TECHNIQUES OF

7.5 STRATEGY FOR INTEGRATION As we have seen, integration is more challenging than differentiation. In finding the deriv-

ative of a function it is obvious which differentiation formula we should apply. But it may not be obvious which technique we should use to integrate a given function.

Until now individual techniques have been applied in each section. For instance, we usually used substitution in Exercises 5.5, integration by parts in Exercises 7.1, and partial fractions in Exercises 7.4. But in this section we present a collection of miscellaneous inte- grals in random order and the main challenge is to recognize which technique or formula to use. No hard and fast rules can be given as to which method applies in a given situation, but we give some advice on strategy that you may find useful.

A prerequisite for strategy selection is a knowledge of the basic integration formulas. In the following table we have collected the integrals from our previous list together with several additional formulas that we have learned in this chapter. Most of them should be memorized. It is useful to know them all, but the ones marked with an asterisk need not be

CHAPTER 7 TECHNIQUES OF INTEGRATION

memorized since they are easily derived. Formula 19 can be avoided by using partial frac- tions, and trigonometric substitutions can be used in place of Formula 20.

TABLE OF INTEGRATION FORMULAS Constants of integration have been omitted.

1. y x n dx 苷 x 1 2. dx 苷 ln x

x 3. a y

4. y a dx 苷

e dx 苷 e

ln a

5. y

6. y cos x dx 苷 sin x

7. 2 y 2 sec x dx 苷 tan x 8. y csc x

9. y sec x tan x dx 苷 sec x

10. y

11. y sec x dx 苷 ln 12. y csc x dx 苷 ln

13. y tan x dx 苷 ln sec x 14. y cot x dx 苷 ln sin x

15. y sinh x dx 苷 cosh x

16. y cosh x dx 苷 sinh x

1 7. y

dx

18. y

2a 2 冟 2 冟 y sx ⱍ

dx

苷 1 *19. dx y 2 2 ln * 20. 苷 ln

sx 2 2

Once you are armed with these basic integration formulas, if you don’t immediately see how to attack a given integral, you might try the following four-step strategy.

1. Simplify the Integrand if Possible Sometimes the use of algebraic manipula- tion or trigonometric identities will simplify the integrand and make the method of integration obvious. Here are some examples:

y sx ( sx ) dx 苷 y ( sx ) dx

y 2 y cos sec 2

y 2 dx 苷

SECTION 7.5 STRATEGY FOR INTEGRATION

2. Look for an Obvious Substitution Try to find some function u苷 in the integrand whose differential du 苷

also occurs, apart from a constant fac- tor. For instance, in the integral

2 dx x

we notice that if u苷x 2 , then du 苷 2x dx . Therefore we use the substitu- tion u苷x 2 instead of the method of partial fractions.

3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led

to the solution, then we take a look at the form of the integrand f . (a) Trigonometric functions. If f is a product of powers of sin x and cos x ,

of tan x and sec x , or of cot x and csc x , then we use the substitutions recom- mended in Section 7.2.

(b) Rational functions. If is a rational function, we use the procedure of Sec- f tion 7.4 involving partial fractions. (c) Integration by parts. If f is a product of a power of (or a polynomial) and x

a transcendental function (such as a trigonometric, exponential, or logarithmic

function), then we try integration by parts, choosing and u d v according to the

advice given in Section 7.1. If you look at the functions in Exercises 7.1, you will see that most of them are the type just described.

(d) Radicals. Particular kinds of substitutions are recommended when certain radicals appear. (i) If

2 2 occurs, we use a trigonometric substitution according to the table in Section 7.3.

(ii) If s occurs, we use the rationalizing substitution u苷 .

s More generally, this sometimes works for s

nn

4. Try Again If the first three steps have not produced the answer, remember that there are basically only two methods of integration: substitution and parts. (a) Try substitution. Even if no substitution is obvious (Step 2), some inspiration

or ingenuity (or even desperation) may suggest an appropriate substitution. (b) Try parts. Although integration by parts is used most of the time on products of the form described in Step 3(c), it is sometimes effective on single func- tions. Looking at Section 7.1, we see that it works on tan x , sin x , and ln x , and these are all inverse functions.

