Page 155 - Calculus Demystified
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SOLUTION CHAPTER 5 Indeterminate Forms
We do this problem by evaluating the limit
−32 5
−32
lim x −1/5 dx = lim x 4/5
M→−∞ M M→−∞ 4 M
5
4/5 4/5
= lim (−32) − M
M→−∞ 4
5
4/5
= lim 16 − M .
M→−∞ 4
This limit equals −∞. Therefore the integral diverges.
∞
You Try It: Evaluate (1 + x) −3 dx.
1
Sometimes we have occasion to evaluate a doubly infinite integral. We do so by
breaking the integral up into two separate improper integrals, each of which can be
evaluated with just one limit.
EXAMPLE 5.23
Evaluate the improper integral
∞
1
dx.
1 + x 2
−∞
SOLUTION
The interval of integration is (−∞, +∞). To evaluate this integral, we break
the interval up into two pieces:
(−∞, +∞) = (−∞, 0]∪[0, +∞).
(The choice of zero as a place to break the interval is not important; any
other point would do in this example.) Thus we will evaluate separately the
integrals
0
+∞ 1 1
dx and dx.
0 1 + x 2 −∞ 1 + x 2
For the first one we consider the limit
N 1 N
lim dx = lim Tan −1
x
N→+∞ 0 1 + x 2 N→+∞ 0
= lim Tan −1 N − Tan −1 0
N→+∞
π
= .
2