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49. y x s4x1 1 dx

1

51. y x s4x 2 1 dx

53. yx 2 sinh mx dx

dx

55. y x xsx

57.yx s3 x c dx

59.ycos x cos3 sin x dx

61.ysx esx dx

sin 2x

63. y 1 cos4 x dx

1

65. y sx 1 sx dx

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS |||| 489

50.

y

1

 

dx

x 2 s

 

 

4x 1

52.

y

dx

 

 

 

x x 4 1

54.

y x sin x 2 dx

56.

y

 

 

 

 

 

 

dx

 

s

x

xs

x

 

 

 

 

 

 

58.

y

 

 

 

x ln x

 

 

 

 

 

 

dx

 

s

 

 

 

 

 

x 2 1

60.

y

 

 

 

 

 

 

 

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x 2s

 

 

 

 

 

 

 

 

 

 

 

4x 2 1

 

 

62.

y

1

 

 

 

 

dx

x

 

3

x

 

 

 

 

 

 

 

 

 

 

s

 

 

 

3

 

ln tan x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

64.

y 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dx

sin x cos x

 

3

 

u 3 1

66.

y2

 

 

 

 

 

 

 

 

 

 

 

du

 

u 3 u 2

67.

y1s

3

s

1 x 2

 

68.

y

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dx

 

dx

 

 

 

x 2

 

 

 

 

1 2e x e x

 

 

 

 

 

 

2x

 

 

 

 

 

 

 

 

 

ln x

1

 

 

 

 

 

 

 

 

 

y

e

dx

70.

y

dx

 

69.

1 e x

x 2

 

 

 

 

 

 

71.

y

 

x arcsin x

72.

y

4 x 10 x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dx

 

 

 

 

 

dx

 

 

s

 

 

 

 

 

 

2 x

 

 

 

 

1 x 2

 

 

 

 

73.

y

 

 

 

 

 

 

 

 

1

 

 

 

 

 

dx

74.

y

 

 

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x 2 x 2 4

s

 

(2 s

 

)4

x

x

 

 

 

 

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

75.

y

 

 

xe

dx

76.

y x 2 bx sin 2x dx

 

s

 

 

 

 

1 e x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sec x cos 2x

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

 

 

77.

y

 

s

 

dx

78.

y

 

 

 

 

 

 

dx

 

1 x 3

sin x sec x

79.

yx sin2 x cos x dx

80.

y

 

sin x cos x

 

 

 

 

 

 

 

 

 

dx

sin4 x cos4 x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

81.

The functions y e x2

and y x 2e x2

don’t have elementary

 

 

antiderivatives, but y 2x 2 1 e x2 does. Evaluate

 

 

x 2x 2 1 e x2 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 antiderivatives of functions like ex2 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 relatively 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 Mathematical 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 arctan x, y 0, and x 1 is rotated about the y-axis. Find the volume of the resulting solid.

SOLUTION Using the method of cylindrical shells, we see that the volume is

V y1 2 x arctan x dx

0

490 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION

N The Table of Integrals appears on Reference Pages 6 –10 at the back of the book.

In the section of the Table of Integrals titled Inverse Trigonometric Forms we locate Formula 92:

 

yu tan 1u du

u2

1

 

tan 1u

 

u

 

 

 

 

 

 

 

 

 

C

 

 

 

 

2

 

 

2

 

 

Thus the volume is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

x2

1

 

 

 

 

 

 

x

1

 

V 2 y0 x tan 1x dx 2

 

 

 

tan 1x

 

0

 

 

 

2

2

 

[ x2 1 tan 1x x]10 2 tan 1 1 1

 

 

2 4 1 21

 

2

 

 

 

 

 

 

 

 

 

M

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

EXAMPLE 2 Use the Table of Integrals to find y

 

 

x

 

 

 

 

dx.

 

 

V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

4x2

 

 

 

 

 

 

 

 

 

 

 

s

 

 

 

 

 

 

 

SOLUTION If we look at the section of the table titled Forms involving sa2 u2 , we see that the closest entry is number 34:

y

 

u2

 

u

 

 

 

a2

sin 1

u

C

 

du

sa2 u2

 

s

 

 

2

2

a

a2 u2

This is not exactly what we have, but we will be able to use it if we first make the substitution u 2x:

 

 

y

 

 

x2

 

 

 

dx y

 

 

u 2 2

 

 

du

 

 

 

1

y

 

 

u2

 

 

 

du

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

s

 

2

 

8

s

 

 

 

 

 

 

 

 

 

 

5 4x2

 

5 u2

5 u2

 

 

 

 

 

 

 

 

 

 

s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Then we use Formula 34 with a2 5 (so a s

 

):

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

y

 

x2

1

 

y

 

 

 

 

 

u2

 

 

 

 

1

 

 

 

u

 

 

 

 

 

 

 

 

 

 

 

5

 

 

 

u

C

 

 

 

 

 

 

 

 

 

 

 

 

s5

 

 

 

 

 

sin 1

 

 

 

 

dx

 

 

 

 

 

 

 

 

 

 

du

 

 

 

 

u2

 

 

 

 

 

8

 

 

 

 

 

 

 

 

 

8

 

2

2

 

 

 

 

5 4x2

 

 

s

5 u2

 

 

 

 

s

5

 

 

s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

5

 

sin 1

 

 

 

 

2x

 

C

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5 4x2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8 s

 

 

 

 

 

16

 

 

 

 

 

 

 

s5

 

 

 

 

 

 

 

 

 

 

 

 

EXAMPLE 3 Use the Table of Integrals to find yx3 sin x dx.