(c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the

denominator or using trigonometric identities) may be useful in transforming the integral into an easier form. These manipulations may be more substantial than in Step 1 and may involve some ingenuity. Here is an example:

dx

dx 苷 y 2 dx

x cos x

2 x dx y sin x

苷 y 2 dx 苷 csc 2

sin

(d) Relate the problem to previous problems. When you have built up some expe-

rience in integration, you may be able to use a method on a given integral that is similar to a method you have already used on a previous integral. Or you may even be able to express the given integral in terms of a previous one. For

CHAPTER 7 TECHNIQUES OF INTEGRATION

instance, x tan 2 x sec x dx is a challenging integral, but if we make use of the iden-

tity tan 2 x苷 sec 2 1 , we can write

y tan 2 x sec x dx 苷 sec 3 y x

y sec x dx

and if x sec 3 x dx has previously been evaluated (see Example 8 in Section 7.2), then that calculation can be used in the present problem. (e) Use several methods. Sometimes two or three methods are required to evalu- ate an integral. The evaluation could involve several successive substitutions of different types, or it might combine integration by parts with one or more substitutions.

In the following examples we indicate a method of attack but do not fully work out the integral.

tan 3 x

EXAMPLE 1 y dx

cos 3 x In Step 1 we rewrite the integral:

tan 3 x

dx 苷 y tan x sec x dx

cos 3

The integral is now of the form tan m

x n x sec x dx with m odd, so we can use the advice in

Section 7.2. Alternatively, if in Step 1 we had written

sin 3 x 1 sin 3 y x dx 苷 y 3 dx 苷 dx

tan 3 x

y cos 6 x

cos 3 x

cos x cos 3 x

then we could have continued as follows with the substitution u苷 cos x :

sin 3 x

6 dx 苷 y sin x dx 苷

cos x

cos 6 x

苷 y 6 du 苷

y du M

e sx dx y

V EXAMPLE 2

According to (ii) in Step 3(d), we substitute u苷

sx 2 . Then x苷u , so dx 苷 2u du and

sx u y e dx 苷 2 y ue du

The integrand is now a product of and the transcendental function u e u so it can be inte-

grated by parts.

SECTION 7.5 STRATEGY FOR INTEGRATION

EXAMPLE 3 y 3 2 dx

x No algebraic simplification or substitution is obvious, so Steps 1 and 2 don’t apply here.

The integrand is a rational function so we apply the procedure of Section 7.4, remember- ing that the first step is to divide.

V EXAMPLE 4 dx y

x sln x Here Step 2 is all that is needed. We substitute u苷 ln x because its differential is

du 苷 dx , which occurs in the integral.

V EXAMPLE 5

dx

Although the rationalizing substitution

u苷

works here [(ii) in Step 3(d)], it leads to a very complicated rational function. An easier method is to do some algebraic manipulation [either as Step 1 or as Step 4(c)]. Multiply- ing numerator and denominator by

, we have

2 y dx

dx 苷 y

2 dx

苷 sin

C AN WE INTEGRATE ALL CONTINUOUS FUNCTIONS? The question arises: Will our strategy for integration enable us to find the integral of every

e x continuous function? For example, can we use it to evaluate 2 x dx ? The answer is No, at least not in terms of the functions that we are familiar with.

The functions that we have been dealing with in this book are called elementary func- tions. These are the polynomials, rational functions, power functions a , exponential functions x , logarithmic functions, trigonometric and inverse trigonometric functions, hyperbolic and inverse hyperbolic functions, and all functions that can be obtained from these by the five operations of addition, subtraction, multiplication, division, and compo- sition. For instance, the function

f sin 2x

is an elementary function.

