2

4= 6 +

ln 2

4= 6 + ln

4√

2 .

[1, Exercise 10, p. 552] 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) .

Discovery Project: Arc Length Contest

C.13.2 §8.2 Area of a Surface of Revolution

We develop a formula for the area of a surface of revolution by observing that an elementof arc of length ∆s will, when rotated about an axis whose distance from the element isr, generate an element of surface whose area is approximately 2πr · ∆s. Rememberingthat

∆s =

√1 +

(∆f(x)

∆x

)2

·∆x =

√1 +

(∆x

∆y

)2

·∆yi .

we can integrate with respect to either x or y as the conditions of the problem demand.

8.2 Exercises

[1, Exercise 22, p. 559] ...Find the area of the surface obtained by rotating the curvey =

√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 48, p. 481]

=π√2[tan θ · sec θ + ln | sec θ + tan θ|]arctan 3

√2

0

= 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. 559] Find the area of the surface obtained by rotating the curvey = 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)2dx

= 2π

∫ 2

1

(u− 1)

√1 +

1

u2du

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

=1

2[tan θ · sec θ + ln | sec θ + tan θ|]arctan 2

π4

by the reduction formula [1, Exercise 48, p. 481]

=1

2

[(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 2π4


=

[(√5 + ln

(√5− 1

2

))−

(√2 + ln(

√2− 1)

)]

etc.

Here again the integral does not require special software or tables, just patience.(I haven’t checked all the computations.)




C.14.1 §8.3 Applications to Physics and Engineering

Omit this section. (But read it if you are majoring in Physics or Engineering!)

C.14.2 §8.4 Applications to Economics and Biology

Omit this section. (But read it if you are majoring into economics or biology.)

C.14.3 §8.5 Probability

Omit this section.

Textbook Chapter 9. DIFFERENTIAL EQUATIONS.

No parts of this chapter are included in the syllabus.

Textbook Chapter 10. PARAMETRIC EQUATIONS ANDPOLAR COORDINATES.

C.14.4 §10.1 Curves Defined by Parametric Equations

A parameter is just a variable. When we call a variable by this term we usually arethinking of a function or set of functions involving the variable as representing a familyof objects. In this first contact, the family will be a set of points . We will be taking thevariable to be a real variable, and so it is natural to consider not only the family of pointson the graphs of a function, but also the way in which the points are generated by theassignment of real numbers to the parameter as representing a point moving along thecurve. We can name the parameter with any available symbol. Often we use the letter t;and a common use for this representation is to treat t as time, so that the curve can bethought of as the trajectory of (i.e., the path traced out by) a moving point. If we adoptthis point of view, and if the parameter values are chosen from an interval a ≤ t ≤ b,then we can speak of the curve (x(t), y(t)), and can think of (x(a), y(a)) and (x(b), y(b))as, respectively, the initial and terminal points. Curves given parametrically in this wayneed not be graphs of functions: a curve may cross vertical lines more than once, andmay even cross itself, possibly more than once.

Graphing Devices This section may be omitted, as, in this course, we shall not beconcerned with the use of graphing devices.


Can the graph of a function, given non-parametrically, be expressed in para-metric form? The curve y = f(x) can be expressed in parametric form as, for example,x = t, y = f(t). But there are infinitely other ways of expressing it parametrically, forexample

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 acurve in parametric form, the parametrization sometimes contains information beyondwhat is available in a non-parametric representation. Sometimes we can see a dynamicway of actually tracing out the curve by allowing the parameter to range through itsdomain. So, for example, the curve

x = cos t

y = sin t

can be thought of as being traced out by a point that moves around the unit circlecentred at the origin, starting at the point (1, 0) at time t = 0, counterclockwise at arate of 1 radian per unit time as t increases (and clockwise at a rate of 1 radian per unittime as t decreases. If we compare the curve with

x = cos 2t

y = sin 2t

we see that both trace out the same circle, covering the curve infinitely often, but thesecond curve moves twice as fast. If we wanted to eliminate the multiple covering, wecould include the inequalities 0 ≤ t < 0 in the first case, or 0 ≤ t < π

2in the second.

The standard way to transform from parametric to non-parametric equations is toeliminate the parameter “between” the equations , which is essentially to solve one equa-tion for the parameter, and to substitute that value into the second equation. However,this operation sometimes “loses information”.

Example C.22 [1, Exercise 13, p. 656] asks you to find a Cartesian equation of thecurve x = cos2 θ, y = sin2 θ,

(−π2

< θ < π2

). We can eliminate θ by solving one equation

for θ and substituting into the second; or, more elegantly, by adding the two equationstogether, obtaining an equation that does not involve θ:

x + y = 1


But this equation does not convey all of the information we started with. One type ofinformation that has been lost is the fact that both x and y, being squares, are non-negative. Thus it is not the whole line x + y = 1 that is equivalent, but only the linesegment joining the points(0, 1) and (1, 0). This feature is essential, since the doubly-infinite line is not the curve represented by the parametric equations. Another type ofinformation that is contained in the parametrization is the way in which this line segmentis traversed. The point is moving back and forth between the point (1, 0) and the point(0, 1), covering the open segment in a parameter interval of length π

4. Distance is not

covered at a constant speed — the point moves fastest in 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 equationx + y = 1 is not enough.

Example C.23 [1, Exercise 17, p. 656] gives the parametric equations

x = cosh t y = sinh t

For every value of t a point with these coordinates can be seen to lie on the curvex2 − y2 = 1, which is a hyperbola with two branches, one opening to the right, andpassing through the point (1, 0), the other opening to the left, and passing throughthe point (−1, 0); the curves are both asymptotic to the lines y = ±x. But, since thehyperbolic cosine is positive, the parametrization applies only to the right branch of thehyperbola: 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 way to describe the curvenon-parametrically is

x2 − y2 = 1

x ≥ 0.

This curve may be parametrized in other ways. For example, we could represent it as

x = sec t

y = tan t

(cf. [1, Exercise 14, p. 456]) but this time we need to restrict the values the parametermay take, for example by

−π

2< θ <

π

2.

The Cycloid In this subsection the textbook describes the construction of this inter-esting curve. You are not expected to remember specific properties of this curve, nor itshistory. However, you should be able to work with this curve if the parametric equationsare given to you, in the same way as you would be expected to be able to work with anyreasonable curve given parametrically.


Families of Parametric Curves Here the author considers a family of curves givenparametrically, in which the various members of the family all have a similar equationobtained by assigning different values to a variable in the parametric equations. Thatvariable is also called a parameter , and the family of curves could be called a parametricfamily of curves represented by parametric equations . Of course, the parameter thatrepresents the family of curves is not the same as the parameter that represents thefamily of points on any specific curve. For example, x2 + y2 = a2 could be considereda parametric family of curves (circles) given in non-parametric form; we could representthe same family in parametric form by

x = a cos t

y = a sin t

where the parameter t represents position on a specific curve, and the parameter arepresents the different curves in the family. We could also interpret the equation byswitching the roles of the parameters: the curve

x = a cos t y = a sin t (23)

with t constant and a variable represents the points on the line through the origin inclinedto the positive x-axis by an angle of t radians. This point of view will become importantin [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 theobvious parametric equations for familiar curves, like the circle).

Example C.24 Equation of a line in the plane.The line in the plane through thepoint (x0, y0) and with slope m has non-parametric equation y = y0 + m(x − x0). Itcan be represented parametrically in infinitely many ways. If we choose to relate theparameter to distance along the line, one can show that the following equations representthe line

x = x0 + t

y = y0 + mt

Check that the line segment joining any two points on this line has slopem

1= m. (Take

two points with coordinates (x0 + a, y0 + ma) and (x0 + b, y0 + mb).)

C.14.5 §10.2 Calculus with Parametric Curves

Tangents If a curve is given parametrically by (x, y) = (x(t), y(t)), then (subjectto certain conditions) we can differentiate y with respect to x by passing through an


intermediate variable t. Recalling that/

dx

dt· dt

dx=

dx

dx= 1 ,

we have

dy

dx=

dy

dt· dt

dx

=dy

dt· 1

dx

dt

so we can determine the first derivativedy

dxin terms of the derivatives of the functions

f and g. This calculation can be extended to the second derivatived2y

dx2, although the

expressions are not as pretty:

d2y

dx2=

d

dx

(dy

dx

)

=d

dt

(dy

dx

)· dt

dx

=d

dt

dy

dtdx

dt

· 1

dx

dt

=

d2y

dt2· dx

dt− d2x

dt2· dy

dt(dx

dt

)2

· 1

dx

dt

=

d2y

dt2· dx

dt− d2x

dt2· dy

dt(dx

dt

)3

Rather than substituting in this formula, memorized, you are advised to be able tocarry out this computation for a specific parametrized curve. It enables us to study theconcavity of a parametrized curve, and to apply the 2nd derivative test if necessary inan optimization problem.

Example C.25 (cf. [1, Exercise 6, p. 666]) (This example from the textbook was chosenat random at the lecture; the intention was not to illustrate clever insights into the


particular example, which is interesting in its own right.) Let’s investigate the curvegiven parametrically by

x = cos θ + sin 2θ

y = sin θ + cos 2θ

Differentiating the parametrizing functions, we obtain33

dx

dθ= − sin θ + 2 cos 2θ

= −4

(sin θ +

1 +√

33

8

)(sin θ +

1−√33

8

)

dy

dθ= 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 withparameter value θ = 0. We find that, when θ = 0,

(dxdθ

, dydθ

)= (1, 2), 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 +1

2(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 that dydθ

= 0. We find that this happens when

cos θ = 0 or sin θ =1

4,

which we could proceed to solve. We can also determine where the curve is vertical, bydetermining where dx

dθ= 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

33Some of these computations were not done at the lecture.


passes through an interval of length 2π, we can see the whole curve by confining θ to theinterval 0 ≤ θ ≤ 2π. In that interval the cosine vanishes when θ = π

2, 3π

2, and sin θ = 1

4

when θ = arcsin 14

and π−arcsin 14. These are the 4 points where the curve is horizontal;

one of those point is the origin, because the curve passes through the origin several times,one of those times with a horizontal tangent. The curve has the shape of a “3-leafedrose” or a “trefoil” (3-leafed clover) centred at the origin, with one petal centred on they-axis. We will see other curves with this shape when we study polar coordinates in [1,§10.3].

We will continue study of this section in the next lecture, March 1st, 2004.


C.15 Supplementary Notes for the Lecture of March 1st, 2004

Release Date: Monday, March 1st, 2004, subject to further revision

Correction to the syllabus. Please note that although the original version of thesyllabus showed that [1, §10.5] was included, this was an error. The only sections ofChapter 10 that are included are §§10.1–10.4. Some of the material in §10.5 is studiedin Math 133.

C.15.1 §10.2 Calculus with Parametric Curves (continued)

Tangents (continued) In the last lecture we considered the curve given in [1, Exercise6, p. 666]):

x = cos θ + sin 2θ

y = sin θ + cos 2θ.

Let’s consider the following question, similar to [1, Exercise 25, p. 666]: “Show that thecurve..has (several)...tangents at (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. Thismeans that either

sin θ = −1

2,

or both34 of the following equations must hold:

cos θ = 0

sin θ = 1 .

For θ between 0 and 2π, these last equations are satisfied when θ =π

2; the first equation

is satisfied when θ =7π

6,11π

6. 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 ratiodydθdxdθ

.

34Note that the second equation implies the first, but the first does not imply the second; we couldthus have shown only the second equation.


Areas Consider the arc between α ≤ t ≤ β of the curve x = f(t), y = g(t). 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 cross vertical linesmore than once. This means that the area could be “folded”, and there might be portionsthat could be counted negatively and cancelling the portions that you are interested in.For that reason you should use this integral only where you are clear about the shape ofthe region whose area you are finding, and there might be situations where the regionshould be broken up into parts and the 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

√(dx

dt

)2

+

(dy

dt

)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)), thisreduces to the formula we derived earlier. Here again, be careful that you are findingthe length of the curve that you intend. In this case there cannot be any cancellation,since the integrand is a square root, which cannot be negative. If you obtain a negativelength, it could simply be a consequence of the direction in which you have parametrizedthe curve, which is harmless; or of an error you have made in you calculations, which isserious.

Surface Area We can also adapt, in the obvious ways, our previous formulæ for areaof surfaces of revolution.

