Page 151 - Calculus Demystified

138 EXAMPLE 5.19 CHAPTER 5 Indeterminate Forms
Analyze the integral
1 1
0 x(1 − x) 1/2 dx.

SOLUTION
The key idea is that we can only handle one singularity at a time. This
integrand is singular at both endpoints 0 and 1. Therefore we divide the domain
of integration somewhere in the middle—at 1/2 say (it does not really matter
where we divide)—and treat the two singularities separately.
First we treat the integral
1/2
1 1/2 dx.
0 x(1 − x)
Since the integrand has a singularity at 0, we consider
1/2 1 1/2Y
lim
(→0 + x(1 − x) dx.
(
This is a tricky integral to evaluate directly. But notice that
1
TEAMFL 1
≥
x(1 − x) 1/2 x · (1) 1/2
when 0 <( ≤ x ≤ 1/2. Thus
1/2 1 1/2 1 1/2 1
x(1 − x) 1/2 dx ≥ x · (1) 1/2 dx = x dx.
( ( (
We evaluate the integral: it equals ln(1/2) − ln (. Finally,
lim − ln ( =+∞.
(→0 +
The first of our integrals therefore diverges.
But the full integral
1 1
0 x(1 − x) 1/2 dx
converges if and only if each of the component integrals

1/2
1 dx
0 x(1 − x) 1/2
and
1 1
1/2 x(1 − x) 1/2 dx

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