If is an elementary function, then f is an elementary function but 2 x f need not

be an elementary function. Consider f x . Since is continuous, its integral exists, f

and if we define the function F by

0 e dt

CHAPTER 7 TECHNIQUES OF INTEGRATION

then we know from Part 1 of the Fundamental Theorem of Calculus that

Thus, f x 2 has an antiderivative , but it has been proved that F F is not an elemen- tary function. This means that no matter how hard we try, we will never succeed in evalu- 2

ating x e x dx in terms of the functions we know. (In Chapter 11, however, we will see how x to express 2 x e dx as an infinite series.) The same can be said of the following integrals:

y sin dx y cos dx

dx

3 1 y sin x sx y dx dx

ln x

In fact, the majority of elementary functions don’t have elementary antiderivatives. You may be assured, though, that the integrals in the following exercises are all elementary functions.

7.5 EXERCISES

1– 80 Evaluate the integral. 3x 2 3x 2

3 25. y 2 dx

y x x 3 dx

1. y cos x

2 sin x

2. y dx

cos x

dx

27. y x

28. y sin sat dt

3. y dx

4. tan 3

tan x

29. y 0 w

d 30. y x dx

2 2t

5. y 0 2 dt

6. y 4 dx

yy

1 e arctan y

y 34.

10. y 0 dx

36. y sin 4x cos 3x dx

37. cos 2 2 38. tan 5 y 3 14. y

15. 2 16. 39. 40. y 3 y 2 dx y sec 0 2 y s4y 2 dy

e 2t

41. y

42. y

2 tan x

x 2 dx e x dx

17. y x sin 2 x dx

18. y 4t dt

44. x 19. y 20. y y y dx

21. 3 arctan sx y

ln x

22. y dx

23. 1 ( ) 8 y x sx

24. 0 2 y ln

dx

47. y x 3 dx

48. y

x 4 4 dx

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS

49. y dx

50. y 2 dx

s3

67. y 1

dx

68. y x

dx

51. y 2 dx

1 dx

52. y 4 e 2x

ln x s4x

69. y x dx

70. y 2 dx

53. y 2 x sinh mx dx

54. y 2 dx

arcsin x

71. y 2 dx

72. y

dx dx

dx

55. y 56. y 1 dx

73. y 2 dx

74. y ( sx sx ) 4

57. y x 3 s dx

x ln x

y x xe

58. dx

sx 2 75. dx

76. y 2 x y

3 59. dx y cos x cos 60. y

s4x sec x cos 2x 2

77. y sx 3 dx

78. y dx

sx 61. 1 y sx e dx 62. y dx

s 3 x 79. sin x cos x x sin y 2 x cos x dx 80. y

sin 4 cos dx 4 x sin 2x

63. y 4 dx

y sin x cos x dx

ln

3 81. The functions y苷e x and y苷x 2 e x 1 don’t have elementary 3 u

antiderivatives, but does. y苷 2 x 65. 2

66. 2 y Evaluate y 2 3 2 du

dx

2 x x dx .

7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS In this section we describe how to use tables and computer algebra systems to integrate

functions that have elementary antiderivatives. You should bear in mind, though, that even the most powerful computer algebra systems can’t find explicit formulas for the antideriv-

x atives of functions like 2 e or the other functions described at the end of Section 7.5.

TABLES OF INTEGRALS Tables of indefinite integrals are very useful when we are confronted by an integral that is

difficult to evaluate by hand and we don’t have access to a computer algebra system. A rel- atively brief table of 120 integrals, categorized by form, is provided on the Reference Pages at the back of the book. More extensive tables are available in CRC Standard Mathe- matical Tables and Formulae, 31st ed. by Daniel Zwillinger (Boca Raton, FL: CRC Press, 2002) (709 entries) or in Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products, 6e (San Diego: Academic Press, 2000), which contains hundreds of pages of integrals. It should be remembered, however, that integrals do not often occur in exactly the form listed in a table. Usually we need to use substitution or algebraic manipulation to transform a given integral into one of the forms in the table.