Example C.26 Let’s determine a formula for the surface area of a doughnut. Supposethat the doughnut is generated by the curve

x = R + r cos θ

y = 0 + r sin θ


where 0 ≤ θ ≤ 2π, where this circle, centred at the point (x, y) = (R, 0), is revolvedaround the y-axis. Assume R ≥ r.

Area of revolution =

∫ 2π

0

2π(R + r cos θ)

√(dx

dt

)2

+

(dy

dt

)2

dt

=

∫ 2π

0

2π(R + r cos θ)√

r2 dt

= 2πr [Rθ + r sin θ]2π0 = 2πr[R · 2π −R · 0] = 4π2rR.

Volumes The textbook appears to say nothing about this, but here again the earlierformulæ can be adapted, to determine, for example, the volume of revolution generatedby a given curve about a given axis.

10.2 Exercises

[1, Exercise 54, p. 667] “ 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 thecentre at the points away from where it crosses the coordinate axes.

Total length =

∫ 2π

0

√(dx

dt

)2

+

(dy

dt

)2

dt

=

∫ 2π

0

√(−3a cos2 θ · sin θ)2 +

(3a sin2 θ · cos θ

)2dt

=

∫ 2π

0

3|a| · | cos θ · sin θ| dt

By symmetry we can find the length of one quarter of the curve:

Total length = 4

∫ π2

0

3|a| · | cos θ · sin θ| dt

= 4|a|∫ π

2

0

3 cos θ · sin θ dt

= 6|a| [sin2 θ]π

2

0= 6|a| .

Laboratory Project: Bezier Curves Omit this subsection.


C.15.2 §10.3 Polar Coordinates

Polar Curves In the polar coordinate system we locate points in the plane by takinga special point, called the pole, a special half-line or ray which emanates from the pole,called the polar axis and a direction for measuring positive angles — usually taken tobe the counter-clockwise direction. Any point can be located if we know its distancefrom the pole, usually denoted by r, and the angle that the line joining them makes withthe polar axis in the positive direction, usually denoted by θ. However, the angle is notunique, since angles of θ and θ + 2nπ will give the same point for any integer n. So herewe have one of the essential differences between polar and cartesian coordinates:

Theorem C.27 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 thatr = 0, any angle θ will give the same point.

Convention permitting negative r It is convenient to broaden the multiplicity ofcoordinates by permitting the distance from the pole to be negative. We do this byagreeing that (r, θ) and (−r, θ+π) represent the same point. This convention permits forcontinuous representation of certain curves, but causes complications at various stages:occasionally additional care is required.

Relations between polar and cartesian coordinates In theory we can set up polarand cartesian systems independently in the plane, placing the pole at any convenientplace. In practice we often place the pole at the origin of a cartesian system, with thepolar axis along the positive x-axis. When the author of the textbook suggests that youare to consider two systems at the same time, and gives no other information, this iswhat he expects you to do.35 When the polar and cartesian systems are placed in this“standard” way, the following relationships hold:

x = r · cos θ

y = r · sin θ

x2 + y2 = r2

y

x= tan θ

Note that, while it is possible to transform from polar to cartesian coordinates withoutambiguity, it is not always possible to move painlessly in the other direction; this is

35However, there is an important application, involving conic sections — ellipses, parabolæ, hyper-bolæ, where we place the origin at a different point; this topic is not on the syllabus of the presentcourse, but you may read about it in [1, §10.6].


because of the non-uniqueness of polar coordinates, about which much more will be said.If you are given an equation in cartesian coordinates, you can transform it to polar bysubstituting the appropriate formulæ for x and y and simplifying — try [1, Problems21-26, p. 678]; if you are given the coordinates of a point in polar coordinates, you cantransform to cartesian in the same way — try [1, Exercises 3-4, p. 677].

Consequences of Non-uniqueness of Polar Coordinates This is a difficult topic,and will require considerable practice before you will become comfortable with it! Someof the problems you will have are related to understanding what is meant by an equationfor a curve: you need to understand that any curve can be represented in multiple ways,even when we use cartesian coordinates. But, when the representation of the pointsthemselves is not unique, the results can be confusing.

We saw with parametric equations that the same point on a curve can appear morethan once on a “curve”. In that context there was a “natural” way of tracing outthe curve, by following increasing values of the parameter. When we come to studypolar coordinates, the situation is much more complicated, because there is no “natural”way of following the generation of the curve, and no one set of coordinates for a pointhas preferential status with respect to others. Polar curves can be expressed by anyrelationship between r and θ, although more often than not the equation will be in theform r = f(θ); however, you will see some equations in the form θ = f(r) — for example,

the equation θ =π

4represents the line through the pole inclined at an angle of

π

4to the

polar axis.

Symmetry When we studied symmetry in the context of cartesian coordinates, weconsidered

• 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 theorigin; and

• periodicity of a graph under a horizontal shift.

These are not the only types of symmetries that a graph can possess, so our study wasselective, and we didn’t investigate thoroughly what principles were involved.

Similarly, when we consider symmetry in the context of polar coordinates, we willnot attempt a thorough study, but will consider types of symmetries which the polarsystem is particularly able to accommodate. Here are the types of symmetry that thetextbook mentions:


• reflective symmetry in the polar axis — exhibited by the invariance of the equationof a curve under the substitution θ 7→ −θ.36

• rotational symmetry under a rotation through an angle of π around the pole —exhibited by the invariance of the equation under the transformation r 7→ −r orunder 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. Becauseof the non-uniqueness of coordinates, curves can have the symmetries listed withoutexhibiting invariance under the transformations listed. For example, the curve θ = π isthe line that extends the polar axis. It certainly has symmetry in the polar axis, butthe equation is not unchanged when we replace θ by −θ. You are not expected to be anexpert in the subject of symmetries.

Tangents to Polar Curves We discuss how to determine the tangents to curves ofthe form r = f(θ). The discussion is based on the observation that, if we superimposepolar and cartesian systems in the most usual way, we can express x and y in terms ofθ by

x = f(θ) · cos θ

y = f(θ) · sin θ

and thereby interpret θ as a parameter in the parametric representation of a curve.

Graphing Polar Curves with Graphing Devices Omit this section — this is adevice-free course.

10.3 Exercises

[1, Exercise 2, p. 677] Plot the point whose polar coordinates are(−1,−π

2

). Then

find two other pairs of polar coordinates of this point, one with r > 0, and onewith r < 0.

Solution: The second coordinate tells us that the point is on the line inclined tothe polar axis at an angle of −π

2radians measured in the positive direction: this

takes us to the ray which is obtained by turning the polar axis in the clockwisedirection through a right angle. But then the negative first polar coordinate tells

36This description is incomplete. See the discussion below of [1, Exercise 37, p. 678].


us to proceed along the opposite ray for 1 unit. If the polar system is superimposedon a cartesian system in the usual way, the point is the unit point on the positivey-axis. This point has the following other sets of polar coordinates:

with positive r:(1, π

2+ 2nπ

)

with negative r:(−1, −π

2+ 2mπ

)or

(−1, 3π2

+ 2`π)

where n, m, and ` are any integers.

[1, Exercise 30, p. 678] Describe the curve with the equation r2 − 3r + 2 = 0. (Thisproblem was not discussed in the lecture, but it is similar to the problem of thecurve r(r − 3) = 0, which was discussed.)

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 pointson either of two concentric circles around the pole.

[1, Exercise 35, p. 678], extended. Describe the curve with equation r = θ.

Solution: This curve is a spiral , turning around the pole. Allow θ to take onnegative values also: the spiral passes through the pole and then turns inside thecurve for positive θ.

[1, Exercise 33, p. 678] Describe the curve with the equation r = 2(1− sin θ).

Solution: Under the transformation θ 7→ π − θ the equation is unchanged, so thecurve has symmetry in the y-axis. This is a “heart-shaped” curve, called a cardiod .The textbook describes another example in [1, Example 9, pp. 675-676], where theauthor shows, as we could show here, that the curve is tangent to the y-axis. Wecall the point of tangency (here the pole) a cusp.


C.16 Supplementary Notes for the Lecture of March 3rd, 2004

Release Date: Wednesday, March 3rd, 2004, subject to further revision

C.16.1 §10.3 Polar Coordinates (continued)

Where is the point with polar coordinates (r, θ)? The ambiguity in polar coordi-nates is not in locating a point with given coordinates — it is only that all points possessmultiple sets of coordinates. If you are given the coordinates (r, θ), you may locate thepoint by:

• First, locating the ray obtained by turning the polar axis (the distinguished raythat emanates from the pole, relative to which we refer all coordinate angles)through an angle of 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 adistance of −r.

10.3 Exercises (continued)

[1, Exercise 12, p. 677] “Sketch the region in the plane consisting of points whosepolar coordinates satisfy the conditions −1 ≤ r ≤ 1, π

4≤ θ ≤ 3π

4.”

Solution: Let’s separate the portion of the region described by positive r, zero r,and negative r:

positive r. 0 < r ≤ 1, π4≤ θ ≤ 3π

4describes a sector of the unit disk centred

at the pole — all points in the region bounded by two perpendicular radiibisecting the first and second quadrants. The points on the radii and thebounding circle are included.

zero r. The region described by r = 0, π4≤ θ ≤ 3π

4consists of the pole alone: the

θ coordinate is irrelevant for the pole.

negative r. −1 ≤ r < 0, π4≤ θ ≤ 3π

4describes the region antipodal to the one

described for positive r — between radii at angles π + π4

and π + 3π4

.

[1, Exercise 18, p. 678] Identify the curve r = 2 cos θ + 2 sin θ.

Solution: The textbook suggests finding a cartesian equation first. This is notalways easy, but is not impractical in the present problem. If we multiply thegiven equation by r, we obtain the equation

r2 − 2r cos θ − 24 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 notethat, in multiplying an equation by a factor — equivalently in multiplying the twoequations

r = 2 cos θ + 2 sin θ (24)

r = 0 (25)

we were, in the second equation, possibly permitting new points to be included inthe “curve”. We must carefully analyze whether the equation r = 0 did that. Butwe know that r = 0 represents only the pole! And the pole was already on thegiven curve: it appears there as

(r, θ) =

(0,

3π

4

).

So multiplying by r = 0 does not introduce any new points.37

[1, Exercise 34, p. 678] Describe the curve with equation r = 1− 3 cos θ.

Solution: It is not practical to represent this curve in cartesian coordinates. Wenote that, when θ is replaced by −θ, the equation is unchanged. This tells us thatthe curve is symmetric about the polar axis. Try tracing it out by starting withthe point with θ = 0. If we superimpose a cartesian system in the usual way, thatpoint is the point with cartesian coordinates (−2, 0). The curve passes through thepole first when θ = arccos 1

3, about 70 degrees. It crosses the y-axis at the point

with cartesian coordinates (x, y) = (0, 1), and then moves to its maximum distanceof 1 + 3 = 4 from the pole when θ = π. Then it returns, crossing the negativey-axis at (x, y) = (0,−1), and passing again through the pole. The curve is one ofthe family called limacon’s , see them sketched in [1, Example 11, p. 677].

[1, 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. While the equation does notremain unchanged when we replace θ by −θ, it changes to r = − sin 2θ, which

37But what would happen if you were to multiply an equation like r = 1 by r = 0?


contains the same points, since it can be rewritten as −r = sin 2(θ + π), which canbe obtained from the original equation by the transformation

(r, θ) 7→ (−rθ + π) .

Thus this curve is symmetric about the polar axis. It is also symmetric about they-axis, since the 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, iflocated in the usual way.

But, when the multiplier of θ is an odd integer, the situation changes. The curver = sin 5θ is again a “rose”, passing through the pole every 72 degrees. It is notsymmetric about the y-axis, nor about the x-axis. (It is symmetric under reflectionsin other axes, and also under rotation through certain angles around the pole.)

C.16.2 §10.4 Areas and Lengths in Polar Coordinates

Areas To find the area bounded by a curve given in polar coordinates we express thearea as the limit of a sum of narrow triangles whose bases are long the bounding curve,and whose upper vertex is at the pole. It will be shown in the lectures that the areasubtended by the arc of the curve r = f(θ) between θ = a and θ = b is then

∫ b

a

1

2· (f(θ))2 dθ .

As with all other formulæ involving polar coordinates, one must use this formula withcare. Be sure that you know precisely what the region looks like; in case of doubt, breakthe region up into parts, and find the areas of the parts separately. The limits must bechosen carefully, to be sure that, for example, you are not computing more area thanyou intend. For example, the curve r = cos θ is a circle of radius 1

2centred at the point(

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

1

2

∫ π

0

cos2 θ dθ =1

4

∫ 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 isswept out as θ ranges from 0 to 2π;

Area =1

2

∫ 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 youcarefully analyze each problem individually.


