ig chapter8

7/21/2019 IG Chapter8

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8 TECHNIQUES OF INTEGRATION

8.1 INTEGRATION BY PARTS

SUGGESTED TIME AND EMPHASIS

112

classes Essential material

POINTS TO STRESS

1. The method of integration by parts; how to chooseu anddvto make the resulting integral simpler.

2. The analogy with u -substitution: u-substitution is undoing the Chain Rule, and integration by parts is

undoing the Product Rule.

QUIZ QUESTIONS

Text Question: Example 1 is an attempt to integrate xsinx . As stated in the subsequent note, it ispossible, using integration by parts, to obtain

xsinx d x = 1

2x2 sinx 1

2

x2 cosx dx . Why is this

equation an indication that we didnt choose our u anddvwisely?

Answer:We are trying to integrate xsinx. If we have to integrate x 2 cosx we have made the problem

morecomplicated, notlesscomplicated.

Drill Question: Compute tln t d t.Answer: 4

3

t3

ln

t 49

t3 + C

MATERIALS FOR LECTURE

Demonstrate how integration by parts works, including heuristics for choosing u anddv. Perhaps notethat the mnemonic LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential)

often helps to determine a good first try for the value ofu . When doing examples, periodically make poor

choices instead of good ones, to illustrate the difference. (For example, compute

x2exd x , attempting

first to do it by letting u= ex

, and then trying again with u= x2

.)

Have students come up with a strategy to compute

x31 x2

dx , which can be solved by partsu= x2,dv= x

1 x2dx

and substitution, or directly by substitution (u= 1 x2).

Compute a volume by cylindrical shells that requires parts, for example, the volume generated by rotatingthe region under y= lnx fromx= 1 to x= eabout thex -axis.

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CHAPTER 8 TECHNIQUES OF INTEGRATION

Draw a function like the one below and have the students try to approximate2

0 xg (x) dx .

0

1

2

3

1 2 3 x

y

Answer:

2

0 xg (x) d x 2g (2) g(2) 0g (0) +g(0) 2 (1) 2 + 0.4 = 0.4

WORKSHOP/DISCUSSION

Compute a definite integral that requires integration by parts (such as /20

xsinx dx).

Solve a problem that requires first a substitution, then integration by parts, such as2xex

2

sin

ln ex2

dx .

Answer:

2xex2

sin

ln ex2

d x= 12

ex2

sinln ex2 cos ln ex2 + C

Work through a non-trivial integration by parts problem with the students, such as x3 ln 2 + x2 dx .Note that it can be solved in two steps, using the substitution u = 2+ x2 and then using parts on1

2

(u 2) ln u du.

GROUP WORK 1: Guess the Method

Divide the students into groups and put problems on the board from the list of examples below (or hand out

the problems, if you prefer). Either have the students integrate the expressions completely, or describe what

method they would use, and what their answer should look like. For closure, do a few problems as a class that

were not covered in the group work.

Examples:

xln 3x dx= 12x2 ln 3x 14x2 + Ce2x sin ex d x= sin ex ex cos ex + C (substitution, then parts)e2x cosx d x= 1

5e2x sinx + 2

5e2x cosx + C (parts twice with a subtraction)

x3 cosx2 dx= 12x2 sin

x2+ 1

2cos

x2+ C (substitution, then parts)

x

2 + x2 ln 2 + x2 dx= 14

2 + x2 ln 2 + x2 1

8

2 + x22 + C (substitution, then parts)

x2 (lnx)2 dx= 13x3 (lnx)2 2

9x3 lnx + 2

27x3 + C (parts twice)

cos

x dx= cos x+ xsin x + C (substitution, then parts)424


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SECTION 8.1 INTEGRATION BY PARTS

GROUP WORK 2: Find the Error

Notice that the answers to the two problems are completely different. Give them the first problem, only

revealing the existence of the second after theyve solved the first.

Answers:

1. The stranger forgot the constant of integration. The last line should read 0 = 1 + C, which is true.One cute hint you can give the students (if you dare) is as follows:

There is something that the stranger failed to C. All you have to do is C it and you will have the

solution to the problem. Do you C what I mean?

2. The penultimate line should read

/4/6 tanx d x= 1|

/4/6 +

/4/6 tanx dx , which gives 0= 0 a true

statement.

HOMEWORK PROBLEMS

Core Exercises: 1, 15, 20, 23, 34, 46, 55, 62

Sample Assignment: 1, 6, 9, 15, 20, 23, 29, 34, 38, 40, 43, 46, 47, 55, 58, 62, 67

Exercise D A N G

1 6 9

15 20 23 29 34

38 40 43 46 47 55 58 62 67

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GROUP WORK 1, SECTION 8.1

Guess the Method

What method(s) could be used to compute the following antiderivatives? Either compute them explicitly, or

describe the best method to use.

1.

xln 3x dx

2.

e2x sin ex d x

3.

e2x cosx d x

4.

x3 cosx2 d x

5.

x

2 + x2 ln 2 + x2 d x

6.

x2 (lnx)2 dx

7.

cos

x dx

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Find the Error

It is a beautiful Spring day. You leave your calculus class feeling sad and depressed. You arent sad because of

the class itself. On the contrary, you have just learned an amazing integration technique: Integration by Parts.

You arent sad because it is your birthday. On the contrary, you are still young enough to actually be happyabout it. You are sad because you know that every time you learn something really wonderful in calculus,

a wild-eyed stranger runs up to you and shows you a proof that it is false. Sure enough, as you cross the

street, he is waiting on the other side.

Good morning, Kiddo, he says.

I just learned integration by parts. Let me have it.

What do you mean? he asks.

Arent you going to run around telling me that all of math is lies?

Well, if you insist, he chuckles... and hands you a piece of paper:

Hey, you say, I dont get it! You did everything right this time!

Yup! says the hungry looking stranger.

But... Zero isnt equal to negative one!

Nope! he says.

You didnt think he could pique your interest again, but he has. Spite him. Find the error in his reasoning.

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Find the Error (The Sequel)

What a wonderful day! You have survived another encounter with the wild-eyed stranger, demolishing his

mischievous pseudo-proof. As you leave his side, you cant resist a taunt.

Didnt your mother tell you never to forget your constants? It seemed a better taunt when you were thinking

it than it did when you said it.

Eh? he says. You come up to him again.

I was just teasing you. Just pointing out that when doing indefinite integration, those constants should not be

forgotten. A simple, silly error, not worthy of you. You look smug. You are the victor.