EXAMPLE 1 The region bounded by the curves y苷 arctan x, y 苷 0 , and x苷 1 is rotated

about the -axis. Find the volume of the resulting solid. y SOLUTION Using the method of cylindrical shells, we see that the volume is

V苷 1 y

CHAPTER 7 TECHNIQUES OF INTEGRATION

N The Table of Integrals appears on Reference

In the section of the Table of Integrals titled Inverse Trigonometric Forms we locate

Pages 6–10 at the back of the book.

2 2 Thus the volume is

0 x tan x

tan

EXAMPLE 2 Use the Table of Integrals to find y 2 dx .

SOLUTION If we look at the section of the table titled Forms involving sa 2 2 , we see

that the closest entry is number 34:

sa 2 2 2 2 sin

2 2 a This is not exactly what we have, but we will be able to use it if we first make the substi-

sa

tution : u苷 2x

x 2 2 du 苷 1 u y 2

2 dx 苷 y 2

du

( so s5 ) :

Then we use Formula 34 with a 2 苷 5 a苷

2 dx 苷 y 2 sin

y 3 x sin x dx .

EXAMPLE 3 Use the Table of Integrals to find

SOLUTION If we look in the section called Trigonometric Forms, we see that none of the entries explicitly includes a u 3 factor. However, we can use the reduction formula

in entry 84 with n苷 3 :

y x 3 3 y x 2 cos x dx

y 2 We now need to evaluate x x cos x dx . We can use the reduction formula in entry 85

85. u n cos u du

u 1 with n苷 2 y , followed by entry 82: sin u du

y x 2 cos x dx 苷 x 2 y x sin x dx

苷x 2

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS

Combining these calculations, we get

where . C苷 3K

y x sx

V EXAMPLE 4 Use the Table of Integrals to find

2 dx . SOLUTION Since the table gives forms involving sa 2 2 , sa 2 2 , and sx 2 2 , but

not sax 2 , we first complete the square:

If we make the substitution

1 (so

1 ), the integrand will involve the

pattern : sa 2 2

y x sx

dx 苷 y su 2 du

苷 u su 2 su 2 y y du

The first integral is evaluated using the substitution t苷u 2 :

2 1 1 2 y 1 u su du 苷 2 y st dt 苷 2 3 t 3 苷 3 2 3

: y sa

2 sa

21. 2 2 du 苷 u 2 2 For the second integral we use Formula 21 with a苷

s3

ln (

2 sa

y su

du 苷 su 2 2 ln ( su 2 )

Thus

y x sx

2 dx

苷 1 2 3 1 2 3 3 sx 2 ln (

sx 2 )

COMPUTER ALGEBRA SYSTEMS We have seen that the use of tables involves matching the form of the given integrand with

the forms of the integrands in the tables. Computers are particularly good at matching pat- terns. And just as we used substitutions in conjunction with tables, a CAS can perform sub- stitutions that transform a given integral into one that occurs in its stored formulas. So it isn’t surprising that computer algebra systems excel at integration. That doesn’t mean that integration by hand is an obsolete skill. We will see that a hand computation sometimes produces an indefinite integral in a form that is more convenient than a machine answer.

To begin, let’s see what happens when we ask a machine to integrate the relatively

simple function y苷 1 . Using the substitution u苷

, an easy calculation

by hand gives

dx 苷 3 ln ⱍ

CHAPTER 7 TECHNIQUES OF INTEGRATION

whereas Derive, Mathematica, and Maple all return the answer

3 ln 共3x ⫺ 2兲

The first thing to notice is that computer algebra systems omit the constant of integra- tion. In other words, they produce a particular antiderivative, not the most general one. Therefore, when making use of a machine integration, we might have to add a constant. Second, the absolute value signs are omitted in the machine answer. That is fine if our

problem is concerned only with values of greater than . But if we are interested in other 2 x

values of , then we need to insert the absolute value symbol. x

In the next example we reconsider the integral of Example 4, but this time we ask a machine for the answer.

EXAMPLE 5 Use a computer algebra system to find x 2 y sx ⫹ 2x ⫹ 4 dx .

