5128_Ch08_434-471

462 Chapter 8 Sequences, L’Hôpital’s Rule, and Improper Integrals
EXAMPLE 5 Evaluating an Integral on (, )
Evaluate
.
1
dxx 2
SOLUTION
According to the definition (part 3) we can write
1
dxx 0
2 1
dxx
2 .
0 1
dxx 2
Next, we evaluate each improper integral on the right-hand side of the equation above.
0
1
dxx 2 lim
a→ 0
a 1
dxx 2
0
lim
a→
x]
tan1 a
0
lim
a→ tan1 0 tan 1 a 0 ( p 2 ) p 2
1
dxx 2 lim
b→
b
0
b
tan 1
b→
x]
0
lim
1
dxx 2
lim
b→
tan 1 b tan 1 0 p 2 0 p 2
Thus,
dx
1 x 2 p 2 p 2 p. Now try Exercise 21.
a
[0, 2] by [–1, 5]
(a)
[0, 2] by [–1, 5]
(b)
Integrands with Infinite Discontinuities
Another type of improper integral arises when the integrand has a vertical asymptote —
an infinite discontinuity— at a limit of integration or at some point between the limits of
integration.
Consider the infinite region in the first quadrant that lies under the curve y 1x
from x 0 to x 1 (Figure 8.17a). First we find the area of the portion from a to 1
(Figure 8.17b).
1
a
dx
x
2x]
Then, we find the limit of this area as a→0 .
1
lim
a→0
a
1
a
2 2a
dx
lim 2 2a 2
x a→0
Figure 8.17 (a) The area under the curve
y 1x from x 0 to x 1 is (b)
1
lim
a→0
a
1x dx.
The area under the curve from 0 to 1 is
1
0
dx
lim
x
a→0 1
a
dx
2.
x