Page 148 - Calculus Demystified

CHAPTER 5
Indeterminate Forms
EXAMPLE 5.15 135
Analyze the integral

3
(x − 3) −2 dx.
2
SOLUTION
This is an improper integral with infinite integrand at 3. We evaluate this
integral by considering
3−( 3−(
lim (x − 3) −2 dx = lim −(x − 3) −1
(→0 + 2 (→0 + 2

−1 −1
= lim ( − 1 .
(→0 +
Thislastlimitis+∞.Wethereforeconcludethattheimproperintegraldiverges.

−1
You Try It: Evaluate the improper integral (dx/(x + 1) 4/5 )dx.
−2
Improper integrals with integrand which is infinite at the left endpoint of
integration are handled in a manner similar to the right endpoint case:

EXAMPLE 5.16
Evaluate the integral
1/2

1
dx.
2
0 x · ln x
SOLUTION
This integral is improper with infinite integrand at 0. The value of the integral
is defined to be
1/2 1

lim dx,
2
(→0 + ( x · ln x
provided that this limit exists.
2
Since 1/(x ln x) is continuous on the interval [(, 1/2] for (> 0, this last
integral can be evaluated directly and will have a finite real value. For clarity,

write ϕ(x) = ln x, ϕ (x) = 1/x. Then the (indefinite) integral becomes

ϕ (x)
dx.
2
ϕ (x)
Clearly the antiderivative is −1/ϕ(x). Thus we see that
1/2
1/2 1 1 1 1
lim dx = lim − = lim − − − .
2
(→0 + ( x·ln x (→0 + lnx ( (→0 + ln(1/2) ln(