Calculus III Unit00 Homework
Section 7.1 pick 10 (review)
Section 7.2 pick 10 (new)
Section 7.3 pick 10 (new)
Section 7.4 pick 10 (Rationalizing substitution only - new)
Section 7.5 pick 15 (use them all)
The techniques of this
chapter enable us to find
the height of a rocket a
minute after liftoff and to
compute the escape
velocity of the rocket.
Techniques of Integration
Because of the Fundamental Theorem of Calculus, we can
integrate a function if we know an antiderivative, that is,
an indefinite integral. We summarize here the most important integrals that we have learned so far.
x n1
C
n1
yx
n
dx ye
x
dx e x C
n 1
y
1
dx ln x C
x
ya
x
dx ax
C
ln a
y sin x dx cos x C
y cos x dx sin x C
y sec x dx tan x C
y csc x dx cot x C
y sec x tan x dx sec x C
y csc x cot x dx csc x C
y sinh x dx cosh x C
y cosh x dx sinh x C
y tan x dx ln sec x C
y cot x dx ln sin x C
2
yx
2
2
1
1
x
dx tan1
2
a
a
a
C
y sa
2
1
x
dx sin1
2
x
a
C
In this chapter we develop techniques for using these basic integration formulas
to obtain indefinite integrals of more complicated functions. We learned the most
important method of integration, the Substitution Rule, in Section 5.5. The other general technique, integration by parts, is presented in Section 7.1. Then we learn
methods that are special to particular classes of functions such as trigonometric functions and rational functions.
Integration is not as straightforward as differentiation; there are no rules that
absolutely guarantee obtaining an indefinite integral of a function. Therefore, in
Section 7.5 we discuss a strategy for integration.
|||| 7.1
Integration by Parts
Every differentiation rule has a corresponding integration rule. For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation. The rule that
corresponds to the Product Rule for differentiation is called the rule for integration by
parts.
The Product Rule states that if f and t are differentiable functions, then
d
f xtx f xtx txf x
dx
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CHAPTER 7 TECHNIQUES OF INTEGRATION
In the notation for indefinite integrals this equation becomes
y f xtx txf x dx f xtx
y f xtx dx y txf x dx f xtx
or
We can rearrange this equation as
y f xtx dx f xtx y txf x dx
1
Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u f x and v tx. Then the differentials are
du f x dx and dv tx dx, so, by the Substitution Rule, the formula for integration
by parts becomes
y u dv uv y v du
2
EXAMPLE 1 Find
y x sin x dx.
SOLUTION USING FORMULA 1 Suppose we choose f x x and tx sin x. Then f x 1
and tx cos x. (For t we can choose any antiderivative of t.) Thus, using Formula
1, we have
y x sin x dx f xtx y txf x dx
xcos x y cos x dx
x cos x y cos x dx
x cos x sin x C
It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as
expected.
SOLUTION USING FORMULA 2 Let
|||| It is helpful to use the pattern:
u
dv du v
Then
and so
ux
dv sin x dx
du dx
v cos x
u
d√
u
√
√
du
y x sin x dx y x sin x dx x cos x y cos x dx
x cos x y cos x dx
x cos x sin x C
SECTION 7.1 INTEGRATION BY PARTS
NOTE
❙❙❙❙
477
Our aim in using integration by parts is to obtain a simpler integral than the one
we started with. Thus, in Example 1 we started with x x sin x dx and expressed it in terms
of the simpler integral x cos x dx. If we had chosen u sin x and dv x dx, then
du cos x dx and v x 22, so integration by parts gives
■
y x sin x dx sin x
x2
1
2
2
yx
2
cos x dx
Although this is true, x x 2 cos x dx is a more difficult integral than the one we started with.
In general, when deciding on a choice for u and dv, we usually try to choose u f x to
be a function that becomes simpler when differentiated (or at least not more complicated)
as long as dv tx dx can be readily integrated to give v.
EXAMPLE 2 Evaluate
y ln x dx.
SOLUTION Here we don’t have much choice for u and dv. Let
u ln x
du Then
dv dx
1
dx
x
vx
Integrating by parts, we get
y ln x dx x ln x y x
dx
x
|||| It’s customary to write x 1 dx as x dx.
x ln x y dx
|||| Check the answer by differentiating it.
x ln x x C
Integration by parts is effective in this example because the derivative of the function
f x ln x is simpler than f .
EXAMPLE 3 Find
2 t
y t e dt.
SOLUTION Notice that t 2 becomes simpler when differentiated (whereas e t is unchanged
when differentiated or integrated), so we choose
u t2
Then
dv e t dt
du 2t dt
v et
Integration by parts gives
3
y t e dt t e
2 t
2 t
2 y te t dt
The integral that we obtained, x te t dt, is simpler than the original integral but is still not
obvious. Therefore, we use integration by parts a second time, this time with u t and
478
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CHAPTER 7 TECHNIQUES OF INTEGRATION
dv e t dt. Then du dt, v e t, and
y te dt te
t
t
y e t dt
te t e t C
Putting this in Equation 3, we get
yt
e dt t 2 e t 2 y te t dt
2 t
t 2 e t 2te t e t C t 2 e t 2te t 2e t C1
EXAMPLE 4 Evaluate
|||| An easier method, using complex numbers,
is given in Exercise 48 in Appendix G.
ye
x
where C1 2C
sin x dx.
SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try choosing
u e x and dv sin x dx anyway. Then du e x dx and v cos x, so integration by
parts gives
ye
4
x
sin x dx e x cos x y e x cos x dx
The integral that we have obtained, x e x cos x dx, is no simpler than the original one, but
at least it’s no more difficult. Having had success in the preceding example integrating
by parts twice, we persevere and integrate by parts again. This time we use u e x and
dv cos x dx. Then du e x dx, v sin x, and
ye
5
x
cos x dx e x sin x y e x sin x dx
At first glance, it appears as if we have accomplished nothing because we have arrived at
|||| Figure 1 illustrates Example 4 by
showing the graphs of f x e x sin x and
1
Fx 2 e xsin x cos x. As a visual check
on our work, notice that f x 0 when F has
a maximum or minimum.
we get
ye
x
sin x dx e x cos x e x sin x y e x sin x dx
This can be regarded as an equation to be solved for the unknown integral. Adding
12
x e x sin x dx to both sides, we obtain
F
f
_3
x e x sin x dx, which is where we started. However, if we put Equation 5 into Equation 4
2 y e x sin x dx e x cos x e x sin x
6
Dividing by 2 and adding the constant of integration, we get
_4
FIGURE 1
ye
x
sin x dx 12 e x sin x cos x C
If we combine the formula for integration by parts with Part 2 of the Fundamental
Theorem of Calculus, we can evaluate definite integrals by parts. Evaluating both sides of
Formula 1 between a and b, assuming f and t are continuous, and using the Fundamental
SECTION 7.1 INTEGRATION BY PARTS
❙❙❙❙
479
Theorem, we obtain
y
6
b
a
EXAMPLE 5 Calculate
y
1
0
b
b
]
f xtx dx f xtx a y txf x dx
a
tan1x dx.
SOLUTION Let
u tan1x
du Then
dv dx
dx
1 x2
vx
So Formula 6 gives
y
1
0
1
]
tan1x dx x tan1x 0 y
1
0
x
dx
1 x2
1 tan1 1 0 tan1 0 y
1
0
1
|||| Since tan x 0 for x 0, the integral in
Example 5 can be interpreted as the area of the
region shown in Figure 2.
1
x
y
2 dx
0 1 x
4
To evaluate this integral we use the substitution t 1 x 2 (since u has another meaning
in this example). Then dt 2x dx, so x dx dt2. When x 0, t 1; when x 1,
t 2; so
y
y=tan–!x
y
0
1
x
dx
1 x2
1
0
x
x
1 2 dt
12 ln t
2 dx 2 y
1 t
1x
]
2
1
12 ln 2 ln 1 12 ln 2
FIGURE 2
Therefore
y
1
0
tan1x dx 1
x
ln 2
y
2 dx 0 1 x
4
4
2
EXAMPLE 6 Prove the reduction formula
|||| Equation 7 is called a reduction formula
because the exponent n has been reduced to
n 1 and n 2.
7
1
y sin x dx n cos x sin
n
n1
x
n1
n
y sin
n2
x dx
where n 2 is an integer.
SOLUTION Let
Then
u sin n1x
dv sin x dx
du n 1 sin n2x cos x dx
v cos x
so integration by parts gives
y sin x dx cos x sin
n
n1
x n 1 y sin n2x cos 2x dx
❙❙❙❙
480
CHAPTER 7 TECHNIQUES OF INTEGRATION
Since cos 2x 1 sin 2x, we have
y sin x dx cos x sin
n
n1
x n 1 y sin n2x dx n 1 y sin n x dx
As in Example 4, we solve this equation for the desired integral by taking the last term
on the right side to the left side. Thus, we have
n y sin n x dx cos x sin n1x n 1 y sin n2x dx
1
y sin x dx n cos x sin
n
or
n1
x
n1
n
y sin
n2
x dx
The reduction formula (7) is useful because by using it repeatedly we could eventually
express x sin n x dx in terms of x sin x dx (if n is odd) or x sin x0 dx x dx (if n is even).
|||| 7.1
Exercises
1–2
|||| Evaluate the integral using integration by parts with the
indicated choices of u and dv.
1.
2.
y x ln x dx;
y sec d ;
■
3–32
5.
u , dv sec 2 d
2
■
3.
u ln x, dv x dx
■
■
■
■
■
■
■
■
■
23.
y
25.
y
y
cos 1x dx
26.
y
27.
y cos x lnsin x dx
28.
y
29.
y cosln x dx
30.
y
31.
y
32.
y
12
0
■
y x cos 5x dx
y re
r2
dr
2
sin x dx
7.
yx
9.
y ln2x 1 dx
4.
6.
x
y xe
y arctan 4t dt
8.
y t sin 2t dt
10.
12.
yx
2
cos mx dx
2
yp
5
y t e dt
15.
y e sin 3 d
16.
ye
17.
y y sinh y dy
18.
y y cosh ay dy
21.
y
0
y
2
1
t sin 3t dt
ln x
dx
x2
20.
22.
■
r3
dr
s4 r 2
1
0
■
■
■
■
t
0
■
arctan1x dx
e s sint s ds
■
■
■
■
■
|||| First make a substitution and then use integration by parts
to evaluate the integral.
33.
y sin sx dx
35.
y
ln p dp
14.
2
s3
33–36
x dx
y ln x dx
2
x 4ln x2 dx
1
x csc 2x dx
x 5 x dx
1
1
y sin
1
0
dx
13.
19.
4
Evaluate the integral.
||||
2
24.
■
11.
y
dy
e 2y
1
0
s
s 2
3 cos 2 d
4
34.
y
36.
yx e
1
e sx dx
5 x2
dx
3 t
y
1
0
y
4
1
■
x
x 1e
■
■
■
■
■
■
■
■
■
■
; 37–40
|||| Evaluate the indefinite integral. Illustrate, and check that
your answer is reasonable, by graphing both the function and its
antiderivative (take C 0).
cos 2 d
2
■
37.
y x cos x dx
38.
yx
39.
y 2x 3e
40.
yx e
32
ln x dx
dx
x
dx
3 x2
dx
st ln t dt
■
■
■
■
■
■
■
■
■
■
■
■
SECTION 7.1 INTEGRATION BY PARTS
41. (a) Use the reduction formula in Example 6 to show that
y sin x dx 2
❙❙❙❙
481
; 53–54
|||| Use a graph to find approximate x-coordinates of the
points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves.
x
sin 2x
C
2
4
53. y x sin x,
(b) Use part (a) and the reduction formula to evaluate
x sin 4x dx.
y x 22
54. y arctan 3x,
■
■
■
■
y x2
■
■
■
■
■
■
■
■
42. (a) Prove the reduction formula
y cos n x dx 1
n1
cos n1x sin x n
n
55–58
|||| Use the method of cylindrical shells to find the volume
generated by rotating the region bounded by the given curves about
the specified axis.
y cos n2x dx
55. y cos x2, y 0, 0 x 1;
(b) Use part (a) to evaluate x cos 2x dx.
(c) Use parts (a) and (b) to evaluate x cos 4x dx.
56. y e , y e , x 1;
43. (a) Use the reduction formula in Example 6 to show that
y
2
0
n1
n
sin n x dx y
58. y e , x 0, y ;
x
2
sin n2x dx
0
■
y
2
0
45–48
||||
y ln x dx x ln x
46.
y x ne x dx x ne x n y x n1e x dx
n
about the x-axis
■
■
vt tt ve ln
■
■
■
■
■
m rt
m
where t is the acceleration due to gravity and t is not too
large. If t 9.8 ms 2, m 30,000 kg, r 160 kgs, and
ve 3000 ms, find the height of the rocket one minute
after liftoff.
Use integration by parts to prove the reduction formula.
45.
■
decreases with time. Suppose the initial mass of the rocket at
liftoff (including its fuel) is m, the fuel is consumed at rate r,
and the exhaust gases are ejected with constant velocity ve
(relative to the rocket). A model for the velocity of the rocket
at time t is given by the equation
1 3 5 2n 1 2 4 6 2n
2
sin 2nx dx n
■
60. A rocket accelerates by burning its onboard fuel, so its mass
44. Prove that, for even powers of sine,
y
■
about x 1
59. Find the average value of f x x 2 ln x on the interval 1, 3.
2 4 6 2n
sin 2n1x dx 3 5 7 2n 1
0
■
about the y-axis
about the y-axis
57. y ex, y 0, x 1, x 0;
where n 2 is an integer.
(b) Use part (a) to evaluate x02 sin 3x dx and x02 sin 5x dx.
(c) Use part (a) to show that, for odd powers of sine,
2
x
x
n y ln xn1 dx
61. A particle that moves along a straight line has velocity
vt t 2et meters per second after t seconds. How far will
it travel during the first t seconds?
47.
y x
2
2na 2
x x 2 a 2 n
2n 1
2n 1
48.
y sec n x dx ■
■
62. If f 0 t0 0 and f and t are continuous, show that
a 2 n dx
■
y x 2 a 2 n1 dx
y sec n2x dx
■
■
■
■
y
a
0
a
f xt x dx f ata f ata y f xtx dx
0
63. Suppose that f 1 2, f 4 7, f 1 5, f 4 3, and
tan x sec n2x
n2
n1
n1
■
(n 12 )
f is continuous. Find the value of x14 x f x dx.
n 1
64. (a) Use integration by parts to show that
■
■
■
■
y f x dx x f x y x f x dx
49. Use Exercise 45 to find x ln x3 dx.
50. Use Exercise 46 to find x x 4e x dx.
51–52
||||
51. y xe0.4x,
52. y 5 ln x,
■
■
(b) If f and t are inverse functions and f is continuous, prove
that
Find the area of the region bounded by the given curves.
■
y 0,
x5
y
y x ln x
■
■
b
a
■
■
■
■
■
■
■
f x dx bf b af a y
f b
f a
t y dy
[Hint: Use part (a) and make the substitution y f x.]
482
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
(c) In the case where f and t are positive functions and
b a 0, draw a diagram to give a geometric interpretation of part (b).
(d) Use part (b) to evaluate x1e ln x dx.
65. We arrived at Formula 6.3.2, V (c) Use parts (a) and (b) to show that
2n 1
I2n1
1
2n 2
I2n
and deduce that lim n l I2n1I2n 1.
(d) Use part (c) and Exercises 43 and 44 to show that
xab 2 x f x dx, by using
cylindrical shells, but now we can use integration by parts to
prove it using the slicing method of Section 6.2, at least for the
case where f is one-to-one and therefore has an inverse function t. Use the figure to show that
lim
nl
This formula is usually written as an infinite product:
d
V b 2d a 2c y t y 2 dy
2 2 4 4 6 6
2
1 3 3 5 5 7
c
Make the substitution y f x and then use integration by
parts on the resulting integral to prove that V xab 2 x f x dx.
y
x=g(y)
2 2 4 4 6 6
2n
2n
1 3 3 5 5 7
2n 1 2n 1
2
and is called the Wallis product.
(e) We construct rectangles as follows. Start with a square of
area 1 and attach rectangles of area 1 alternately beside or
on top of the previous rectangle (see the figure). Find the
limit of the ratios of width to height of these rectangles.
y=ƒ
d
x=b
c
x=a
0
66. Let In a
b
x
x02 sin n x dx.
(a) Show that I2n2 I2n1 I2n.
(b) Use Exercise 44 to show that
2n 1
I2n2
I2n
2n 2
|||| 7.2
Trigonometric Integrals
In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine.
EXAMPLE 1 Evaluate
3
y cos x dx.
SOLUTION Simply substituting u cos x isn’t helpful, since then du sin x dx. In order
to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of
sine would require an extra cos x factor. Thus, here we can separate one cosine factor
and convert the remaining cos2x factor to an expression involving sine using the identity
sin 2x cos 2x 1:
cos 3x cos 2x cos x 1 sin 2x cos x
We can then evaluate the integral by substituting u sin x, so du cos x dx and
y cos x dx y cos x cos x dx y 1 sin x cos x dx
3
2
y 1 u 2 du u 13 u 3 C
sin x 13 sin 3x C
2
SECTION 7.2 TRIGONOMETRIC INTEGRALS
❙❙❙❙
483
In general, we try to write an integrand involving powers of sine and cosine in a form
where we have only one sine factor (and the remainder of the expression in terms of
cosine) or only one cosine factor (and the remainder of the expression in terms of sine).
The identity sin 2x cos 2x 1 enables us to convert back and forth between even powers
of sine and cosine.
EXAMPLE 2 Find
5
2
y sin x cos x dx
SOLUTION We could convert cos 2x to 1 sin 2x, but we would be left with an expression in
terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and
rewrite the remaining sin 4x factor in terms of cos x :
sin 5x cos 2x sin2x2 cos 2x sin x 1 cos 2x2 cos 2x sin x
|||| Figure 1 shows the graphs of the integrand
sin 5x cos 2x in Example 2 and its indefinite integral (with C 0). Which is which?
Substituting u cos x, we have du sin x dx and so
y sin x cos x dx y sin x
5
2
2
2
cos 2x sin x dx
0.2
y 1 cos 2x2 cos 2x sin x dx
_π
π
y 1 u 2 2 u 2 du y u 2 2u 4 u 6 du
_0.2
u3
u5
u7
2
3
5
7
C
13 cos 3x 25 cos 5x 17 cos 7x C
FIGURE 1
In the preceding examples, an odd power of sine or cosine enabled us to separate a
single factor and convert the remaining even power. If the integrand contains even powers
of both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix D):
sin 2x 12 1 cos 2x
|||| Example 3 shows that the area of the region
shown in Figure 2 is 2.
EXAMPLE 3 Evaluate
y
0
and
cos 2x 12 1 cos 2x
sin 2x dx.
SOLUTION If we write sin 2x 1 cos 2x, the integral is no simpler to evaluate. Using the
half-angle formula for sin 2x, however, we have
1.5
y=sin@ x
y
0
sin 2x dx 12 y 1 cos 2x dx 0
[ (x 1
2
1
2
0
]
sin 2x)
12 ( 12 sin 2) 12 (0 12 sin 0) 12 0
_0.5
π
Notice that we mentally made the substitution u 2x when integrating cos 2x. Another
method for evaluating this integral was given in Exercise 41 in Section 7.1.