Finding the intersections of curves in polar coordinates The textbook [1, p.681] discusses the difficulties that of finding intersections which are a consequence ofthe multiple sets of coordinates for points. However, the text suggests that finding theintersections can only be done visually, and that is not true. In fact, intersections can befound algebraically, but one must be careful and thorough. I attach below a discussion Iprepared for a class some years ago, where another textbook had made a false statementthat I felt obliged to correct; the error is continued in the current edition of the sametextbook.

Example C.28 In an example in another textbook [21, Example 8, p. 579], [20, Exam-ple 8, p. 635] the objective is to find the points where the curves

r = 1 + sin θ (26)

r2 = 4 sin θ (27)

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 equationsare graphed”. We show here all intersection points can be found algebraically! We neverresort to calculations on a sketch: all procedures can be justified theoretically — thesketch serves only to help visualize a situation that can be adequately described verballyand/or with mathematical formulæ.

–2

–1

0

1

2

–1 –0.5 0.5 1


Had the curves been given in cartesian coordinates, we could have found all intersectionsby solving the equations simultaneously. Why can’t we solve the polar equations inthe same way? The difficulty derives from the fact that any point has infinitely manydifferent polar representations. More precisely, a point that can be represented by polarcoordinates (r, θ) also has coordinates ((−1)nr, θ+nπ), where n is any integer — positiveor 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 samepoint appears with different coordinates .

Solve the given equations algebraically: By eliminating sin θ between the two equa-tions, we obtain 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 ofthe equations the transformation

(r, θ) 7→ (−r, θ + π) (28)

and solve it with the original form of the other equation. Repeat this process untilthe equations transform to a pair already solved. Equation (26) transforms to

−r = 1 + sin(θ + π) (29)

which is equivalent tor = −1 + sin θ (30)

which equation we solve with (27). Eliminating sin θ yields r = 2 ± 2√

2, sosin θ = 3 ± 2

√2. The upper sign is inadmissible, as a sine cannot exceed 1 in

magnitude. Hence r = 2− 2√

2 and

sin θ = 3− 2√

2 . (31)

The solutions to (31) 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 coor-dinate in these cases is negative, we could equally well represent the points as

(2√

2− 2, sin−1(3− 2

√2)

+ π)


and (2√

2− 2,− sin−1(3− 2

√2))

.

A second application of (28), to (29), restores the original equation; hence thereare no other intersection points, except possibly the pole.

Check whether the pole is on both curves: On (26) the pole appears as

(0,

3π

2

),

etc.; on (27) it appears as (0, 0), etc. Thus the pole is also a point of intersection.The reason we did not find it when we solved pairs of equations is that it appearson the two curves only with different sets of coordinates, no two related by (28).


C.17 Supplementary Notes for the Lecture of March 8th, 2004

Release Date: Monday, March 8th, 2004, subject to further revision

C.17.1 §10.4 Areas and lengths in polar coordinates (continued)

Example C.29 [1, Exercise 29, p., 683] “Find the area of the region that lies insideboth curves: r = sin θ, r = cos θ.”Solution: The curves are circles through the pole. To see this, multiply the first by r = 0(which, of course, could be bringing the pole into the curve). The resulting equation isr2 = 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 multiplying by 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 circleof the same radius centred at the point (r, θ) =

(12, 0

); polar coordinates for the centre

of the first circle are, for example,(

12, π

4

).

Where do the curves meet? (This is [1, Exercise 37, p. 683], which should havepreceded the present problem in the exercises!) We can begin this investigation bysolving their equations, which imply that tan θ = 1, so θ = π

4+ nπ, where n is any

integer; that implies that r = ± 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, π

4+ 2mπ

)and

(r, θ) =(− 1√

2, π

4+ (2m + 1)π

)are all the same point — having cartesian coordinates(

12, 1

2

).

But the curves also meet at the pole, even though this did not show up when wesolved the equations. This is because the pole appears on the first curve with θ = nπ,and on the second curve with θ = (n+ 1

2)π. The only practical way to determine whether

curves intersect at the pole 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 underthe transformation (r, θ) 7→ (−r, θ+π), we find that the equation does not change (exceptfor a sign change on both sides); the same applies to the second equation. Thus, in thepresent example, there are no intersections other than the pole in which a point willappear on the two curves with different coordinates. We can comfortably proceed tothe integration part of the problem, confident that we have not missed any points ofintersection.

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 by the arc 0 ≤ θ ≤ π4

of the circle r = sin θ; hence

Area = 2 · 1

2

∫ π4

0

sin2 θ dθ

=1

2

∫ π4

0

(1− cos 2θ) dθ

1

2

[θ − sin 2θ

2

]π4

0

=π

8− 1

4.

(You could have solved this problem using Cartesian coordinates.)

The preceding example is straightforward. The next one is not.

Example C.30 [1, Exercise 26, p. 483] Find the area of the region that lies inside thecurve

r = 2 + sin θ (32)

and outside the curver = 3 sin θ . (33)

Solution: The first step is to determine where the curves intersect. We can start bysolving equations (32) and (33) algebraically. We find that r = 3 and θ = π

2+2nπ, where

n is any integer. The second curve is a circle with centre at the point (r, θ) =(

32, π

2

). 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, π

2

). Could there be any other points

of intersection? There don’t appear to be any, from the sketch, but we can resolve thisquestion algebraically. The pole does not lie on first curve, since that would entail that0 = 2 + sin θ, i.e., that the sine of an angle exceeds 1 in magnitude. Since that is notpossible, the pole cannot lie on the curve. If there were to be any other intersections,they would have to have different sets of coordinates on the 2 curves. We investigatewhat happens to (33) 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 (32) changes to−r = 2 + sin(θ + π) = 2− sin θ, or

r = −2 + sin θ. (34)

When we solve this equation with (33), we find that −2 = 0, a contradiction that tellsus that there cannot be any solutions. 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 theouter curve, and then subtract from it the area of the disk inside. The outer curve istraced out as θ ranges from 0 to 2π, so the area of the region it bounds will be

1

2

∫ 2π

0

(2 + sin θ)2 dθ =1

2

∫ 2π

0

(4 + 4 sin θ + sin2 θ) dθ

=1

2

∫ 2π

0

(4 + 4 sin θ +

1− cos 2θ

2

)dθ

=1

2

[9θ

2− 4 cos θ − 1

4sin 2θ

]2π

0

=9π

2

The inner curve bounds an area of

1

2

∫ π

0

9 sin2 θ dθ =9

4

∫ π

0

(1− cos 2θ) dθ

=9

4

[θ − 1

2sin 2θ

]π

0

=9

4· 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

9π

2− 9π

4=

9π

4.

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. 681]? We could do this for theportion of the area above the polar axis and its extension, where we would obtain

1

2

∫ π

0

((2 + sin θ)2 − (3 sin θ)2) dθ

=1

2

∫ π

0

(4 + 4 sin θ − 4(1− cos 2θ)) dθ

=1

2[−4 cos θ + 2 sin 2θ]π0 = (2 + 0)− (−2 + 0) = 4

For the remainder of the area only curve (32) and the extended polar axis serve as aboundary:

1

2

∫ 2π

π

(2 + sin θ)2 dθ =1

2

[9θ

2− 4 cos θ − 1

4sin 2θ

]2π

π

=1

2

[(9π − 4− 0)−

(9π

2− (−4)− 0

)]=

9π

2− 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

range of the integration? We would have been subtracting the area of the inside disktwice!

And what would have happened if we had taken (34) as the equation of the outercurve? If we used it only to find the area of the (larger) region bounded, we would obtainthe correct value there. But, if we had taken the difference of squares under the integralsign, we would obtain values for regions that do not correspond to the one whose are weare trying to find.

Before finding an area in polar coordinates, (or, indeed, in cartesian coordinates aswell), you are urged to make a sketch and study the element of area that you are summingin the limit, to ensure that the integral represents the area that you are seeking.

Arc Length To develop a formula for the arc length of a curve given in polar coor-dinates as r = f(θ), we can apply the theory of curves given in parametric form to thecurve

x = f(θ) · cos θ

y = f(θ) · sin θ

We find that (dx

dθ

)2

+

(dy

dθ

)2

=

(dr

dθ

)2

+ r2

so the length is given by the integral

L =

∫ b

a

√(dr

dθ

)2

+ r2 dθ

where the limits θ = a and θ = b need to be determined from the parametrization of thecurve. As usual, you should be careful to determine the appropriate values of θ to definethe portion of the curve that interest you.

10.4 Exercises

[1, Exercise 22, p. 683] “Find the area enclosed by the loop of the strophoid r =2 cos θ − sec θ.”

Solution: First we observe that the curve is entirely traced out when θ passesthrough an interval of length 2π; so, without limiting generality, we may confineourselves to such an interval, say −π

2< θ < π

2, and thereby remove the some


possible ambiguity of coordinates. We must exclude the points where sec θ isundefined. For convenience, let’s confine θ to the union of intervals

−π

2< θ <

π

2and

π

2< θ <

3π

2.

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 seethe whole curve, it suffices to investigate angles θ between −π

2and 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: thistells us that 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 θ comes from −π2, we find that r is negative, and the point is in the 2nd

quadrant. It passes through the pole first when θ = −π4, and is then in the fourth

quadrant, striking the polar axis when θ = 0, at the point (r, θ) = (1, 0) and itthen moves into the first quadrant, following the mirror image of the portion in thefourth quadrant, passing through the pole again when θ = π

4, and moving into the

3rd quadrant. If we consider the cartesian coordinates of the curve in parametricform, we have

x = 2 cos2 θ − 1

y = sin 2θ − tan θ

and see that, as cos θ → 0, x → −1, and y → ±∞, so the curve is asymptotic tothe vertical line x = −1. Our present problem is to determine the area of the loopfrom θ = −π

4to θ = π

4, or, by symmetry,

2

(1

2

) ∫ 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 46, p. 684] Find the exact length of the arc of the polar curve r = e2θ

from θ = 0 to θ = 2π.

Solution:

Length =

∫ 2π

0

√(dr

dθ

)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. 684] Graph the curve r = cos2 θ2, and find its length.

Solution: The equation can be simplified to read

r =1 + cos

(2 · θ

2

)

2

which we recognize to be a cardioid. As θ ranges over the interval 0 ≤ θ ≤ 2π theentire curve is traced out once.

Length =

∫ 2π

0

√(dr

dθ

)2

+ r2 dθ

=1

2

∫ 2π

0

√(− sin θ)2 + (1 + cos θ)2 dθ

=1

2

∫ 2π

0

√2(1 + cos θ) dθ

=1

2

∫ 2π

0

√2

(2 cos2

θ

2

)dθ

=

∫ 2π

0

∣∣∣∣cos

(θ

2

)∣∣∣∣ dθ

=

[2 sin

θ

2

]π

0

−[2 sin

θ

2

]2π

π

= 2− (−2) = 4 .


C.17.2 §10.5 Conic Sections

Omit this section.

Review

Problems Plus



Release Date: Thursday, March 11th, 2004, subject to further revision

Textbook Chapter 11. INFINITE SEQUENCES AND SERIES.

C.18.1 §11.1 Sequences

A sequence is an ordered set of objects, usually labelled with either the non-negativeintegers 0, 1, . . . , n, . . . or by the positive integers 1, 2, . . . , n, . . . . In this chapterwe shall be interested in sequences of real numbers or of real functions . Technically,such a sequence is a function that maps either the non-negative integers or the positiveintegers on to the set of real numbers (or to the set of real functions). We usually denotea sequence by a single letter, e.g. a, and show the labelling by either a parenthesizedvalue, as a(n), or a subscripted values, as an; when we wish to talk about a sequence withspecific terms, we may just list the terms with a general term that shows the pattern weare 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 thetwo entries 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 togeneralize a pattern that appears to hold between a few terms at the beginning of thesequence.

The following definition is included for completeness, but you are not expected to beable to work with it, in the same way that the formal definition of the limit for a functionof a continuous real variable was discussed only for completeness.

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. Othernotations are 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

of limx→∞

f(x) = L; another way of writing the same definition is

an → L as n →∞.

The precise definition will be discussed in the lectures, and is to be found as [1, Definition2, p. 703], 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 thebraces and when we speak of the limit of a sequence.