SOLUTION Maple responds with the answer

1 3 s3

3 共x ⫹ 2x ⫹ 4 兲 ⫺ 4 共2x ⫹ 2兲 sx 2 ⫹ 2x ⫹ 4 ⫺ arcsinh 共1 ⫹ x兲

2 3 This looks different from the answer we found in Example 4, but it is equivalent because

the third term can be rewritten using the identity

This is Equation 3.11.3.

arcsinh x 苷 ln ( x⫹ sx 2 ⫹1 )

Thus

s3 s3

3 冋 3 3 共1 ⫹ x兲 册

arcsinh

共1 ⫹ x兲 苷 ln 1 共1 ⫹ x兲 ⫹ s |

苷 ln

[ 1⫹x⫹ s共1 ⫹ x兲 2 ⫹3 ]

s3

苷 ln

⫹ ln ( x⫹ 1⫹ sx 2 ⫹ 2x ⫹ 4 )

s3

The resulting extra term 3 ⫺

2 ln ( 1 兾 s3 ) can be absorbed into the constant of integration.

Mathematica gives the answer

6 6 2 冉 3 冊 冉 s3 冊

sx ⫹ 2x ⫹ 4 ⫺ arcsinh

x 2 2 3 1⫹x

Mathematica combined the first two terms of Example 4 (and the Maple result) into a single term by factoring.

Derive gives the answer

6 sx 2 ⫹ 2x ⫹ 4 2 ⫹x⫹5 3 2 ln ( sx 2 ⫹ 2x ⫹ 4 ⫹x⫹1 )

The first term is like the first term in the Mathematica answer, and the second term is identical to the last term in Example 4.

y 2 x ⫹5 8 dx .

EXAMPLE 6 Use a CAS to evaluate

SOLUTION Maple and Mathematica give the same answer:

1 x 18 ⫹ 5 x 16 1750

18 2 ⫹ 50x 14 ⫹ 3 x 12 ⫹ 4375x 10 ⫹ 21875x 8 ⫹ 3 x 6 ⫹ 156250x 4 ⫹ 2 x 2

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS

It’s clear that both systems must have expanded 2 ⫹5 共x 8 兲 by the Binomial Theorem and

then integrated each term.

If we integrate by hand instead, using the substitution u苷x 2 ⫹5 , we get

Derive and the TI-89/92 also give this answer. 1 8 y

x 2 共x ⫹5 兲 dx 苷

18 共x ⫹5 9 兲 ⫹C

For most purposes, this is a more convenient form of the answer.

EXAMPLE 7 Use a CAS to find y sin 5 x cos 2 x dx .

SOLUTION In Example 2 in Section 7.2 we found that

1 5 1 2 y 1 sin x cos 2 x dx 苷 ⫺ 5 3 7 cos 3 x⫹ 5 cos x⫺ 7 cos x⫹C

Derive and Maple report the answer

7 sin x cos x⫺

35 sin x cos x⫺ 105 cos x

whereas Mathematica produces

64 cos x ⫺ 192 cos 3x ⫹ 320 cos 5x ⫺ 448 cos 7x We suspect that there are trigonometric identities which show these three answers are

equivalent. Indeed, if we ask Derive, Maple, and Mathematica to simplify their expres- sions using trigonometric identities, they ultimately produce the same form of the answer as in Equation 1.

7.6 EXERCISES

1– 4 Use the indicated entry in the Table of Integrals on the

11. 0 t 2 e y ⫺t

12. x y 2 csch 共x 3 ⫹1 兲 dx

dt

Reference Pages to evaluate the integral.

tan s7 ⫺ 2x 3 2 3x 共1兾z兲

1. y 2 dx ; entry 33 2. y dx ; entry 55

13. y z 2 dz

14. y sin sx dx

s3 ⫺ 2x

3. y sec 共␲x兲 dx ; entry 71

4. y e sin 3␪ d␪ ; entry 98

15. e 2x arctan 共e x 兲 dx

y x sin 共x 兲 cos共3x 兲 dx

2 17. dx y y s6 ⫹ 4y ⫺ 4y dy 18. y

2x 3 ⫺ 3x 2 5–30 Use the Table of Integrals on Reference Pages 6 –10 to evalu-

sin 2␪ ate the integral.

