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

532 Chapter 7 Techniques of Integration
y
y= 1
œ„„„„ x-2
endpoint of f2, 5g, we use part (b) of Definition 3:
y 5
2
dx
− lim
sx 2 2 t l2 1
y 5
t
dx
sx 2 2
− lim
t l2 1 2sx 2 2 g t
5
0
FIGURE 10
area=2œ„3
1 2 3 4 5
x
− lim
t l2 1 2(s3 2 st 2 2 ) − 2s3
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. n
Example 6 Determine whether y y2
sec x dx converges or diverges.
0
SOLUtion Note that the given integral is improper because lim x lsy2d
2 sec x − `.
Using part (a) of Definition 3 and Formula 14 from the Table of Integrals, we have
y y2
sec x dx −
0
−
lim y t
t lsy2d 2 0
sec x dx − lim
t l sy2d 2 ln | sec x 1 tan x |g 0
t
lim flnssec t 1 tan td 2 ln 1g − `
t lsy2d2 because sec t l ` and tan t l ` as t l sy2d 2 . Thus the given improper integral is
divergent.
Example 7 Evaluate y 3 dx
if possible.
0 x 2 1
SOLUtion Observe that the line x − 1 is a vertical asymptote of the integrand. Since
it occurs in the middle of the interval f0, 3g, we must use part (c) of Definition 3 with
c − 1:
y 3
0
where y 1
0
dx
x 2 1 − y 1
0
dx
x 2 1 − lim y t
t l1 2 0
dx
x 2 1 1 y 3
1
dx
x 2 1
dx
x 2 1 − lim
t l1 2 ln | x 2 1 |g 0
t
n
− lim
(ln
t
| t 2 1 | 2 ln | 21 |) − lim
l1 2 t l1
2
lns1 2 td − 2`
because 1 2 t l 0 1 as t l 1 2 . Thus y 1 dxysx 2 1d is divergent. This implies that
0
y 3 dxysx 2 1d is divergent. [We do not need to evaluate 0 y3 dxysx 2 1d.]
n
1
Warning 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:
y 3
0
dx
x 2 1 − ln | x 2 1 |g 3 − ln 2 2 ln 1 − ln 2
0
This is wrong because the integral is improper and must be calculated in terms of limits.
From now on, whenever you meet the symbol y b f sxd dx you must decide, by looking at
a
the function f on fa, bg, whether it is an ordinary definite integral or an improper integral.
Example 8 y 1
ln x dx.
0
SOLUtion We know that the function f sxd − ln x has a vertical asymptote at 0 since
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.