James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

530 Chapter 7 Techniques of Integration
Example 3 Evaluate y`
2`
1
1 1 x 2 dx.
SOLUtion It’s convenient to choose a − 0 in Definition 1(c):
y`
2`
1
1 1 x 2 dx − y 0 2`
1
1 1 x dx 1 y`
2 0
We must now evaluate the integrals on the right side separately:
y`
0
y 0 2`
1
2
dx − lim
1 1 x t l ` yt 0
1
2
dx − lim
1 1 x t l 2` y0 t
dx
1 1 x − lim
t
xg 2 t l ` tan21 0
1
1 1 x 2 dx
− lim
t l ` stan21 t 2 tan 21 0d − lim
t l ` tan21 t − 2
dx
1 1 x − lim
0
xg 2 t l 2` tan21 t
− lim
t l 2` stan21 0 2 tan 21 td − 0 2S2 2D− 2
y
1
y=
1+≈
area=π
Since both of these integrals are convergent, the given integral is convergent and
y`
2`
1
1 1 x 2 dx − 2 1 2 −
FIGURE 6
0
x
Since 1ys1 1 x 2 d . 0, the given improper integral can be interpreted as the area of
the infinite region that lies under the curve y − 1ys1 1 x 2 d and above the x-axis (see
Figure 6).
n
Example 4 For what values of p is the integral
convergent?
y`
1
1
x p dx
SOLUtion We know from Example 1 that if p − 1, then the integral is divergent, so
let’s assume that p ± 1. Then
y`
1
1
p
dx − lim
x
t l ` yt 1
− lim
t l `
x 2p dx − lim
t l `
1
1 2 pF 1
t p21
x 2p11
x−t
2p 1 1Gx−1
G 2 1
If p . 1, then p 2 1 . 0, so as t l `, t p21 l ` and 1yt p21 l 0. Therefore
y`
1
1
x dx − 1
p p 2 1
if p . 1
and so the integral converges. But if p , 1, then p 2 1 , 0 and so
1
t p21 − t 12p l ` as t l `
and the integral diverges.
n
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