19. y 2 sin x cos x ln 共sin x兲 dx

20. y d␪

s5 ⫺ sin ␪

5. 2x cos y ⫺1 0 x dx

6. y 2

x 2 dx

22. x 3 2 ⫺x y 4 dx

y 0 s4x

3⫺e 2x

s4x 2 ⫺7 21. dx

ln ( 1⫹ )

24. sx 6 y sin 2x dx

7. 8. y sx tan 3 共␲x兲 dx y dx

23. sec y 5 x dx

dx 2 ⫺3 2 1

y x s4x 2 ⫹9

9. 2 10. s2y

s4 ⫹ 共ln x兲

y y 2 25. y dx

dy

26. y x e

4 ⫺x

0 dx

CHAPTER 7 TECHNIQUES OF INTEGRATION

27. se 2x ⫺ 1 dx

28. y e t sin

y CAS 共␣t ⫺ 3兲 dt 43. (a) Use the table of integrals to evaluate F 共x兲 苷 x f 共x兲 dx ,

where

x 4 dx

29. y ␪ tan

sx 10 ⫺2 30. y 2 ␪ d␪

What is the domain of and ? f F 31. Find the volume of the solid obtained when the region under 2 (b) Use a CAS to evaluate F 共x兲 the curve . What is the domain of the y苷x s4 ⫺ x , 0艋x艋2 , is rotated about the

function that the CAS produces? Is there a discrepancy F y -axis.

between this domain and the domain of the function F

32. The region under the curve y苷 tan 2 x from 0 to ␲ 兾4 is

that you found in part (a)?

rotated about the -axis. Find the volume of the resulting x solid. CAS 44. Computer algebra systems sometimes need a helping hand

from human beings. Try to evaluate 33. Verify Formula 53 in the Table of Integrals (a) by differentia-

y 共1 ⫹ ln x兲 s1 ⫹ 共x ln x兲 2 dx

tion and (b) by using the substitution t 苷 a ⫹ bu .

34. Verify Formula 31 (a) by differentiation and (b) by substi- tuting . u苷a sin ␪

with a computer algebra system. If it doesn’t return an

answer, make a substitution that changes the integral into one 35– 42 Use a computer algebra system to evaluate the integral.

CAS

that the CAS can evaluate.

Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent.

CAS 45– 48 Use a CAS to find an antiderivative F of such f

4 5 that F 共0兲 苷 0 . Graph and f F 35. and locate approximately the y sec x dx 36. y csc x dx

x -coordinates of the extreme points and inflection points of . F

dx

x 2 ⫺1

37. 2 y f x sx 2 ⫹4 dx 38. 45. 共x兲 苷 4 y 2

e x 共3e x ⫹2 兲

x ⫹x ⫹1 46. f 共x兲 苷 xe ⫺x sin x, ⫺5 艋 x 艋 5

39. y x

s1 ⫹ 2x dx 4 40. y sin x dx

47. f 共x兲 苷 sin 4 x cos 6 x , 0艋x艋␲

41. y tan x dx

42. y 3 dx

PATTERNS IN INTEGRALS

PROJECT In this project a computer algebra system is used to investigate indefinite integrals of families of

functions. By observing the patterns that occur in the integrals of several members of the family, you will first guess, and then prove, a general formula for the integral of any member of the family.

1. (a) Use a computer algebra system to evaluate the following integrals.

(i) y dx

(ii)

y 共x ⫹ 1兲共x ⫹ 5兲

dx

共x ⫹ 2兲共x ⫹ 3兲

(iii) y dx

y 共x ⫹ 2兲 2

(iv)

dx

共x ⫹ 2兲共x ⫺ 5兲

(b) Based on the pattern of your responses in part (a), guess the value of the integral

共x ⫹ a兲共x ⫹ b兲 dx if . a苷b What if ? a苷b

(c) Check your guess by asking your CAS to evaluate the integral in part (b). Then prove it

using partial fractions.