Yup. Indefinite integrals always have those pesky constants. For some reason he isnt looking defeated. He

is looking crafty.

Right. Well, Im going to be going now...

Of course, Kiddo,definiteintegrals dont have constants, sure as elephants dont have exoskeletons.Yes. Well, I really must be going.

Surprisingly quickly, he snatches the paper out of your hand, and adds to it. This is what it now looks like.

No constants missing here! Happy Birthday! The stranger leaves, singing the Happy Birthday song in a

minor key. Now there are no constants involved in the argument. But the conclusion is the same: 0 = 1. Isthe stranger right? Has he finally demonstrated that all that youve learned is suspect and contradictory? Or

can you, using your best mathematical might, find the error in this new version of his argument?

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8.2 TRIGONOMETRIC INTEGRALS


1 class Recommended material

POINTS TO STRESS

1. Integration of powers of the sine and cosine functions.

2. Integration of powers of the tangent and secant functions.

QUIZ QUESTIONS

Text Question:Whenm is odd, we can integrate sinm x dx by lettingu= cosx . Why doesm have tobe odd for this trick to work?

Answer: When m is odd, we can write sinm x dx as

1 cos2x

(m1)/2

sinx d x , and then the

u-substitution works. Ifm is not odd, then(m

1) /2 is not an integer. Less formal answers that correctly

address the issue of parity should be given credit.

Drill Question: Compute

sin2xcos3x d x .

Answer: 15

sin5x + 13

sin3x + C


Give several examples, such as

sin4xcos3x dx ,

sin7x 3

cosxd x,

sin2xcos4x d x , and sinx(cosx)1 dx to review the strategies for evaluating

sinm xcosn x d x:

Ifm or n is odd, peel off one power of sin x or cosx and use sin2x + cos2x= 1.

Ifm andn are both even, use the half-angle identities, as done in the text.Answers:

sin4xcos3 x d x= sin4x 1 sin2x cosx dx . Lettingu= sinx givesu4

1 u2

du= 1

5sin5x 1

7sin7x + C

sin7x 3

cosx dx=

1 cos2x

3cos1/3xsinx d x . Lettingu= cosx gives

1 u2

3u1/3 du= 3

22(cosx)22/3 + 9

16(cosx)16/3 9

10(cosx)10/3 + 3

4(cosx)4/3 + C

sin2xcos4 x d x= 1

cos2x

22 1 +

cos2x

24 d x

= 164

1 + 2cos2x cos2 2x 4cos3 2x cos4 2x + 2cos5 2x + cos6 2x dx

The odd powers of cos 2x can now be integrated by the previous method. The even powers require further

use of the half-angle identities.sinx(cosx)1 dx= tanx d x= ln |secx | + C

Give a couple of examples such as tanxsec4x d x and tanxsec3.28x dx to illustrate thestraightforward cases of

tanm xsecn x d x .

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Answers:

tanxsec4x d x =

tanx

tan2x + 1 sec2x dx . Letting u = tanx givesu1/2

u2 + 1 du= 2

3(tanx)3/2 + 2

7tanx7/2 + C.

tanxsec3.28x d x= sec2.28 (tanxsecx) d x . Lettingu= secx gives u2.28 du= sec3.28x3.28

+ C.

Derive the equation sin mxcos nx dx= 0 in two ways, first by computing sin mxcos mx dx usingFormula 2 and then by simply noting that sin mxcos nx is an odd function.

WORKSHOP/DISCUSSION

Derive the equation

secx d x= ln |secx + tanx | + C. Use this equation to compute

tan4xsecx d x .

Show how the computation oftan5x d x is quite different from the previous computation. Use the double-angle formula cos2 = 1 + cos2

2 to compute

d1 + cos2 .

Have the students find the volume generated by rotating the region under y= 1 + sin2x , 0 x aboutthex -axis.

GROUP WORK 1: An Equality Tester

This activity thoroughly explores a family of integrals that are interesting in their own right, using a compu-

tation that comes in handy in the study of Fourier series.

It is best to pose Problem 1 before handing out the sheet, because the students may disagree on the relative

areas of the two functions before they see Problem 2.

For Problem 2, the students may need the hint to consider the cases m= nandm= nseparately.Answers:

1. (a)

x

y

1

0 2

One has thrice the period of the other.2

0 sin2 3x dx=

2

0 sin2x d x=

(b)2

0 sin2 mx dx= ifm is an integer not equal to zero; 2

0 sin2 mx dx= 0 ifm= 0.

2. (a) 2

0 sin mxsin nx dx = 0 if m and n are positive integers with m = n. (This can be proven by

computation, and illustrated by graphical analysis.)2

0 sin mxsin nx dx= ifm andn are positive nonzero integers withm= n, by Problem 1(b).

(b) Again, this can be seen by direct computation, or using the hint and the fact that

cos mxcos nx sin mxsin nx= cos (m + n)x(c)

2

0 cos mxcos nx dx= 0 ifm andn are positive integers withm= n;

2

0 cos mxcos nx dx= if

m andn are positive nonzero integers withm= n.430


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SECTION 8.2 TRIGONOMETRIC INTEGRALS

GROUP WORK 2: Find the Error

Introduce this activity by writing

A = BC= D

on the blackboard and asking, If A= C, can we conclude that B= D? Then hand out the exercise. Ifstudents answer the problem by simply saying, He forgot the + C, make sure that they understand theimplication of the strangers computations, namely, that the functions y= cos2x andy= 2cos2x differ bya constant.

HOMEWORK PROBLEMS

Core Exercises: 3, 8, 13, 31, 52, 55, 57, 62

Sample Assignment: 3, 8, 13, 26, 31, 46, 47, 52, 55, 57, 59, 62, 66, 67

Exercise D A N G3 8

13 26 31 46 47 52 55 57 59

62 66 67

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An Equality Tester

1. (a) Graph sin2x and sin2 3x for 0 x 2. What is the relationship between these two functions?What do you think is the relationship between the areas bounded by these two functions from 0 to

2?

(b) Letm 0 be an integer. Compute 20

sin2 mx dx.

2. Letm andn be nonnegative integers.

(a) Compute2

0 sin mxsin nx dx .

(b) Show that2

0 sin mxsin nx dx= 2

0 cos mxcos nx dx .

Hint:Consider2

0 (cos mxcos nx sin mxsin nx) d x .

