6.5. IMPROPER INTEGRALS
383
1
2
3
y
x
y =
1
(x−2) 2
Figure 6.22: The function f (x) =
1
(x−2) 2 on an interval including x = 2.
vertical asymptote in f (x) =
1
(x−2) 2 at x = 2 by writing
3
1
1
(x − 2) 2 dx = lim
a→2 −
a
1
1
(x − 2) 2 dx + lim
b→2 +
3
b
1
(x − 2) 2 dx
and then evaluate two separate limits of proper integrals. For instance, doing so for the
integral with a approaching 2 from the left, we find
2
1
1
(x − 2) 2 dx = lim
a→2 −
a
1
1
(x − 2) 2 dx
= lim
a→2 −
−
1
(x − 2)
a
1
= lim
a→2 −
−
1
(a − 2)
+
1
1 − 2
= ∞,
since
1
a−2 → −∞ as a approaches 2 from the left. Thus, the improper integral
2
1
1
(x−2) 2 dx
diverges; similar work shows that
3
2
1
(x−2) 2 dx also diverges. From either of these two
results, we can conclude that that the original integral,
3
1
1
(x−2) 2 dx diverges, too.
Activity 6.15.
For each of the following definite integrals, decide whether the integral is improper or
not. If the integral is proper, evaluate it using the First FTC. If the integral is improper,
determine whether or not the integral converges or diverges; if the integral converges,
find its exact value.
(a)
1
0
1
x 1/3 dx
(b)
2
0
e −x dx