5128_Ch08_434-471

464 Chapter 8 Sequences, L’Hôpital’s Rule, and Improper Integrals
3
1
dx
x 1 23 lim
c→1 3
lim
c→1
c
dx
x 1 23
3
3x 113]
c
We conclude that
3
0
lim
c→1 33 113 3c 1 13 3 3 2
dx
x 1 23 3 3 3 2.
Now try Exercise 25.
EXAMPLE 7 Infinite Discontinuity at an Endpoint
Evaluate 2
dx
.
1 x 2
SOLUTION
The integrand has an infinite discontinuity at x 2 and is continuous on 1, 2.
Thus,
2
dx
lim
1 x 2 c→2 c
dx
1 x 2
c
lim ln x
c→2
2]
1
lim ln c 2 ln 1
c→2
.
The original integral diverges and has no value. Now try Exercise 29.
Test for Convergence and Divergence
When we cannot evaluate an improper integral directly (often the case in practice) we first
try to determine whether it converges or diverges. If the integral diverges, that’s the end of
the story. If it converges, we can then use numerical methods to approximate its value. In
such cases the following theorem is useful.
THEOREM 6 Comparison Test
Let f and g be continuous on a, with 0 f x gx for all x a. Then
1.
f x dx converges if
gx dx converges.
a
a
2.
gx dx diverges if
f x dx diverges.
a
a
EXAMPLE 8 Investigating Convergence
Does the integral 1
ex2 dx converge?
SOLUTION
Solve Analytically By definition,
e x2 dx lim
1
b→
b
1
e x2 dx.