(c) Compute2

0 cos mxcos nx dx .

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Find the Error

It is a beautiful Spring morning. Everywhere you look, people are happily going to their classes, or coming

from their classes. High School is fun! calls out one student, and about twenty more yell Sure is!

in unison. Someone else calls out, I love History! A bunch of other students call Great subject! inresponse. Swept up in the spirit of things, you call out, Calculus is wonderful! Lies! Lies! calls out a

lone, familiar voice. You wheel around and directly behind you is a wild-eyed hungry-looking stranger.

Oh, dont be silly, you say. I just learned about trigonometric integration. It wasnt that hard a section, and

there isnt a single lie in it.

He looks up at you and says, Oh, really? Perhaps you can take a quick true/false quiz, and see how easy the

section is. The stranger then whips out a sheet of paper with this on it:

Both are clearly true! he shouts, before you have a chance to think. AND we know that

2sin2x= 2 (2sinxcosx) = 4sinxcosx ! Thus cos 2x= 2cos2x! Ho ho!Ho ho? you ask.

Ho ho, I say; ho, ho, I mean! Because at x = 0, cos2x = 1, and 2cos2x = 2! Once again, yourCalculus gets you into trouble! Two equals one, two equals one! sings the stranger, to the tune of Nyah,

nyah, nyah nyah, nyah, as he skips off into the distance.

Consider the strangers test. Are the answers true to both questions? And if so, then could the strangerbe correct? If 1= 2, then how can you tell odd numbers from even ones? Would one still be the loneliestnumber? How many turtle doves would your true love give to you on the second day of Christmas? Or is

there a possibility that there is an error somewhere in the strangers reasoning? Find the error.

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8.3 TRIGONOMETRIC SUBSTITUTION


1

2 1 class Recommended material

POINTS TO STRESS

1. The basic trigonometric substitutions and when to use them.

2. The use of trigonometric identities and right-triangle trigonometry to convert antiderivatives back to

expressions in the original variable, for example,

dx1 + x23/2= sin

tan1x

= x1 + x2

.

QUIZ QUESTIONS

Text Question: The book states that when doing an integral where the term 1 +x2 occurs, it often helpsto use the substitution x = tan . How could introducing a trigonometric function possibly make thingssimpler?

Answer:This substitution allows us to use the simplifying identity 1 + tan2 = sec2 .

Drill Question: Compute1

0

4 x2 d x using the substitution x= 2sin tand the fact that

cos2 t d t= 12

cos tsin t+ 12

t+ CAnswer: 1

2

3 + 1

3


Go over the table of trigonometric substitutions listed below, emphasizing when to use the different forms,and the restrictions that need to be placed on for each.

Expression Substitution Identitya2 x2 x= asin ,

2

2 1 sin2 = cos2

a2 + x2 x= atan , 2

2 1 + tan2 = sec2

x2

a2 x

=asec , 0

2 or

3

2 sec2

1

=tan2

Examples:

d x

25 x2= arcsin 1

5x + C,

dx

1 + 9x2= 1

3arctan 3x + C

Show how to derive identities such as sintan1x = xx2 + 1

by setting up a right triangle as in Figure 1.

Have the students evaluate 1 x x2 d x in two ways: first by completing the square, and then usingthe trigonometric substitutionx + 1

2=

5

2 sin .

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SECTION 8.3 TRIGONOMETRIC SUBSTITUTION

WORKSHOP/DISCUSSION

Compute 10

x2

4 x2 d x using a trigonometric substitution. Point out that because this is a definiteintegral, we dont need to use trigonometric identities at the end to find the antiderivative in terms of the

original variablex .

Answer:10 x

24 x2 d x = 16 /60 sin2 ucos2 u du using the substitution 2 sin u = x , and16

/60

sin2 ucos2 u d u= 14

3 + 1

3.

Evaluate

x2

x2 a2 d xin two different ways and compare the computations: first use the trigonometricsubstitution x= asec , then use the hyperbolic substitution x= acosh t.

GROUP WORK 1: Pizza for Three

The introduction to this exercise is very important. The goal is to slice a

14pizza, with two parallel lines across the entire pizza, to create threepieces of equal area.

Draw the figure on the board and explain that they need to find the value of

c. (If the class is particularly quick, the introduction can be abbreviated, but

it is better to say too much here than too little.)

This problem is also a good excuse to order pizza for a hard-working class.

Note that this is Problem 1 from Problems Plus after Chapter 8. A complete

solution to this problem can be found in the Solutions Manual.

x

y

7

_7

_c c_7 7

Answer:c 1.855

GROUP WORK 2: Look Before You Compute

The goal of this activity is to show students that it sometimes pays to look at the geometry of a problem before

immediately applying techniques.

Answers:

1.

12 + 4x x2 =

42 (x 2)2

2. 4sin = x 2 gives

/2/216 cos

2 d = 8.

3.

2

4

_2 2 4 6

y

x0

This is a semicircle of radius 4 and center(0, 2)with equation(x 2)2 + y2 = 16.435


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HOMEWORK PROBLEMS

Core Exercises: 1, 12, 22, 32, 36, 40

Sample Assignment: 1, 4, 12, 19, 22, 25, 32, 33, 34, 36, 37, 39, 40, 42

Exercise D A N G1 4

12 19 22 25 32 33 34 36 37

39 40 42

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Pizza for Three

How do you cut a 14pizza into three pieces of equal area, using just two parallel cuts?

x

y

7

_7

_c c_7 7

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Look Before You Compute

Consider the definite integral

6212 + 4x x2dx

1. Rewrite the integrand in the form

b2 (x a)2.

2. Use a trigonometric substitution to evaluate the integral.

3. Graph the original integrand over the range[2, 6]. Evaluate the integral directly by interpreting it as anarea.

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8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS


1

2 1 1

2classes Optional material

POINT TO STRESS

The idea that a given rational function can be broken down into a set of standard integrals, each of which

can be computed routinely.

QUIZ QUESTIONS

Text Question: Why would one want to write 1(x + 2) (x + 3) as the sum of two fractions?

Answer:It is much easier to find dx

x

+2

and d x

x

+3

than it is to find d x

(x

+2) (x

+3)

directly.

Drill Question: Compute

d x

x2 3x + 2 .Answer:ln |x 1| + ln |x 2| + C


As a warm-up, remind students how to compute

A

x + a dx and

B

(x + a)2 d x . If partial fractions are to

be covered in depth, also compute

d x

x2 + 4x + 8=

dx

(x + 2)2 + 4.