Since you bring to this course some understanding of limits of functions, we willoccasionally appeal to the following

Theorem C.31 [1, Theorem 3, p. 704] If f is a function, and an such that f(n) = an

for all n, and if limx→∞

f(x) = L, then limn→∞

an = L.

We also generalize our definition of the limit of a sequence to permit limits to be ±∞;the definition is again analogous to the corresponding definition for the meaning oflim

x→∞f(x) = ±∞.

A sequence that has limit L is said to converge to L; if there is no limit, the sequenceis said to diverge. We do not permit L to be ±∞ in this usage, so a sequence whoselimit is ±∞ is said to diverge.

Limit Laws We can prove limit laws for sequences analogous to the limit laws we sawin the previous course for functions. I shall not give the details in these notes. Rememberthe usual restriction that we cannot divide by 0, so there must be a restriction on thequotient law. The limit laws can be extended to infinite limits whenever the operationscan be justified; so, we can think of ∞ +∞ = ∞, but we cannot attach a meaning to∞ + (−∞), nor of 0 · ∞, as it is not possible to assign to these expressions a meaningthat will be consistent with the algebraic operations on real numbers.

Sequences of powers For a fixed real number a we know the behavior of a functionax as x →∞, and how it depends on a; this permits us to study the behavior of an whenthe exponent is 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

Example C.32 (This example was mentioned in the lecture, but I did not show there

how to find its limit.) 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 + 1

n

)n →1

e


as n → ∞. Thus, for sufficiently large n, every term is less than half of the one beforeit, which is positive. Compared to the nth term, the n + 10th will be less than 1

1000of

the size; the n + 20th will be less than one-millionth the size, etc. The sequence is thusapproaching 0.

Increasing and Decreasing Sequences We defined the concepts of increasing anddecreasing in connection with real functions of a real variable (cf. [1, p. 21]. The identicaldefinitions apply for sequences, where we consider the domain of the function to be thepositive or non-negative integers. When a sequence is either increasing or decreasingwe say that it is monotonic or monotone. Sometimes we find it convenient to workwith a slightly weaker property than increasing , and may speak of a sequence as beingnon-decreasing , which permits the function to remain constant for a while.

A function is bounded above if there exists a number M which is greater than allvalues of the function. For example, the sine function is bounded above, since sin x ≤ 1for all x in its domain. We could also have observed that sin x < 10000, and concludethat such a statement justifies the conclusion that the function is bounded above: wedon’t care how high above the graph of the function the bounding line appears, onlythat such a line exists. But the function tan x is not bounded above. In the same waywe can define bounded below and bounded — meaning bounded both above and below.An important theorem we shall need in this chapter is

Theorem C.33 1. A sequence that is bounded above and monotonely increasing (oreven monotonely non-decreasing) is convergent.

2. A sequence that is bounded below and monotonely decreasing (or even monotonelynon-increasing) is convergent.

(Note that the theorem stated here is stronger than [1, Monotonic Sequence Theorem,p., 709].)

11.1 Exercises

[1, Exercise 28, p. 711] Determine whether the sequence

ln n

ln 2n

converges.

Solution:

ln n

ln 2n=

ln n

ln 2 + ln n

= 1− ln 2

ln 2n

The subtracted fraction has a constant numerator, but the denominator is increas-ing, so the fraction is decreasing as n increases; hence 1 minus the fraction is


increasing ; in fact, we can see that the subtracted fraction is approaching 0, so thegiven fraction is approaching 1.

[1, Exercise 32, p. 711] Determine whether the sequence an = ln(n + 1) − ln n con-verges.

Solution: We might be tempted to analyze a difference by using the difference lawfor limits. But we find that each of the terms approaches +∞, and we cannot givea meaning to ∞−∞. So that approach will not work.

However, if we observe that the difference of logarithms is

lnn + 1

n= ln

(1 +

1

n

)

we can reason as follows. As n → ∞, 1n→ 0, 1 + 1

n→ 1 + 0 = 1. By continuity

of the logarithm, the sequence will approach ln 1 = 0.




Release Date: Monday, March 15th, 2004, subject to further revision

C.19.1 §11.2 Series

The sum of the terms in a sequence. Having defined what we mean by convergenceof a sequence 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 mathemati-cian, the operation of addition is a binary operation on numbers, that is, it maps tworeal numbers on to one real number. This operation has a number of properties thatmathematicians 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 propertythat x + 0 = x for every real number x;

and other properties that we will not discuss here. The second property permits us totalk of more general sums than of just 2 numbers. If we have three real numbers x, y,z, then the second 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 not consider the details of this definition in this calculus course.38 Butthe definition we obtain in this way applies only to a finite sequence of numbers. Inthis section we consider a broader definition, to permit us to speak of the sum of all theterms in a sequence. In this context we no 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 bothsingular and plural in English: there is no word “serie” in English39. Our definition

38This is a topic that would likely appear in a first Abstract Algebra course, like MATH 235.39But the singular word for “series” in French is serie, and the plural is series.


proceeds from the partial 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 adifferent letter — m — for the general term in this sequence, because the letter n was“busy”40.) We say that

∞∑n=0

an = L

if L is the limit of the sequence of partial sums of the series. The series is then said toconverge, or to be convergent ; if it does not converge, it diverges or is divergent . Here,as with series and functions, we generalize to write that a sum = ∞ or = −∞, but stilldescribe such series as divergent. We will often consider series which we will know to beconvergent, without being able to specify the precise value of the limit.

“Geometric” series. A geometric series is one, each of whose terms after the first isa constant multiple of its predecessor. We will often represent such series as

a + ar + ar2 + . . . + arn + . . .

where a 6= 0 and the “common ratio” is the ratio of each term to its predecessor, heredenoted by r. In the lecture of March 10th, 2004 we derived the value of the partialsums as

a + ar + . . . + arm =m∑

n=0

arn =

a(1−rn)

1−rif r 6= 1

(n + 1)a if r = 1.

This ratio can be seen to have the following limit properties:

a + ar + . . . + arm =m∑

n=0

arn =

= a1−r

if −1 < r < 1

= +∞ if r ≥ 1 and a > 0= −∞ if r ≥ 1 and a < 0diverges if r ≤ −1

.

40Technically, we call n a bound variable in this case.


Divergence of the “harmonic” series. The harmonic series is

1 +1

2+

1

3+ . . . +

1

n+ . . . .

In the next section we will see a proof, using an improper integral, that this seriesdiverges. Here we shall see a simpler proof without integrals. We need only observe thefollowing inequalities:

12≥ 1

2

13

+ 14

> 14

+ 14

= 24

= 12

15

+ 16

+ 17

+ 18

> 18

+ 18

+ 18

+ 18

= 48

= 12

. . .1

2n−1+1+ 1

2n−1+2+ . . . + 1

2n−1+ 1

2n > 2n−1 · 12n = 1

2

Thus we can make the partial sums as large as we like by proceeding out sufficiently farin the series. It follows that ∞∑

n=1

1

n= +∞ ,

so the harmonic series diverges.

“Telescoping” series. Sometimes the partial sums can be interpreted as sums wherethere is 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

)(cf. [1, Example 6, p. 717]).

Changes of variable in a sum. We have already seen the concept of changing avariable in connection with definite integrals. We can carry out similar changes in sums —both finite and infinite, although we will not investigate all the details. So, for example,

we can change the 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 definiteintegral.


“Necessary” and “sufficient” conditions. Suppose that A and B are sentencesthat may be true or false. If A cannot be true except when B is true, we say that thetruth of A entails the truth 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 andy are members 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 thatB is a necessary condition for A, since A cannot be true unless B is true.

Mathematicians usually search for conditions that are both necessary and sufficient,since these can characterize a situation. But often we have to be satisfied with conditionsthat are either of one type or the other.

A “necessary condition” for convergence of a series The implication that inter-ests use here is 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 canprove, sometimes, is that a certain series is not convergent, since no series that “failsthe test” can be convergent. Many textbooks call this test the “nth term test”; yourtextbook calls it “The” test for divergence. This is not a standard term, and you mightwish to avoid using it, since a listener who has not read Stewart’s books might not knowwhat you are referring to.41 In practice you should “internalize” this test and alwaysapply it, since you might otherwise waste time trying to prove that a divergent series isconvergent.

41You will meet other tests that could be given such a name, although this is the most likely oneto distinguish in this way. Stewart’s name for the test may be no more objectionable than the namein general use: it is not good form to use the symbol n in the name of a test, since the test does notdepend on giving that particular name to the variable that indexes the members of the sequence; myobjections to the name Stewart assigns are that (1) it is not universally accepted; and (2) the definitearticle suggests uniqueness, and this is not the only test that exists.


Operations on series. Convergent series may be added term by term, or multipliedby a constant:

∞∑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 rearrangea series; we shall see later that rearrangement of the terms can result in a series havinga different sum, or even in a convergent series being rendered divergent.

11.2 Exercises

[1, Exercise 38, p. 720] Express the “repeating decimal expansion” 6.254 as a ratioof integers.

Solution: By a “repeating decimal expansion”, we intend that the digits under thehorizontal 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+

54

1000+

54

100000+

54

10000000+ . . .

which can be seen as a constant added to a geometric series:

(6 +

2

10

)+

54

1000

(1 +

1

100+

(1

100

)2

+

(1

100

)3

+ . . .

)

whose sum is

6 +2

10+

54

1000· 1

1− 1100

= 6 +1

5+

3

55=

344

55.

(In this way we can show that any “repeating decimal” is a rational number.)



Distribution Date: Wednesday, March 17th, 2004, subject to further revision

C.20.1 §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 whetherthe series converges. That is the test Stewart calls “The Test for Divergence”, and itgives negative information: 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. That is the only test we shall have thatcan be applied to “general” series, where the signs of the terms do not follow a specificpattern.

Tests of Positive Series for convergence. Most of the tests we shall meet requirea specific arrangement of signs for the terms. In fact, all but one of them will requirethat all the signs are the same. We call such series “positive” series, thinking of all thesigns as being +; but similar results hold when all signs are −. The first test of positiveseries will be discussed in this section. I do not think of this as the most elementary ofthe tests, and would have preferred to discuss the material of the next sections first; butI will grudgingly follow the order of topics in the textbook.

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 widthis 1, and whose height is an. The test will be restricted to sequences where the terms aremonotonely decreasing. Suppose that we know of a function f defined on the interval[1,∞), whose graph passes through the left upper end-points of the rectangles. Thenthe area under the curve will be less than the sum of the areas of the rectangles. If wecan show that the area under the curve is infinite, we will be able to conclude that thesum of the series is divergent. In a similar way, if we can pass the graph of a function fthrough the right upper endpoints of the rectangles, then the area under the curve willexceed the sum of the areas of the rectangles: if the area under the curve is convergent,then the same can be said about the series. In these ways we can shown that

the series converges precisely when the integral under the curve converges!

We call this result


Theorem C.34 (The Integral Test) If f is a continuous, positive, decreasing42 func-tion defined on the interval [1,∞), and if the sequence an has the property thatan = 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 dependon the lower limits of the sum and the definite integral. (However, when, below, weinvestigate bounds for the actual value of the sum of the series, then our bounds willdepend on the specific values we choose for the limits of sum and integral.)

Application: the “p-series”. The “p-series” are the series∑ 1

np, where p is a pos-

itive constant. The integral test shows that

∑ 1

npis convergent if and only if p > 1 .

The case p = 1 is the harmonic series, which we have already shown to be divergent,using a different proof. (This application is sometimes called the “p-series test”.) Sincethe tests we will meet in [1, §11.4] involve comparing a given series with series that weknow to be convergent and divergent, the p-series are particularly important as theyprovide us with a family of series that can be used for comparison purposes.

Estimating the Sum of a Series In the proof of the Integral Test we comparedthe sum of a series of decreasing positive terms with the area under the graph of afunction that is positive and decreasing. This comparison can be refined, and gives riseto inequalities that bound the value of the sum from below and above. Let us, extendingthe notation used in the enunciation of the theorem, define

Rm =∞∑

m=n+1

am;

that is, Rm is the sum of the “tail” of the series, starting with the (n + 1)st term. Then

∫ ∞

n+1

f(x) dx ≤ Rn ≤∫ ∞

n

f(x) dx ,

42Actually, it suffices for the function to be non-increasing : we can permit the function values to stayat the same level so long as they eventually decrease again and eventually approach 0.


or, alternatively, ∫ ∞

n+1

f(x) dx ≤ Rn ≤ an+1 +

∫ ∞

n+1

f(x) dx ,

or

0 ≤ Rn −∫ ∞

n+1

f(x) dx ≤ an+1 .