FIGURE 2
EXAMPLE 4 Find
4
y sin x dx.
SOLUTION We could evaluate this integral using the reduction formula for
x sin n x dx
(Equation 7.1.7) together with Example 1 (as in Exercise 41 in Section 7.1), but a better
484
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
method is to write sin 4x sin 2x2 and use a half-angle formula:
y sin x dx y sin x dx
4
2
y
2
1 cos 2x
2
2
dx
14 y 1 2 cos 2x cos 2 2x dx
Since cos 2 2x occurs, we must use another half-angle formula
cos 2 2x 12 1 cos 4x
This gives
y sin x dx y 1 2 cos 2x 1
4
4
14 y
1
2
1 cos 4x dx
( 32 2 cos 2x 12 cos 4x) dx
14 ( 32 x sin 2x 18 sin 4x) C
To summarize, we list guidelines to follow when evaluating integrals of the form
x sin mx cos nx dx, where m 0 and n 0 are integers.
Strategy for Evaluating
y sin
m
x cos nx dx
(a) If the power of cosine is odd n 2k 1, save one cosine factor and use
cos 2x 1 sin 2x to express the remaining factors in terms of sine:
y sin
m
x cos 2k1x dx y sin m x cos 2xk cos x dx
y sin m x 1 sin 2xk cos x dx
Then substitute u sin x.
(b) If the power of sine is odd m 2k 1, save one sine factor and use
sin 2x 1 cos 2x to express the remaining factors in terms of cosine:
y sin
2k1
x cos n x dx y sin 2xk cos n x sin x dx
y 1 cos 2xk cos n x sin x dx
Then substitute u cos x. [Note that if the powers of both sine and cosine are
odd, either (a) or (b) can be used.]
(c) If the powers of both sine and cosine are even, use the half-angle identities
sin 2x 12 1 cos 2x
cos 2x 12 1 cos 2x
It is sometimes helpful to use the identity
sin x cos x 12 sin 2x
SECTION 7.2 TRIGONOMETRIC INTEGRALS
❙❙❙❙
485
We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since
ddx tan x sec 2x, we can separate a sec 2x factor and convert the remaining (even)
power of secant to an expression involving tangent using the identity sec 2x 1 tan 2x.
Or, since ddx sec x sec x tan x, we can separate a sec x tan x factor and convert the
remaining (even) power of tangent to secant.
EXAMPLE 5 Evaluate
6
4
y tan x sec x dx.
SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x factor in
terms of tangent using the identity sec 2x 1 tan 2x. We can then evaluate the integral
by substituting u tan x with du sec 2x dx :
y tan x sec x dx y tan x sec x sec x dx
6
4
6
2
2
y tan 6x 1 tan 2x sec 2x dx
y u 61 u 2 du y u 6 u 8 du
u7
u9
C
7
9
17 tan 7x 19 tan 9x C
EXAMPLE 6 Find
y tan sec d.
5
7
SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left with
a sec 5 factor, which isn’t easily converted to tangent. However, if we separate a
sec tan factor, we can convert the remaining power of tangent to an expression
involving only secant using the identity tan 2 sec 2 1. We can then evaluate the
integral by substituting u sec , so du sec tan d :
y tan 5
sec 7 d y tan 4 sec 6 sec tan d
y sec 2 12 sec 6 sec tan d
y u 2 12 u 6 du y u 10 2u 8 u 6 du
u 11
u9
u7
2
C
11
9
7
111 sec 11 29 sec 9 17 sec 7 C
The preceding examples demonstrate strategies for evaluating integrals of the form
x tan mx sec nx dx for two cases, which we summarize here.
486
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
Strategy for Evaluating
y tan
m
x sec nx dx
(a) If the power of secant is even n 2k, k 2, save a factor of sec 2x and use
sec 2x 1 tan 2x to express the remaining factors in terms of tan x :
y tan
m
x sec 2kx dx y tan m x sec 2xk1 sec 2x dx
y tan m x 1 tan 2xk1 sec 2x dx
Then substitute u tan x.
(b) If the power of tangent is odd m 2k 1, save a factor of sec x tan x and
use tan 2x sec 2x 1 to express the remaining factors in terms of sec x :
y tan
2k1
x sec n x dx y tan 2xk sec n1x sec x tan x dx
y sec 2x 1k sec n1x sec x tan x dx
Then substitute u sec x.
For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to
integrate tan x by using the formula established in (5.5.5):
y tan x dx ln sec x C
We will also need the indefinite integral of secant:
1
y sec x dx ln sec x tan x C
We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x :
sec x tan x
y sec x dx y sec x sec x tan x dx
y
sec 2x sec x tan x
dx
sec x tan x
If we substitute u sec x tan x, then du sec x tan x sec 2x dx, so the integral
becomes x 1u du ln u C. Thus, we have
y sec x dx ln sec x tan x C
SECTION 7.2 TRIGONOMETRIC INTEGRALS
EXAMPLE 7 Find
❙❙❙❙
487
3
y tan x dx.
SOLUTION Here only tan x occurs, so we use tan 2x sec 2x 1 to rewrite a tan 2x factor in
terms of sec 2x :
y tan x dx y tan x tan x dx
3
2
y tan x sec 2x 1 dx
y tan x sec 2x dx y tan x dx
tan 2x
ln sec x C
2
In the first integral we mentally substituted u tan x so that du sec 2x dx.
If an even power of tangent appears with an odd power of secant, it is helpful to express
the integrand completely in terms of sec x. Powers of sec x may require integration by
parts, as shown in the following example.
EXAMPLE 8 Find
3
y sec x dx.
SOLUTION Here we integrate by parts with
u sec x
du sec x tan x dx
Then
dv sec 2x dx
v tan x
y sec x dx sec x tan x y sec x tan x dx
3
2
sec x tan x y sec x sec 2x 1 dx
sec x tan x y sec 3x dx y sec x dx
Using Formula 1 and solving for the required integral, we get
y sec x dx (sec x tan x ln sec x tan x ) C
3
1
2
Integrals such as the one in the preceding example may seem very special but they
occur frequently in applications of integration, as we will see in Chapter 8. Integrals of
the form x cot m x csc n x dx can be found by similar methods because of the identity
1 cot 2x csc 2x.
Finally, we can make use of another set of trigonometric identities:
2 To evaluate the integrals (a) x sin mx cos nx dx, (b) x sin mx sin nx dx, or
(c) x cos mx cos nx dx, use the corresponding identity:
|||| These product identities are discussed in
Appendix D.
(a) sin A cos B 12 sinA B sinA B
(b) sin A sin B 12 cosA B cosA B
(c) cos A cos B 12 cosA B cosA B
❙❙❙❙
488
CHAPTER 7 TECHNIQUES OF INTEGRATION
EXAMPLE 9 Evaluate
y sin 4x cos 5x dx.
SOLUTION This integral could be evaluated using integration by parts, but it’s easier to use
the identity in Equation 2(a) as follows:
y sin 4x cos 5x dx y
1
2
sinx sin 9x dx
12 y sin x sin 9x dx
12 (cos x 19 cos 9x C
|||| 7.2
1–47
1.
3.
5.
7.
9.
11.
13.
15.
||||
Exercises
Evaluate the integral.
3
2
y sin x cos x dx
34
y
2
sin 5x cos 3x dx
5
4
y cos x sin x dx
y
2
0
y
0
cos2 d
y 1 cos 0
d
sin 4x cos 2x dx
3
y sin x scos x dx
2
3
17.
y cos x tan x dx
19.
y
21.
y sec x tan x dx
23.
y tan x dx
25.
27.
29.
1 sin x
dx
cos x
2
2
6
y sec t dt
y
3
0
5
4
tan x sec x dx
3
6.
10.
2
y
4.
8.
sin 43t dt
4
2.
y tan x sec x dx
12.
14.
6
y tan ay dy
34.
y tan x sec x dx
cot 2x dx
36.
y
37.
y cot csc d
38.
y csc
39.
y csc x dx
40.
y
41.
y sin 5x sin 2x dx
42.
y sin 3x cos x dx
43.
y cos 7 cos 5 d
44.
y
45.
y
1 tan 2x
dx
sec 2x
46.
y cos x 1
47.
y t sec t
y tan x dx
33.
y cos d
y
2
0
35.
y
6
3
y sin x cos x dx
tan 3
4
2
cos 5x dx
y sin mx dx
2
6
2
4
cot 3x dx
3
y
2
0
y
0
3
3
sin 2 2 d
cos6 d
3
6
4
x cot 6 x dx
csc 3x dx
2
y x cos x dx
y
2
0
sin 2x cos 2x dx
16.
y cos cos sin d
18.
y cot sin d
20.
y cos x sin 2x dx
22.
y
24.
y tan x dx
5
5
2
2
0
sec 4t2 dt
4
4
y tan 2x sec 2x dx
3
0
tan 4t 2 dt
■
■
■
■
■
■
■
■
■
■
■
■
48. If x04 tan 6 x sec x dx I , express the value of
x04 tan 8 x sec x dx in terms of I.
; 49–52
|||| Evaluate the indefinite integral. Illustrate, and check that
your answer is reasonable, by graphing both the integrand and its
antiderivative (taking C 0.
5
4
4
49.
y sin x dx
50.
y sin x cos x dx
51.
y sin 3x sin 6x dx
52.
y sec
4
28.
3
y
2
dx
sec tan d
4
y
0
2
cos x sin x
dx
sin 2x
4
26.
30.
32.
5
31.
5
tan 5x sec6x dx
■
■
■
■
■
■
■
■
4
x
dx
2
■
■
■
53. Find the average value of the function f x sin 2x cos 3x on
the interval , .
■
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
54. Evaluate x sin x cos x dx by four methods: (a) the substitution
||||
current that varies from 155 V to 155 V with a frequency
of 60 cycles per second (Hz). The voltage is thus given by
the equation
Et 155 sin120 t
Find the area of the region bounded by the given curves.
55. y sin x,
y sin 3x,
56. y sin x,
y 2 sin 2x,
■
■
■
■
■
x 2
x 0,
■
where t is the time in seconds. Voltmeters read the RMS (rootmean-square) voltage, which is the square root of the average
value of Et 2 over one cycle.
(a) Calculate the RMS voltage of household current.
(b) Many electric stoves require an RMS voltage of 220 V.
Find the corresponding amplitude A needed for the voltage
Et A sin120 t.
x 2
x 0,
■
■
■
■
■
■
; 57–58
|||| Use a graph of the integrand to guess the value of the
integral. Then use the methods of this section to prove that your
guess is correct.
57.
y
2
0
■
■
cos 3x dx
■
58.
■
■
■
y
■
2
0
■
65–67
sin 2 x cos 5 x dx
■
■
■
■
|||| Find the volume obtained by rotating the region bounded
by the given curves about the specified axis.
about the x-axis
60. y tan 2x, y 0, x 0, x 4;
about the x-axis
61. y cos x, y 0, x 0, x 2;
about y 1
62. y cos x, y 0, x 0, x 2;
about y 1
■
■
■
■
■
■
■
■
■
66.
y sin mx sin nx dx 67.
y cos mx cos nx dx |||| 7.3
■
■
■
0
0
■
if m n
if m n
if m n
if m n
■
■
■
■
■
■
■
68. A finite Fourier series is given by the sum
N
f x a
n
sin nx
n1
■
63. A particle moves on a straight line with velocity function
vt sin t cos 2 t. Find its position function s f t if
f 0 0.
Prove the formula, where m and n are positive integers.
y sin mx cos nx dx 0
■
■
||||
65.
59–62
59. y sin x, x 2, x , y 0;
489
64. Household electricity is supplied in the form of alternating
u cos x, (b) the substitution u sin x, (c) the identity
sin 2x 2 sin x cos x, and (d) integration by parts. Explain the
different appearances of the answers.
55–56
❙❙❙❙
■
a 1 sin x a 2 sin 2x a N sin Nx
Show that the mth coefficient a m is given by the formula
am 1
y
f x sin mx dx
Trigonometric Substitution
In finding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 dx arises,
where a 0. If it were x xsa 2 x 2 dx, the substitution u a 2 x 2 would be effective
but, as it stands, x sa 2 x 2 dx is more difficult. If we change the variable from x to by
the substitution x a sin , then the identity 1 sin 2 cos 2 allows us to get rid of the
root sign because
sa 2 x 2 sa 2 a 2 sin 2 sa 21 sin 2 sa 2 cos 2 a cos Notice the difference between the substitution u a x (in which the new variable is
a function of the old one) and the substitution x a sin (the old variable is a function of
the new one).
In general we can make a substitution of the form x tt by using the Substitution
Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is, t is one-to-one. In this case, if we replace u by x and x by t in the Substitution
Rule (Equation 5.5.4), we obtain
2
y f x dx y f ttt
t dt
This kind of substitution is called inverse substitution.
2
490
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
We can make the inverse substitution x a sin provided that it defines a one-to-one
function. This can be accomplished by restricting to lie in the interval 2, 2.
In the following table we list trigonometric substitutions that are effective for the given
radical expressions because of the specified trigonometric identities. In each case the restriction on is imposed to ensure that the function that defines the substitution is one-to-one.
(These are the same intervals used in Appendix D in defining the inverse functions.)
Table of Trigonometric Substitutions
Expression
Substitution
Identity
sa 2 x 2
x a sin ,
2
2
1 sin 2 cos 2
sa 2 x 2
x a tan ,
2
2
1 tan 2 sec 2
sx 2 a 2
x a sec ,
0
EXAMPLE 1 Evaluate
y
3
or 2
2
sec 2 1 tan 2
s9 x 2
dx.
x2
SOLUTION Let x 3 sin , where 2 2. Then dx 3 cos d and
s9 x 2 s9 9 sin 2 s9 cos 2 3 cos 3 cos (Note that cos 0 because 2 2.) Thus, the Inverse Substitution Rule
gives
3 cos s9 x 2
y x 2 dx y 9 sin 2 3 cos d
y
cos 2
d y cot 2 d
sin 2
y csc 2 1 d
cot C
3
x
¨
œ„„„„„
9-≈
Since this is an indefinite integral, we must return to the original variable x. This can be
done either by using trigonometric identities to express cot in terms of sin x3 or
by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle.
Since sin x3, we label the opposite side and the hypotenuse as having lengths x and 3.
Then the Pythagorean Theorem gives the length of the adjacent side as s9 x 2, so we
can simply read the value of cot from the figure:
FIGURE 1
cot x
sin ¨=
3
s9 x 2
x
(Although 0 in the diagram, this expression for cot is valid even when 0.)
Since sin x3, we have sin1x3 and so
y
x
s9 x 2
s9 x 2
dx
sin1
2
x
x
3
C
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
❙❙❙❙
491
EXAMPLE 2 Find the area enclosed by the ellipse
x2
y2
1
2 a
b2
SOLUTION Solving the equation of the ellipse for y, we get
y
y2
x2
a2 x2
2 1 2 b
a
a2
(0, b)
y
or
b
sa 2 x 2
a
(a, 0)
0
FIGURE 2
¥
≈
+ =1
b@
a@
x
Because the ellipse is symmetric with respect to both axes, the total area A is four times
the area in the first quadrant (see Figure 2). The part of the ellipse in the first quadrant is
given by the function
b
y sa 2 x 2
0xa
a
1
4
and so
Ay
a
0
b
sa 2 x 2 dx
a
To evaluate this integral we substitute x a sin . Then dx a cos d. To change the
limits of integration we note that when x 0, sin 0, so 0; when x a,
sin 1, so 2. Also
sa 2 x 2 sa 2 a 2 sin 2 sa 2 cos 2 a cos a cos since 0 2. Therefore
A4
b
a
y
a
0
4ab y
sa 2 x 2 dx 4
2
0
b
a
cos 2 d 4ab y
y
2 1
2
0
[
2ab 12 sin 2
2
0
]
2
0
2ab
a cos a cos d
1 cos 2 d
00
2
ab
We have shown that the area of an ellipse with semiaxes a and b is ab. In particular,
taking a b r, we have proved the famous formula that the area of a circle with
radius r is r 2.
NOTE Since the integral in Example 2 was a definite integral, we changed the limits of
integration and did not have to convert back to the original variable x.
■
EXAMPLE 3 Find
1
y x sx
2
2
4
dx.
SOLUTION Let x 2 tan , 2 2. Then dx 2 sec 2 d and
sx 2 4 s4tan 2 1 s4 sec 2 2 sec 2 sec Thus, we have
dx
y x sx
2
2
4
y
2 sec 2 d
1
2
4 tan 2 sec 4
sec y tan d
2
492
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
To evaluate this trigonometric integral we put everything in terms of sin and cos :
sec
1
cos 2
cos 2
2
tan cos sin sin 2
Therefore, making the substitution u sin , we have
y
dx
1
2
x sx 4
4
2
œ„„„„„
≈+4
x
¨
2
1
u
y
C
du
u2
1
C
4 sin csc C
4
We use Figure 3 to determine that csc sx 2 4x and so
FIGUR E 3
tan ¨=
1
4
cos 1
d 2
sin 4
y
dx
x
2
y x sx
2
EXAMPLE 4 Find
y sx
2
4
sx 2 4
C
4x
x
dx.
4
2
SOLUTION It would be possible to use the trigonometric substitution x 2 tan here (as in
Example 3). But the direct substitution u x 2 4 is simpler, because then du 2x dx
and
y
NOTE
x
1
dx 2
sx 4
2
du
y su
su C sx 2 4 C
Example 4 illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution. You should look for a simpler method first.
■
EXAMPLE 5 Evaluate
y sx
dx
, where a 0.
a2
2
SOLUTION 1 We let x a sec , where 0 dx a sec tan d and
2 or 32. Then
sx 2 a 2 sa 2sec 2 1 sa 2 tan 2 a tan a tan Therefore
y sx
dx
a sec tan y
d
a2
a tan 2
≈-a@
œ„„„„„
The triangle in Figure 4 gives tan sx 2 a 2a, so we have
¨
a
FIGU RE 4
x
sec ¨=
a
y sec d ln sec tan C
x
y sx
dx
x
sx 2 a 2
ln
C
2
a
a
a
2
ln x sx 2 a 2 ln a C
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
❙❙❙❙
493
Writing C1 C ln a, we have
y sx
1
dx
ln x sx 2 a 2 C1
a2
2
SOLUTION 2 For x 0 the hyperbolic substitution x a cosh t can also be used. Using the
identity cosh 2 y sinh 2 y 1, we have
sx 2 a 2 sa 2 cosh 2 t 1 sa 2 sinh 2 t a sinh t
Since dx a sinh t dt, we obtain
y sx
dx
a sinh t dt
y
y dt t C
a2
a sinh t
2
Since cosh t xa, we have t cosh1xa and
dx
x
cosh1
a
sx 2 a 2
y
2
C
Although Formulas 1 and 2 look quite different, they are actually equivalent by
Formula 3.9.4.
NOTE
As Example 5 illustrates, hyperbolic substitutions can be used in place of trigonometric substitutions and sometimes they lead to simpler answers. But we usually use
trigonometric substitutions because trigonometric identities are more familiar than hyperbolic identities.