Remind students of the process of polynomial division, perhaps by rewriting 2x3

+ 3x2

+ 7x + 42x + 1 as

x2 + x + 3 + 12x + 1 , and then computing

2x3 + 3x2 + 7x + 4

2x + 1 dx .

Be sure to indicate that in order to use partial fractions, we need the degree of the numerator less than

the degree of the denominator. So to compute

x4 + 2x2 1d x , we first use long division to rewrite it as

x2 + 1 + 3x2 1

dx .

Find the coefficients for the partial fraction decomposition for x + 3

(x 2) (x 1)d x in two different ways:

first using two linear equations, and then using the method of creating zeros [setting x = 1 and thenx= 2 in x + 3 = A (x + 2) + B(x 1)].

Go over the process of partial fractions for quadratic terms, using 3x2 + 2

(x 1) =

1

x 1x + 1

x2 + 2 ,

and (if the subject is to be covered exhaustively) 1

x2 + 12x =1

x x

x2 + 1x

x2 + 12 .439


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SECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS

GROUP WORK 2: Finding Coefficients

Answers:

1. (a) 15 (x + 2)+

1

180 (x 3)+1

6 (x + 3)2+ 7

36 (x + 3) (b) 3

56

1 + 5xx2 + x + 2

3

56

26 + 5xx2 4x 4

2.1

5 (x + 2)+1

180 (x 3)+1

6 (x + 3)2+7

36 (x + 3) 3. 3

56 1

+5x

x2 + x + 23

56 26

+5x

x2 4x 4

HOMEWORK PROBLEMS

Core Exercises: 2, 7, 23, 35, 44, 51, 54, 55, 62

Sample Assignment: 2, 5, 7, 12, 16, 23, 34, 35, 41, 44, 48, 51, 54, 55, 57, 62, 64, 65

Exercise D A N G

2 5

7 12 16 23 34 35 41 44 48 51 54 55

57 62 64 65

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Partial Fractions (Version 1)

1. Compute the following integrals:

(a) dx

x 1

(b)

dx

x + 3

2. Factorx 2 + 2x 3.

3. Compute

x d x

x2 + 2x 3 .

4. Compute 5x + 5x2 + 2x 3d x .

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(a) dx

x + 1

(b)

dx

x + 2

2. Factor 2x3 + 6x2 + 4x .

3. Compute

2x + 3

2x3 + 6x2 + 4x dx .

4. Compute 3x2 + 6x + 22x3 + 6x2 + 4x dx .

443


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(a) dx

x + 1

(b)

dx

x + 2

(c)

d x

x2 + 4

2. Factorx 4 + 3x3 + 6x2 + 12x + 8.

3. Compute

20x2 d x

x4 + 3x3 + 6x2 + 12x + 8 .

4. Compute

x4 + 3x3 + 26x2 + 12x + 8

x4 + 3x3 + 6x2 + 12x + 8 d x .

444


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Finding Coefficients

1. Write the following rational functions as a product of powers of linear terms and irreducible quadratic

terms.

(a) 1x2 x 6 x2 + 6x + 9

(b) 3

x2 + x 2 x2 4x 4

2. Find the partial fraction decomposition for the function in Problem 1(a) using a linear system.

3. Find the partial fraction decomposition for the function in Problem 1(b) using the method of creating

zeros.

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8.5 STRATEGY FOR INTEGRATION


1 class Optional material

POINTS TO STRESS

1. The four-step strategy suggested in the text.

2. If at first you dont succeed, try again with a different method.

3. There are elementary functions that do not have elementary antiderivatives


This section gives the instructor a good opportunity to work a variety of examples with the students. Thefollowing challenging integrals provide opportunities to use the various techniques and strategies:

x3

1 x2 d x

ex

1/3d x

(xlnx)2 dx

tan1x d x

cos

x d x

cosxsinx d x

1 + sin4x

3d xx1/2

x3/2 x1/2 ln (lnx)x

2

dx

1e exd x

x5 cosx3 dx

Answers: x3

1 x2 dx=

1

5x2 + 2

15

1 x2

3/2 + C, ex1/3 dx= 3ex1/3x2/3 2x1/3 + 2+ C,(xlnx )2 d x= 1

27x3

9 ln2x 6 lnx + 2

+ C, tan1x dx= xarctanx 12

lnx2 + 1

+ C,cos

x d x= 2cos x + 2xsin x + C,

cosxsinx

1 + sin4xd x= 1

2arctan

sin2x

+ C,

3 d x

x2 x = 3 lnx + 3 ln (x 1) + C,

ln (lnx)x

2d x= ln2 (lnx) 2 ln (lnx) + 2 lnx + C,

dxe ex=

1

ex ln (e

x e)+ C, x5 cosx3 dx= 13cosx3 + 13x3 sinx3 + C

Discuss integrating functions with parameters. For example, compute the antiderivative

d x

x2 + A bybreaking it into cases. Also examine parameters as part of the limits of integration, as in solving the

equation

a3

x d x

x2 8= 2.

Go through a few integrals that require special approaches, such as secx d x .Answer:

secx dx=

(secx + tanx) secx

secx + tanx d x= ln (secx + tanx) + C

WORKSHOP/DISCUSSION

Find the area under the curve f (x) = 1ex + ex fromx= 1 to x= 2. (See Exercise 74.)

Answer:

2

1

dx

ex + ex= arctan e2 arctan e 0.218

If the velocity of a particle is given by v(t)= tln tt2 1

, determine the distance the particle has traveled

fromt= 2 to t= 5. (See Exercise 56.)446


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SECTION 8.5 STRATEGY FOR INTEGRATION

Answer: We let u = lnx , dv = xx2 1

, followed by a trigonometric substitution to obtain tln t

t2 1dt=

t2 1 ln t

t2 1 arcsec t+ C. Our final answer is thus

2

6 l n 5 2

6 + arcsec5

3 ln 2 +

3 13

3.839.

If the rate of change of population growth with respect to time is given byb (t)= t3

+ 1t3 t2 , find the total

change from year 1 to year 3. (See Exercise 66.)

GROUP WORK 1: Putting It All Together

The students may not understand the idea of taking one step, and describing a strategy. Perhaps do Problem 1

(or a similar problem) for them as an example. In Problem 3, students need to realize that ln x = lnx .Answers:

1. (a) Substituteu= cosx . (b) e5 cosx sinxcos2x d x= u

2e5udu

(c) Integrate by parts (twice).