In particular, when n = 0, this gives the inequalities

0 ≤∞∑

n=1

an −∫ ∞

1

f(x) dx ≤ a1 .

Example C.35 We proved earlier that the telescoping series∞∑

n=1

1

n(n + 1)converges,

and we found the sum. The integral test could be used to prove that it converges, butthe test would not give the exact value of the sum, although it would provide boundswhich compare the sum to the improper integral

∫ ∞

1

1

x(x + 1)dx = lim

a→∞((ln a− ln 1)− (ln(a + 1)− ln 2))

= lima→∞

(ln

a

a + 1+ ln 2

)

= ln 1 + ln 2 = ln 2.

so the sum of the series is bounded between ln 2 = 0.6931471806 and ln 2 + 12

=1.193147181, which is a weaker result than the fact we already know that the sum isequal to 1.

11.3 Exercises

[1, Exercise 18] To investigate the convergence of∞∑

n=1

1

n2 − 4n + 5.

Solution: We know that

∫1

x2 − 4x + 5dx =

∫1

(x− 2)2 + 1dx = arctan(x− 2) + C ,

and might be tempted to compare the integral

∫ ∞

1

1

x2 − 4x + 5dx 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

1

x2 − 4x + 5,

0 ≤∞∑

n=2

1

n2 − 4n + 5≤ lim

x→∞arctan(x− 2)− arctan(0)

=π

2

and so the sum of the series is bounded between1

2and

1 + π

2.


C.21 Supplementary Notes for the Lecture of March 22nd,2004

Distribution Date: Monday, March 22nd, 2004, subject to further revision

C.21.1 §11.4 The Comparison Tests

We continue to study series in a “reverse-logical” order. That is, as we proceed, we willbe encountering properties that are more and more basic. In this section we will meetanother theorem that — like the Integral Test — applies only to series whose terms allhave the same sign, the so-called “positive series”, or “series of positive terms”. Finally,in [1, §11.6] we will see a theorem [1, Theorem 3, p., 741] that will justify our continuedinvestigation of series with positive terms. When we have covered all the sections on thesyllabus, I hope to return to the various tests.

The comparison tests are designed so that we can infer the convergence or divergenceof a given positive series by comparison of its terms with those of another series thatwe know to be convergent or divergent. So, in order to use this test, we need to haveavailable as large as possible a family of positive series whose convergence or divergenceare known. Let’s remember 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 thebeginning (have the same convergence properties as the original series)

• series obtained from a given series by multiplying by a positive constant (have thesame convergence properties as the original series)

Theorem C.36 (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 alln sufficiently large”, which means, in mathematical jargon, that all we require isthat the inequality holds after some integer, which could be as large as you like; ifthe inequality fails a finite number of times, even an enormously large finite numberof times, that is sufficient 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 forcomparison purposes 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.37 (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 is delicate, and difficult for students to remember, so we usually do not discuss itin this course.)

Estimating Sums


11.4 Exercises

[1, Exercise 20, p. 735] Determine whether the series∞∑

n=1

1 + 2n

1 + 3nconverges or diverges.

Solution: This problem was investigated at length at the lecture. It was shown

that we could not compare naively with2n

3n, which is a convergent geometric series

because the terms here are slightly larger. However a number of other possibilitiessuggested themselves. For example, we have

1 + 2n

1 + 3n≤ 1 + 2n

3n=

(1

3

)n

+

(2

3

)n

< 2

(2

3

)n

which is the general term of a geometric series with common ratio 23

< 1, whichis convergent. Other comparisons were shown in the lecture, some of them com-plicated to prove. For the purpose of proving only convergence (and not boundingthe series) one tries to find the simplest possible comparison.


n=1

n!

nnconverges or diverges.

Solution: In this case we showed that the general term was

n!

nn=

1

n· 2

n· 3

n· . . . · n− 1

n· n

n

≤ 1

n· 2

n· 1 =

2

n2

which is the general term in a convergent p-series. Hence the given series is con-vergent. 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, dependingon which series is in the numerator; our version of this test does not permit thelimit to be either 0 or ∞.


n=1

sin1

nconverges 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 haveused the Comparison Test? Unfortunately, the geometric argument that we usedto prove the result quoted earlier showed [1, p. 212] that

0 < sin1

n<

1

n


win which the inequality on the right is the opposite of what we need to infer fromthe divergence of the Harmonic Series that the given series diverges by using theComparison Test.


n=1

1

n1+ 1n

converges or diverges.

Solution: It appears that the series “resembles” the Harmonic Series, and we willtry to compare it in the limit. The guess is a fortunate one, since the limit doesexist.

limn→∞

1

n1+ 1n

1n

= limn→∞

1

n1n

= limn→∞

1

eln nn

=1

elim

n→∞ln nn

by continuity of exponential

=1

e0= 1 by L’Hospital’s Rule

Since the Harmonic Series diverges and the ratio of the terms of the given series tothose of this divergent series approaches a non-zero positive real number, the givenseries 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 larger than those of the series wewere given.



Distribution Date: Wednesday, March 24th, 2004, subject to further revision

C.22.1 §11.5 Alternating Series

As with the preceding lecture, we are investigating a specialized topic concerning seriesprior to meeting a more general theorem in [1, §11.6] at the next lecture. The presentsection will broader the class of series that we can work with.

An alternating series is one in which the signs alternate signs. Our main interest, asin the preceding 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 you offer a “good” upperbound for the error — i.e., for the difference between the sum of the series and thegiven partial sum.

Our interest in alternating series derives partly from some specific series that will ariselater in [1, Chapter 11], in the portion of the chapter assigned to Calculus III.

By the “Test for Divergence”, an alternating series cannot converge unless the limitof the sequence of its terms is 0. For these series, and these series alone, we have a partialconverse of that test:

Theorem C.38 (Leitniz’s Alternating Series Test) An alternating series∞∑

n=1

(−1)n−1bn

for which:

(i) 0 ≤ bn+1 ≤ bn for all n; and

(11) limn→∞

bn = 0

converges.

Estimating Sums While the estimation of errors can be complicated for other series,the following very simple bounds hold for alternating series:

Corollary C.39 (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


at the Nth term cannot exceed the very next term; more precisely,∣∣∣∣∣

∞∑n=N+1

(−1)n−1bn

∣∣∣∣∣ < |bN+1| .

11.5 Exercises

[1, Exercise 17, p. 739] Test for convergence or divergence the series∞∑

n=1

(−1)n sinπ

n.

Solution: This problem should be compared with [1, Exercise 31, p. 735] (cf. thesenotes, p. 2131). In the earlier problem we proved the divergence of a series whoseterms resemble those of the present series, except that it was not alternating.But the present series satisfies the conditions of the alternating series test, as theterms alternate in sign, are monotonely decreasing (because the sine function is anincreasing function to the right of 0), and the limit of the terms is 0. Thus theseries converges. When we have covered [1, §11.6] we will be able to describe thepresent series as being Conditionally Convergent — it is convergent, but loses thatproperty if all terms are replaced by their absolute value.


n=1

(−1)n

√n

1 + 2√

n.

Solution: The general term of the given series is,

(−1)n

√n

1 + 2√

n= (−1)n 1

1√n

+ 2

whose magnitude → 1

2as n → ∞. Thus the sequence of terms of this series does

not have a limit as n → ∞, and the series cannot converge, by the “Test forDivergence”. In fact, the other condition of the Leibniz Theorem, that the termsdecrease in magnitude, also fails to hold! But here we can say more: not only canwe state that the Leibniz Test yields no information, but we can state that theTest for Divergence shows that the series actually diverges.


n=1

(−1)n(−n

5

)n

.

Solution: This series also diverges, because the terms to not have limit 0.

(At the end of the lecture some comments were made about the topics to be seen in [1,§11.5] in the next lecture.)



Distribution Date: Sunday, March 28th, 2004, subject to further revision

C.23.1 §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 positiveseries, we see a reason for our interest in this type of series. To state the theorem werequire two definitions:

Definition C.2 1. A series∑

an is absolutely convergent if the series∑ |an| is con-

vergent.

2. A series∑

an is conditionally convergent if∑

an is convergent but not absolutelyconvergent, i.e., if

∑an converges but

∑ |an| diverges.

Note that, thus far, you must not interpret the word absolutely as an adverb modifyingthe participle convergent ; until the following theorem is available, you should treat ab-solutely convergent as a two-word name for a property, nothing more. Now we state thetheorem that will permit a broader interpretation of the name.

Theorem C.40 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 yourtextbook.)

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 thata conditionally convergent series is convergent and . . . .

While investigation of the convergence of general series, i.e., series whose terms haveboth plus and minus signs, can be difficult, we can begin by considering the series ofabsolute values: if the resulting series converges, we can infer that the original seriesis convergent. But, if the series of absolute values diverges, or if we are unable to sayanything about it, then we can make no inference at all.

The Ratio Test. The Ratio Test is a test for positive series which considers the limitof the ratio of a term to its predecessor. The textbook states the test in terms of theabsolute values of terms of a general series, so we will present it in that variant:

Theorem C.41 (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 infor-

mation concerning the possible convergence of the series∑

an.

The Root Test. The Root Test is another test for positive series; here also the text-book presents it in a form that applies to all series.

Theorem C.42 (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 infor-

mation concerning the possible convergence of the series∑

an.

You will need to practice on many problems in order to become comfortable with all thetests you have met, and to know the limitations of each of them. You should not besurprised if some problems are amenable to the use of more than one test.

Rearrangements The author reports on the result of Bernhardt Riemann that theterms of a conditionally convergent series may be written in another order to create seriesthat will converge to any given real number, and even to diverge.

11.6 Exercises

[1, Exercise 3, p. 745] I modify the problem: Determine whether the series

∞∑n=0

(−1)n 1000000n

n!

is absolutely convergent, conditionally convergent, or divergent.

Solution: Apply the Ratio Test to the series of absolute values. The ratio of then + 1th term to the nth is 100000

n+1→ 0 as n → ∞. Thus, by the Ratio Test, the

series is absolutely convergent.


[1, Exercise 12, p. 745] Determine whether the series

∞∑n=0

sin 4n

4n


Solution: Compare the series absolute values with the geometric series∑

14n . Since

the latter converges, so does the former. Thus the given series is absolutely con-vergent.


∞∑n=0

(−1)n

(ln n)n


Solution: Consider the series of absolute values. The ratio of the n + 1th termto its predecessor is complicated, so we will look at the nth root of the nth term,obtaining

1

ln n→ 0

as n →∞. The given series is absolutely convergent.


∞∑n=0

(−1)n

n ln n


Solution: Here the ratio of the absolute value of the n + 1th term to that of thenth is (

1 +1

n

)· ln(n + 1)

ln n

in which the first factor approaches 1, and the second (by L’Hospital’s Rule) alsoapproaches 1. Thus the Ratio Test fails to give useful information about thisseries. We are not helpless, however. Let’s first consider the series absolute values.The terms are decreasing and approaching 0, so we may use the Integral Test [1,Exercise 21, p. 729] to show that it is divergent. On the other hand, the originalseries is alternating, and the terms are decreasing and approaching 0; hence, bythe Leibniz Test, that series converges; consequently it is conditionally convergent.



∞∑n=0

3− cos n

n23 − 1


Solution: For n > 2 this is a positive series; the numerator behaves erratically, andwe will not be able to calculate a limit using the Ratio or Root Tests. But since, forlarge n, the denominator is larger than n

23 , we would expect this series to diverge

by comparison with the appropriate p-series. More precisely, the divergence of thep-series with p = 2

3implies the divergence of the given series, since

3− cos n

n23 − 2

>3− 1

n23 − 2

>3− 1

n23

=2

n23

.


C.24 Supplementary Notes for the Lecture of March 31st, 2004

Distribution Date: Wednesday, March 31st, 2004Corrected Monday, April 5th, 2004, subject to further revision

C.24.1 §11.7 Strategy for Testing Series

This lecture was based on problems that I chose at random at the lecture.

11.7 Exercises


n=1

(n√

2− 1).

Solution: I first complete the method I was using at the lecture, then show how itmay be simplified. I started with the observation that the following factorizationholds:

(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 smallerthan

(n√

2)n

, so the denominator is smaller than 2n. Thus the terms are greaterthan terms in a multiple of the harmonic series, so the series must diverge, by theComparison Test.

