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

SECTION 7.8 Improper Integrals 533
lim x l 0
1 ln x − 2`. Thus the given integral is improper and we have
y 1
ln x dx − lim y 1
ln x dx
0
t l0 1 t
Now we integrate by parts with u − ln x, dv − dx, du − dxyx, and v − x:
y
y 1
t
ln x dx − x ln xg t 1 2
y 1
t
dx
− 1 ln 1 2 t ln t 2 s1 2 td − 2t ln t 2 1 1 t
0
1
area=1
x
To find the limit of the first term we use l’Hospital’s Rule:
ln t
lim t ln t − lim
t l01 t l0 1 1yt − lim
t l0 1
1yt
21yt 2
− lim
t l0 1 s2td − 0
y=ln x
FIGURE 11
Therefore y 1
0
ln x dx − lim s2t ln t 2 1 1 td − 20 2 1 1 0 − 21
t l01 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.
n
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.
y
g
f
Comparison Theorem Suppose that f and t are continuous functions with
f sxd > tsxd > 0 for x > a.
(a) If y`
a
f sxd dx is convergent, then y`
tsxd dx is convergent.
a
(b) If y`
a
tsxd dx is divergent, then y`
a
f sxd dx is divergent.
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 sxd is finite, then so is the area under the
bottom curve y − tsxd. And if the area under y − tsxd is infinite, then so is the area
under y − f sxd. [Note that the reverse is not necessarily true: If y`
a
y`
a
f sxd dx may or may not be convergent, and if y`
a
or may not be divergent.]
f sxd dx is divergent, y`
a
tsxd dx is convergent,
tsxd dx may
y
0
y=e _x
FIGURE 13 13
y=e _x2
7et070813
10/13/09
1
x
Example 9 Show that y` 2 0 e2x dx is convergent.
SOLUtion We can’t evaluate the integral directly because the antiderivative of e 2x 2 is
not an elementary function (as explained in Section 7.5). We write
y`
0 e2x 2 dx − y 1
0
e 2x 2 dx 1 y`
1 e2x 2 dx
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 2x 2 < 2x
and therefore e 2x 2 < e 2x . (See Figure 13.) The integral of e 2x is easy to evaluate:
y`
1 e2x dx − lim
t l ` yt 1
e 2x dx − lim
t l ` se21 2 e 2t d − e 21
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