2. (a) Substitutex= u3. (b) d x

x2/3 + 3x1/3 + 2= 3 u2 du

u2 + 3u + 2(c) Use long division and then partial fractions.

3. (a) Note that ln (x) = lnx . (b) x5 ln (x) d x= x5 lnx dx(c) Integrate by parts withu= lnx .

4. (a) Integrate by parts withu= ln (1 + ex),dv= e2xd x .

(b)

e

2x

ln (1 + ex

) d x= 1

2e

2x

ln (1 + ex

) 1

2 e2x

1 + exex

d x

(c) Substituteu= 1 + ex oru= ex .5. (a) Expand(ex + cosx )2 = e2x + 2ex cosx + cos2x .

(b)

(ex + cosx)2 dx=

e2x dx + 2

ex cosx d x +

cos2x d x

(c) The first and third integrals are simple, and the second can be integrated by parts (twice).

6. (a) Substituteu= lnx . (b)

ln (lnx)

xd x=

ln u d u (c) Integrate by parts.

GROUP WORK 2: Integration Jeopardy

This activity, designed to last for sixty to ninety minutes, is meant as a computational review of integration

techniques. Technology should not be permitted.

The Integration Jeopardy game board should be put on an overhead projector, or copied onto the blackboard.

The game is played in two rounds, each consisting of 20 questions, followed by a third round with a final

question. Students should be put into at most six mixed-ability teams of between three and seven players per

team. (It is most fun for the students if they get to name their team, but this process can take time!)

447


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27/59

SECTION 8.5 STRATEGY FOR INTEGRATION

HOMEWORK PROBLEMS

Core Exercises: 2, 9, 25, 30, 41, 48, 52

Sample Assignment: 2, 5, 9, 13, 19, 25, 30, 33, 35, 41, 48, 52, 54, 59, 62, 66, 70, 75, 79

Exercise D A N G2 5 9

13 19 25 30 33 35 41 48

52 54 59 62 66 70 75 79

449


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29/59


30/59


31/59

Integration Jeopardy (Questions and Answers)

Fun with Trigonometry

Value Question Answer

100

sinx3

cosxdx 3

2

3

cos2x + C

200 sinxcos (cosx) d x sin (cosx) + C300

tan2 sec4 d 1

5tan3 sec2 + 2

15tan3 + C

400

1 + cosx

sinxdx ln (cscx cotx) + ln sinx + C

500

tan2 d tan 12

2 1

2ln

1 + tan2 + C

Potpourri


100

x3

e2tdt

d x

e2x

4 e

6

2x + c

200

d

d xf(x)

d x f (x) + C

300 d

dr

r23

sin

x2

d x 2

sin r4

r+ C

400

cos t d x (cos t)x + C500

ln

1 + x2

d x xln

1 + x2

2x + 2 arctanx + C

453


32/59


Round 2: Double Jeopardy

Integration by Substitution


200 sin (2x +3) d x

1

2cos (2x

+3)

+C

400

x 2

cosx3 + 1 dx 1

3sin

x3 + 1+ C

600

1 + x

9

xdx

1 + x

105

+ C

800

e2x

1 + ex d x ex ln (1 + ex) + C

1000

tanxsec3x dx 13

sec3x + C

Integration by Parts


200

xex dx xex ex + C400

x3 lnx dx 1

4x4 lnx 1

16x4 + C

600

x2 sinx d x x2 cosx + 2cosx + 2xsinx + C800

arcsinx d x xarcsinx +

1 x2+ C

1000

e2x cosx d x 25

e2x cosx + 15

e2x sinx + C

Definite Integrals


200

2

1

x2 + 1x

d x 185

2 12

5

400

/3

/4

sinx d x 12+ 1

2

2

600 ln 6

ln 3

8ex dx 24

800

/3

0

sin

cos2 d 1

1000

4

1

tln t d t 32

3 ln 2 28

9

454


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Fun with Trigonometry


200

sin (2x + 3) d x 12

cos (2x + 3) + C400 x

2 cos x3 + e2 d x

1

3sin x

3 + e2+ C600

tanysec3 y d y

1

3sec3 y + C800

(x + sinx)2 d x 1

3x3 + 1

2x + 2sinx 2xcosx 1

2cosxsinx + C

1000

ln (tanx)

sinxcosxd x 1

2[ln (tanx)]2 + C

Potpourri


20011x4 sinx d x 0

400

cosx

1 + sin2x d x arctan (sinx )+ C

600

d

dt

sin4 t

et

2

dt

sin4 t

et2

+ C

800

ex+ex

d x eex + C

1000

cos (lnx) dx 12x(sin (lnx) + cos (lnx)) + C

Round 3: Final Jeopardy

Choose one of the following:

Question Answer x4

x10 + 4d x 1

10arctan

1

2x5

+ C cotxln (sinx ) d x

ln (sinx)2

2 + C

e 3xd x 3e

3x 3

x2 6

3

xe3x

+6e

3x

+C

(Hint:Substituteu= x1/3 and then use parts twice.)

Thanks to Ben Nicholson

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35/59


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HOMEWORK PROBLEMS

Core Exercises: 1, 7, 14, 31, 37

Sample Assignment: 1, 7, 8, 14, 17, 26, 31, 34, 37, 40, 41, 44, 45

Exercise D A N G1 7 8

14 17 26 31 34 37 40 41

44 45

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DISCOVERY PROJECT Patterns in Integrals

This project gives students the opportunity to discover some of the table formulas by going over several exam-

ples using a computer algebra system. Students can learn general formulas for partial fractions,

sin (ax ) cos (bx ), xn lnx , and xnex . Each problem is self-contained and instructors may wish to choose

selected problems, or break the class into four groups and have each student tackle a particular problem.

Hopefully students will see that by writing down a few examples, a general pattern can be determined.

More advanced students could also be given the task, using the text examples as a model, of finding patterns

in the integrals of a class of functions that they come up with on their own.

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8.7 APPROXIMATE INTEGRATION

TRANSPARENCY AVAILABLE

#15 (Figures 1 and 2)


1 class Optional material (essential for science/engineering majors)

POINTS TO STRESS

1. Left and right endpoint approximations.

2. The Midpoint Rule and the Trapezoidal Rule.

3. The geometry behind the Midpoint and Trapezoidal rules, and how to use technology to do numerical

integration.