We could have used the Limit Comparison Test also, if we know that

limn→∞

n√

2 = limn→∞

eln 2n = 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 .

By the Limit Comparison Test, the divergence of the harmonic series implies thedivergence of the given series.

Note that, if we alternate the signs in this last problem, we obtain a series whose termsare decreasing and approaching 0, so the Alternating Series Test tells us that that seriesconverges, i.e., that it is conditionally convergent.


C.25 Supplementary Notes for the Lecture of April 5th, 2004

Distribution Date: Thursday, April 8th, 2004, subject to further revision

Estimates of Remainders after truncation of series The definition of the state-ment that ∞∑

n=1

an = S

may be paraphrased as saying that the sum∞∑

n=N+1

an of the “tail” of the series after the

partial sum of the first N terms may be made arbitrarily small as N → ∞. Thus farmost of our attention has been devoted to simply proving or disproving that a seriesconverges. But, in some cases, we are able to accurately estimate how many terms weneed to add before the partial sum is within a certain tolerance, i.e., before the sum ofthe “tail” is within a given tolerance. This is an area of estimation where often a moreclever practitioner can obtain results better than the average. In the lecture I discussedonly the basic ideas, and omitted some refinements that are in the textbook.

The “tail” of an infinite series is also an infinite series. So, when we are findingbounds for the error on truncation of a series, we are actually finding bounds for the sumof another series.

Remember that, unless a series consists only of terms of all the same sign — theso-called “positive” series — you do not have the right to “rearrange” the terms: theymust be summed in precisely the same order in which they are given originally.43 As withthe results we have seen hitherto, most of our results are restricted to positive series; theexception is the specific bound that can be used for series satisfying the “AlternatingSeries Test”.

43Changes in order that affect only a finite number of terms will not cause any problem, however.

Information for Students in MATH 141 2003 01 3001

D Problem Assignments from Previous Years

D.1 1998/1999

The problem numbers listed below refer to the textbook in use at that time, [21], [23].For many of the problems there are answers in the textbook or in the Student SolutionManual [24].

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”problems in the textbook. In most of the following problems there is a reference to a“similar” problem in the textbook. You should always endeavour to show as much ofyour work as possible, and to reduce your solution to “simplest terms”. Remember thatthe main reason for submitting this assignment is to have an opportunity for your tutorto grade your work; the actual grade obtained should be of lesser significance.

In Exercises 1-5 below, evaluate the indefinite integral, and verify by differentiation:

1. (cf. [21, Exercise 5.2.5, p. 294])

∫ (3

x4− 5x

12 − x2 + 4x−3

)dx

2.

∫ (3

x− 2

1 + x2

)dx

3. (cf. [21, Exercise 5.2.13, p. 294])

∫ (xex2 − e4x

)dx

4. (cf. [21, Exercise 5.2.19, p. 294])

∫(1−√x)(2x + 3)2 dx

5. (cf. [21, Exercise 5.2.27, p. 294])

∫(4 cos 8x− 2 sin πx + cos 2πx− (sin 2π)x) dx

6. (cf. [21, Example 5.2.8, p. 289]) Determine the differentiable function y(x) such

thatdy

dx=

1√1− x2

and y(2−

12

)=

π

2.

7. (This is [21, Exercise 5.2.51, p. 295] written in purely mathematical terminology.)

Solve the initial value problem:d

dx

(dy

dx

)= sin x, where y = 0 and

dy

dx= 0 when

x = 0. [Hint: First use one of the initial values to determine the general value

ofdy

dxfrom the given “differential equation”; then use the second initial value to

determine y(x) completely.]

8. ([21, Exercise 5.3.4, p. 306]) Write the following in “expanded notation”, i.e. with-

out using the symbol∑

:6∑

j=1

(2j − 1).


9. (cf. [21, Exercise 5.3.18, p. 306]) Write the following sum in “summation notation”:

x− x3

3+

x5

5− x7

7+ ...± x999

999

where the signs are alternating +, −, +, −, ... The sign of the last term has notbeen given — you should determine it.

10. (cf. [21, Example 5.3.6, p. 302]) Given that

n∑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

0

sin 4x dx.

2. Interpreting the following integral as the area of a region, evaluate it using knownarea formulas: ∫ 6

0

√36− x2 dx.

3. Use properties of integrals to establish the following inequality without evaluatingthe integral: ∫ 1

0

1

1 +√

xdx ≤

∫ 1

0

1

1 + x3dx.

4. Deduce the Second Comparison Property of integrals from the First ComparisonProperty [21, p. 325, §5.5].

5. Apply the Fundamental Theorem of Calculus [21, p. 331, §5.6] to find the derivativeof the given function: ∫ x

−1

(t2 + 2)15 dt.


6. Differentiate the functions

(a)

∫ x3

0

cos t dt

(b)

∫ 3x

1

sin t2 dt.

7. Solve the initial value problemdy

dx=√

1 + x2 , y(1) = 5 . Express your answer

in terms of a definite integral (which you need not attempt to evaluate). Thisproblem can be solved using the methods of [21, Chapter 5].

8. Evaluate the indefinite integrals:

(a)

∫2x√

3− 2x2 dx

(b)

∫x2 sin(3x3) dx

(c)

∫x + 3

x2 + 6x + 3dx

9. Evaluate the definite integrals:

(a)

∫ 8

0

t√

t + 2 dt

(b)

∫ π/2

0

(1 + 3 sin η)3/2 cos η dη

(c)

∫ π

0

sin2 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

be solved using the methods of [21, Chapter 5]. It is not necessary to use methodsof [21, Chapter 9].



D.2.3 Assignment 3

In all of these problems you are expected to show all your work neatly. (This assignmentis only a sampling. Your are advised to try other problems from your textbook; solutionsto some can be found in the Student Solution Manual [22].)

1. [21, Exercise 6.1.6, p. 382] As n → ∞, the interval [2, 4] is to be subdivided inton subintervals of equal length ∆x by n − 1 equally spaced points x1, x2, ..., xn−1

(where x0 = 2, xn = 4). Evaluate limn→∞

n∑i=1

1

xi

∆x by computing the value of the

appropriate related integral.

2. (a) [21, Exercise 6.2.6, p. 391] Use the method of cross-sections to find the volumeof the solid that is generated by rotating the plane region bounded by y =9− x2 and y = 0 about the x-axis.

(b) (cf. Problem 2a) Use the method of cylindrical shells to find the volume of thesolid that is generated by rotating the plane region bounded by y = 9 − x2

and y = 0 about the x-axis.

(c) Use the method of cross-sections to find the volume of the solid that is gen-erated by rotating the plane region bounded by y = 9 − x2 and y = 0 aboutthe y-axis.

(d) (cf. Problem 2c) Use the method of cylindrical shells to find the volume of thesolid that is generated by rotating the plane region bounded by y = 9 − x2

and y = 0 about the y-axis.

3. (a) [21, 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, andx = 1.

(b) (cf. Problem 3a) Set up an integral that would be obtained if the methodof cylindrical shells were used to represent the volume of the solid that isgenerated by rotating around the line y = −1 the region bounded by y = 2e−x,y = 2, and x = 1. YOU ARE NOT EXPECTED TO EVALUATE THEINTEGRAL.

4. (cf. [21, Exercise 6.2.40, p. 392]) The base of a certain solid is a circular diskwith diameter AB of length 2a. Find the volume of the solid if each cross sectionperpendicular to AB is an equilateral triangle.

5. (a) [21, Exercise 6.3.26, p. 401] Use the method of cylindrical shells to find thevolume of 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 thesolid generated by rotating around the y-axis the region bounded by the curves

y =1

1 + x2, y = 0, x = 0, x = 2.

6. (cf. [21, 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) [21, 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 thesurface of revolution generated by revolving the arc of the curve y = x3 fromx = 1 to x = 2 around the y-axis.

8. [21, Exercise 7.2.44, p. 442] Evaluate the indefinite integral

∫x + 1

x2 + 2x + 3dx

9. (cf. [21, Exercise 7.2.36, p. 442]) Determine the value of the function f(x) =∫ x

−1

t2

8− t3dt for any point x < 2.

10. (cf. [21, 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)

∫dx

2√

x(1 + x)


(c)

∫ex

1 + e2xdx

(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

ASSIGNMENT 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

ASSIGNMENT 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 2x

cos2 2xdx

(d)

∫sec6 2t dt


D.2.5 Assignment 5

1. [21, 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. [21, Exercise 9.5.8, p. 540] Find

∫1

(x + 1)(x2 + 1)dx.

3. (a) [21, Exercise 9.5.23] Find

∫x2

(x + 2)3dx.

(b) Find the volume of the solid of revolution generated by the region bounded

by y =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

by y =x

(x + 2)32

, y = 0, x = 1, and x = 2 about the y-axis.

4. [21, Exercise 9.5.38, p. 540] Make a preliminary substitution before using themethod of partial fractions:

∫cos θ

sin2 θ(sin θ − 6)dθ

5. [21, Exercise 9.6.6, p. 547] Use trigonometric substitutions to evaluate the integral∫x2

√9− 4x2

dx.

6. [21, Exercise 9.6.26, p. 547] Use trigonometric substitutions to evaluate the integral∫1

9 + 4x2dx.

7. [21, Exercise 9.6.35, p. 547] Use trigonometric substitutions to evaluate the integral∫ √x2 − 5

x2dx.

8. [21, Exercise 9.7.14, p. 553] Evaluate the integral

∫x√

8 + 2x− x2 dx.

9. [21, Exercise 9.8.17, p. 561] Determine whether the following improper integral

converges; if it does converge, evaluate it:

∫ ∞

−∞

x

x2 + 4dx.


10. [21, Exercise 9.8.27, p. 561] Determine whether the following improper integral


∫ ∞

0

cos x dx.

11. (cf. [21, Exercise 9.8.14, p. 561]) Determine whether the following improper integral


∫ +8

−8

1

(x + 4)23

dx.

12. [21, Exercise 10.2.2, p. 580] Find two polar coordinate representations, one withr ≥ 0, and the other with r ≤ 0 for the points with the following rectangularcoordinates:

(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 — equationsin both rectangular and polar coordinates:

(a) [21, Exercise 10.2.20, p. 580] The horizontal line through (1, 3).

(b) [21, Exercise 10.2.26, p. 580] The circle with centre (3, 4) and radius 5.

14. (a) [21, Exercise 10.2.56, p. 581] Showing all your work, find all points of inter-section of 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 polarequations r2 = 4 sin θ and r2 = −4 sin θ.

[Note: The procedure sketched in the solution of [21, Example 10.2.8, p. 579] forfinding points of intersection is incomplete. Your instructor will discuss a system-atic procedure in 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)) as t 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 + 7

2n3 + n2,

(b) an =1 + (−1)n

√n

(3/2)n,

(c) an = n sin πn,

(d) an =

(n− 1

n + 1

)n

.

8. Determine, for each of the following infinite series, whether it converges or diverges.If it converges, find its sum.

(a) 1 + 3 + 5 + 7 + . . . + (2n− 1) + . . . ,

(b) 4 + 43

+ . . . + 43n + . . . ,

(c)∞∑

n=1

(5−n − 7−n),


(d)∞∑

n=1

( e

π

)n

.

9. Find the set of all those values of x for which the series∞∑

n=1

(x

3

)n

is a convergent

geometric 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 givenvalues of 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 forex.

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 seriesconverge or diverge.

(a)∞∑

n=1

1

1 + 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, althoughthe experiment had to be terminated in mid-term because of technical problems.)

D.4 2001/2002

This was the first time WeBWorK assignments were used exclusively in this course.


D.5 MATH 141 2003 01

WeBWorK assignments were used exclusively for assignments. The questions are notavailable for publication.


E Final Examinations from Previous Years

E.1 Final Examination in Mathematics 189-121B (1996/1997)

1. [4 MARKS] Find the derivative of the function F defined by

F (x) =

∫ x4

x2

sin√

t dt .

2. [4 MARKS] Evaluate

∫ π

−π2

f(x) dx , where

f(x) =

cos x, −π

2≤ x ≤ π

33πx + 1, π

3< x ≤ π

.


∫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

and x− 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 solidthat is generated.


∫x ln x dx .


∫sin2 x cos5 x dx .

9. [6 MARKS] Determine the partial fraction decomposition of the following ratio ofpolynomials:

x5 + 2

x2 − 1.

10. [4 MARKS] Determine whether or not the following sequence converges as n →∞ . If it does, find the limit:

(1 +

x

n

)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−k2

converges 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 withinthe circle 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) .