■
EXAMPLE 6 Find
y
3 s32
0
x3
dx.
4x 2 932
SOLUTION First we note that 4x 2 932 s4x 2 9
)3 so trigonometric substitution
is appropriate. Although s4x 9 is not quite one of the expressions in the table of
trigonometric substitutions, it becomes one of them if we make the preliminary substitution u 2x. When we combine this with the tangent substitution, we have x 32 tan ,
which gives dx 32 sec 2 d and
2
s4x 2 9 s9 tan 2 9 3 sec When x 0, tan 0, so 0; when x 3s32, tan s3, so 3.
y
3 s32
0
27
3
x3
3 8 tan dx
y
0
4x 2 932
27 sec3
163 y
3
0
163 y
3
0
3
2
sec 2 d
3
tan 3
3 sin d 163 y
d
0
sec cos2
1 cos 2
sin d
cos 2
Now we substitute u cos so that du sin d. When 0, u 1; when
3, u 12.
494
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
Therefore
y
3 s32
0
2
x3
12 1 u
12
3
du 163 y 1 u 2 du
32 dx 16 y
2
1
1
4x 9
u
2
EXAMPLE 7 Evaluate
1
u
u
3
16
x
y s3 2x x
2
12
163 [( 12 2) 1 1] 323
1
dx.
SOLUTION We can transform the integrand into a function for which trigonometric substitution is appropriate by first completing the square under the root sign:
3 2x x 2 3 x 2 2x 3 1 x 2 2x 1
4 x 12
This suggests that we make the substitution u x 1. Then du dx and x u 1, so
x
y s3 2x x
2
dx y
u1
du
s4 u 2
We now substitute u 2 sin , giving du 2 cos d and s4 u 2 2 cos , so
|||| Figure 5 shows the graphs of the integrand
in Example 7 and its indefinite integral (with
C 0 ). Which is which?
x
y s3 2x x
2
dx y
2 sin 1
2 cos d
2 cos 3
y 2 sin 1 d
2
_4
2 cos C
s4 u 2 sin1
u
2
_5
s3 2x x 2 sin1
FIGURE 5
|||| 7.3
4–30
|||| Evaluate the integral using the indicated trigonometric
substitution. Sketch and label the associated right triangle.
2.
3.
■
yx
2
1
dx ;
sx 2 9
y x 3 s9 x 2 dx ;
y sx
■
x1
2
C
Exercises
1–3
1.
C
x3
dx ;
2 9
■
■
x 3 sec 4.
y
5.
y
■
0
2
■
■
■
■
■
■
x3
dx
s16 x 2
1
dt
t 3 st 2 1
1
dx
s25 x 2
7.
yx
9.
y sx
x 3 tan ■
2 s3
s2
x 3 sin Evaluate the integral.
||||
2
dx
2 16
6.
y
8.
y
10.
2
0
x 3 sx 2 4 dx
sx 2 a 2
dx
x4
y st
t5
dt
2 2
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
11.
y s1 4x
13.
y
15.
y a
17.
y sx
1
12.
y
sx 2 9
dx
x3
14.
y u s5 u
x2
dx
x 2 32
16.
yx
x
dx
7
18.
y ax
2
2
s1 x 2
dx
x
19.
y
21.
y
23.
y s5 4x x
25.
y s9x
27.
y x
29.
y x s1 x 4 dx
23
0
x 3s4 9x 2 dx
2
2
2
dx
1
dx
6x 8
dx
2x 22
0
du
2
y
dx
b 2 32
t
20.
y s25 t
22.
y
24.
y st
26.
y s4x x
28.
y 5 4x x
1
2
¨
O
dt
2
y
dt
6t 13
2
2
0
■
■
■
■
■
■
■
38. A charged rod of length L produces an electric field at point
2 52
Pa, b given by
cos t
dt
s1 sin 2 t
EP y
■
■
■
31. (a) Use trigonometric substitution to show that
y sx
dx
x 4 sx 2 2
; 37. Use a graph to approximate the roots of the equation
x 2 s4 x 2 2 x. Then approximate the area bounded by
the curve y x 2 s4 x 2 and the line y 2 x.
dx
■
x
Graph the integrand and its indefinite integral on the same
screen and check that your answer is reasonable.
La
a
■
R
; 36. Evaluate the integral
dx
y
Q
sx 2 1 dx
x2
30.
P
2
dx
s16x 2 9
2
0
495
equation x 2 y 2 r 2. Then A is the sum of the area of the
triangle POQ and the area of the region PQR in the figure.]
x sx 2 4 dx
dx
2
❙❙❙❙
dx
ln ( x sx 2 a 2 ) C
a2
2
b
dx
4 0 x 2 b 2 32
where is the charge density per unit length on the rod and 0
is the free space permittivity (see the figure). Evaluate the integral to determine an expression for the electric field EP.
y
P (a, b)
(b) Use the hyperbolic substitution x a sinh t to show that
y sx
dx
x
sinh1
a2
a
2
0
L
x
C
These formulas are connected by Formula 3.9.3.
39. Find the area of the crescent-shaped region (called a lune)
bounded by arcs of circles with radii r and R. (See the figure.)
32. Evaluate
y x
2
x2
dx
a 2 32
(a) by trigonometric substitution.
(b) by the hyperbolic substitution x a sinh t.
r
R
33. Find the average value of f x sx 2 1x, 1 x 7.
34. Find the area of the region bounded by the hyperbola
9x 4y 36 and the line x 3.
2
2
35. Prove the formula A 2 r 2 for the area of a sector of a circle
1
with radius r and central angle . [Hint: Assume 0 2
and place the center of the circle at the origin so it has the
40. A water storage tank has the shape of a cylinder with diameter
10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage of the
total capacity is being used?
41. A torus is generated by rotating the circle x 2 y R2 r 2
about the x-axis. Find the volume enclosed by the torus.
496
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
|||| 7.4
Integration of Rational Functions by Partial Fractions
In this section we show how to integrate any rational function (a ratio of polynomials) by
expressing it as a sum of simpler fractions, called partial fractions, that we already know
how to integrate. To illustrate the method, observe that by taking the fractions 2x 1
and 1x 2 to a common denominator we obtain
2
1
2x 2 x 1
x5
2
x1
x2
x 1x 2
x x2
If we now reverse the procedure, we see how to integrate the function on the right side of
this equation:
yx
2
x5
dx x2
y
2
1
x1
x2
dx
2 ln x 1 ln x 2 C
To see how the method of partial fractions works in general, let’s consider a rational
function
f x Px
Qx
where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions
provided that the degree of P is less than the degree of Q. Such a rational function is called
proper. Recall that if
Px a n x n a n1 x n1 a 1 x a 0
where a n 0, then the degree of P is n and we write degP n.
If f is improper, that is, degP degQ, then we must take the preliminary step
of dividing Q into P (by long division) until a remainder Rx is obtained such that
degR degQ. The division statement is
f x 1
Px
Rx
Sx Qx
Qx
where S and R are also polynomials.
As the following example illustrates, sometimes this preliminary step is all that is
required.
EXAMPLE 1 Find
≈+x +2
x-1 ) ˛
+x
˛-≈
≈+x
≈-x
2x
2x-2
2
y
x3 x
dx.
x1
SOLUTION Since the degree of the numerator is greater than the degree of the denominator,
we first perform the long division. This enables us to write
y
x3 x
dx x1
y
x2 x 2 2
x1
dx
x3
x2
2x 2 ln x 1 C
3
2
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
❙❙❙❙
497
The next step is to factor the denominator Qx as far as possible. It can be shown that
any polynomial Q can be factored as a product of linear factors (of the form ax b) and
irreducible quadratic factors (of the form ax 2 bx c, where b 2 4ac 0). For instance, if Qx x 4 16, we could factor it as
Qx x 2 4x 2 4 x 2x 2x 2 4
The third step is to express the proper rational function RxQx (from Equation 1) as
a sum of partial fractions of the form
A
ax bi
or
Ax B
ax bx c j
2
A theorem in algebra guarantees that it is always possible to do this. We explain the details
for the four cases that occur.
CASE I
■
The denominator Qx is a product of distinct linear factors.
This means that we can write
Qx a 1 x b1 a 2 x b 2 a k x bk where no factor is repeated (and no factor is a constant multiple of another). In this case
the partial fraction theorem states that there exist constants A1, A2 , . . . , Ak such that
2
Rx
A1
A2
Ak
Qx
a 1 x b1
a2 x b2
a k x bk
These constants can be determined as in the following example.
EXAMPLE 2 Evaluate
x 2 2x 1
dx.
3
3x 2 2x
y 2x
SOLUTION Since the degree of the numerator is less than the degree of the denominator,
we don’t need to divide. We factor the denominator as
2x 3 3x 2 2x x2x 2 3x 2 x2x 1x 2
Since the denominator has three distinct linear factors, the partial fraction decomposition
of the integrand (2) has the form
3
|||| Another method for finding A, B, and C
is given in the note after this example.
x 2 2x 1
A
B
C
x2x 1x 2
x
2x 1
x2
To determine the values of A, B, and C, we multiply both sides of this equation by the
product of the denominators, x2x 1x 2, obtaining
4
x 2 2x 1 A2x 1x 2 Bxx 2 Cx2x 1
Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get
5
x 2 2x 1 2A B 2Cx 2 3A 2B Cx 2A
498
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
|||| Figure 1 shows the graphs of the integrand
in Example 2 and its indefinite integral (with
K 0). Which is which?
2
The polynomials in Equation 5 are identical, so their coefficients must be equal. The
coefficient of x 2 on the right side, 2A B 2C, must equal the coefficient of x 2 on the
left side—namely, 1. Likewise, the coefficients of x are equal and the constant terms are
equal. This gives the following system of equations for A, B, and C:
2A B 2C 1
_3
3
3A 2B C 2
2A 2B 2C 1
_2
FIGURE 1
Solving, we get A 12 , B 15 , and C 101 , and so
x 2 2x 1
dx 3
3x 2 2x
y 2x
y
1 1
1
1
1
1
2 x
5 2x 1
10 x 2
dx
12 ln x 101 ln 2x 1 101 ln x 2 K
|||| We could check our work by taking the terms
to a common denominator and adding them.
In integrating the middle term we have made the mental substitution u 2x 1, which
gives du 2 dx and dx du2.
NOTE
We can use an alternative method to find the coefficients A, B, and C in
Example 2. Equation 4 is an identity; it is true for every value of x. Let’s choose values of
x that simplify the equation. If we put x 0 in Equation 4, then the second and third terms
on the right side vanish and the equation then becomes 2A 1, or A 12 . Likewise,
x 12 gives 5B4 14 and x 2 gives 10C 1, so B 15 and C 101 . (You may object
that Equation 3 is not valid for x 0, 12 , or 2, so why should Equation 4 be valid for those
values? In fact, Equation 4 is true for all values of x, even x 0, 12 , and 2. See Exercise 67
for the reason.)
■
EXAMPLE 3 Find
yx
2
dx
, where a 0.
a2
SOLUTION The method of partial fractions gives
1
1
A
B
2 x a
x ax a
xa
xa
2
and therefore
Ax a Bx a 1
Using the method of the preceding note, we put x a in this equation and get
A2a 1, so A 12a. If we put x a, we get B2a 1, so B 12a.
Thus
y
dx
1
2 x a
2a
2
y
1
1
xa
xa
dx
1
(ln x a ln x a
2a
) C
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
❙❙❙❙
499
Since ln x ln y lnxy, we can write the integral as
y
6
dx
1
xa
ln
C
x2 a2
2a
xa
See Exercises 53–54 for ways of using Formula 6.
CASE II
■
Qx is a product of linear factors, some of which are repeated.
Suppose the first linear factor a 1 x b1 is repeated r times; that is, a 1 x b1 r occurs in
the factorization of Qx. Then instead of the single term A1a 1 x b1 in Equation 2, we
would use
A1
A2
Ar
2 a 1 x b1
a 1 x b1 a 1 x b1 r
7
By way of illustration, we could write
x3 x 1
A
B
C
D
E
2 2
3 2 x x 1
x
x
x1
x 1
x 13
but we prefer to work out in detail a simpler example.
EXAMPLE 4 Find
y
x 4 2x 2 4x 1
dx.
x3 x2 x 1
SOLUTION The first step is to divide. The result of long division is
x 4 2x 2 4x 1
4x
x1 3
3
2
2
x x x1
x x x1
The second step is to factor the denominator Qx x 3 x 2 x 1. Since Q1 0,
we know that x 1 is a factor and we obtain
x 3 x 2 x 1 x 1x 2 1 x 1x 1x 1
x 12x 1
Since the linear factor x 1 occurs twice, the partial fraction decomposition is
4x
A
B
C
2
2 x 1 x 1
x1
x 1
x1
Multiplying by the least common denominator, x 12x 1, we get
8
4x Ax 1x 1 Bx 1 Cx 12
A Cx 2 B 2Cx A B C
|||| Another method for finding the coefficients:
Put x 1 in (8): B 2.
Put x 1: C 1.
Put x 0: A B C 1.
Now we equate coefficients:
AB C0
A B 2C 4
A B C 0
500
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
Solving, we obtain A 1, B 2, and C 1, so
y
x 4 2x 2 4x 1
dx x3 x2 x 1
CASE III
■
y
x1
1
2
1
2 x1
x 1
x1
dx
x2
2
x ln x 1 ln x 1 K
2
x1
x2
2
x1
x
ln
K
2
x1
x1
Qx contains irreducible quadratic factors, none of which is repeated.
If Qx has the factor ax 2 bx c, where b 2 4ac 0, then, in addition to the partial
fractions in Equations 2 and 7, the expression for RxQx will have a term of the form
Ax B
ax 2 bx c
9
where A and B are constants to be determined. For instance, the function given by
f x xx 2x 2 1x 2 4
has a partial fraction decomposition of the form
x
A
Bx C
Dx E
2
2
2
x 2x 1x 4
x2
x 1
x 4
2
The term given in (9) can be integrated by completing the square and using the formula
y
10
EXAMPLE 5 Evaluate
y
dx
1
x
tan1
x2 a2
a
a
C
2x 2 x 4
dx.
x 3 4x
SOLUTION Since x 3 4x xx 2 4 can’t be factored further, we write
2x 2 x 4
A
Bx C
2
xx 2 4
x
x 4
Multiplying by xx 2 4, we have
2x 2 x 4 Ax 2 4 Bx Cx
A Bx 2 Cx 4A
Equating coefficients, we obtain
AB2
C 1
Thus A 1, B 1, and C 1 and so
y
2x 2 x 4
dx x 3 4x
y
4A 4
1
x1
2
x
x 4
dx
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
❙❙❙❙
501
In order to integrate the second term we split it into two parts:
y
x1
x
1
dx y 2
dx y 2
dx
x2 4
x 4
x 4
We make the substitution u x 2 4 in the first of these integrals so that du 2x dx.
We evaluate the second integral by means of Formula 10 with a 2:
y
2x 2 x 4
1
x
1
dx y dx y 2
dx y 2
dx
2
xx 4
x
x 4
x 4
ln x 12 lnx 2 4 12 tan1x2 K
EXAMPLE 6 Evaluate
4x 2 3x 2
dx.
2
4x 3
y 4x
SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain
4x 2 3x 2
x1
1
2
2
4x 4x 3
4x 4x 3
Notice that the quadratic 4x 2 4x 3 is irreducible because its discriminant is
b 2 4ac 32 0. This means it can’t be factored, so we don’t need to use the
partial fraction technique.
To integrate the given function we complete the square in the denominator:
4x 2 4x 3 2x 12 2
This suggests that we make the substitution u 2x 1. Then, du 2 dx and
x u 12, so
y
4x 2 3x 2
dx 4x 2 4x 3
y
1
x 12 y
x 14 y
x1
4x 2 4x 3
1
2
dx
u 1 1
u1
du x 14 y 2
du
2
u 2
u 2
u
1
du 14 y 2
du
u2 2
u 2
x 18 lnu 2 2 x 18 ln4x 2 4x 3 NOTE
■
1
1
u
tan1
4 s2
s2
C
1
2x 1
tan1
4 s2
s2
C
Example 6 illustrates the general procedure for integrating a partial fraction of
the form
Ax B
ax bx c
2
where b 2 4ac 0
502
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
We complete the square in the denominator and then make a substitution that brings the
integral into the form
y
Cu D
u
1
du
2
2 du C y
2
2 du D y
2
u a
u a
u a2
Then the first integral is a logarithm and the second is expressed in terms of tan1.
CASE IV
■
Qx contains a repeated irreducible quadratic factor.
If Qx has the factor ax 2 bx cr, where b 2 4ac 0, then instead of the single
partial fraction (9), the sum
A1 x B1
A2 x B2
Ar x Br
ax 2 bx c
ax 2 bx c2
ax 2 bx cr
11
occurs in the partial fraction decomposition of RxQx. Each of the terms in (11) can be
integrated by first completing the square.
|||| It would be extremely tedious to work out by
hand the numerical values of the coefficients in
Example 7. Most computer algebra systems,
however, can find the numerical values very
quickly. For instance, the Maple command
convertf, parfrac, x
or the Mathematica command
Apart[f]
gives the following values:
A 1,
E
15
8
,
B 18 ,
C D 1,
F ,
G H 34 ,
1
8
I 12 ,
EXAMPLE 7 Write out the form of the partial fraction decomposition of the function
x3 x2 1
xx 1x 2 x 1x 2 13
SOLUTION
x3 x2 1
xx 1x 2 x 1x 2 13
J 12
A
B
Cx D
Ex F
Gx H
Ix J
2
2
2
2 x
x1
x x1
x 1
x 1
x 2 13
EXAMPLE 8 Evaluate
y
1 x 2x 2 x 3
dx.
xx 2 12
SOLUTION The form of the partial fraction decomposition is
1 x 2x 2 x 3
A
Bx C
Dx E
2
2
2
2
xx 1
x
x 1
x 12
Multiplying by xx 2 12, we have
x 3 2x 2 x 1 Ax 2 12 Bx Cxx 2 1 Dx Ex
Ax 4 2x 2 1 Bx 4 x 2 Cx 3 x Dx 2 Ex
A Bx 4 Cx 3 2A B Dx 2 C Ex A
If we equate coefficients, we get the system
AB0
C 1
2A B D 2
C E 1
which has the solution A 1, B 1, C 1, D 1, and E 0. Thus
A1
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
y
1 x 2x 2 x 3
dx xx 2 12
y
y
1
x1
x
2
2
x
x 1
x 12
503
dx
dx
x
dx
x dx
y 2
dx y 2
y 2
x
x 1
x 1
x 12
ln x 12 lnx 2 1 tan1x |||| In the second and fourth terms we made the
mental substitution u x 2 1.
❙❙❙❙
1
K
2x 2 1
We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral
y
x2 1
dx
xx 2 3
could be evaluated by the method of Case III, it’s much easier to observe that if
u xx 2 3 x 3 3x, then du 3x 2 3 dx and so
y
Useful, Do this.
x2 1
dx 13 ln x 3 3x C
xx 2 3
Rationalizing Substitutions
Some nonrational functions can be changed into rational functions by means of appropriate substitutions. In particular, when an integrand contains an expression of the form
n
n
tx, then the substitution u s
tx may be effective. Other instances appear in the
s
exercises.