85.yu n cos u du

u n sin u n yu n 1 sin u du

SOLUTION If we look in the section called Trigonometric Forms, we see that none of the entries explicitly includes a u3 factor. However, we can use the reduction formula in entry 84 with n 3:

yx3 sin x dx x3 cos x 3 yx2 cos x dx

We now need to evaluate xx2 cos x dx. We can use the reduction formula in entry 85 with n 2, followed by entry 82:

yx2 cos x dx x2 sin x 2 yx sin x dx

x2 sin x 2 sin x x cos x K

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS |||| 491

Combining these calculations, we get

yx3 sin x dx x3 cos x 3x2 sin x 6x cos x 6 sin x C

where C 3K.

M

V EXAMPLE 4 Use the Table of Integrals to find yxsx2 2x 4 dx.

SOLUTION Since the table gives forms involving sa2 x2 , sa2 x2 , and sx2 a2 , but not sax2 bx c , we first complete the square:

x2 2x 4 x 1 2 3

If we make the substitution u x 1 (so x u 1), the integrand will involve the pattern sa2 u2 :

yxsx2 2x 4 dx y u 1 su2 3 du

21. ysa 2 u 2 du u sa 2 u 2

2

a 2 ln(u sa 2 u 2 ) C

2

yusu2 3 du ysu2 3 du

The first integral is evaluated using the substitution t u2 3:

yusu2 3 du 12 yst dt 12 23 t3 2 13 u2 3 3 2

For the second integral we use Formula 21 with a s3 :

ysu2 3 du u su2 3 32 ln(u su2 3 )

2

Thus

yxsx2 2x 4 dx

13 x2 2x 4 3 2 x 1 sx2 2x 4 32 ln(x 1 sx2 2x 4 ) C

2

M

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 patterns. And just as we used substitutions in conjunction with tables, a CAS can perform substitutions 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 1 3x 2 . Using the substitution u 3x 2, an easy calculation by hand gives

y 3x 1 2 dx 13 ln 3x 2 C

492 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION

whereas Derive, Mathematica, and Maple all return the answer

13 ln 3x 2

The first thing to notice is that computer algebra systems omit the constant of integration. 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 x greater than 23 . But if we are interested in other values of x, then we need to insert the absolute value symbol.

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 yxsx2 2x 4 dx.

SOLUTION Maple responds with the answer

31 x2 2x 4 3 2 41 2x 2 s

 

 

3

arcsinh

s3

1 x

x2 2x 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

arcsinh x ln(x s

 

 

 

)

 

 

 

 

 

 

 

 

 

 

 

 

N This is Equation 3.11.3.

 

 

 

 

 

 

 

 

 

x2 1

 

 

 

 

 

 

 

 

 

 

 

 

Thus

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

s

 

 

 

 

 

 

 

 

 

 

 

 

s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

x

1

 

 

 

 

2

 

1

arcsinh

3

 

 

 

x ln 3 1

s|3 1 x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

1

 

[1 x s

 

 

 

 

]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 x 2 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

s3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ln

 

1

 

 

 

ln(x 1 s

 

 

 

 

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x2 2x 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

s3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The resulting extra term 23 ln(1 s

 

) can be absorbed into the constant of integration.

3

Mathematica gives the answer

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

 

 

 

 

 

x

 

 

x2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

1 x

 

 

 

 

 

 

 

 

 

x2

2x

4

arcsinh

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6

 

6

 

 

3 s

2

 

s3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Mathematica combined the first two terms of Example 4 (and the Maple result) into a

single term by factoring.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Derive gives the answer

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

61 s

x2 2x 4

2x2 x 5 23 ln(s

x2 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.

M

EXAMPLE 6 Use a CAS to evaluate yx x2 5 8 dx.

 

SOLUTION Maple and Mathematica give the same answer:

 

 

1

x18 25 x16 50x14 17503 x12 4375x10 21875x8 2187503 x6 156250x4 3906252

x2

 

18

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS |||| 493

It’s clear that both systems must have expanded x2 5 8 by the Binomial Theorem and then integrated each term.

If we integrate by hand instead, using the substitution u x2 5, we get

N Derive and the TI-89/92 also give this answer.

yx x2 5 8 dx

1

x2 5 9 C

 

18

 

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

M

EXAMPLE 7 Use a CAS to find ysin5x cos2x dx.