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 =3

2− x

2, and y = 0

is revolved around the line y = 0. Determine the volume of the solid of revolutionwhich is 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 − 8

x4 − 16.

6. [5 MARKS] Showing all your work, evaluate limx→0+

xx .


∫x3e−x2

dx .


∫x3 − 1

x3 + xdx .


∫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 θ and outside the circle r = 2 .

11. [5 MARKS] Showing all your work, obtain a second-degree Taylor polynomial for

f(x) =

∫ x

0

et(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 seriesconverges:

∞∑n=1

21n

n2

14. [5 MARKS] Showing all your work, determine whether the following series con-verges:

∞∑n=1

1

n · 2n


E.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 =3

2− x

2, and y = 0

is revolved around the line y = 0. Determine the volume of the solid of revolutionwhich is generated.

3. [10 MARKS] Find the area of the surface of revolution generated by revolving thecurve

y =1

2

(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 πx

x2 − 4.

6. [5 MARKS] Evaluate limx→∞

(cos

1

x2

)x4

.


∫e2x

1 + e4xdx .


∫x2 cos x dx .


∫x3 − 1

x3 + xdx .


∫ √a2 − u2 du , where |u| < a.


11. [10 MARKS] Find the area of the region that lies within the limacon r = 1 +2 cos θ and outside the circle r = 2 .

12. [5 MARKS] Showing all your work, obtain a second-degree Taylor polynomial for

f(x) =

∫ x

0

es(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 seriesconverges.

∞∑n=1

ln n

n

15. [5 MARKS] Showing all your work, determine whether the following series con-vereges.

∞∑n=1

n2 + 1

en(n + 1)2


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 aboutthe line x = 1 the region bounded by the curve (x − 1)2 = 5 − 4y and theline y = 1 .

3. [8 MARKS] Find the volume of the solid generated by revolving about the linex = 0 the region bounded by the curves

y = sin x

y = −2

x = 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 follow-ing inequalities are true.

e < F (e) < ee3+1 .

(b) [4 MARKS] Determine the value ofd

dxF (x3) at each of the

following points:

i. at x = 0 .

ii. at x = 2 .

6. [4 MARKS] Showing all your work, evaluate

∫sin3 πx dx .


∫x2e−x dx .


∫x− 1

x3 − x2 − 2xdx .


∫x3 + x2 + x− 1

x2 + 2x + 2dx .

10. [8 MARKS] Find the area of the region inside the curve r = 3 sin θ and outside thecurve r = 2− cos θ.

11. Showing all your work, determine whether each of the following integrals is con-vergent or divergent:


(a) [4 MARKS]

∞∫

0

sin x dx .

(b) [4 MARKS]

2∫

0

dx

1− x2.

12. Showing all your work, determine whether each of the following sequences is con-vergent or divergent.

(a) [4 MARKS]

n sinπ

n

(b) [4 MARKS](2n + 1) e−n

13. Showing all your work, determine whether each of the following infinite series isconvergent or divergent:

(a) [4 MARKS]∞∑

n=1

1

4n3.

(b) [4 MARKS]∞∑

n=1

(1

n+

1

n2

).

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 + 2

n(n + 1).

(b) [4 MARKS]∞∑

n=1

(−1)n cos n

n2.


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 revolvingabout the line x = 0 the region bounded by the curve y = 4 − x2 and thelines x = 0 and y = 0 .


3. [8 MARKS] Find the volume of the solid generated by revolving about the linex = 0 the region bounded by the curves

y = sin x

y = 2

x = 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 follow-ing inequalities are true.

0 < F (e) < e .

(b) [4 MARKS] Determine the value ofd

dxF (x) at each of the

following points:

i. at x = 0 .

ii. at x =π

2.


∫x5e−x2

dx .


∫x3 − x2 + x + 1

x2 − 2x + 2dx .

8. [8 MARKS] Find the area of the region inside the curve r = 6 sin θ and outside thecurve r = 4− 2 sin θ.

9. Showing all your work, determine whether each of the following integrals is con-vergent or divergent:


(a) [4 MARKS]

∞∫

0

cos x dx .

(b) [4 MARKS]

4∫

0

dx

4− x2.

10. Showing all your work, determine whether each of the following sequences is con-vergent or divergent.

(a) [4 MARKS]

n sinπ

n

(b) [4 MARKS](2n + 1) e−n

11. Showing all your work, determine whether each of the following infinite series isconvergent or divergent:

(a) [4 MARKS]∞∑

n=1

1

4n5.

(b) [4 MARKS]∞∑

n=1

(1

n− 1

n3

).


1. [11 MARKS] Find the area of the region bounded by the curves x = y2 andx = −y2 + 12y − 16 .

2. [11 MARKS] Let C denote the arc of the curve y = cosh x for −1 ≤ x ≤ 1 .Find the 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 it is 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 n

ln (n2).


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 regionbounded by 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

0

e2x 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 andoutside r = 2 .

7. [11 MARKS] Determine whether the following improper integral converges:

∫ 1

0

ln x

x2dx .

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 isconvergent, divergent, conditionally convergent and/or absolutely convergent.

(a) [6 MARKS]∞∑

n=1

(−1)n ln n

n.

(b) [6 MARKS]∞∑

n=1

cos nπ

n.


1. Showing all your work, determine, for each of the following infinite series, whetheror not it converges.

(a) [3 MARKS]∞∑i=1

n

n3 + 1.


(b) [3 MARKS]∞∑

n=1

ln

(n

3n + 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 .

5. [14 MARKS] Evaluate the integral

∫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 areconstants;

(b) [6 MARKS] the value ofd2y

dx2at 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 − 4and y = −2x2 + 5x− 2 .


1. (a) [6 MARKS] Showing all your work, find F ′(1) when

F (t) =

∫ 2t

1

x

x3 + x + 7dx .

(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]

∫1

2x3 + xdx;

(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 =1

2, and above by the curve y =

1

1 + x2.

4. [15 MARKS] Showing all your work, find the volume generated by revolving aboutthe y-axis the smaller region bounded by the circle x2 + y2 = 25 and the linex = 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

given by r ≥ 0 , −π

2≤ θ ≤ π

2;

(c) [5 MARKS] find the coordinates of the points on the curve where the tangentline is parallel to the line θ = 0 .

6. For each of the following integrals, determine whether it is convergent or divergent;if it is convergent, you are expected to determine its value. Show all your work.

(a) [7 MARKS]

∫ 2

−1

1

x3dx ;

(b) [7 MARKS]

∫ ∞

1

xe−x2

dx .

7. Showing all your work, determine, for each of the following series, whether or notit converges:

(a) [5 MARKS]∞∑

n=2

1

n(ln n)2;

(b) [5 MARKS]∞∑

n=1

(−1)n

(n2 − 1

n2 + 1

);

(c) [5 MARKS]∞∑

n=1

n + 1

3n.


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 u

1 + sec u· tan u du

(d) [3 MARKS]

∫et

1 + e2tdt


2. Let K denote the curvey = x2 , (0 ≤ x ≤ 1) .

(a) [6 MARKS] Determine the area of the surface of revolution generated byrevolving K about the y-axis.

(b) [6 MARKS] Determine the volume of the solid of revolution formed by re-volving about the line y = 0 the region bounded by K and the linesx = 1 and y = 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 theintegral.

(b) [4 MARKS] Determine the area of the region subtended by C at the pole —i.e. of the region bounded by the arc C and the line θ = 0.

(c) [4 MARKS] The given curve can be represented in cartesian coordinates para-metrically 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).

4. [12 MARKS] Showing all your work, evaluate the integral

∫40− 16x2

(1− 4x2) (1 + 2x)dx .

5. [12 MARKS] Showing all your work, determine the area of the region bounded bythe curves y = arctan x and 4y = π x in the first quadrant.

6. (a) [4 MARKS] Showing all your work, determine the value of


(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, whetherit is absolutely convergent, conditionally convergent, or divergent.

(a) [4 MARKS]∞∑

n=5

(−1)n n2 − 1

6n2 + 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 + 5

2n.


1. Showing all your work, evaluate each of the following, always simplifying youranswer as much as possible:

(a) [3 MARKS]

∫ex sin x dx

(b) [3 MARKS]

∫ 12

0

sin−1 y√1− y2

dy

(equivalently,

∫ 12

0

arcsin y√1− y2

dy

).

(c) [3 MARKS]

∫(u2 + 2u)e−u du

(d) [3 MARKS]

∫1 + cos t

sin tdt

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 ofthe surface of revolution generated by revolving K about the x-axis.

(b) [6 MARKS] Determine the volume of the solid of revolution formed by re-

volving about the line y =√

32

the region bounded by K and the lines

x = 0 and y =√

32

.

(You may assume that

∫ √2x− x2 dx =

x− 1

2

√2x− x2 +

1

2arccos(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 wherethe tangent is horizontal.

(b) [6 MARKS] By determining the value ofd2y

dx2as a function of t, determine all

points (x, y) on C at which the ordinate (y-coordinate) is a (local) maximum,and all points at which the ordinate is a (local) minimum.

4. [12 MARKS] Showing all your work, evaluate the indefinite integral

∫4x3

(x2 − 9) (3x + 9)dx .

5. [12 MARKS] Showing all your work, determine the area of the region bounded bythe curves x− 2y + 7 = 0 and y2 − 6y − x = 0 .

6. (a) [6 MARKS] Showing all your work, evaluate


(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

) · 1

n.

9. [10 MARKS] Prove or disprove the following statement: The point with polarcoordinates

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 tosketch the curves.


E.12 Final Examination in MATH 141 2003 01

1. [10 MARKS] Find the area of the region bounded in the first quadrant by thecurves

y = ex, y = e−x, y = e2x−3 .

Simplify your answer as much as possible. (Your instructors are aware that you donot have the use of a calculator.)

2. Showing all your work, evaluate each of the following indefinite integrals:

(a) [5 MARKS]

∫1

x2 + 2x + 17dx

(b) [5 MARKS]

∫ln(ln x)

xdx

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 ofrevolution obtained by rotating Ω about the y-axis.

(b) [7 MARKS] Showing all your work, determine the area of the surface of rev-olution obtained by rotating the arc

x =√

16 + y2, (0 ≤ y ≤ 3)

about the y-axis. You may assume that

d

dθ(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 =3π

2.

(b) [6 MARKS] Find the length of C.

6. Give, for each of the following statements, a specific example to show that thestatement is not a theorem:

(a) [3 MARKS] If an∞n=0 is a sequence such that limn→∞

an = 0, then∞∑

n=0

an con-

verges.

(b) [3 MARKS] If the series∞∑

n=0

an and∞∑

n=0

bn are both divergent, then∞∑

n=0

(an+bn)

is divergent.

(c) [3 MARKS] If a series∞∑

n=0

an 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.


E.13 Supplemental/Deferred Examination in MATH 141 200301

1. [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]

∫e

1x

x2dx

(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 showthat the integral does not converge. (The examiners are aware that you donot have access 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 ofrevolution obtained by rotating Ω about the y-axis.

(b) [6 MARKS] Showing all your work, determine the volume of the solid ofrevolution 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 parametervalue t = −1

2.


(b) [6 MARKS] Find the area of the surface obtained by rotating the curve aboutthe x-axis.

6. Give, for each of the following statements, a specific example to show that thestatement is not a theorem:

(a) [3 MARKS] If bn∞n=0 is a sequence such that limn→∞

bn = 1, then∞∑

n=0

bn con-

verges.

(b) [3 MARKS] If a series∞∑

n=0

b2n converges, then

∞∑n=0

bn converges.

(c) [3 MARKS] If the series∞∑

n=0

an and∞∑

n=0

bn are both divergent, then∞∑

n=0

(anbn)

is divergent.


(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 thecurve r = 2− 2 sin θ and outside the curve r = 3 .


F WeBWorK

F.1 Frequently Asked Questions (FAQ)

F.1.1 Where is WeBWorK?

WeBWorK is located on Web servers of the Department of Mathematics and Statistics,and is accessible at the following URL’s:

http://msr01.math.mcgill.ca/webwork/m141w04or

http://msr02.math.mcgill.ca/webwork/m141w04

If your student number ends with 1, 3, 5, 7, or 9, you should use the URL

http://msr01.math.mcgill.ca/webwork/m141w04;

if your student number ends with 0, 2, 4, 6, or 8, you should use

http://msr02.math.mcgill.ca/webwork/m141w04.