EXAMPLE 9 Evaluate
y
sx 4
dx.
x
SOLUTION Let u sx 4. Then u 2 x 4, so x u 2 4 and dx 2u du.
Therefore
y
u
u2
sx 4
dx y 2
2u du 2 y 2
du
x
u 4
u 4
2
y
1
4
u2 4
du
We can evaluate this integral either by factoring u 2 4 as u 2u 2 and using
partial fractions or by using Formula 6 with a 2:
y
du
sx 4
dx 2 y du 8 y 2
x
u 4
2u 8 1
u2
ln
C
22
u2
2sx 4 2 ln
sx 4 2
C
sx 4 2
❙❙❙❙
504
CHAPTER 7 TECHNIQUES OF INTEGRATION
|||| 7.4
Exercises
1–6
|||| Write out the form of the partial fraction decomposition
of the function (as in Example 7). Do not determine the numerical
values of the coefficients.
2x
x 33x 1
1. (a)
(b)
x1
x3 x2
2. (a)
(b)
1
x 3 2x 2 x
x1
x3 x
x2
(b)
x 1x 2 x 1
2
3. (a) 2
x 3x 4
3
34.
yx
dx
4
x2
36.
y
38.
y x x
y
35.
yx
37.
y x
■
3
3
x3
dx
2x 42
2
■
■
■
■
■
0
1
0
■
x4
x4 1
(b)
t4 t2 1
t 2 1t 2 42
39.
y x sx 1 dx
6. (a)
x4
x xx 2 x 3
(b)
1
x6 x3
41.
y
43.
y sx
45.
1
y sx sx dx [Hint: Substitute u ■
7–38
■
■
■
x
7.
y x 6 dx
9.
y x 5x 2 dx
x9
■
■
■
8.
■
r2
y r 4 dr
1
1
16
9
17.
19.
21.
23.
25.
27.
29.
1
y
2
1
44.
y
1
0
3
x3
dx
1
2
3
3
y
x1
dx
x 2 3x 2
47.
ye
ax
dx
2
bx
14.
y x ax b dx
16.
x 4x 10
dx
x2 x 6
2x 3
dx
x 12
4y 7y 12
dy
y y 2 y 3
3
y
1
0
3
y
x 2x 1
dx
x3 x
x2
dx
x 3x 22
y
20.
y
y
5x 2 3x 2
dx
x 3 2x 2
22.
y s s 1
y
x2
dx
x 13
y
10
dx
x 1x 2 9
y
x 3 x 2 2x 1
dx
x 2 1x 2 2
2
x4
dx
2x 5
2x
24.
26.
28.
ds
2
■
■
■
48.
■
■
6
.]
sx
sx .]
12
cos x
dx
sin 2x sin x
y
■
■
■
■
■
49–50
|||| Use integration by parts, together with the techniques of
this section, to evaluate the integral.
49.
y lnx
2
■
■
2
x 2 dx
■
■
■
50.
■
■
1
y x tan
■
x dx
■
■
■
3
2
; 52. Graph both y 1x 2x and an antiderivative on the
same screen.
y
y
x2 x 6
dx
x 3 3x
53.
y
x 2 2x 1
dx
x 12x 2 1
y
x 2x x 1
dx
x 4 5x 2 4
55. The German mathematician Karl Weierstrass (1815–1897)
2
■
2
; 51. Use a graph of f x 1x 2x 3 to decide whether
2
x0 f x dx is positive or negative. Use the graph to give a rough
estimate of the value of the integral and then use partial
fractions to find the exact value.
x3
dx
x 13
3
30.
e 2x
dx
3e x 2
■
2
1
dx
x 52x 1
yx
4
sx
dx
x2 x
1
2
18.
1
dx
3
1s
x
13
12.
0
■
■
■
0
y
1
dx
x2 1
yx
y
42.
1
y sx sx dx [Hint: Substitute u 13.
15.
sx
dx
x4
y x sx 2 dx
46.
1
■
1
40.
y t 4t 1 dt
y
■
||||
10.
11.
2
■
■
function and then evaluate the integral.
Evaluate the integral.
||||
3
■
2x 3 5x
dx
x 4 5x 2 4
Useful,
do these.
39–48
Make a substitution to express the integrand as a rational
2x 1
x 1 3x 2 4 2
5. (a)
■
x3
dx
1
■
(b)
3
3
x4 1
dx
2
12
x
x 2 4x 3
4. (a)
x
dx
x 2 4x 13
x 2 2x
dx
x 3x 2 4
33.
2
1
y
yx
5
1
dx
1
32.
31.
53–54
|||| Evaluate the integral by completing the square and using
Formula 6.
■
yx
■
2
dx
2x
■
54.
■
■
■
■
y 4x
■
2
2x 1
dx
12x 7
■
■
■
noticed that the substitution t tanx2 will convert any
■
SECTION 7.5 STRATEGY FOR INTEGRATION
x
2
1
s1 t 2
and
sin
x
2
ty
(b) Show that
cos x 1 t2
1 t2
2t
1 t2
sin x and
64. Factor x 4 1 as a difference of squares by first adding and
subtracting the same quantity. Use this factorization to evaluate
x 1x 4 1 dx.
(c) Show that
dx 56–59
2
dt
1 t2
CAS
decomposition of the function
Use the substitution in Exercise 55 to transform the integrand into a rational function of t and then evaluate the integral.
dx
y 3 5 sin x
58.
y
2
3
■
1
dx
1 sin x cos x
■
60–61
■
||||
■
■
■
f x 1
57.
y 3 sin x 4 cos x dx
59.
y 2 sin x sin 2x
■
1
■
■
■
dx
CAS
■
■
1
,
x 6x 8
61. y x1
,
x1
■
■
12x 5 7x 3 13x 2 8
100x 80x 116x 4 80x 3 41x 2 20x 4
6
5
(b) Use part (a) to find x f x dx and graph f and its indefinite
integral on the same screen.
(c) Use the graph of f to discover the main features of the
graph of x f x dx.
a 5, b 10
2
66. (a) Find the partial fraction decomposition of the function
f x Find the area of the region under the given curve from
60. y 4x 3 27x 2 5x 32
30x 13x 4 50x 3 286x 2 299x 70
5
(b) Use part (a) to find x f x dx (by hand) and compare with
the result of using the CAS to integrate f directly.
Comment on any discrepancy.
a to b.
■
65. (a) Use a computer algebra system to find the partial fraction
||||
56.
PS
dP
Pr 1P S
Suppose an insect population with 10,000 females grows at a
rate of r 0.10 and 900 sterile males are added. Evaluate the
integral to give an equation relating the female population to
time. (Note that the resulting equation can’t be solved explicitly for P.)
t
s1 t 2
a 2, b 3
67. Suppose that F, G, and Q are polynomials and
■
■
■
■
■
■
■
■
■
Fx
Gx
Qx
Qx
62. Find the volume of the resulting solid if the region under the
curve y 1x 2 3x 2 from x 0 to x 1 is rotated
about (a) the x-axis and (b) the y-axis.
63. One method of slowing the growth of an insect population
without using pesticides is to introduce into the population a
number of sterile males that mate with fertile females but produce no offspring. If P represents the number of female insects
in a population, S the number of sterile males introduced each
generation, and r the population’s natural growth rate, then the
|||| 7.5
505
female population is related to time t by
rational function of sin x and cos x into an ordinary rational
function of t.
(a) If t tanx2, x , sketch a right triangle or use
trigonometric identities to show that
cos
❙❙❙❙
for all x except when Qx 0. Prove that Fx Gx for
all x. [Hint: Use continuity.]
68. If f is a quadratic function such that f 0 1 and
y
f x
dx
x 2x 13
is a rational function, find the value of f 0.
Strategy for Integration
As we have seen, integration is more challenging than differentiation. In finding the derivative 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 integrals 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.
506
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
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
memorized since they are easily derived. Formula 19 can be avoided by using partial fractions, and trigonometric substitutions can be used in place of Formula 20.
Table of Integration Formulas Constants of integration have been omitted.
x n1
n1
1.
yx
n
dx 3.
ye
x
dx e x
5.
n 1
1
dx ln x
x
2.
y
4.
ya
y sin x dx cos x
6.
y cos x dx sin x
7.
y sec x dx tan x
8.
y csc x dx cot x
9.
y sec x tan x dx sec x
10.
y csc x cot x dx csc x
11.
y sec x dx ln sec x tan x 12.
y csc x dx ln csc x cot x 13.
y tan x dx ln sec x 14.
y cot x dx ln sin x 15.
y sinh x dx cosh x
16.
y cosh x dx sinh x
17.
y
dx
1
x
tan1
x2 a2
a
a
18.
y
dx
x
sin1
a
sa 2 x 2
*19.
y
dx
1
xa
ln
2 x a
2a
xa
*20.
y
dx
ln x sx 2 a 2
sx a 2
2
2
x
dx ax
ln a
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 manipulation or
trigonometric identities will simplify the integrand and make the method of integration obvious. Here are some examples:
y sx (1 sx ) dx y (sx x) dx
tan sin y sec d y cos cos d
2
2
y sin cos d 12 y sin 2 d
SECTION 7.5 STRATEGY FOR INTEGRATION
❙❙❙❙
507
y sin x cos x dx y sin x 2 sin x cos x cos x dx
2
2
2
y 1 2 sin x cos x dx
2. Look for an Obvious Substitution Try to find some function u tx in the inte-
grand whose differential du tx dx also occurs, apart from a constant factor.
For instance, in the integral
y
x
dx
x2 1
we notice that if u x 2 1, then du 2x dx. Therefore, we use the substitution u x 2 1 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 x.
(a) Trigonometric functions. If f x 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 recommended in Section 7.2.
(b) Rational functions. If f is a rational function, we use the procedure of Section 7.4 involving partial fractions.
(c) Integration by parts. If f x is a product of a power of x (or a polynomial)
and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv 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 sx 2 a 2 occurs, we use a trigonometric substitution according to the
table in Section 7.3.
n
n
(ii) If s
ax b occurs, we use the rationalizing substitution u s
ax b.
n
More generally, this sometimes works for stx.
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 functions. Looking at Section 7.1, we see that it works on tan1x, sin1x, 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
y 1 cos x y
y
1
1 cos x
1 cos x
dx y
dx
1 cos x 1 cos x
1 cos 2x
1 cos x
dx sin 2x
y
csc 2x cos x
sin 2x
dx
508
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
(d) Relate the problem to previous problems. When you have built up some experience 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
instance, x tan 2x sec x dx is a challenging integral, but if we make use of the
identity tan 2x sec 2x 1, we can write
y tan x sec x dx y sec x dx y sec x dx
2
3
and if x sec 3x 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 evaluate 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.
EXAMPLE 1
tan 3x
y cos x dx
3
In Step 1 we rewrite the integral:
y
tan 3x
dx y tan 3x sec 3x dx
cos 3x
The integral is now of the form x tan m x sec n x dx with m odd, so we can use the advice in
Section 7.2.
Alternatively, if in Step 1 we had written
y
tan 3x
sin 3x 1
sin 3x
dx
dx
dx
y
y
cos 3x
cos 3x cos 3x
cos 6x
then we could have continued as follows with the substitution u cos x :
y
sin 3x
1 cos 2x
1 u2
dx
sin
x
dx
du
y
y
cos 6x
cos 6x
u6
y
EXAMPLE 2
ye
sx
u2 1
du y u 4 u 6 du
u6
dx
According to Step 3(d)(ii) we substitute u sx. Then x u 2, so dx 2u du and
ye
sx
dx 2 y ue u du
The integrand is now a product of u and the transcendental function e u so it can be integrated by parts.
SECTION 7.5 STRATEGY FOR INTEGRATION
❙❙❙❙
509
x5 1
dx
3x 2 10x
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, remembering that the first step is to divide.
EXAMPLE 3
yx
EXAMPLE 4
y xsln x
3
dx
Here Step 2 is all that is needed. We substitute u ln x because its differential is
du dxx, which occurs in the integral.
1x
dx
1x
Although the rationalizing substitution
EXAMPLE 5
y
u
1x
1x
works here [Step 3(d)(ii)], 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)]. Multiplying numerator and denominator by s1 x, we have
y
1x
1x
dx y
dx
1x
s1 x 2
y
1
x
dx y
dx
2
s1 x
s1 x 2
sin1x s1 x 2 C
Can We Integrate All Continuous Functions?
The question arises: Will our strategy for integration enable us to find the integral of every
2
continuous function? For example, can we use it to evaluate x e 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 functions. These are the polynomials, rational functions, power functions x a , exponential
functions a 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 composition. For instance, the function
f x x2 1
lncosh x xe sin 2x
x 3 2x 1
is an elementary function.
If f is an elementary function, then f is an elementary function but x f x dx need not
2
be an elementary function. Consider f x e x . Since f is continuous, its integral exists,
and if we define the function F by
x
2
Fx y e t dt
0
❙❙❙❙
510
CHAPTER 7 TECHNIQUES OF INTEGRATION
then we know from Part 1 of the Fundamental Theorem of Calculus that
Fx e x
2
2
Thus, f x e x has an antiderivative F , but it has been proved that F is not an elementary function. This means that no matter how hard we try, we will never succeed in evalu2
ating x e x dx in terms of the functions we know. (In Chapter 11, however, we will see how
2
to express x e x dx as an infinite series.) The same can be said of the following integrals:
y
ex
dx
x
y sx
3
y sinx
1 dx
y
2
dx
1
dx
ln x
y cose
y
x
dx
sin x
dx
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
1–80
1.
3.
5.
7.
9.
Evaluate the integral.
||||
sin x sec x
dx
tan x
y
y
2t
dt
t 32
2
0
y
e
dy
1 y2
1
3
1
r 4 ln r dr
11.
y sin 13.
y 1 x
15.
y
17.
y x sin x dx
19.
ye
21.
yt e
23.
y (1 sx ) dx
3
cos d
5
dx
12
0
2 32
x
dx
s1 x 2
2
xe x
dx
3 2t
1
0
4.
6.
8.
x1
dx
x 2 4x 5
y
2.
dt
8
10.
25.
yx
3x 2 2
dx
2
2x 8
26.
yx
27.
y cot x lnsin x dx
28.
y sin sat dt
29.
y
30.
y x
31.
y
32.
y
33.
y s3 2x x
34.
y
35.
y
36.
y sin 4x cos 3x dx
37.
y
38.
y
39.
y
40.
y s4y
41.
y tan d
42.
yx
43.
y e s1 e
44.
y s1 e
45.
yx e
46.
y 1e
47.
yx
48.
y
y tan d
3
x
dx
s3 x 4
y
arctan y
1
y
Exercises
y x csc x cot x dx
y
x1
dx
x 2 4x 5
4
0
x
dx
x4 x2 1
y
12.
y sin x coscos x dx
14.
y
16.
y
18.
y 1e
20.
ye
22.
y x sin
24.
y lnx
s1 ln x
dx
x ln x
x2
dx
s1 x 2
s22
0
e 2t
3
sx
4t
dt
dx
1
2
x dx
1 dx
3w 1
dw
w2
5
0
1
1x
dx
1x
2
dx
x 8 sin x dx
1
4
0
cos2 tan2 d
x
dx
1 x 2 s1 x 2
2
x
x
dx
5 x 3
dx
xa
dx
2
a2
3x 2 2
dx
3
2x 8
2
2
2
4x dx
s2x 1
dx
2x 3
2
4
4
0
2
1 4 cot x
dx
4 cot x
tan 5 sec 3 d
2
1
dy
4y 3
tan1x dx
1 ex
x
x
dx
dx
x
dx
x4 a4
❙❙❙❙
SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS
49.
1
y x s4x 1 dx
1
51.
y x s4x
53.
yx
55.
50.
yx
1
dx
s4x 1
y
69.
y 1e
71.
yx
x ln x
dx
2
1
73.
y x 2x
ln1 x dx
75.
y sin x sin 2x sin 3x dx
77.
y 1x
79.
y x sin
dx
1
52.
y x x
54.
y x sin x dx
y x 4 4 sx 1 dx
56.
y sx
57.
y x sx c dx
58.
yx
59.
ye
3x
1
dx
ex
60.
y x sx dx
61.
yx
10
x4
dx
16
62.
y x 1
63.
y sxe
64.
y
65.
2
2
1
dx
sinh mx dx
1
3
y
sx
dx
1
dx
sx 1 sx
|||| 7.6
66.
4
2
2
1
3
x3
10
3
4
y
3
2
arctanst
dt
st
3
67.
2
dx
lntan x
dx
sin x cos x
u3 1
du
u3 u2
■
1
e 2x
4
x
dx
x
dx
4x 2 3
1
sx
■
2
3
2
4
dx
dx
x cos x dx
■
■
■
■
2
1
68.
y 1 2e
70.
y
72.
y 1 st dt
74.
ye
76.
y x
78.
y sin x sec x dx
80.
y sin
x
ex
511
dx
lnx 1
dx
x2
st
3
x
dx
e x
2
bx sin 2x dx
sec x cos 2x
sin x cos x
dx
4
x cos 4 x
■
■
■
■
■
■
2
81. The functions y e x and y x 2e x don’t have elementary
2
antiderivatives, but y 2x 2 1e x does. Evaluate
2
x 2x 2 1e x dx.
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 antideriv2
atives of functions like e x 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, 30th ed. by Daniel Zwillinger (Boca Raton, FL: CRC
Press, 1995) (581 entries) or in Gradshteyn and Ryzhik’s Table of Integrals, Series,
and Products, 6e (New York: 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
1
V y 2 x arctan x dx
0
512
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
|||| The Table of Integrals appears on the
Reference Pages at the back of the book.
In the section of the Table of Integrals entitled Inverse Trigonometric Forms we locate
Formula 92:
y u tan
1
u du u2 1
u
tan1u C
2
2
Thus, the volume is
V 2 y
1
0
x2 1
x
x tan x dx 2
tan1x 2
2
1
1
0
x 1 tan x x 2 tan 1 1
1
2
1
1
0
24 1 12 2 EXAMPLE 2 Use the Table of Integrals to find
x2
y s5 4x
dx.
2
SOLUTION If we look at the section of the table entitled Forms involving sa 2 u 2, we see
that the closest entry is number 34:
u2
u
a2
u
du sa 2 u 2 sin1
2 u2
2
2
a
sa
y
C
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
u22 du
1
dx y
2
2
8
s5 4x
s5 u 2
y
u2
du
s5 u 2
Then we use Formula 34 with a 2 5 (so a s5 ):
x2
y s5 4x
2
dx 1
8
u2
y s5 u
2
du 1
8
u
5
u
s5 u 2 sin1
2
2
s5
x
5
2x
sin1
s5 4x 2 8
16
s5
EXAMPLE 3 Use the Table of Integrals to find
yx
3
C
C
sin x dx.
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 3:
yx
85.
yu
n
cos u du
u n sin u n y u n1 sin u du
3
sin x dx x 3 cos x 3 y x 2 cos x dx
We now need to evaluate x x 2 cos x dx. We can use the reduction formula in entry 85
with n 2, followed by entry 82:
yx
2
cos x dx x 2 sin x 2 y x sin x dx
x 2 sin x 2sin x x cos x K
SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS
❙❙❙❙
513
Combining these calculations, we get
yx
3
sin x dx x 3 cos x 3x 2 sin x 6x cos x 6 sin x C
where C 3K .
