Page 149 - Calculus Demystified

CHAPTER 5
Indeterminate Forms
136
Now as ( → 0 we have ln ( →−∞ hence 1/ ln ( → 0. We conclude that
+
the improper integral converges to 1/ ln 2.
0 −1/2
You Try It: Evaluate the improper integral 1/(x + 2) dx.
−2
Many times the integrand has a singularity in the middle of the interval of inte-
gration. In these circumstances we divide the integral into two pieces for each of
which the integrand is infinite at one endpoint, and evaluate each piece separately.
EXAMPLE 5.17
Evaluate the improper integral
4
4(x + 1) −1/5 dx.
−4
SOLUTION
The integrand is unbounded as x tends to −1. Therefore we evaluate
separately the two improper integrals
−1 4
4(x + 1) −1/5 dx and 4(x + 1) −1/5 dx.
−4 −1
The first of these has the value
−1−(
−1−(
lim 4(x + 1) −1/5 dx = lim 5(x + 1) 4/5
(→0 + −4 (→0 + −4
4/5 4/5
= lim 5 (−() − (−3)
(→0 +
=−5 · 3 4/5
The second integral has the value

4
4
lim 4(x + 1) −1/5 dx = lim 5(x + 1) 4/5
(→0 + −1+( (→0 + −1+(
4/5 4/5
= lim 5 5 − (
(→0 +
= 5 9/5 .
We conclude that the original integral converges and
4 −1 4

4(x + 1) −1/5 dx = 4(x + 1) −1/5 dx + 4(x + 1) −1/5 dx
−4 −4 −1
=−5 · 3 4/5 + 5 9/5 .
3
You Try It: Evaluate the improper integral x −1 dx.
−4