SECTION 7.7 APPROXIMATE INTEGRATION

2. (a) Use a computer algebra system to evaluate the following integrals.

(i) y sin x cos 2x dx

(ii) y sin 3x cos 7x dx

(iii) y sin 8x cos 3x dx

(b) Based on the pattern of your responses in part (a), guess the value of the integral

y sin ax cos bx dx

(c) Check your guess with a CAS. Then prove it using the techniques of Section 7.2. For what values of and is it valid? a b

3. (a) Use a computer algebra system to evaluate the following integrals. (i)

x y 2 y y ln x dx

(iv) x 3 ln x dx

x y 7 y ln x dx

(v)

(b) Based on the pattern of your responses in part (a), guess the value of

x y n ln x dx

(c) Use integration by parts to prove the conjecture that you made in part (b). For what values of is it valid? n

4. (a) Use a computer algebra system to evaluate the following integrals.

(i) y xe x dx

x 2 e y x dx

x 3 e y x dx

(ii)

(iii)

(iv) x 4 e x dx

x 5 y x y e dx

(v)

(b) Based on the pattern of your responses in part (a), guess the value of x 6 e x dx x . Then

use your CAS to check your guess. (c) Based on the patterns in parts (a) and (b), make a conjecture as to the value of the integral

y n x e x dx

when is a positive integer. n (d) Use mathematical induction to prove the conjecture you made in part (c).

7.7 APPROXIMATE INTEGRATION There are two situations in which it is impossible to find the exact value of a definite

integral.

a f 共x兲 dx using the Fundamental Theorem of Calculus we need to know an antiderivative of . Sometimes, f

The first situation arises from the fact that in order to evaluate b x

however, it is difficult, or even impossible, to find an antiderivative (see Section 7.5). For example, it is impossible to evaluate the following integrals exactly:

s1 ⫹ x y 3 dx ⫺1

0 e dx

CHAPTER 7 TECHNIQUES OF INTEGRATION

The second situation arises when the function is determined from a scientific experi- ment through instrument readings or collected data. There may be no formula for the func- tion (see Example 5).

In both cases we need to find approximate values of definite integrals. We already know one such method. Recall that the definite integral is defined as a limit of Riemann sums, so any Riemann sum could be used as an approximation to the integral: If we divide 关a, b兴

into subintervals of equal length n ⌬x 苷 共b ⫺ a兲兾n , then we have

a f 共x兲 dx ⬇

i苷 兺 1

f 共x *兲 ⌬x i

where x * i is any point in the th subinterval i 关x i⫺ 1 ,x i 兴 . If x i * is chosen to be the left endpoint

of the interval, then x i *苷x i⫺ 1 and we have

b 0 n x¸ ⁄ ¤ ‹ x¢

1 f 共x兲 dx ⬇ L n y 苷

a 兺 f 共x i⫺ 1 i苷 兲 ⌬x 1

(a) Left endpoint approximation

If f 共x兲 艌 0 , then the integral represents an area and (1) represents an approximation of this

area by the rectangles shown in Figure 1(a). If we choose x i * to be the right endpoint, then x *苷x i i and we have

a f 共x兲 dx ⬇ R n 苷 兺 f 共x i i苷 兲 ⌬x 1

[See Figure 1(b).] The approximations L n and R n defined by Equations 1 and 2 are called

0 x¸ ⁄ ¤ ‹ x¢

x the left endpoint approximation and right endpoint approximation, respectively. In Section 5.2 we also considered the case where x i * is chosen to be the midpoint x i of

(b) Right endpoint approximation

the subinterval 关x i⫺ 1 ,x i 兴 . Figure 1(c) shows the midpoint approximation M n , which appears

to be better than either L n or R n .