EXAMPLE 4 Use the Table of Integrals to find
y xsx
2
2x 4 dx.
SOLUTION Since the table gives forms involving sa 2 x 2, sa 2 x 2, and sx 2 a 2, but
not sax 2 bx c , we first complete the square:
x 2 2x 4 x 12 3
If we make the substitution u x 1 (so x u 1), the integrand will involve the
pattern sa 2 u 2 :
y xsx
2
2x 4 dx y u 1 su 2 3 du
y usu 2 3 du y su 2 3 du
The first integral is evaluated using the substitution t u 2 3:
y usu
21.
y sa
2
u 2 du u
sa 2 u 2
2
2
3 du 12 y st dt 12 23 t 32 13 u 2 332
For the second integral we use Formula 21 with a s3 :
2
a
ln (u sa 2 u 2 ) C
2
y su
2
3 du u
3
su 2 3 2 ln(u su 2 3 )
2
Thus
y xsx
2
2x 4 dx
13x 2 2x 432 x1
3
sx 2 2x 4 2 ln( x 1 sx 2 2x 4 ) C
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 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 13x 2. Using the substitution u 3x 2, an easy calculation
by hand gives
1
y 3x 2 dx 13 ln 3x 2 C
514
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
whereas Derive, Mathematica, and Maple all return the answer
1
3
ln3x 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
y xsx
2
2x 4 dx.
SOLUTION Maple responds with the answer
1
3
3
s3
arcsinh
1 x
2
3
x 2 2x 432 14 2x 2sx 2 2x 4 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 sx 2 1 )
|||| This is Equation 3.9.3.
Thus
arcsinh
s3
s3
1 x ln
1 x s 13 1 x2 1
3
3
|
ln
1
1 x s1 x2 3
s3
ln
1
ln( x 1 sx 2 2x 4 )
s3
[
]
The resulting extra term 32 ln(1s3 ) can be absorbed into the constant of integration.
Mathematica gives the answer
5
x
x2
6
6
3
sx 2 2x 4 3
1x
arcsinh
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
1
6
3
sx 2 2x 4 2x 2 x 5 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.
EXAMPLE 6 Use a CAS to evaluate
y xx
2
58 dx.
SOLUTION Maple and Mathematica give the same answer:
1
18
12
x 18 52 x 16 50x 14 1750
4375x 10 21875x 8 218750
x 6 156250x 4 390625
x2
3 x
3
2
SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS
❙❙❙❙
515
It’s clear that both systems must have expanded x 2 58 by the Binomial Theorem and
then integrated each term.
If we integrate by hand instead, using the substitution u x 2 5, we get
y xx
|||| Derive and the TI-89/92 also give this answer.
2
58 dx 181 x 2 59 C
For most purposes, this is a more convenient form of the answer.
EXAMPLE 7 Use a CAS to find
5
2
y sin x cos x dx.
SOLUTION In Example 2 in Section 7.2 we found that
y sin x cos x dx 5
1
1
3
2
cos 3x 25 cos 5x 17 cos7x C
Derive and Maple report the answer
8
17 sin 4x cos 3x 354 sin 2x cos 3x 105
cos 3x
whereas Mathematica produces
10
1
3
1
645 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 expressions using trigonometric identities, they ultimately produce the same form of the answer
as in Equation 1.
F
5
0
EXAMPLE 8 If f x x 60 sin 4x cos 5x, find the antiderivative F of f such that F0 0.
FIGURE 1
Graph F for 0 x 5. Where does F have extreme values and inflection points?
SOLUTION The antiderivative of f produced by Maple is
10
fª
F
32
2
Fx 12 x 2 203 sin 3x cos 6x 207 sin x cos 6x 47 cos 4x sin x 16
21 cos x sin x 21 sin x
f
0
and we note that F0 0. This expression could probably be simplified, but there’s no
need to do so because a computer algebra system can graph this version of F as easily as
any other version. A graph of F is shown in Figure 1. To locate the extreme values of F,
we graph its derivative F f in Figure 2 and observe that F has a local maximum
when x 2.3 and a local minimum when x 2.5. The graph of F f in Figure 2
shows that F has inflection points when x 0.7, 1.3, 1.8, 2.4, 3.3, and 3.9.
5
_7
FIGURE 2
|||| 7.6
Exercises
1–4
|||| Use the indicated entry in the Table of Integrals on the
Reference Pages to evaluate the integral.
1.
y
s7 2x 2
dx ; entry 33
x2
2.
3x
y s3 2x dx ;
entry 55
■
3.
y sec x dx ;
entry 71
4.
y e sin 3 d ;
entry 98
3
2
■
■
■
■
■
■
■
■
■
■
■
516
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
5–30 |||| Use the Table of Integrals on the Reference Pages to
evaluate the integral.
5.
7.
9.
11.
y
1
2x cos1x dx
0
3x
ye
yx
y
0
1
2
6.
8.
cos 4x dx
y
1
dx
x 2 s4x 2 7
3
2
dx
s4x 2 9
10.
12.
tan 1z
dz
z2
13.
y
15.
ye
17.
y y s6 4y 4y
19.
y sin x cos x lnsin x dx
23.
seche x dx
2
dy
2
y
ex
dx
3 e 2x
5
s2y 3
dy
y2
y
y
0
1
y sin
16.
y x sinx
18.
yx
20.
y e 1 2e
2
y
2
0
2
39.
y x s1 2x dx
40.
y sin 4x dx
41.
y tan5x dx
42.
y x 5sx
■
■
■
CAS
1
0
2
sec 4x dx
■
■
■
■
■
■
2
1 dx
■
■
■
■
43. Computer algebra systems sometimes need a helping hand
from human beings. Ask your CAS to evaluate
x
1 dx
2x
1 dx
6
instead. Why do you think it was successful with this form of
the integrand?
2x dx
CAS
44. Try to evaluate
x 4ex dx
t
y 1 ln x s1 x ln x
2
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.
sin t 3 dt
sec 2 tan 2
■
■
y 2 x s2
y s9 tan d
CAS
■
■
31. Find the volume of the solid obtained when the region under
the curve y x s4 x 2, 0 x 2, is rotated about the
y-axis.
32. The region under the curve y tan 2x 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 .
y tan x
4
30.
x 4 dx
10
2
■
38.
x s4x x dx
3
ye
■
cos 2x dx
3
28.
■
y sin x
If it doesn’t return an answer, ask it to try
x
1 dx
y sx
37.
dx
y
29.
y x 21 x 3 4 dx
y 2 x s4
26.
2x
36.
2
x5
dx
s2
s4 ln x
dx
x
y se
dx
■
cos3x 2 dx
x
2
■
2
sx dx
y sin
27.
y x 2s5 x
x 2 cos 3x dx
24.
y
35.
3
14.
22.
y sec x dx
25.
|||| 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.
y csc x2 dx
3
21.
35–42
2
t 2et dt
x
CAS
45–48
|||| Use a CAS to find an antiderivative F of f such
that F0 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
x x2 1
4
46. f x xex sin x,
5 x 5
0x
47. f x sin 4x cos 6x,
48. f x ■
■
x3 x
x6 1
■
■
■
■
■
■
■
■
■
■
DISCOVERY PROJECT PATTERNS IN INTEGRALS
❙❙❙❙
517
DISCOVERY PROJECT
CAS
Patterns in Integrals
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.
1
(i)
y x 2x 3 dx
(iii)
y x 2x 5 dx
1
1
(ii)
y x 1x 5 dx
(iv)
y x 22 dx
1
(b) Based on the pattern of your responses in part (a), guess the value of the integral
1
y x ax 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.
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 a and b is it valid?
3. (a) Use a computer algebra system to evaluate the following integrals.
(i)
(iv)
y ln x dx
(ii)
y x ln x dx
y x 3 ln x dx
(v)
y x 7 ln x dx
(iii)
y x 2 ln x dx
(b) Based on the pattern of your responses in part (a), guess the value of
y x n ln x dx
(c) Use integration by parts to prove the conjecture that you made in part (b). For what
values of n is it valid?
4. (a) Use a computer algebra system to evaluate the following integrals.
(i)
y xe x dx
(ii)
y x 2e x dx
(iv)
y x 4e x dx
(v)
y x 5e x dx
(iii)
y x 3e x dx
(b) Based on the pattern of your responses in part (a), guess the value of x x 6e x dx. 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
n x
yx e
dx
when n is a positive integer.
(d) Use mathematical induction to prove the conjecture you made in part (c).
518
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
|||| 7.7
Approximate Integration
There are two situations in which it is impossible to find the exact value of a definite
integral.
The first situation arises from the fact that in order to evaluate xab f x dx using the
Fundamental Theorem of Calculus we need to know an antiderivative of f . Sometimes,
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:
y
1
2
e x dx
0
y
1
s1 x 3 dx
1
The second situation arises when the function is determined from a scientific experiment through instrument readings or collected data. There may be no formula for the function (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 a, b
into n subintervals of equal length x b an, then we have
y
y
b
a
n
f x dx f x* x
i
i1
where x *i is any point in the ith subinterval x i1, x i . If x *i is chosen to be the left endpoint
of the interval, then x *i x i1 and we have
0
x¸
⁄
x™
x£
x¢
x
y
1
f x dx L n a
(a) Left endpoint approximation
f x
i1
x
i1
If f x 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 *i x i and we have
y
y
2
0
n
b
x¸
⁄
x™
x£
x¢
x
(b) Right endpoint approximation
y
b
a
n
f x dx Rn f x x
i
i1
[See Figure 1(b).] The approximations L n and Rn defined by Equations 1 and 2 are called
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 xi of
the subinterval x i1, x i . Figure 1(c) shows the midpoint approximation Mn , which appears
to be better than either L n or Rn.
Midpoint Rule
y
b
a
where
0
–x¡
–
x™
–x£
–x¢
(c) Midpoint approximation
FIGURE 1
x
and
f x dx Mn x f x1 f x2 f xn x ba
n
xi 12 x i1 x i midpoint of x i1, x i SECTION 7.7 APPROXIMATE INTEGRATION
❙❙❙❙
519
Another approximation, called the Trapezoidal Rule, results from averaging the approximations in Equations 1 and 2:
y
b
a
f x dx 1
2
n
n
f x i1 x i1
f x x
i
i1
x
2
n
f x i1 f x i i1
x
f x 0 f x 1 f x 1 f x 2 f x n1 f x n 2
x
f x 0 2 f x 1 2 f x 2 2 f x n1 f x n 2
y
Trapezoidal Rule
y
b
a
f x dx Tn x
f x0 2 f x1 2 f x2 2 f xn1 f x n 2
where x b an and xi a i x.
0
x¸
⁄
x™
x£
x¢
x
The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates
the case f x 0. The area of the trapezoid that lies above the ith subinterval is
FIGURE 2
x
Trapezoidal approximation
y=
f x i1 f x i 2
x
f x i1 f x i 2
and if we add the areas of all these trapezoids, we get the right side of the Trapezoidal
Rule.
1
x
EXAMPLE 1 Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n 5 to
approximate the integral x12 1x dx.
SOLUTION
(a) With n 5, a 1, and b 2, we have x 2 15 0.2, and so the Trapezoidal Rule gives
y
1
2
FIGURE 3
2
1
1
0.2
dx T5 f 1 2 f 1.2 2 f 1.4 2 f 1.6 2 f 1.8 f 2
x
2
0.1
1
2
2
2
2
1
1
1.2
1.4
1.6
1.8
2
0.695635
y=
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
2 1
y1 x dx x f 1.1 f 1.3 f 1.5 f 1.7 f 1.9
1
x
1
FIGURE 4
2
1
5
1
1
1
1
1
1.1
1.3
1.5
1.7
1.9
0.691908
This approximation is illustrated in Figure 4.
❙❙❙❙
520
CHAPTER 7 TECHNIQUES OF INTEGRATION
In Example 1 we deliberately chose an integral whose value can be computed explicitly so that we can see how accurate the Trapezoidal and Midpoint Rules are. By the
Fundamental Theorem of Calculus,
y
2
1
y
b
a
f x dx approximation error
1
2
dx ln x]1 ln 2 0.693147 . . .
x
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 5 are
ET 0.002488
and
EM 0.001239
In general, we have
b
ET y f x dx Tn
a
Module 5.1/5.2/7.7 allows you to compare approximation methods.
Approximations to y
2
1
and
b
EM y f x dx Mn
a
The following tables show the results of calculations similar to those in Example 1, but
for n 5, 10, and 20 and for the left and right endpoint approximations as well as the
Trapezoidal and Midpoint Rules.
1
dx
x
Corresponding errors
n
Ln
Rn
Tn
Mn
5
10
20
0.745635
0.718771
0.705803
0.645635
0.668771
0.680803
0.695635
0.693771
0.693303
0.691908
0.692835
0.693069
n
EL
ER
ET
EM
5
10
20
0.052488
0.025624
0.012656
0.047512
0.024376
0.012344
0.002488
0.000624
0.000156
0.001239
0.000312
0.000078
We can make several observations from these tables:
1. In all of the methods we get more accurate approximations when we increase the
2.
|||| It turns out that these observations are true
in most cases.
3.
4.
5.
value of n. (But very large values of n result in so many arithmetic operations that
we have to beware of accumulated round-off error.)
The errors in the left and right endpoint approximations are opposite in sign and
appear to decrease by a factor of about 2 when we double the value of n.
The Trapezoidal and Midpoint Rules are much more accurate than the endpoint
approximations.
The errors in the Trapezoidal and Midpoint Rules are opposite in sign and appear
to decrease by a factor of about 4 when we double the value of n.
The size of the error in the Midpoint Rule is about half the size of the error in the
Trapezoidal Rule.
Figure 5 shows why we can usually expect the Midpoint Rule to be more accurate than
the Trapezoidal Rule. The area of a typical rectangle in the Midpoint Rule is the same as
the trapezoid ABCD whose upper side is tangent to the graph at P. The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid AQRD used in
SECTION 7.7 APPROXIMATE INTEGRATION
C
B
D
x i-1
x–i
521
the Trapezoidal Rule. [The midpoint error (shaded red) is smaller than the trapezoidal error
(shaded blue).]
These observations are corroborated in the following error estimates, which are proved
in books on numerical analysis. Notice that Observation 4 corresponds to the n 2 in each
denominator because 2n2 4n 2. The fact that the estimates depend on the size of the
second derivative is not surprising if you look at Figure 5, because f x measures how
much the graph is curved. [Recall that f x measures how fast the slope of y f x
changes.]
P
A
❙❙❙❙
xi
3 Error Bounds Suppose f x K for a x b. If ET and EM are the errors
in the Trapezoidal and Midpoint Rules, then
C
R
E T
P
Kb a3
12n 2
E and
M
Kb a3
24n 2
B
Let’s apply this error estimate to the Trapezoidal Rule approximation in Example 1. If
f x 1x, then f x 1x 2 and f x 2x 3. Since 1 x 2, we have 1x 1, so
Q
A
f x D
FIGURE 5
2
2
2
3 x
13
Therefore, taking K 2, a 1, b 2, and n 5 in the error estimate (3), we see that
|||| K can be any number larger than all the
values of f x , but smaller values of K
give better error bounds.
E T
22 13
1
0.006667
2
125
150
Comparing this error estimate of 0.006667 with the actual error of about 0.002488, we see
that it can happen that the actual error is substantially less than the upper bound for the
error given by (3).
EXAMPLE 2 How large should we take n in order to guarantee that the Trapezoidal and
Midpoint Rule approximations for x12 1x dx are accurate to within 0.0001?
SOLUTION We saw in the preceding calculation that f x 2 for 1 x 2, so we can
take K 2, a 1, and b 2 in (3). Accuracy to within 0.0001 means that the size of
the error should be less than 0.0001. Therefore, we choose n so that
213
0.0001
12n 2
Solving the inequality for n, we get
n2 |||| It’s quite possible that a lower value for n
would suffice, but 41 is the smallest value for
which the error bound formula can guarantee us
accuracy to within 0.0001.
or
n
2
120.0001
1
40.8
s0.0006
Thus, n 41 will ensure the desired accuracy.
522
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
For the same accuracy with the Midpoint Rule we choose n so that
213
0.0001
24n 2
n
which gives
1
29
s0.0012
EXAMPLE 3
y
2
(a) Use the Midpoint Rule with n 10 to approximate the integral x01 e x dx.
(b) Give an upper bound for the error involved in this approximation.
y=e x
SOLUTION
2
(a) Since a 0, b 1, and n 10, the Midpoint Rule gives
y
1
0
2
e x dx x f 0.05 f 0.15 f 0.85 f 0.95
0.1
e 0.0025 e 0.0225 e 0.0625 e 0.1225 e 0.2025 e 0.3025
e 0.4225 e 0.5625 e 0.7225 e 0.9025
0
1
1.460393
x
FIGURE 6
Figure 6 illustrates this approximation.
2
2
2
(b) Since f x e x , we have f x 2xe x and f x 2 4x 2 e x . Also, since
0 x 1, we have x 2 1 and so
2
0 f x 2 4x 2 e x 6e
Taking K 6e, a 0, b 1, and n 10 in the error estimate (3), we see that an upper
bound for the error is
6e13
e
0.007
24102
400
|||| Error estimates are upper bounds for
the error. They give theoretical, worst-case
scenarios. The actual error in this case turns
out to be about 0.0023.
Simpson’s Rule
Another rule for approximate integration results from using parabolas instead of straight
line segments to approximate a curve. As before, we divide a, b into n subintervals of
equal length h x b an, but this time we assume that n is an even number. Then
on each consecutive pair of intervals we approximate the curve y f x 0 by a parabola
as shown in Figure 7. If yi f x i , then Pi x i , yi is the point on the curve lying above x i.
A typical parabola passes through three consecutive points Pi , Pi1 , and Pi2 .
y
y
P¸
P¡
P∞
P¸ (_h, y¸)
Pß
P¡ (0, › )
P™
P£
0
a=x¸
FIGURE 7
⁄
x™
x£
P™ (h, fi )
P¢
x¢
x∞
xß=b
x
_h
FIGURE 8
0
h
x
SECTION 7.7 APPROXIMATE INTEGRATION
❙❙❙❙
523
To simplify our calculations, we first consider the case where x 0 h, x 1 0, and
x 2 h. (See Figure 8.) We know that the equation of the parabola through P0 , P1 , and P2
is of the form y Ax 2 Bx C and so the area under the parabola from x h to
x h is
|||| Here we have used Theorem 5.5.7. Notice
that Ax 2 C is even and Bx is odd.
y
h
h
h
Ax 2 Bx C dx 2 y Ax 2 C dx
0
2 A
2 A
x3
Cx
3
3
h
0
h
h
Ch 2Ah 2 6C
3
3
But, since the parabola passes through P0h, y0 , P10, y1 , and P2h, y2 , we have
y0 Ah2 Bh C Ah 2 Bh C
y1 C
y2 Ah 2 Bh C
and therefore
y0 4y1 y2 2Ah 2 6C
Thus, we can rewrite the area under the parabola as
h
y0 4y1 y2 3
Now, by shifting this parabola horizontally we do not change the area under it. This means
that the area under the parabola through P0 , P1 , and P2 from x x 0 to x x 2 in Figure 7
is still
h
y0 4y1 y2 3
Similarly, the area under the parabola through P2 , P3 , and P4 from x x 2 to x x 4 is
h
y2 4y3 y4 3
If we compute the areas under all the parabolas in this manner and add the results, we get
y
b
a
f x dx h
h
h
y0 4y1 y2 y2 4y3 y4 yn2 4yn1 yn 3
3
3
h
y0 4y1 2y2 4y3 2y4 2yn2 4yn1 yn 3
Although we have derived this approximation for the case in which f x 0, it is a reasonable approximation for any continuous function f and is called Simpson’s Rule after
the English mathematician Thomas Simpson (1710–1761). Note the pattern of coefficients: 1, 4, 2, 4, 2, 4, 2, . . . , 4, 2, 4, 1.