 

SOLUTION

In Example 2 in Section 7.2 we found that

 

1

ysin5x cos2x dx 31 cos3x 52 cos5x 71 cos7x C

 

Derive and Maple report the answer

 

 

71 sin4x cos3x

4

sin2x cos3x

8

cos3x

 

 

35

105

 

whereas Mathematica produces

645 cos x 1921 cos 3x 3203 cos 5x 4481 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 expressions using trigonometric identities, they ultimately produce the same form of the answer as in Equation 1. M

7.6E X E R C I S E S

1– 4 Use the indicated entry in the Table of Integrals on the Reference Pages to evaluate the integral.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1.

y

s7

2x2

 

dx; entry 33

2.

y

 

 

3x

 

 

dx;

entry 55

x2

 

 

s

 

 

 

 

3

 

2x

 

3.

ysec3 x dx; entry 71

4.

ye2 sin 3 d ;

entry 98

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5–30 Use the Table of Integrals on Reference Pages 6–10 to evaluate the integral.

 

1

 

 

 

 

3

 

 

1

 

 

 

 

 

 

 

5.

y0

2x cos 1x dx

6.

y2

 

 

 

 

dx

 

x2 s

 

 

 

4x2 7

 

 

 

 

 

 

 

 

 

ln(1

 

)

 

 

 

 

 

 

 

 

 

 

 

 

x

7.

ytan3 x dx

8.

y

 

 

 

 

 

s

 

 

 

 

dx

 

 

 

s

 

 

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

9.

y

 

 

10.

y

s

2y2 3

 

 

dy

 

x2s

 

 

 

 

y2

 

 

 

 

4x2 9

 

 

 

 

 

 

11.

y01 t2e t dt

 

 

 

 

 

 

 

3

 

 

 

 

 

 

13.

y

tan 1 z

dz

z 2

15.

ye2x arctan ex dx

 

yys

 

dy

6 4y 4y2

17.

 

 

ysin2x cos x ln sin x dx

19.

 

 

y

e

x

 

 

 

 

21.

 

dx

3

e2x

23.

ysec5x dx

 

 

 

 

 

 

s

 

 

 

 

 

 

25.

y

4 ln x 2

dx

 

 

 

x

 

 

12. yx2 csch x3 1 dx

14.ysin 1sx dx

16.yx sin x2 cos 3x2 dx

18.

y

dx

 

 

 

2x3 3x2

 

 

20.

y

 

 

sin 2

d

 

 

 

 

 

 

s

 

 

 

 

5 sin

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

22.

y0

x3s4x2 x4 dx

24.

ysin6 2x dx

 

 

 

 

1

 

 

 

 

 

 

 

 

26.

y0

x4e x dx

 

 

 

 

 

494 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION

27.

ys

e2x 1

dx

28.

yet sin t 3 dt

 

 

4

 

 

 

 

 

sec

2

 

tan

2

 

 

 

29.

y

 

 

x dx

30.

y

 

 

 

d

 

 

 

 

 

 

 

s

 

 

 

 

s

 

 

x10 2

 

9 tan2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

31.Find the volume of the solid obtained when the region under the curve y xs4 x2 , 0 x 2, is rotated about the y-axis.

32. The region under the curve y tan2x from 0 to 4 is rotated about the x-axis. Find the volume of the resulting solid.

33.Verify Formula 53 in the Table of Integrals (a) by differentiation and (b) by using the substitution t a bu.

34.Verify Formula 31 (a) by differentiation and (b) by substituting u a sin .

CAS 35– 42 Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent.

35.

ysec4x dx

36.

ycsc5x dx

 

 

 

 

 

 

 

 

 

 

 

dx

 

 

37.

yx2sx2 4 dx

38.

y

 

 

 

ex 3ex 2

39.

yxs

 

 

dx

40.

ysin4x dx

 

 

1 2x

 

 

41.

ytan5x dx

42.

y

 

1

 

 

 

dx

s

1

3

 

 

 

 

 

 

 

 

 

 

 

 

 

sx

CAS 43. (a) Use the table of integrals to evaluate F x x f x dx, where

f x

 

 

1

 

 

 

 

 

xs1

x2

 

What is the domain of f and F?

(b)Use a CAS to evaluate F x . What is the domain of the function F that the CAS produces? Is there a discrepancy between this domain and the domain of the function F that you found in part (a)?

CAS 44. Computer algebra systems sometimes need a helping hand from human beings. Try to evaluate

y 1 ln x s1 x ln x 2 dx

with a computer algebra system. If it doesn’t return an answer, make a substitution that changes the integral into one that the CAS can evaluate.

CAS 45– 48 Use a CAS to find an antiderivative F of f such that F 0 0. Graph f and F and locate approximately the

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

45.

f x

 

x2 1

 

x4 x2 1

 

46.

f x xe x sin x,

5 x 5

47.

f x sin4x cos6x,

0 x

48.

f x

x3 x

 

x6 1

 

 

 

 

 

 

 

 

 

D I S C O V E R Y

CAS

PATTERNS IN INTEGRALS

 

 

 

 

 

 

P R O J E C T

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

1

 

dx

(ii)

y

1

 

dx

 

 

 

 

 

 

 

 

x 2 x 3

x 1 x 5

 

 

 

(iii)

y

1

 

dx

(iv)

y

1

dx

 

 

 

x 2 x 5

 

x 2 2

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

1

y x a x b dx

if a b. What if a b?

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

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