4. (for science/engineering majors) Comparisons of the accuracies of the different approximating techniques,

and error bounds.

QUIZ QUESTIONS

Text Question: When would we need to use these techniques in the real world?Answer: We would need these techniques when we are given discrete rate-of-change data, with no

underlying function. (Other answers are also acceptable.)

Drill Question: The function fis continuous on the closed interval [2, 10] and has values given in the

table below. Using the subintervals [2, 6], [6, 7], and [7, 10], what is the trapezoidal approximation of10

2 f (x) dx ?

x 2 6 7 10

f(x) 5 15 17 11

(A) 16 (B) 32 (C) 49 (D) 72 (E) 144

Answer:(C)


Describe how numerical integration methods can be used to construct a model of a quantity whose

derivative matches a collection of experimental data points. For example, there are situations in whichwe are interested in velocity, but acceleration is easier to measure. These methods are particularly useful

when, as is often the case, there is no elementary underlying function.

Illustrate the geometry behind the Trapezoidal and Midpoint Rules, perhaps using Figure 5. Formulate intuitive comparisons of the accuracy of Rn, Ln , and Mn . Discuss the special cases of

monotone functions briefly to prepare students for Group Work 2. TECallows some visual and numerical

experimentation on several different functions.

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40/59


GROUP WORK 1: Position from Samples (Part 2)

Have the class revisit their automobile data previously collected in the group work Position from Samples

(Part 1) in Section 5.1. (If this activity was not previously assigned, it may be assigned at this time.) With

the new tools that they have at their disposal, especially the Trapezoidal Rule, the students should be able to

improve their estimates of the area under the velocity curve.GROUP WORK 2: Comparison of Methods

The purpose of this exercise is to show that if f is increasing or decreasing, the left and right endpoint

approximations bound the integral, and the Trapezoidal Rule gives an in-between value which is usually

closer to the actual value. In addition, if a curve is concave up (or concave down) then one can tell whether

the Trapezoidal Rule gives an over- or an underestimate.

Answers:

1. 3.178, 4.787, 3.983. The left endpoint approximation is an underestimate, the right endpoint

approximation is an overestimate, and the Midpoint Rule gives an underestimate.

2. 2.565, 2.565, 2.565. There is not enough information to tell.

3. 0.8546, 0.6080, 0.7313. The left endpoint approximation is an over estimate, the right endpoint

approximation is an underestimate, and the Midpoint Rule gives an underestimate.

4. (a) If the function is increasing, the right endpoint approximation gives an overestimate.

(b) If the function is decreasing, the left endpoint approximation gives an overestimate.

(c) If the function is concave up, the Trapezoidal Rule gives an overestimate.

HOMEWORK PROBLEMS

Core Exercises: 1, 2, 10, 19, 26, 29, 32, 37

Sample Assignment: 1, 2, 3, 10, 15, 16, 19, 22, 23, 26, 27, 29, 30, 32, 34, 35, 37, 38, 42, 43

Exercise D A N G

1 2 3

10 15 16 19 22

23 26

Exercise D A N G

27 29 30 32 34 35 37 38

42 43

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Comparison of Methods

For each of the integrals in Problems 13, first sketch the corresponding area, and then approximate the area

using right and left endpoint approximations and the Trapezoid Rule, all withn= 4. From your sketch alone,determine if each approximation is an overestimate, an underestimate, or if there is not enough informationto tell.

1.5

1 lnx d x

2.11secx dx

3.1

0 cos (tanx) dx

4. (a) What condition on a function guarantees that the right endpoint approximation is an overestimate?

(b) What condition on a function guarantees that the left endpoint approximation is an overestimate?

(c) What condition on a function guarantees that the Trapezoidal Rule gives an overestimate?

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8.8 IMPROPER INTEGRALS


1 112

classes Essential material: Infinite limits of integration, comparisons

Optional material: Vertical asymptotes

POINTS TO STRESS

1. A careful definition of convergence and divergence as related to improper integrals of typea

f (x) d x

(including an interpretation of the integral as the area under a curve).

2. A careful definition of convergence and divergence as related to integrals of discontinuous functions.

3. The Comparison Theorem for improper integrals, including result 2 from Example 4.

QUIZ QUESTIONS

Text Question: Why is

3

1

dx

x2 2 an improper integral? What two improper integrals will tell us if 3

1

dx

x2 2converges?

Answer:The integrand is undefined forx=

2;

2

1

d x

x2 2and

3

2

d x

x2 2 .

Drill Question: Compute0

t e2tdt.

(A) 12

(B) 14

(C) 14

(D) 12

(E) Divergent

Answer:(C)


Illustrate the geometric interpretation of improper integrals as areas under infinite curves, stressingExample 4. Give examples of functions that enclose finite and infinite areas, including functions with

vertical asymptotes.

Discuss the Comparison Theorem (including a geometric justification) and work several examples. Forexample, show why

1

d x

x3 + 7x2 + 2x + 1converges, without resorting to computing the antiderivativeby partial fractions. Stress that for convergence at infinity, only the tail part of the integral matters;

convergence is independent of the value of the integral over any finite interval. Point out that when

a f(x) d x diverges, it is possible that lim

tta

f(x) d x is not infinite, but rather

simply does not exist. (For example,0

cosx dx .)

Explain the following paradox: The curve y= 1/x to the right ofx = 1 encloses infinite area, thatis,

1

1

xd x diverges. But if we rotate it about the x -axis and look at the corresponding infinite solid, that

solid has finite volume.

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SECTION 8.8 IMPROPER INTEGRALS

Answer: There is no paradox. It is possible for

a shape to have finite volume and, at the same

time, have infinite cross-sectional area. A two-

dimensional analogue is easier to visualize. One

can enclose a finite area with a curve of infinite

perimeter. For example, the curve at right has

infinite perimeter, but fits inside a circle of radius

3

2, showing that it has finite area.

1

1

_1

_1 0 x

y

r=sin 1

2

+ 12

WORKSHOP/DISCUSSION

Show that

2

0

d x2 x2

is an improper integral that converges to 2

.

Show why

3

1

d x

x2 2 converges if and only if

2

1

dx

x2 2 and

3

2

dx

x2 2 converge, and then why 3

1

dx

x2 2diverges.

Ask if we can find p such that 1

0

1

xpd x= 100. Now find p such that

1

0

1

xpd x= 500, and then such

that1

0

1xp

dx= 1,000,000 = 106.