524
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
Simpson’s Rule
|||| Thomas Simpson was a weaver who taught
himself mathematics and went on to become one
of the best English mathematicians of the 18th
century. What we call Simpson’s Rule was
actually known to Cavalieri and Gregory in the
17th century, but Simpson popularized it in his
best-selling calculus textbook, entitled A New
Treatise of Fluxions.
y
b
a
f x dx Sn x
f x 0 4 f x 1 2 f x 2 4 f x 3 3
2 f xn2 4 f xn1 f xn where n is even and x b an.
EXAMPLE 4 Use Simpson’s Rule with n 10 to approximate
x12 1x dx.
SOLUTION Putting f x 1x, n 10, and x 0.1 in Simpson’s Rule, we obtain
y
2
1
1
dx S10
x
x
f 1 4 f 1.1 2 f 1.2 4 f 1.3 2 f 1.8 4 f 1.9 f 2
3
0.1
3
1
4
2
4
2
4
2
4
2
4
1
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
0.693150
Notice that, in Example 4, Simpson’s Rule gives us a much better approximation
S10 0.693150 to the true value of the integral ln 2 0.693147. . . than does the
Trapezoidal Rule T10 0.693771 or the Midpoint Rule M10 0.692835. It turns out
(see Exercise 48) that the approximations in Simpson’s Rule are weighted averages of
those in the Trapezoidal and Midpoint Rules:
S2n 13 Tn 23 Mn
(Recall that ET and EM usually have opposite signs and EM is about half the size of ET .)
In many applications of calculus we need to evaluate an integral even if no explicit formula is known for y as a function of x. A function may be given graphically or as a table
of values of collected data. If there is evidence that the values are not changing rapidly,
then the Trapezoidal Rule or Simpson’s Rule can still be used to find an approximate value
for xab y dx, the integral of y with respect to x.
EXAMPLE 5 Figure 9 shows data traffic on the link from the United States to SWITCH,
the Swiss academic and research network, on February 10, 1998. Dt is the data
throughput, measured in megabits per second Mbs. Use Simpson’s Rule to estimate
the total amount of data transmitted on the link up to noon on that day.
D
8
6
4
2
FIGURE 9
0
3
6
9
12
15
18
21
24 t (hours)
SECTION 7.7 APPROXIMATE INTEGRATION
❙❙❙❙
525
SOLUTION Because we want the units to be consistent and Dt is measured in megabits
per second, we convert the units for t from hours to seconds. If we let At be the
amount of data (in megabits) transmitted by time t, where t is measured in seconds, then
At Dt. So, by the Net Change Theorem (see Section 5.4), the total amount of data
transmitted by noon (when t 12 60 2 43,200) is
A43,200 y
43,200
0
Dt dt
We estimate the values of Dt at hourly intervals from the graph and compile them in
the table.
t hours
t seconds
Dt
t hours
t seconds
Dt
0
1
2
3
4
5
6
0
3,600
7,200
10,800
14,400
18,000
21,600
3.2
2.7
1.9
1.7
1.3
1.0
1.1
7
8
9
10
11
12
25,200
28,800
32,400
36,000
39,600
43,200
1.3
2.8
5.7
7.1
7.7
7.9
Then we use Simpson’s Rule with n 12 and t 3600 to estimate the integral:
y
43,200
0
At dt t
D0 4D3600 2D7200 4D39,600 D43,200
3
3600
3.2 42.7 21.9 41.7 21.3 41.0
3
21.1 41.3 22.8 45.7 27.1 47.7 7.9
143,880
Thus, the total amount of data transmitted up to noon is about 144,000 megabits, or
144 gigabits.
In Exercises 27 and 28 you are asked to demonstrate, in particular cases, that the error
in Simpson’s Rule decreases by a factor of about 16 when n is doubled. That is consistent
with the appearance of n 4 in the denominator of the following error estimate for Simpson’s
Rule. It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules, but
it uses the fourth derivative of f .
f
4
x K for a x b.
If ES is the error involved in using Simpson’s Rule, then
4 Error Bound for Simpson’s Rule Suppose that
ES Kb a5
180n 4
526
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
EXAMPLE 6 How large should we take n in order to guarantee that the Simpson’s Rule
approximation for x12 1x dx is accurate to within 0.0001?
SOLUTION If f x 1x, then f 4x 24x 5. Since x 1, we have 1x 1 and so
f
|||| Many calculators and computer algebra systems have a built-in algorithm that computes an
approximation of a definite integral. Some of
these machines use Simpson’s Rule; others use
more sophisticated techniques such as adaptive
numerical integration. This means that if a function fluctuates much more on a certain part of
the interval than it does elsewhere, then that
part gets divided into more subintervals. This
strategy reduces the number of calculations
required to achieve a prescribed accuracy.
4
x 24
24
x5
Therefore, we can take K 24 in (4). Thus, for an error less than 0.0001 we should
choose n so that
2415
0.0001
180n 4
n4 This gives
n
or
24
1800.0001
1
6.04
s0.00075
4
Therefore, n 8 (n must be even) gives the desired accuracy. (Compare this with
Example 2, where we obtained n 41 for the Trapezoidal Rule and n 29 for the
Midpoint Rule.)
EXAMPLE 7
2
(a) Use Simpson’s Rule with n 10 to approximate the integral x01 e x dx.
(b) Estimate the error involved in this approximation.
SOLUTION
(a) If n 10, then x 0.1 and Simpson’s Rule gives
|||| Figure 10 illustrates the calculation in
Example 7. Notice that the parabolic arcs are
2
so close to the graph of y e x that they are
practically indistinguishable from it.
y
y
1
0
e x dx 2
x
f 0 4 f 0.1 2 f 0.2 2 f 0.8 4 f 0.9 f 1
3
0.1 0
e 4e 0.01 2e 0.04 4e 0.09 2e 0.16 4e 0.25 2e 0.36
3
4e 0.49 2e 0.64 4e 0.81 e 1 1.462681
2
(b) The fourth derivative of f x e x is
y=e ≈
f 4x 12 48x 2 16x 4 e x
2
and so, since 0 x 1, we have
0 f 4x 12 48 16e 1 76e
0
FIGURE 10
1
x
Therefore, putting K 76e, a 0, b 1, and n 10 in (4), we see that the error is at
most
76e15
0.000115
180104
(Compare this with Example 3.) Thus, correct to three decimal places, we have
y
1
0
2
e x dx 1.463
❙❙❙❙
SECTION 7.7 APPROXIMATE INTEGRATION
|||| 7.7
1. Let I Exercises
x04 f x dx, where
(Round your answers to six decimal places.) Compare your results
to the actual value to determine the error in each approximation.
f is the function whose graph is
shown.
(a) Use the graph to find L 2 , R2, and M2.
(b) Are these underestimates or overestimates of I ?
(c) Use the graph to find T2. How does it compare with I ?
(d) For any value of n, list the numbers L n , Rn , Mn , Tn , and I in
increasing order.
y
3
5.
x 2 sin x dx, n 8
■
■
■
■
6.
■
y
1
0
■
esx dx, n 6
■
■
■
■
■
7–18 |||| Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and
(c) Simpson’s Rule to approximate the given integral with the
specified value of n. (Round your answers to six decimal places.)
f
7.
y
1
9.
y
11.
y
13.
y
15.
y
17.
y
1
2
3
4 x
were used to estimate x02 f x dx, where f is the function whose
graph is shown. The estimates were 0.7811, 0.8675, 0.8632,
and 0.9540, and the same number of subintervals were used in
each case.
(a) Which rule produced which estimate?
(b) Between which two approximations does the true value of
x02 f x dx lie?
■
y
2
4
1 x 2 dx,
s
0
ln x
dx,
1x
2
1
12
0
2
n 10
sine t2 dt, n 8
e 1x dx,
1
n4
cos x
dx,
x
5
1
n8
1
dy,
1 y5
3
0
n8
■
■
n6
■
■
■
8.
y
10.
y
12.
y
14.
y
16.
y
18.
y
■
12
0
3
0
4
0
4
0
6
4
4
2
■
sinx 2 dx, n 4
dt
,
1 t2 t4
n6
s1 sx dx, n 8
sx sin x dx, n 8
lnx 3 2 dx,
ex
dx,
x
■
n 10
n 10
■
■
■
19. (a) Find the approximations T10 and M10 for the integral
x02 ex
2
dx.
(b) Estimate the errors in the approximations of part (a).
(c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to
within 0.00001?
1
y=ƒ
2
20. (a) Find the approximations T8 and M8 for x01 cosx 2 dx.
x
(b) Estimate the errors involved in the approximations of
part (a).
(c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to
within 0.00001?
1
2
; 3. Estimate x0 cosx dx using (a) the Trapezoidal Rule and
(b) the Midpoint Rule, each with n 4. From a graph of the
integrand, decide whether your answers are underestimates or
overestimates. What can you conclude about the true value of
the integral?
21. (a) Find the approximations T10 and S10 for x01 e x dx and the
corresponding errors ET and ES.
(b) Compare the actual errors in part (a) with the error estimates given by (3) and (4).
(c) How large do we have to choose n so that the approximations Tn , Mn , and Sn to the integral in part (a) are accurate to
within 0.00001?
2
; 4. Draw the graph of f x sinx 2 in the viewing rectangle
0, 1
by 0, 0.5
and let I x01 f x dx.
(a) Use the graph to decide whether L 2 , R2 , M2, and T2 underestimate or overestimate I .
(b) For any value of n, list the numbers L n , Rn , Mn , Tn , and I in
increasing order.
(c) Compute L 5 , R5 , M5, and T5. From the graph, which do you
think gives the best estimate of I ?
22. How large should n be to guarantee that the Simpson’s Rule
2
approximation to x01 e x dx is accurate to within 0.00001?
CAS
Use (a) the Midpoint Rule and (b) Simpson’s Rule to
approximate the given integral with the specified value of n.
||||
0
■
2. The left, right, Trapezoidal, and Midpoint Rule approximations
0
y
2
0
5–6
527
23. The trouble with the error estimates is that it is often very diffi-
cult to compute four derivatives and obtain a good upper bound
K for f 4x by hand. But computer algebra systems have no
528
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
problem computing f 4 and graphing it, so we can easily find
a value for K from a machine graph. This exercise deals
with approximations to the integral I x02 f x dx, where
f x e cos x.
(a) Use a graph to get a good upper bound for f x .
(b) Use M10 to approximate I .
(c) Use part (a) to estimate the error in part (b).
(d) Use the built-in numerical integration capability of your
CAS to approximate I .
(e) How does the actual error compare with the error estimate
in part (c)?
(f) Use a graph to get a good upper bound for f 4x .
(g) Use S10 to approximate I .
(h) Use part (f) to estimate the error in part (g).
(i) How does the actual error compare with the error estimate
in part (h)?
(j) How large should n be to guarantee that the size of the
error in using Sn is less than 0.0001?
CAS
Use Simpson’s Rule to estimate the area of the pool.
6.2
|||| Find the approximations L n , Rn , Tn , and Mn for n 4, 8,
and 16. Then compute the corresponding errors EL , ER, ET , and EM.
(Round your answers to six decimal places. You may wish to use
the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when
n is doubled?
25–26
x 3 dx
0
■
■
26.
■
■
■
■
y
2
0
■
e x dx
■
■
■
■
■
|||| Find the approximations Tn , Mn , and Sn for n 6 and 12.
Then compute the corresponding errors ET, EM, and ES. (Round
your answers to six decimal places. You may wish to use the sum
command on a computer algebra system.) What observations can
you make? In particular, what happens to the errors when n is
doubled?
27–28
27.
■
y
4
28.
sx dx
1
■
■
■
■
■
■
y
2
1
■
f x
x
f x
0.0
0.4
0.8
1.2
1.6
6.8
6.5
6.3
6.4
6.9
2.0
2.4
2.8
3.2
7.6
8.4
8.8
9.0
32. A radar gun was used to record the speed of a runner during
the first 5 seconds of a race (see the table). Use Simpson’s
Rule to estimate the distance the runner covered during those
5 seconds.
t (s)
v (ms)
t (s)
v (ms)
0
0.5
1.0
1.5
2.0
2.5
0
4.67
7.34
8.86
9.73
10.22
3.0
3.5
4.0
4.5
5.0
10.51
10.67
10.76
10.81
10.81
33. The graph of the acceleration at of a car measured in fts2 is
shown. Use Simpson’s Rule to estimate the increase in the
velocity of the car during the 6-second time interval.
a
12
xe x dx
■
x
(b) If it is known that 4 f x 1 for all x, estimate the
error involved in the approximation in part (a).
1
1
4.8
value of the integral x03.2 f x dx.
1
y
5.6 5.0 4.8
31. (a) Use the Midpoint Rule and the given data to estimate the
24. Repeat Exercise 23 for the integral y s4 x 3 dx.
25.
7.2
6.8
■
■
■
29. Estimate the area under the graph in the figure by using (a) the
Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s
Rule, each with n 4.
8
4
0
2
4
6 t
y
34. Water leaked from a tank at a rate of rt liters per hour, where
the graph of r is as shown. Use Simpson’s Rule to estimate the
total amount of water that leaked out during the first six hours.
r
4
1
0
1
2
3
4 x
30. The widths (in meters) of a kidney-shaped swimming pool
were measured at 2-meter intervals as indicated in the figure.
2
0
2
4
6 t
SECTION 7.7 APPROXIMATE INTEGRATION
t
P
t
P
0:00
0:30
1:00
1:30
2:00
2:30
3:00
1814
1735
1686
1646
1637
1609
1604
3:30
4:00
4:30
5:00
5:30
6:00
1611
1621
1666
1745
1886
2052
529
3
1 x 3, y 0, x 0 ,
39. The region bounded by the curves y s
35. The table (supplied by San Diego Gas and Electric) gives the
power consumption in megawatts in San Diego County from
midnight to 6:00 A.M. on December 8, 1999. Use Simpson’s
Rule to estimate the energy used during that time period. (Use
the fact that power is the derivative of energy.)
❙❙❙❙
and x 2 is rotated about the x-axis. Use Simpson’s Rule with
n 10 to estimate the volume of the resulting solid.
CAS
40. The figure shows a pendulum with length L that makes a maxi-
mum angle 0 with the vertical. Using Newton’s Second Law
it can be shown that the period T (the time for one complete
swing) is given by
T4
L
t
y
2
0
dx
s1 k 2 sin 2x
where k sin( 12 0 ) and t is the acceleration due to gravity.
If L 1 m and 0 42, use Simpson’s Rule with n 10 to
find the period.
36. Shown is the graph of traffic on an Internet service provider’s
T1 data line from midnight to 8:00 A.M. D is the data throughput, measured in megabits per second. Use Simpson’s Rule to
estimate the total amount of data transmitted during that time
period.
¨¸
D
0.8
41. The intensity of light with wavelength traveling through
a diffraction grating with N slits at an angle is given by
I N 2 sin 2kk 2, where k Nd sin and d is the
distance between adjacent slits. A helium-neon laser with
wavelength 632.8 109 m is emitting a narrow band of
light, given by 106 106, through a grating with
10,000 slits spaced 104 m apart. Use the Midpoint Rule with
10
n 10 to estimate the total light intensity x10
I d emerging from the grating.
0.4
6
0
2
4
6
8 t (hours)
6
37. If the region shown in the figure is rotated about the y-axis to
form a solid, use Simpson’s Rule with n 8 to estimate the
volume of the solid.
x020 cos x dx. Compare your result to the actual value. Can
you explain the discrepancy?
43. Sketch the graph of a continuous function on 0, 2
for which
the Trapezoidal Rule with n 2 is more accurate than the
Midpoint Rule.
y
4
44. Sketch the graph of a continuous function on 0, 2
for which
2
0
42. Use the Trapezoidal Rule with n 10 to approximate
the right endpoint approximation with n 2 is more accurate
than Simpson’s Rule.
2
4
6
10 x
8
45. If f is a positive function and f x 0 for a x b, show
that
b
38. The table shows values of a force function f x where x is
measured in meters and f x in newtons. Use Simpson’s Rule
to estimate the work done by the force in moving an object a
distance of 18 m.
Tn y f x dx Mn
a
46. Show that if f is a polynomial of degree 3 or lower, then
Simpson’s Rule gives the exact value of xab f x dx.
x
0
3
6
9
12
15
18
47. Show that 2 Tn Mn T2n.
f x
9.8
9.1
8.5
8.0
7.7
7.5
7.4
48. Show that 3 Tn 3 Mn S2n.
1
1
2
530
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
|||| 7.8
Improper Integrals
In defining a definite integral xab f x dx we dealt with a function f defined on a finite interval a, b
and we assumed that f does not have an infinite discontinuity (see Section 5.2).
In this section we extend the concept of a definite integral to the case where the interval is
infinite and also to the case where f has an infinite discontinuity in a, b
. In either case
the integral is called an improper integral. One of the most important applications of this
idea, probability distributions, will be studied in Section 8.5.
Type I: Infinite Intervals
Consider the infinite region S that lies under the curve y 1x 2, above the x-axis, and to
the right of the line x 1. You might think that, since S is infinite in extent, its area must
be infinite, but let’s take a closer look. The area of the part of S that lies to the left of the
line x t (shaded in Figure 1) is
Try painting a fence that never ends.
Resources / Module 6
/ How To Calculate
/ Start of Improper Integrals
At y
t
1
1
1
dx x2
x
t
1
1
1
t
Notice that At 1 no matter how large t is chosen.
y
y=
1
≈
area=1 -
x=1
0
FIGURE 1
t
1
We also observe that
lim At lim 1 tl
1
t
tl
1
t
x
1
The area of the shaded region approaches 1 as t l (see Figure 2), so we say that the area
of the infinite region S is equal to 1 and we write
y
1
y
y
y
area= 21
0
1
2
x
y
area= 45
area= 23
0
1
t 1
dx 1
y
2 dx tlim
l
1
x
x2
1
3
x
0
1
area=1
5 x
0
1
x
FIGURE 2
Using this example as a guide, we define the integral of f (not necessarily a positive
function) over an infinite interval as the limit of integrals over finite intervals.