MIDPOINT RULE

(c) Midpoint approximation

and 1 x

i 苷 2 共x i⫺ 1 ⫹x i 兲 苷 midpoint of 关x i⫺ 1 ,x i 兴

FIGURE 1

Another approximation, called the Trapezoidal Rule, results from averaging the approx- imations in Equations 1 and 2:

2 冋 i苷 兺 1 i苷 兺 1 册 2 冋 i苷 兺 1 册

苷 ⌬x y n

( f 共x i⫺ 1 兲 ⫹ f 共x i 兲 )

⌬x

[ ( f 共x 0 兲 ⫹ f 共x 1 兲 ) ⫹ ( f 共x 1 兲 ⫹ f 共x 2 兲 ) ⫹⭈⭈⭈⫹ ( f 共x n⫺ 1 兲 ⫹ f 共x n 兲 ) ]

2 ⌬x

关 f 共x 0 兲 ⫹ 2 f 共x 1 兲 ⫹ 2 f 共x 2 兲 ⫹ ⭈ ⭈ ⭈ ⫹ 2 f 共x n⫺ 1 兲 ⫹ f 共x n 兲兴

SECTION 7.7 APPROXIMATE INTEGRATION

TRAPEZOIDAL RULE

b 苷 y ⌬x

a f 共x兲 dx ⬇ T n

关 f 共x 0 兲 ⫹ 2 f 共x 1 兲 ⫹ 2 f 共x 2 兲 ⫹ ⭈ ⭈ ⭈ ⫹ 2 f 共x n⫺ 1 兲 ⫹ f 共x n 兲兴

where and ⌬x 苷 共b ⫺ a兲兾n x i 苷 a⫹i ⌬x .

The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates the case f 共x兲 艌 0 . The area of the trapezoid that lies above the th subinterval is i

and if we add the areas of all these trapezoids, we get the right side of the Trapezoidal Trapezoidal approximation

Rule.

EXAMPLE 1 Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n苷 5 to

approximate the integral 2 x

1 共1兾x兲 dx .

y= 1 x SOLUTION

(a) With n苷

5, a 苷 1 , and b苷 2 , we have ⌬x 苷 共2 ⫺ 1兲兾5 苷 0.2 , and so the Trape-

zoidal Rule gives

2 1 y 0.2

1 dx ⬇T 5 苷

关 f 共1兲 ⫹ 2 f 共1.2兲 ⫹ 2 f 共1.4兲 ⫹ 2 f 共1.6兲 ⫹ 2 f 共1.8兲 ⫹ f 共2兲兴

This approximation is illustrated in Figure 3.

(b) The midpoints of the five subintervals are 1.1 , 1.3 , 1.5 , 1.7 , and 1.9 , so the Midpoint

Rule gives

y= 1 y 1 x dx ⬇ ⌬x 关 f 共1.1兲 ⫹ f 共1.3兲 ⫹ f 共1.5兲 ⫹ f 共1.7兲 ⫹ f 共1.9兲兴 x

This approximation is illustrated in Figure 4.

In Example 1 we deliberately chose an integral whose value can be computed explicitly

1 2 so that we can see how accurate the Trapezoidal and Midpoint Rules are. By the Funda-

FIGURE 4

mental Theorem of Calculus,

1 dx 苷 ln x ] 1 苷 x ln 2 苷 0.693147 . . .

a f 共x兲 dx 苷 approximation ⫹ error

The error in using an approximation is defined to be the amount that needs to be added to the approximation to make it exact. From the values in Example 1 we see that the errors

in the Trapezoidal and Midpoint Rule approximations for n苷 5 are

E T ⬇ ⫺0.002488

and

E M ⬇ 0.001239

CHAPTER 7 TECHNIQUES OF INTEGRATION

In general, we have

T 苷 y a f 共x兲 dx ⫺ T n and E M 苷 y a f 共x兲 dx ⫺ M n

TEC Module 5.2 / 7.7 allows you to The following tables show the results of calculations similar to those in Example 1, but

compare approximation methods.

for n苷

5, 10 , and 20 and for the left and right endpoint approximations as well as the

Trapezoidal and Midpoint Rules.