Show that

2

0

d x2 x2

is an improper integral that converges to

2.

GROUP WORK 1: Convergence with a Parameter

Have the students prove that

e

d x

x(lnx)pconverges only for p >1. Perhaps just ask them for what values

of p the integral converges, and see if they figure it out for themselves. Be sure that they check the cases

p = 0 andp = 1 carefully. If time permits, determine the values ofpfor which e1

dx

x(lnx)pconverges.

Answer:Substitution with u = lnx will help prove thate

d x

x(lnx)p converges for p > 1, and that e

1

d x

x(lnx)pconverges for p < 1.

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GROUP WORK 2: Whats Wrong?

For students, the hardest part of the comparison test is often figuring out which way the implications go. This

activity helps them to really understand this test.

Answers:

1. Positive numbers are greater than negative numbers.

2. limn

n1

d x

x2= 1 2.It diverges. lim

n

n1

dx

x

is infinite.

4. They do not. The Comparison Test only applies if the integrands are positive.

GROUP WORK 3: Deceptive Visions

Things are not always what they seem in the realm of improper integrals, as this exercise attempts to illustrate.

Emphasize to the students that theyshouldntbe able to predict just by looking in Part A that one of the three

curves has infinite area under it, and that the other two have finite areas. Make sure, during closure, to try to

come to some understanding of how such a thing can happen: that the very subtle differences in the curves

values add up when we go off to infinity. Part B is an integral which occurs in aerodynamics and has a

surprising value.

Answers:

Part A

1. There isnt an obvious difference because the functions are very similar.

2.

10

1

d x

x= ln 2 + ln 5 2.302585,

10

1

d x

x1.01 2.276278,

10

1

d x

x1.001 2.299936. The functions are

similar, as are the areas.3. Answers will vary.

4.

100

1

dx

x= ln 100 4.605170,

100

1

d x

x1.01= 100 1000.01+ 100 4.500741,

100

1

dx

x1.001= 1000 1000.001 + 1000 4.594582. The first and third are closest together, which

makes sense since the integrands are most similar.

5. The first diverges, the second converges to 100, and the third converges to 1000. The students should

notice that while the first was closest to the third for x= 100, the first diverges while the third does not.

Part B

1. It is not defined at x= 1.2. Multiply both numerator and denominator by

1 + x .

3.

1 + x

1 x2d x=

d x

1 x2+ x dx

1 x2= arcsinx

1 x2

4. limn1

arcsin n

1 n2

arcsin (1)

1 (1)2

= arcsin1 arcsin(1) = 466


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SECTION 8.8 IMPROPER INTEGRALS

GROUP WORK 4: Improper Integration Jeopardy

This is an extension of Integration Jeopardy (see Group Work 2 in Section 8.5) emphasizing improper inte-

grals. It can be used in addition to integration jeopardy, or combined with the former, allowing the improper

integration categories to replace one or more of the others.

HOMEWORK PROBLEMS

Core Exercises: 2, 4, 6, 21, 35, 47, 50, 57, 67

Sample Assignment: 2, 4, 6, 13, 19, 21, 30, 35, 36, 42, 47, 50, 54, 57, 65, 67, 71

Exercise D A N G

2 4 6

13 19

21 30 35 36 42 47 50 54 57 65 67 71

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Whats Wrong?

Let f (x) = 1x2

andg(x) = 1x

.

1. Show that, forx

1, f (x)

g(x).

2. Show that

1

1

x2d x= 1.

3. Does

1 1

x d x converge or diverge?

4. Do your answers to Problems 2 and 3 contradict the Comparison Test? Why or why not?

468


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Deceptive Visions

Part A

Graph the following three functions, on the same set of axes, from x

=1 to x

=10:

f (x) = 1x

g(x ) = 1x1.01

h(x) = 1x1.001

1. Can you see a significant difference among the three graphs? If so, what is it? If not, why not?

2. Compute the areas under the three curves from x = 1 tox = 10. Do you get different results? Why or

why not?

3. Do you expect a big difference in the areas under the three curves betweenx= 1 andx= 20? How aboutbetweenx= 1 andx= 50?

4. Compute the areas under the three curves from x= 1 to x= 100. Do you get very different results?Which two of the three functions areas are the closest? Why?

5. What do you think will happen to the area under each of the three functions if you take x from 1 out to

infinity? Try it.

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Deceptive Visions

Part B

We want to evaluate

1

1

1 + x1 x d x .

1. Why is this an improper integral?

2. Show that

1 + x1 x =

1 + x1 x2

if1< x


49/59


Improper Integration Jeopardy (Game Boards)

Round 1: Jeopardy

Infinite

Limits

Vertical

Asymptotes Potpourri

100 100 100

200 200 200

300 300 300

400 400 400


Infinite

Limits

Vertical

Asymptotes Potpourri

200 200 200

400 400 400

600 600 600

800 800 800

471


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Improper Integration Jeopardy (Questions and Answers)

Round 1: Jeopardy

Infinite LimitsValue Question Answer

1001

e2x dx 12

e2

200

1

4x

1 + x2 dx

3001

xex d x 0

400

4

dxx + 2

500 Does

ex

2

dx converge? Yes (compare withex and use symmetry)

Vertical Asymptotes


100

9

0

dxx

6

200

3

2

dx

x4 Divergent

300

1

0

lnxx

4

400 0 secx d x Divergent500

2

0

x 32x 3d x Divergent

Potpourri


100 Does

1

dx

1 + x3 converge? Yes. Compare with 1

x3.

200

1

dx

x p, p> 1

1

p 1

300 10

dxxp

, 0 < p< 1 11 p

400 Does

0

d x

x pconverge for any value ofp? No.

1

0

d x

xpdiverges for p 1;

1

d x

xpdiverges for p 1

500 Does

1

0

d x

1 + x3 converge? No. Compare with x2

1 +x3 .

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Improper Integration Jeopardy (Questions and Answers)


Infinite Limits


200 0 ex/2 d x 2400

0

1

1 + x2 dx

2

600

x

1 + x2 d x Divergent

800k

ex/ndx nek/n

1000

0

d x

1 + ex ln 2

Vertical Asymptotes


200

1

1

d x

x2 Divergent

400

1

0

1

(4y 1)2 d y Divergent

6002

0 r2 ln r dr 8

3ln 2 8

9

800

2

1

1

x4d x

1000 1

0 lnx d x 1

Potpourri


200 Does7

2sinx2

dx converge? Yes. Its not even improper!