SECTION 7.8 IMPROPER INTEGRALS
1
❙❙❙❙
531
Definition of an Improper Integral of Type 1
(a) If xat f x dx exists for every number t a, then
y
a
t
f x dx lim y f x dx
tl
a
provided this limit exists (as a finite number).
(b) If xtb f x dx exists for every number t b, then
y
b
f x dx lim
t l
y
t
b
f x dx
provided this limit exists (as a finite number).
b
The improper integrals xa f x dx and x
f x dx are called convergent if the
corresponding limit exists and divergent if the limit does not exist.
a
(c) If both xa f x dx and x
f x dx are convergent, then we define
y
f x dx y
a
f x dx y f x dx
a
In part (c) any real number a can be used (see Exercise 74).
Any of the improper integrals in Definition 1 can be interpreted as an area provided that
f is a positive function. For instance, in case (a) if f x 0 and the integral xa f x dx
is convergent, then we define the area of the region S x, y x a, 0 y f x in
Figure 3 to be
AS y f x dx
a
y
y=ƒ
S
FIGURE 3
0
a
x
This is appropriate because xa f x dx is the limit as t l of the area under the graph of
f from a to t.
EXAMPLE 1 Determine whether the integral
x1 1x dx is convergent or divergent.
SOLUTION According to part (a) of Definition 1, we have
y
1
1
t 1
dx lim y
dx lim ln x
tl 1 x
tl
x
]
t
1
lim ln t ln 1 lim ln t tl
tl
The limit does not exist as a finite number and so the improper integral x1 1x dx is
divergent.
532
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
y
Let’s compare the result of Example 1 with the example given at the beginning of this
section:
1
y=
≈
y
1
finite area
0
x
1
FIGURE 4
y=
1
x
y
1
dx diverges
x
1
Geometrically, this says that although the curves y 1x 2 and y 1x look very similar
for x 0, the region under y 1x 2 to the right of x 1 (the shaded region in Figure 4)
has finite area whereas the corresponding region under y 1x (in Figure 5) has infinite
area. Note that both 1x 2 and 1x approach 0 as x l but 1x 2 approaches 0 faster than
1x. The values of 1x don’t decrease fast enough for its integral to have a finite value.
EXAMPLE 2 Evaluate
y
1
dx converges
x2
y
0
xe x dx.
SOLUTION Using part (b) of Definition 1, we have
y
0
infinite area
xe x dx lim
t l
y
t
0
xe x dx
We integrate by parts with u x, dv e x dx so that du dx, v e x :
0
1
x
y
0
t
FIGURE 5
]
0
0
xe x dx xe x t y e x dx
t
te t 1 e t
We know that e t l 0 as t l , and by l’Hospital’s Rule we have
lim te t lim
t l
t l
t
1
t t lim
l et
e
lim e t 0
t l
Therefore
y
0
xe x dx lim te t 1 e t t l
0 1 0 1
EXAMPLE 3 Evaluate
y
1
dx.
1 x2
SOLUTION It’s convenient to choose a 0 in Definition 1(c):
y
1
0
1
1
dx y
2 dx y
2 dx
1 x
0 1 x
1 x2
We must now evaluate the integrals on the right side separately:
y
0
t
1
dx
dx lim y
lim tan1x
tl 0 1 x2
tl
1 x2
]
t
0
lim tan 1t tan1 0 lim tan1t tl
tl
2
SECTION 7.8 IMPROPER INTEGRALS
y
0
1
0
dx
lim tan1x
y
2 dx t lim
l t 1 x 2
t l
1x
]
❙❙❙❙
533
0
t
lim tan 1 0 tan 1t
t l
0 2
2
Since both of these integrals are convergent, the given integral is convergent and
y=
1
1+≈
y
y
area=π
0
FIGURE 6
x
1
2 dx 1x
2
2
Since 11 x 2 0, the given improper integral can be interpreted as the area of
the infinite region that lies under the curve y 11 x 2 and above the x-axis (see
Figure 6).
EXAMPLE 4 For what values of p is the integral
y
1
convergent?
1
dx
xp
SOLUTION We know from Example 1 that if p 1, then the integral is divergent, so let’s
assume that p 1. Then
1
t
y1 x p dx tlim
y x p dx
l 1
xp1
lim
t l p 1
lim
tl
xt
x1
1
1
1
1 p t p1
If p 1, then p 1 0, so as t l , t p1 l and 1t p1 l 0. Therefore
y
1
1
1
p dx x
p1
if p 1
and so the integral converges. But if p 1, then p 1 0 and so
1
t
p1
t 1p l as t l and the integral diverges.
We summarize the result of Example 4 for future reference:
2
y
1
1
dx is convergent if p 1 and divergent if p 1.
xp
Type 2: Discontinuous Integrands
Suppose that f is a positive continuous function defined on a finite interval a, b but has
a vertical asymptote at b. Let S be the unbounded region under the graph of f and above
the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a
534
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
y
horizontal direction. Here the region is infinite in a vertical direction.) The area of the part
of S between a and t (the shaded region in Figure 7) is
y=ƒ
x=b
t
At y f x dx
a
0
a
x
t b
If it happens that At approaches a definite number A as t l b, then we say that the
area of the region S is A and we write
FIGURE 7
y
b
a
t
f x dx lim ya f x dx
tlb
We use this equation to define an improper integral of Type 2 even when f is not a positive function, no matter what type of discontinuity f has at b.
3
|||| Parts (b) and (c) of Definition 3 are illustrated
in Figures 8 and 9 for the case where f x 0
and f has vertical asymptotes at a and c,
respectively.
Definition of an Improper Integral of Type 2
(a) If f is continuous on a, b and is discontinuous at b, then
y
b
a
t
f x dx lim ya f x dx
tlb
if this limit exists (as a finite number).
(b) If f is continuous on a, b and is discontinuous at a, then
y
y
b
a
b
f x dx lim y f x dx
tla
t
if this limit exists (as a finite number).
0
a t
b
The improper integral xab f x dx is called convergent if the corresponding limit
exists and divergent if the limit does not exist.
x
(c) If f has a discontinuity at c, where a c b, and both xac f x dx and
xcb f x dx are convergent, then we define
FIGURE 8
y
y
b
a
EXAMPLE 5 Find
0
a
5
2
FIGURE 9
b
c
1
dx.
sx 2
y
5
2
y
y=
a
SOLUTION We note first that the given integral is improper because f x 1sx 2
has the vertical asymptote x 2. Since the infinite discontinuity occurs at the left endpoint of 2, 5, we use part (b) of Definition 3:
b x
c
y
c
f x dx y f x dx y f x dx
1
œ„„„„
x-2
dx
5
dx
lim y
t l2
t sx 2
sx 2
lim 2sx 2
t l2
]
5
t
lim 2(s3 st 2 )
t l2
2s3
3
area=2 œ„
0
1
FIGURE 10
2
3
4
5
x
Thus, the given improper integral is convergent and, since the integrand is positive, we
can interpret the value of the integral as the area of the shaded region in Figure 10.
SECTION 7.8 IMPROPER INTEGRALS
y
EXAMPLE 6 Determine whether
/2
0
❙❙❙❙
535
sec x dx converges or diverges.
SOLUTION Note that the given integral is improper because lim x l /2 sec x . Using
part (a) of Definition 3 and Formula 14 from the Table of Integrals, we have
y
/2
0
sec x dx lim
t l /2
y
t
0
sec x dx
lim ln sec x tan x
t l /2
]
t
0
lim lnsec t tan t ln 1
t l /2
because sec t l and tan t l as t l 2. Thus, the given improper integral is
divergent.
EXAMPLE 7 Evaluate
y
dx
if possible.
x1
3
0
SOLUTION Observe that the line x 1 is a vertical asymptote of the integrand. Since it
occurs in the middle of the interval 0, 3, we must use part (c) of Definition 3 with
c 1:
3
dx
1
dx
3
dx
y0 x 1 y0 x 1 y1 x 1
y
where
1
0
dx
t
dx
lim y
lim ln x 1
t
l1
t l1
0
x1
x1
lim (ln t 1 ln 1
t l1
]
t
0
)
lim ln1 t t l1
because 1 t l 0 as t l 1. Thus, x01 dxx 1 is divergent. This implies that
x03 dxx 1 is divergent. [We do not need to evaluate x13 dxx 1.]
|
If we had not noticed the asymptote x 1 in Example 7 and had instead
confused the integral with an ordinary integral, then we might have made the following
erroneous calculation:
WARNING
■
y
3
0
dx
ln x 1
x1
]
3
0
ln 2 ln 1 ln 2
This is wrong because the integral is improper and must be calculated in terms of limits.
From now on, whenever you meet the symbol xab f x dx you must decide, by looking
at the function f on a, b, whether it is an ordinary definite integral or an improper
integral.
EXAMPLE 8 Evaluate
y
1
0
ln x dx.
SOLUTION We know that the function f x ln x has a vertical asymptote at 0 since
lim x l 0 ln x . Thus, the given integral is improper and we have
y
1
0
1
ln x dx lim y ln x dx
tl0
t
536
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
Now we integrate by parts with u ln x, dv dx, du dxx, and v x :
y
1
t
]
1
1
ln x dx x ln x t y dx
t
1 ln 1 t ln t 1 t
t ln t 1 t
To find the limit of the first term we use l’Hospital’s Rule:
lim t ln t lim
t l 0
tl0
ln t
1t
y
lim
tl0
0
1
1t
1t 2
lim t 0
x
tl0
area=1
y
Therefore
1
0
ln x dx lim t ln t 1 t
tl0
0 1 0 1
y=ln x
Figure 11 shows the geometric interpretation of this result. The area of the shaded region
above y ln x and below the x-axis is 1.
FIGURE 11
A Comparison Test for Improper Integrals
Sometimes it is impossible to find the exact value of an improper integral and yet it
is important to know whether it is convergent or divergent. In such cases the following theorem is useful. Although we state it for Type 1 integrals, a similar theorem is true for
Type 2 integrals.
Comparison Theorem Suppose that f and t are continuous functions with
f x tx 0 for x a.
(a) If xa f x dx is convergent, then xa tx dx is convergent.
(b) If xa tx dx is divergent, then xa f x dx is divergent.
y
f
g
0
a
FIGURE 12
x
We omit the proof of the Comparison Theorem, but Figure 12 makes it seem plausible.
If the area under the top curve y f x is finite, then so is the area under the bottom curve
y tx. And if the area under y tx is infinite, then so is the area under y f x.
[Note that the reverse is not necessarily true: If xa tx dx is convergent, xa f x dx may
or may not be convergent, and if xa f x dx is divergent, xa tx dx may or may not be
divergent.]
EXAMPLE 9 Show that
y
0
2
e x dx is convergent.
2
SOLUTION We can’t evaluate the integral directly because the antiderivative of ex is not
an elementary function (as explained in Section 7.5). We write
y
0
2
1
2
2
ex dx y ex dx y ex dx
0
1
SECTION 7.8 IMPROPER INTEGRALS
y
y=e _x
537
and observe that the first integral on the right-hand side is just an ordinary definite integral. In the second integral we use the fact that for x 1 we have x 2 x, so x 2 x
2
and therefore ex ex. (See Figure 13.) The integral of ex is easy to evaluate:
2
y=e_x
y
1
0
❙❙❙❙
t
ex dx lim y ex dx
tl
lim e1 et e1
x
1
1
tl
FIGURE 13
2
Thus, taking f x ex and tx ex in the Comparison Theorem, we see that
2
2
x1 ex dx is convergent. It follows that x0 ex dx is convergent.
2
In Example 9 we showed that x0 ex dx is convergent without computing its value. In
Exercise 70 we indicate how to show that its value is approximately 0.8862. In probability theory it is important to know the exact value of this improper integral, as we will see
in Section 8.5; using the methods of multivariable calculus it can be shown that the exact
value is s2. Table 1 illustrates the definition of an improper integral by showing how
2
the (computer-generated) values of x0t ex dx approach s2 as t becomes large. In fact,
2
these values converge quite quickly because ex l 0 very rapidly as x l .
TABLE 1
x0t ex
t
1
2
3
4
5
6
2
dx
0.7468241328
0.8820813908
0.8862073483
0.8862269118
0.8862269255
0.8862269255
EXAMPLE 10 The integral
because
y
1
1 ex
dx is divergent by the Comparison Theorem
x
1 ex
1
x
x
and x1 1x dx is divergent by Example 1 [or by (2) with p 1].
Table 2 illustrates the divergence of the integral in Example 10. It appears that the
values are not approaching any fixed number.
TABLE 2
|||| 7.8
(a)
y
(c)
y
1
2
0
x1t 1 ex x dx
2
5
10
100
1000
10000
0.8636306042
1.8276735512
2.5219648704
4.8245541204
7.1271392134
9.4297243064
Exercises
1. Explain why each of the following integrals is improper.
t
4 x 4
dx
(b)
y
x
dx
x 5x 6
(d)
y
x e
2
2
0
0
sec x dx
1
dx
x 5
2
2. Which of the following integrals are improper? Why?
1
dx
2x 1
sin x
(c) y
2 dx
1 x
(a)
y
2
1
(b)
y
(d)
y
1
0
2
1
1
dx
2x 1
lnx 1 dx
❙❙❙❙
538
CHAPTER 7 TECHNIQUES OF INTEGRATION
3. Find the area under the curve y 1x 3 from x 1 to x t
and evaluate it for t 10, 100, and 1000. Then find the total
area under this curve for x 1.
; 4. (a) Graph the functions f x 1x
and tx 1x in the
viewing rectangles 0, 10 by 0, 1 and 0, 100 by 0, 1.
(b) Find the areas under the graphs of f and t from x 1 to
x t and evaluate for t 10, 100, 10 4, 10 6, 10 10, and 10 20.
(c) Find the total area under each curve for x 1, if it exists.
5–40
1.1
0.9
Determine whether each integral is convergent or
divergent. Evaluate those that are convergent.
||||
5.
y
7.
y
1
1
dx
3x 12
6.
y
1
dw
s2 w
8.
y
1
9.
y
11.
y
13.
y
y2
1
y
x
dx
1 x2
12.
y
xex dx
14.
y
15.
y sin d
16.
y
17.
y
x1
dx
x 2 2x
18.
y
19.
y
se 5s ds
20.
y
21.
y
ln x
dx
x
22.
y
23.
y
24.
y
25.
27.
29.
31.
2
2
1
0
1
y
1
y
3
0
y
0
1
y
3
2
33.
y
35.
y
37.
y
33
0
0
1
1
x2
dx
9 x6
ln x
dx
x2
1
dx
sx
26.
28.
1
dx
x2
30.
1
dx
x4
32.
e
0
6
y
0
y
3
0
y
9
1
y
1
0
y
sec x dx
36.
y
38.
y
1
0
4
0
2
0
; 44.
; 45.
; 46.
■
||||
■
■
ln x
dx
sx
■
■
■
■
■
x/2
2
2
2
■
■
■
■
■
■
■
■
■
■
■
; 47. (a) If tx sin xx , use your calculator or computer to
2
make a table of approximate values of x1t tx dx for t 2,
5, 10, 100, 1000, and 10,000. Does it appear that x1 tx dx
is convergent?
(b) Use the Comparison Theorem with f x 1x 2 to show
that x1 tx dx is convergent.
(c) Illustrate part (b) by graphing f and t on the same screen
for 1 x 10. Use your graph to explain intuitively why
x1 tx dx is convergent.
; 48. (a) If tx 1(sx 1), use your calculator or computer to
make a table of approximate values of x2t tx dx for t 5,
10, 100, 1000, and 10,000. Does it appear that x2 tx dx is
convergent or divergent?
(b) Use the Comparison Theorem with f x 1sx to show
that x2 tx dx is divergent.
(c) Illustrate part (b) by graphing f and t on the same screen
for 2 x 20. Use your graph to explain intuitively why
x2 tx dx is divergent.
49–54
|||| Use the Comparison Theorem to determine whether the
integral is convergent or divergent.
cos 2x
dx
1 x2
50.
y
dx
x e 2x
52.
y
dx
x sin x
54.
y
y
1
dx
xsx
51.
y
1
dx
3
x9
s
53.
y
dx
s1 x 2
55. The integral
x3
dx
2x 3
■
1
0
S x, y x 2, 0 y e S x, y 0 y 2x 9
S x, y x 0, 0 y xx 9
S x, y 0 x 2, 0 y sec x
S {x, y 2 x 0, 0 y 1sx 2 }
49.
1
dx
x2 x 6
■
y
Sketch the region and find its area (if the area is finite).
x arctan x
dx
1 x 2 2
1
dy
4y 1
40.
2
e x dx
ln x
dx
x3
■
■
re r3 dr
1
; 43.
dz
z 2 3z 2
42.
3
0
z 2 ln z dz
41. S x, y x 1, 0 y e x dt
cos 2
d
2
0
41–46
x 2ex dx
y
■
2 v 4 dv
34.
ex
dx
e 1
2t
x 1 15 dx
x
2
0
10.
e
x
dx
x 2 2
dy
4
1
dx
2x 5
0
39.
■
1
1
/2
0
■
■
■
■
■
y
0
■
1
1
1
0
■
2 e x
dx
x
x
dx
s1 x 6
ex
dx
sx
■
■
■
1
dx
sx 1 x
is improper for two reasons: The interval 0, is infinite and
the integrand has an infinite discontinuity at 0. Evaluate it by
expressing it as a sum of improper integrals of Type 2 and
Type 1 as follows:
1
1
1
1
y0 sx 1 x dx y0 sx 1 x dx y1 sx 1 x dx
■
SECTION 7.8 IMPROPER INTEGRALS
56. Evaluate
y
2
1
dx
x sx 2 4
57–59
|||| Find the values of p for which the integral converges and
evaluate the integral for those values of p.
y
59.
y
■
1
dx
xp
1
0
1
58.
y
e
1
dx
x ln x p
■
■
ys y
R
s
2r
x r dr
sr 2 s 2
If the actual density of stars in a cluster is x r 12 R r2,
find the perceived density ys.
67. A manufacturer of lightbulbs wants to produce bulbs that last
x p ln x dx
0
539
a photograph. Suppose that in a spherical cluster of radius R
the density of stars depends only on the distance r from the
center of the cluster. If the perceived star density is given by
ys, where s is the observed planar distance from the center of
the cluster, and x r is the actual density, it can be shown that
by the same method as in Exercise 55.
57.
❙❙❙❙
■
■
■
■
■
■
■
■
60. (a) Evaluate the integral x0 x nex dx for n 0, 1, 2, and 3.
(b) Guess the value of x0 x nex dx when n is an arbitrary positive integer.
(c) Prove your guess using mathematical induction.
x dx is divergent.
61. (a) Show that x
about 700 hours but, of course, some bulbs burn out faster than
others. Let Ft be the fraction of the company’s bulbs that burn
out before t hours, so Ft always lies between 0 and 1.
(a) Make a rough sketch of what you think the graph of F
might look like.
(b) What is the meaning of the derivative rt Ft?
(c) What is the value of x0 rt dt ? Why?