If you access WeBWorK through WebCT, the link on your page will have been pro-grammed to take you to the correct WeBWorK server automatically.

F.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. Your initial password will be your 9-digit student ID number. Youwill be able to change this password after you sign on to WeBWorK.44

Your e-mail address. The WeBWorK system requires each user to have an e-mailaddress. After signing on to WeBWorK, you should verify that the e-mail addressshown is the one that you prefer. You should endeavour to keep your e-mail address upto date, since the instructors may send messages to the entire class through this route.

We suggest that you use either your UEA45 or your po-box address. You may forwardyour mail from these addresses to any other convenient address, (cf. §1.8.1.)

44If you forget your password you will have to send a message to your instructor so that the systemadministrator may be instructed to reset the password at its initial value.

45Uniform E-mail Address



F.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.

F.1.4 When will assignments be available on WeBWorK?

Each assignment will have a begin date and a due date. The assignment is available toyou after the begin date; solutions will be made available soon after the due date.

F.1.5 Do WeBWorK assignments cover the full range of problems that Ishould be able to solve in this course?

WeBWorK problems are typically short, and only very brief answers are expected.The questions on the “Regular” WeBWorK assignments (##R1 through R10) are asampling of some types of problem you should be able to solve after successfully com-pleting this course. Some types of calculus problems do not lend themselves to this kindof treatment, and may not appear on the WeBWorK assignments. Use of WeBWorKdoes not replace studying the textbook — including the worked examples,attending lectures and tutorials, and working exercises from the textbook— using the Student Solutions Manual [3] to check your work. Students arecautioned not to draw conclusions from the presence, absence, or relative frequencies ofproblems of particular types, or from particular sections of the textbook. Certain sec-tions of the textbook remain examination material even though no problems are includedin the WeBWorK assignments.

F.1.6 May I assume that the distribution of topics on quizzes and finalexaminations will parallel the distribution of topics in the WeBWorKassignments?

No!

F.1.7 WeBWorK provides for different kinds of “Display Mode”. Whichshould I use?

“Display mode” is the mode that you enter when you first view a problem; and, later,when you submit your answer. You may wish to experiment with the different formats.The “best” is usually typeset2 mode, which should look similar to the version thatyou print out (cf. next question); of intermediate quality is formatted text , but somecharacters may fail to print properly; the lowest quality is text mode, which is essentiallythe way the author of the problem entered his data into the system. If your computer


has difficulty displaying in typeset2 46 mode, you may have to use one of the other modes.Typeset mode is related to the TEX and systems that mathematicians use in typesettingtheir documents; the notes that you are reading here were prepared using ; it containsformatting instructions in a “markup” language, and is difficult for inexperienced readers.

F.1.8 WeBWorK provides for printing assignments in “Portable DocumentFormat” (.pdf) or “PostScript” (.ps) form. Which should I use?

Most newer home computers have already been loaded with the Acrobat Reader for .pdffiles; if the Reader has not been installed on your computer47, you will find instructionsfor downloading this (free) software in §1.5.5 of these notes. If you are not happy with.pdf files, and wish to print and view PostScript files, you may require such (free) softwareas Ghostscript and Ghostview, available at

http://www.cs.wisc.edu/∼ghost/gsview/index.html

Most computers available to you on campus should be capable of printing in eitherof .pdf and PostScript formats.

F.1.9 What is the relation between WeBWorK and WebCT?

WebCT is the proprietary system of Web Course Tools that has been implementedby McGill University. You may access the web page for this course, and WeBWorKthrough your WebCT account48, and WebCT will link you to the appropriate serverfor WeBWorK. If you follow this route to WeBWorK, you will still have to log inwhen you reach the WeBWorK site. At the present time we will be using WebCTprimarily for the posting of grades, and as a convenient repository for links to notes andannouncements in the course. We are not planning to use the potential WebCT sitesthat exist for the tutorial sections: use only the site for the lecture section in which youare registered.

F.1.10 Which browser should I use for WeBWorK?

We recommend that you use Internet Explorer or Netscape. While other browsers maygive satisfactory results, your instructors and tutors do not have time to correct errors inyour WeBWorK records that could be attributed to idiosyncracies in another browser.Information about browsers supported by WebCT may be obtained at

http://ww2.mcgill.ca/icc/webct/browserCheck/browser.html#recommend

46or typeset, which is another mode currently available47At the time these notes were written, the latest version of the Reader was 6.0, but recent, earlier

versions will also work properly.48http://webct.mcgill.ca


F.1.11 What do I have to do on WeBWorK?

After you sign on to WeBWorK, and click on “Begin Problem Sets”, you will see a listof Assignments, each with a due date. You may print out a copy of your assignmentby clicking on “Get hard copy”. This is your version of the assignment, and it willdiffer from the assignments of other students in the course. You should spend some timeworking on the assignment away from the computer. When you are ready to submityour solutions, sign on again, and select the same assignment. This time click on “Doproblem set”. You can expect to become more comfortable with the system as youattempt several problems; but, in the beginning, there are likely to be situations whereyou cannot understand what the system finds wrong with some of your answers. Itis useful to click on the Preview Answers button to see how the system interprets ananswer that you have typed in. As the problems may become more difficult, you mayhave to refer to the “Help” page, and also to the “List of functions” which appears onthe page listing the problems. Don’t submit an answer until you are happy with theinterpretation that the Preview Answers button shows that the system will be taking ofyour answer.

F.1.12 How can I learn how to use WeBWorK?

As soon as your instructor announces that the WeBWorK accounts are ready, signon and try assignments P1 or R1. Neither of these will have limits on the number ofattempts at problems. The system is self-instructive, so we will not burden you with along list of instructions.

You will need to learn how to enter algebraic expressions into WeBWorK as it is codedto read what you type in a way that may different from what you expect. For exam-ple, the symbol ^ is used for writing exponents (powers). If you type 2^3, WeBWorKwill interpret this as 23 = 8. However, if you type 2^3+x, WeBWorKwill interpretit as 23 + x, i.e. as 8 + x; if you wish to write 23+x, you have to type 2^(3+x). Youmay obtain more information from the List of Available Functions, available online, or athttp://msr01.math.mcgill.ca/webwork_system_html/docs/available_functions.html

F.1.13 Where should I go if I have difficulties with WeBWorK ?

If you have difficulties signing on to WeBWorK, or with the viewing or printing func-tions on WeBWorK, or with the specific problems on your version of an assignment,you may send an e-mail distress message directly from WeBWorK by clicking on theFeedback button. You may also report the problem to your instructor and/or your

tutor, but the fastest way of resolving your difficulty is usually the Feedback . Please


give as much information as you can. (All of the instructors and tutors are able to viewfrom within WeBWorK the answers that 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, notonly to the hours of the tutor of the section in which you are registered.

F.1.14 Can the WeBWorK system ever break down or degrade?

Like all computer systems, WeBWorK can experience technical problems. The systemsmanager is continually monitoring its performance. If you experience a difficulty whenonline, please click on the Feedback button and report it. If that option is not availableto you, please communicate with either instructor by e-mail.

If you leave your WeBWorK assignment until the hours close to the due time onthe due date, you should not be surprised if the system is slow to respond. This isnot a malfunction, but is simply a reflection of the fact that other students have alsobeen procrastinating! To benefit from the speed that the system can deliver under normalconditions, do not delay your WeBWorK until the last possible day! If a systems failureinterferes with the due date of an assignment, arrangements will be made to change thatdate, and an e-mail message will be broadcast to all users (to the e-mail addresses onrecord).49

F.1.15 How many attempts may I make to solve a particular problem onWeBWorK?

Regular Assignments ##R1, R3, R5, R7, R9 are intended to prepare you for the evennumbered regular assignments, and there will be no limit on the numbers of attemptson the former. It is planned to limit the number of attempts at problems on the evennumbered assignments; information about the number of attempts permitted shouldappear at the top of each assignment. The odd numbered assignment is intended toprepare you for a painless completion of the subsequent even numbered assignment.

F.1.16 Will all Regular WeBWorK assignments have the same length? thesame value?

The numbers of problems on the various assignments may not be the same, and theindividual problems may vary in difficulty. But all 10 Regular assignments will countequally in the 10% of the final grade allocated to them. TO REITERATE: You willreceive credit for both the grade you obtain on the odd-numbered assignment (with

49But slowness of the system just before the due time will not normally be considered a systemsfailure.


unlimited attempts) and your grade on following even-numbered assignment (with re-stricted attempts).

F.1.17 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

examination.


G Contents of the DVD disks for

Larson/Hostetler/Edwards

These excellent disks were produced to accompany the textbook, Calculus of a SingleVariable: Early Transcendental Functions, 3rd Edition[18] (called LHE in the chartsbelow). The correspondences shown to sections of [10] are only approximate.

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 Numeri-

cally25 2.2, 2.4

1 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 Prob-

lem68 2.1

1 2.2 Basic Differentiation Rules and Rates ofChange

34 2.3

1 2.3 The Product and Quotient Rules andHigher Order 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 The-

orem15 4.2

2 3.3 Increasing and Decreasing Functions andthe First 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 4.103 4.2 Area 503 4.3 Riemann Sums and Definite Integrals 373 4.4 The Fundamental Theorem of Calculus 493 4.5 Integration by Substitution 473 4.6 Numerical Integration 283 4.7 The Natural Logarithmic Functions: Inte-

gration34

3 4.8 Inverse Trigonometric Functions: Integra-tion

25

3 4.9 Hyperbolic Functions 31


DVD LHE Stewart# Section Subject Minutes Section4 6 Chapter 6: Applications of Integration4 6.2 The Disc and Washer Methods 224 7.2 Integration by Parts 174 7.4 Trigonometric Substitution 314 7.5 Partial Fractions 334 7.7 L’Hopital’s Rule 22 4.4

(The coverage extends to part of the material for Math 141 as well.)


H References

H.1 Stewart Calculus Series

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

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

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

[4] J. Stewart, Single Variable Calculus (Early Transcendentals), Fifth Edition.Thomson*Brooks/Cole (2003); bundled with Student Solutions Manual for Stew-art’s Single Variable Calculus (Early Transcendentals), Fifth Edition. Thom-son*Brooks/Cole (2003). ISBN 0-534-42976-9.

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

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

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

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

[9] H. Keynes, J. Stewart, D. Clegg, Tools for Enriching Calculus , CD to accompany[1] and [2]. Thomson*Brooks/Cole (2003). ISBN 0-534-39731-X.

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

[11] J. Stewart, Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole (1999).ISBN 0-534-36298-2.

[12] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’sSingle Variable Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole(1999). ISBN 0-534-36301-6.


H.2 Other Calculus Textbooks

H.2.1 R. A. Adams

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

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

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

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

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

H.2.2 Larson, Hostetler, et al.

[18] Calculus Instructional DVD Program, for use with (inter alia) Lar-son/Hostetler/Edwards, Calculus of a Single Variable: Early Transcendental Func-tions, Third Edition [19]. Houghton Mifflin (2003). ISBN 0-618-25097-2.

[19] R. Larson, R. P. Hostetler, B. H. Edwards, D. E. Heyd, Calculus, Early Transcen-dental Functions, Third Edition. Houghton Mifflin Company, Boston (2003). ISBN0-618-22307-X.

H.2.3 Edwards and Penney

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

[21] C. H. Edwards, Jr., and D. E. Penney, Calculus with Analytic Geometry, EarlyTranscendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997).ISBN 0-13-793076-3.

[22] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Calculus withAnalytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall,Englewood Cliffs, NJ (1997). ISBN 0-13-079875-4.


[23] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus with AnalyticGeometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, EnglewoodCliffs, NJ (1997). ISBN 0-13-793092-5.

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

H.2.4 Others, not “Early Transcendentals”

[25] G. H. Hardy, A Course of Pure Mathematics, 10th edition. Cambridge UniversityPress (1967).

H.3 Other References

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

[27] McGill Undergraduate Programs Calendar 2002/2003. Also accessible athttp://www.mcgill.ca/courses/#UGRAD.

[28] Notes Distributed to Students in Section 1 of Math 189-141B (1999/2000).

[29] G. N. Berman, A Problem Book in Mathematical Analysis. Mir Publishers, Moscow(1977 translation from 1975 Russian edition).

[30] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual, Single VariableCalculus, Early Transcendentals, Sixth Edition. Prentice Hall, Englewood Cliffs,NJ (2002). ISBN 0-13-066155-4.

information for students in math 141 2004 01

Documents