400 Does

2

0

sinxx

dx converge? Yes

6000

ddt

sin4 t

et

2

dt limt

sin4 t

et

2

= 0

800 Show that limt

t

tx dx= 0. 12 t2 12 t2 = 0

1000 Show thatx dx is divergent.

x dx=

0

x d x +0

x d x; both diverge.

Round 3: Final Jeopardy

Does0

dxx + x4

converge or diverge? Prove your answer.

Answer:It converges. Compare with 1

xnear 0, and with

1

x2for largex .

473


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8 SAMPLE EXAM

Problems marked with an asterisk (*) are particularly challenging and should be given careful consideration.

1. For each integral below, indicate a technique that can be used to evaluate the integral, and then apply the

technique to rewrite the integral as a simpler one. Finally, without actually solving the integral, indicate

how to proceed from there.

(a)

d x

9 x2

(b)

x 7

9 x2dx

(c)

3x

x2 93/2 d x

(d)

x3ex2

dx

(e)

3x + 2

x2 + 6x + 8d x

(f)

3x + 2

(x + 2) (x + 1) (x 1)d x

(g)

x3

2 + x2

5/2

d x

2. Consider the regionSbounded by the curves y= x + 1x2

andy= x 1x2

forx 1.

(a) Is the area ofSfinite or infinite? If finite, find the area.

(b) Now suppose we rotateSaround the x axis. Is the volume of the resulting solid finite or infinite? If

finite, find the volume.

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9. (a) Show that limtttsinx d x= 0.

(b) Show that sinx dx is divergent.

(c) Do the answers to parts (a) and (b) contradict one another? Explain your reasoning. Include a graph

in your explanation.

10. Which area below is equal to

2

0

4 d x4 + x2

3/2 ?(a)

y

0

1

2 u

/20

cos u d u

(b)

0

1

1 u

y

_4

/40

sin u d u

(c)

0

1

1 u

y

_4

/40

cos u d u

(d)

0

1

1 u

y

_4

/40

sec u d u

11. (a) Show that the following formula is valid for any differentiable function f.x f (x) d x= x f (x) f (x) dx

(b) Compute x3x 1

d x using the above formula.

(c) Suppose that f(x) is continuous and differentiable, f(2)= 2, f (6)= 6, and6

2 f (x) dx=

10. Compute 62

x f (x) d x .

12. Give a direct argument, without any computations, to show that

sinx(1 + cosx)2 d x= 0.

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CHAPTER 8 SAMPLE EXAM

13. Let f(x)be the function graphed below.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x

y

0

We wish to approximate1

0 f(x) d x. Which of the following gives the best approximation in this case

and why?

(a) The Midpoint Rule with 2 subintervals

(b) The Midpoint Rule with 4 subintervals

(c) The Left Endpoint Rule with 4 subintervals

14. Let f(x)be the function graphed below.

x

y

Four students approximated the area under f(x) from 0 to 1. They all used the same number of

subintervals, but they each used a different method. Here are their results:

George 2.453Vicki 2.638Todd 2.555

Pat 2.178

Which student used which method? Explain.

Left Endpoint Approximation: ______________

Right Endpoint Approximation:______________

Midpoint Rule:________________

Trapezoid Rule:________________

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15. Consider the three functions f(x),g(x)and 1/x2, graphed below.

0

1

2

3

4

5

6

1 2 3 x

g(x)1/x@

f(x)

y

Which of the following must be true, might be true, or cannot be true?

(a)1

f (x) dx converges.

(b)1

0 g(x) d x converges.

(c)

1

g(x) d x diverges.

16. Ifx= tan , show that sin 2 = 2x1 + x2 .

17. Show that the areas of the two shaded regions below are the same:

0

1

1 x

y

f (x) = 11 + x22

0

1

1

y

_4

g() = cos2

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8 SAMPLE EXAM SOLUTIONS

1. Answers may vary. The following are sample correct answers:

(a) Use the trigonometric substitutionx= 3sin , to get d.Or use algebraic simplification, followed by the substitutionu= x

3, to get

du

1 u2.

(b)

2

1

x 79 x2

d x=

2

1

x9 x2

dx 7

2

1

d x9 x2

. The first integral can be done using the sub-

stitution u= 9 x2 to get

8

5

du

2

u. For the second integral, the substitutionx = 3sin yields

2

1

d x

9 x2

= arcsin2/3

arcsin 1/3

d, while the substitutionv

= 1

3x yields

2

1

d x

9 x2

= 2/3

1/3

dv

1 v2

.

(c) u= x2 9

3

2u3/2du

(d) u= x2 12

ueu du

(e) Partial fractions expansion:

5

x + 42

x + 2

dx

(f) Partial fractions expansion:

4

3 (x + 2)+1

2 (x + 1)+5

6 (x 1)

d x

(g) u= 2 + x2 12

(u 2) u5/2 du

2. (a) Area =1

x + 1

x2

x 1x2

dx=

1

2

x2d x= 2

(b) Volume =1

x + 1

x2

2

x 1x2

2d x=

1

4

xd x , which is divergent.

3. (a) Vp=1

x2pdx This integral converges for p > 1

2.

(b) Bp= 10

2x 1

xpd x.. This integral converges for p > 2.

(c) Any value ofpsuch that 12

< p 1. For these values ofp, Vp=

2p 1 .

4. (a)

1

4 u2 du

(b) Use partial fractions to get 14

ln |sinx 2| + 14

ln |sinx + 2| + C.479


58/59


59/59

CHAPTER 8 SAMPLE EXAM SOLUTIONS

14. George 2.453 Trapezoid Rule (underestimate)

Vicki 2.638 Right Endpoint Approximation (largest)

Todd 2.555 Midpoint Rule (overestimate)

Pat 2.178 Left Endpoint Approximation (smallest)

15. (a) f (x) < 1/x2 for x

1 and 11/x2 d x converges, so

1

f(x) d x must converge and the

statement must be true.

(b) g(x) >1/x2 for 0< x 1/x2 on(1, ), so the convergence of1

1/x2

d x

tells us nothing.

16.

x

1

1+x@

sin2 = 2sin cos = 2

x1 + x2

11 + x2

= 2x

1 + x2

17. The first area is

1

0

11 + x22 . Lettingx= tan gives

/4

0

sec2 d

sec4 =

/4

0

cos2 d, which is the

second area.

ig chapter8

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