68. As we will see in Section 9.4, a radioactive substance decays
(b) Show that
exponentially: The mass at time t is mt m0e kt, where
m0 is the initial mass and k is a negative constant. The mean
life M of an atom in the substance is
t
lim y x dx 0
tl
■
t
This shows that we can’t define
y
M k y te kt dt
0
t
f x dx lim y f x dx
tl
t
62. The average speed of molecules in an ideal gas is
v
4
s
M
2RT
32
y
0
2
v 3eMv 2RT dv
where M is the molecular weight of the gas, R is the gas
constant, T is the gas temperature, and v is the molecular
speed. Show that
v
8RT
M
63. We know from Example 1 that the region
x, y x 1, 0 y 1x has infinite area. Show
that by rotating about the x-axis we obtain a solid with
finite volume.
64. Use the information and data in Exercises 29 and 30 of Sec-
tion 6.4 to find the work required to propel a 1000-kg satellite
out of Earth’s gravitational field.
65. Find the escape velocity v0 that is needed to propel a rocket
of mass m out of the gravitational field of a planet with mass
M and radius R. Use Newton’s Law of Gravitation (see Exercise 29 in Section 6.4) and the fact that the initial kinetic
energy of 12 mv 20 supplies the needed work.
66. Astronomers use a technique called stellar stereography to
determine the density of stars in a star cluster from the
observed (two-dimensional) density that can be analyzed from
For the radioactive carbon isotope, 14 C, used in radiocarbon
dating, the value of k is 0.000121. Find the mean life of a
14
C atom.
69. Determine how large the number a has to be so that
y
a
1
dx 0.001
x2 1
2
70. Estimate the numerical value of x0 ex dx by writing it as
4 x 2
0
x 2
4
the sum of x e dx and x e dx. Approximate the first integral by using Simpson’s Rule with n 8 and show that the
second integral is smaller than x4 e4x dx, which is less than
0.0000001.
71. If f t is continuous for t 0, the Laplace transform of f is
the function F defined by
Fs y f test dt
0
and the domain of F is the set consisting of all numbers s for
which the integral converges. Find the Laplace transforms of
the following functions.
(a) f t 1
(b) f t e t
(c) f t t
72. Show that if 0 f t Me at for t 0, where M and a are
constants, then the Laplace transform Fs exists for s a.
73. Suppose that 0 f t Me at and 0 f t Ke at for t 0,
where f is continuous. If the Laplace transform of f t is Fs
❙❙❙❙
540
CHAPTER 7 TECHNIQUES OF INTEGRATION
and the Laplace transform of f t is Gs, show that
Gs sFs f 0
77. Find the value of the constant C for which the integral
sa
y
f x dx is convergent and a and b are real numbers,
74. If x
0
show that
y
a
f x dx y f x dx y
1
C
x2
sx 2 4
dx
0
x
2
b
a
2
1
2
x0 ex
2
f x dx y f x dx
78. Find the value of the constant C for which the integral
b
dx.
y
1
0
7 Review
■
0
dx x sln y dy by interpreting the
integrals as areas.
76. Show that x e
x
C
x2 1
3x 1
dx
converges. Evaluate the integral for this value of C.
CONCEPT CHECK
1. State the rule for integration by parts. In practice, how do you
use it?
2. How do you evaluate x sin mx cos nx dx if m is odd? What if n is
odd? What if m and n are both even?
■
5. State the rules for approximating the definite integral xab f x dx
with the Midpoint Rule, the Trapezoidal Rule, and Simpson’s
Rule. Which would you expect to give the best estimate? How
do you approximate the error for each rule?
6. Define the following improper integrals.
3. If the expression sa 2 x 2 occurs in an integral, what sub-
stitution might you try? What if sa 2 x 2 occurs? What if
sx 2 a 2 occurs?
4. What is the form of the partial fraction expansion of a rational
function PxQx if the degree of P is less than the degree of
Q and Qx has only distinct linear factors? What if a linear
factor is repeated? What if Qx has an irreducible quadratic
factor (not repeated)? What if the quadratic factor is repeated?
■
(a)
x x 2 4
B
A
can be put in the form
.
2
x 4
x2
x2
x2 4
B
C
A
2.
can be put in the form .
2
x x 4
x
x2
x2
y
a
f x dx
(b)
y
b
f x dx
(c)
y
f x dx
7. Define the improper integral xab f x dx for each of the follow-
ing cases.
(a) f has an infinite discontinuity at a.
(b) f has an infinite discontinuity at b.
(c) f has an infinite discontinuity at c, where a c b.
8. State the Comparison Theorem for improper integrals.
TRUE-FALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
1.
converges. Evaluate the integral for this value of C.
75. Show that x0 x 2ex dx ||||
■
8. The Midpoint Rule is always more accurate than the
Trapezoidal Rule.
9. (a) Every elementary function has an elementary derivative.
(b) Every elementary function has an elementary antiderivative.
10. If f is continuous on 0, and x1 f x dx is convergent, then
x0 f x dx is convergent.
3.
x2 4
B
A
can be put in the form 2 .
x x 4
x
x4
11. If f is a continuous, decreasing function on 1, and
4.
A
x 4
B
can be put in the form 2
.
x x 2 4
x
x 4
12. If xa f x dx and xa tx dx are both convergent, then
2
2
5.
y
6.
y
4
0
1
x
dx 12 ln 15
2
x 1
1
dx is convergent.
x s2
t
7. If f is continuous, then x
f x dx lim t l xt
f x dx.
lim x l f x 0 , then x1 f x dx is convergent.
xa f x tx dx is convergent.
13. If xa f x dx and xa tx dx are both divergent, then
xa f x tx dx is divergent.
14. If f x tx and x0 tx dx diverges, then x0 f x dx also
diverges.
CHAPTER 7 REVIEW
■
EXERCISES
Note: Additional practice in techniques of integration is provided in
Exercises 7.5.
1–40
1.
3.
y
x
dx
x 10
5
0
y
2
0
cos d
1 sin 7
y tan x sec x dx
7.
sinln t
dt
t
9.
y
y
4
1
4.
3
5.
6.
8.
x 32 ln x dx
10.
sx 2 1
dx
x
2
11.
y
13.
yx
15.
y sin 17.
y x sec x tan x dx
1
3
dx
x
2
cos 5 d
y
x1
dx
9x 2 6x 5
21.
y
dx
2 4x
sx
23.
y csc
25.
y
19.
4
4x dx
3x 3 x 2 6x 4
dx
x 2 1x 2 2
2
cos 3x sin 2x dx
y
29.
y
31.
y
33.
y 4 x
1
1
x 5 sec x dx
ln 10
0
y
5
0
0
y
dt
2t 1 3
4
1
yy
1
dy
4y 12
dx
x 2s1 x 2
y
y
2
sarctan x
dx
1 x2
1
0
sin x
dx
1 x2
1
12.
y
14.
y
x2 2
dx
x2
16.
y
sec 6
d
tan 2
18.
y
x 2 8x 3
dx
x 3 3x 2
y
dt
sin 2t cos 2t
22.
y
x3
dx
x 110
24.
ye
26.
y 1e
20.
1
x
■
41–50
ye0.6y dy
41.
y
43.
y
45.
y
47.
y
49.
y
■
||||
3
■
y
ln x
dx
sx
46.
y
dx
x2 x 2
48.
y
50.
y
dx
4x 4x 5
2
stan d
sin 2
■
■
■
■
■
■
Evaluate the integral or show that it is divergent.
44.
4
■
dx
x ln x
0
■
y
2
4
■
3
y
42.
1
0
■
40.
1
dx
2x 13
■
■
■
■
■
■
■
1
0
6
2
1
0
1
1
1
t2 1
dt
t2 1
y
dy
sy 2
1
dx
2 3x
x1
dx
3
x4
s
tan1x
dx
x2
■
■
■
; 51–52
|||| Evaluate the indefinite integral. Illustrate and check that
your answer is reasonable by graphing both the function and its
antiderivative (take C 0).
51.
y lnx
■
■
2
2x 2 dx
■
■
■
52.
■
■
y sx
x3
dx
2 1
■
■
■
■
■
2
3
; 53. Graph the function f x cos x sin x and use the graph to
guess the value of the integral x02 f x dx. Then evaluate the
integral to confirm your guess.
CAS
54. (a) How would you evaluate x x 5e2x dx by hand?
(Don’t actually carry out the integration.)
(b) How would you evaluate x x 5e2x dx using tables?
(Don’t actually do it.)
(c) Use a CAS to evaluate x x 5e2x dx.
(d) Graph the integrand and the indefinite integral on the same
screen.
cos x dx
dx
x
■
sx 1
xe 2x
dx
1 2x 2
12
■
■
■
■
■
■
■
■
■
■
■
3
27.
0
y
■
2.
541
■
39.
Evaluate the integral.
||||
❙❙❙❙
e xse x 1
dx
ex 8
x2
28.
y sx 1 dx
30.
y e s1 e
32.
y
34.
y arcsin x dx
36.
y
1 tan d
1 tan 38.
y x tan
3
55–58
|||| Use the Table of Integrals on the Reference Pages to
evaluate the integral.
dx
2 32
1
dx
sx x 32
35.
y
37.
y cos x sin x
2
cos 2x dx
s1 e 2x dx
2x
55.
ye
4
x sin x
dx
cos 3 x
57.
y sx
0
■
dx
x
x
■
2
x 1 dx
■
■
■
■
5
56.
y csc t dt
58.
y s1 2 sin x dx
■
cot x
■
■
■
■
2
1
x2 dx
59. Verify Formula 33 in the Table of Integrals (a) by differentia-
tion and (b) by using a trigonometric substitution.
60. Verify Formula 62 in the Table of Integrals.
61. Is it possible to find a number n such that x0 x n dx is
convergent?
■
542
❙❙❙❙
CHAPTER 7 TECHNIQUES OF INTEGRATION
62. For what values of a is x0 e ax cos x dx convergent? Evaluate the
integral for those values of a.
point to be 53 cm. The circumference 7 cm from each end is
45 cm. Use Simpson’s Rule to make your estimate.
63–64
|||| Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and
(c) Simpson’s Rule with n 10 to approximate the given integral.
Round your answers to six decimal places.
63.
y
1
s1 x 4 dx
0
64.
y
2
0
ssin x dx
28 cm
■
■
■
■
■
■
■
■
■
■
■
■
65. Estimate the errors involved in Exercise 63, parts (a) and (b).
How large should n be in each case to guarantee an error of
less than 0.00001?
66. Use Simpson’s Rule with n 6 to estimate the area under the
curve y e xx from x 1 to x 4.
71. Use the Comparison Theorem to determine whether the integral
x3
dx
1 x 2
is convergent or divergent.
y
1-minute intervals and recorded in the chart. Use Simpson’s
Rule to estimate the distance traveled by the car.
72. Find the area of the region bounded by the hyperbola
t (min)
v (mih)
t (min)
v (mih)
0
1
2
3
4
5
40
42
45
49
52
54
6
7
8
9
10
56
57
57
55
56
73. Find the area bounded by the curves y cos x and y cos 2x
between x 0 and x .
74. Find the area of the region bounded by the curves
y 1(2 sx ), y 1(2 sx ), and x 1.
75. The region under the curve y cos 2x, 0 x 2, is rotated
about the x-axis. Find the volume of the resulting solid.
76. The region in Exercise 75 is rotated about the y-axis. Find the
volume of the resulting solid.
77. If f is continuous on 0, and lim x l f x 0, show that
y
0
68. A population of honeybees increased at a rate of rt bees per
week, where the graph of r is as shown. Use Simpson’s Rule
with six subintervals to estimate the increase in the bee population during the first 24 weeks.
r
f x dx f 0
78. We can extend our definition of average value of a continuous
function to an infinite interval by defining the average value of
f on the interval a, to be
1
t
lim
y f x dx
tl t a a
(a) Find the average value of y tan1x on the interval 0, .
(b) If f x 0 and xa f x dx is divergent, show that the average value of f on the interval a, is lim x l f x, if this
limit exists.
(c) If xa f x dx is convergent, what is the average value of f
on the interval a, ?
(d) Find the average value of y sin x on the interval 0, .
12000
8000
4000
0
4
8
12
16
20
t
24
(weeks)
79. Use the substitution u 1x to show that
y
0
69. (a) If f x sinsin x, use a graph to find an upper bound
5
y 2 x 2 1 and the line y 3.
67. The speedometer reading (v) on a car was observed at
CAS
for f 4x .
(b) Use Simpson’s Rule with n 10 to approximate
x0 f x dx and use part (a) to estimate the error.
(c) How large should n be to guarantee that the size of the
error in using Sn is less than 0.00001?
70. Suppose you are asked to estimate the volume of a football.
You measure and find that a football is 28 cm long. You use a
piece of string and measure the circumference at its widest
ln x
dx 0
1 x2
80. The magnitude of the repulsive force between two point charges
with the same sign, one of size 1 and the other of size q, is
q
F
4 0 r 2
where r is the distance between the charges and 0 is a
constant. The potential V at a point P due to the charge q is
defined to be the work expended in bringing a unit charge to P
from infinity along the straight line that joins q and P. Find a
formula for V .
PROBLEMS
PLUS
Cover up the solution to the example and try it yourself first.
EXAMPLE
(a) Prove that if f is a continuous function, then
y
a
0
a
f x dx y f a x dx
0
(b) Use part (a) to show that
y
2
0
sin n x
n
n dx sin x cos x
4
for all positive numbers n.
SOLUTION
|||| The principles of problem solving are
discussed on page 80.
(a) At first sight, the given equation may appear somewhat baffling. How is it possible
to connect the left side to the right side? Connections can often be made through one of
the principles of problem solving: introduce something extra. Here the extra ingredient is
a new variable. We often think of introducing a new variable when we use the Substitution Rule to integrate a specific function. But that technique is still useful in the present
circumstance in which we have a general function f .
Once we think of making a substitution, the form of the right side suggests that it
should be u a x. Then du dx. When x 0, u a; when x a, u 0. So
y
a
0
0
a
f a x dx y f u du y f u du
a
0
But this integral on the right side is just another way of writing x0a f x dx. So the given
equation is proved.
(b) If we let the given integral be I and apply part (a) with a 2, we get
Iy
2
0
|||| The computer graphs in Figure 1 make it
seem plausible that all of the integrals in the
example have the same value. The graph of each
integrand is labeled with the corresponding
value of n.
sin n x
2
sin n2 x
dx
n dx y
n
0
sin x cos x
sin 2 x cos n2 x
n
A well-known trigonometric identity tells us that sin2 x cos x and
cos2 x sin x, so we get
Iy
2
0
1
3
4
2
Notice that the two expressions for I are very similar. In fact, the integrands have the
same denominator. This suggests that we should add the two expressions. If we do so,
we get
1
2I y
2
0
0
FIGURE 1
cos n x
dx
cos x sin n x
n
π
2
Therefore, I 4.
sin n x cos n x
2
dx y 1 dx 0
sin n x cos n x
2
; 1. Three mathematics students have ordered a 14-inch pizza. Instead of slicing it in the tradi-
P RO B L E M S
tional way, they decide to slice it by parallel cuts, as shown in the figure. Being mathematics
majors, they are able to determine where to slice so that each gets the same amount of pizza.
Where are the cuts made?
1
dx.
x7 x
The straightforward approach would be to start with partial fractions, but that would be brutal.
Try a substitution.
2. Evaluate y
1
3
7
3. Evaluate y (s
1 x7 s
1 x 3 ) dx.
0
4. A man initially standing at the point O walks along a pier pulling a rowboat by a rope of
14 in
length L. The man keeps the rope straight and taut. The path followed by the boat is a curve
called a tractrix and it has the property that the rope is always tangent to the curve (see the
figure).
(a) Show that if the path followed by the boat is the graph of the function y f x, then
FIGURE FOR PROBLEM 1
y
pier
f x dy
sL 2 x 2
dx
x
(b) Determine the function y f x.
L
(x, y)
5. A function f is defined by
(L, 0)
O
FIGURE FOR PROBLEM 4
f x y cos t cosx t dt
x
0
0 x 2
Find the minimum value of f .
6. If n is a positive integer, prove that
y
1
0
ln xn dx 1n n!
7. Show that
y
1
0
2 2n n!2
2n 1!
1 x 2 n dx Hint: Start by showing that if In denotes the integral, then
Ik1 2k 2
Ik
2k 3
; 8. Suppose that f is a positive function such that f is continuous.
(a) How is the graph of y f x sin nx related to the graph of y f x? What happens
as n l ?
(b) Make a guess as to the value of the limit
lim
nl
y
1
0
f x sin nx dx
based on graphs of the integrand.
(c) Using integration by parts, confirm the guess that you made in part (b). [Use the fact that,
since f is continuous, there is a constant M such that f x M for 0 x 1.]
9. If 0 a b, find lim
tl0
y
1
0
1t
bx a1 x t dx
.
t1
x
; 10. Graph f x sine and use the graph to estimate the value of t such that xt f x dx is a
maximum. Then find the exact value of t that maximizes this integral.
y
11. The circle with radius 1 shown in the figure touches the curve y 2x twice. Find the area
of the region that lies between the two curves.
12. A rocket is fired straight up, burning fuel at the constant rate of b kilograms per second. Let
v vt be the velocity of the rocket at time t and suppose that the velocity u of the exhaust
gas is constant. Let M Mt be the mass of the rocket at time t and note that M decreases
as the fuel burns. If we neglect air resistance, it follows from Newton’s Second Law that
y=| 2x |
0
FIGURE FOR PROBLEM 11
FM
x
dv
ub
dt
where the force F Mt. Thus
1
M
dv
ub Mt
dt
Let M1 be the mass of the rocket without fuel, M2 the initial mass of the fuel, and
M0 M1 M2 . Then, until the fuel runs out at time t M2 b, the mass is M M0 bt.
(a) Substitute M M0 bt into Equation 1 and solve the resulting equation for v. Use the
initial condition v 0 0 to evaluate the constant.
(b) Determine the velocity of the rocket at time t M2 b. This is called the burnout velocity.
(c) Determine the height of the rocket y yt at the burnout time.
(d) Find the height of the rocket at any time t.
13. Use integration by parts to show that, for all x 0,
0y
0
sin t
2
dt ln1 x t
ln1 x
; 14. The Chebyshev polynomials Tn are defined by
Tnx cosn arccos x
n 0, 1, 2, 3, . . .
(a) What are the domain and range of these functions?
(b) We know that T0x 1 and T1x x. Express T2 explicitly as a quadratic polynomial
and T3 as a cubic polynomial.
(c) Show that, for n 1,
Tn1x 2x Tnx Tn1x
(d) Use part (c) to show that Tn is a polynomial of degree n.
(e) Use parts (b) and (c) to express T4 , T5 , T6 , and T7 explicitly as polynomials.
(f) What are the zeros of Tn ? At what numbers does Tn have local maximum and minimum
values?
(g) Graph T2 , T3 , T4 , and T5 on a common screen.
(h) Graph T5 , T6 , and T7 on a common screen.
(i) Based on your observations from parts (g) and (h), how are the zeros of Tn related to the
zeros of Tn1 ? What about the x-coordinates of the maximum and minimum values?
1
(j) Based on your graphs in parts (g) and (h), what can you say about x1
Tnx dx when n is
odd and when n is even?
(k) Use the substitution u arccos x to evaluate the integral in part (j).
(l) The family of functions f x cosc arccos x are defined even when c is not an integer
(but then f is not a polynomial). Describe how the graph of f changes as c increases.