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

SECTION 7.8 Improper Integrals 529
The limit does not exist as a finite number and so the improper integral y`
1
divergent.
s1yxd dx is
n
Let’s compare the result of Example 1 with the example given at the beginning of this
section:
1
1
y`
1
2
dx converges y`
dx diverges
x 1 x
Geometrically, this says that although the curves y − 1yx 2 and y − 1yx look very similar
for x . 0, the region under y − 1yx 2 to the right of x − 1 (the shaded region in Figure 4)
has finite area whereas the corresponding region under y − 1yx (in Figure 5) has infinite
area. Note that both 1yx 2 and 1yx approach 0 as x l ` but 1yx 2 approaches 0 faster than
1yx. The values of 1yx don’t decrease fast enough for its integral to have a finite value.
y
y
y= 1 ≈
1
y=
x
finite area
infinite area
0
1
x
0
1
x
FIGURE 4
y`
1 s1yx 2 d dx converges
FIGURE 5
s1yxd dx diverges
y`
1
Example 2 Evaluate y 0 xe x dx.
2`
SOLUtion Using part (b) of Definition 1, we have
y 0 2`
xe x dx − lim
t l2` y0 xe x dx
t
We integrate by parts with u − x, dv − e x dx so that du − dx, v − e x :
y 0
t
xe x dx − xe x g t
0
2 y 0
t
e x dx
− 2te t 2 1 1 e t
TEC In Module 7.8 you can investigate
visually and numerically whether
several improper integrals are convergent
or divergent.
We know that e t l 0 as t l 2`, and by l’Hospital’s Rule we have
lim te t − lim
t l2` t l2`
t
e 2t − lim
t l2`
1
2e 2t
Therefore
− lim
t l2` s2e t d − 0
y 0 2`
xe x dx − lim
t l2` s2te t 2 1 1 e t d
− 20 2 1 1 0 − 21 n
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.