6.5. IMPROPER INTEGRALS
381
nonnegative. Further, we note that our primary interest is in functions f for which
lim x→∞ f (x) = 0, for if the function f does not approach 0 as x → ∞, then it is
impossible for
∞
a
f (x) dx to converge.
Activity 6.14.
Determine whether each of the following improper integrals converges or diverges. For
each integral that converges, find its exact value.
(a)
∞
1
1
x 2 dx
(b)
∞
0
e
−x/4 dx
(c)
∞
2
9
(x + 5) 2/3 dx
(d)
∞
4
3
(x + 2) 5/4 dx
(e)
∞
0
xe
−x/4 dx
(f)
∞
1
1
x p dx, where p is a positive real number
⊳
Improper Integrals Involving Unbounded Integrands
It is also possible for an integral to be improper due to the integrand being unbounded on
the interval of integration. For example, if we consider
1
0
1
√
x
dx,
we see that because f (x) =
1
√
x
has a vertical asymptote at x = 0, f is not continuous on
[0, 1], and the integral is attempting to represent the area of the unbounded region shown
at right in Figure 6.21.
Just as we did with improper integrals involving infinite limits, we address the problem
of the integrand being unbounded by replacing such an improper integral with a limit
of proper integrals. For example, to evaluate
1
0
1
√
x
dx, we replace 0 with a and let a
approach 0 from the right. Thus,
1
0
1
√
x
dx = lim
a→0 +
1
a
1
√
x
dx,
381
nonnegative. Further, we note that our primary interest is in functions f for which
lim x→∞ f (x) = 0, for if the function f does not approach 0 as x → ∞, then it is
impossible for
∞
a
f (x) dx to converge.
Activity 6.14.
Determine whether each of the following improper integrals converges or diverges. For
each integral that converges, find its exact value.
(a)
∞
1
1
x 2 dx
(b)
∞
0
e
−x/4 dx
(c)
∞
2
9
(x + 5) 2/3 dx
(d)
∞
4
3
(x + 2) 5/4 dx
(e)
∞
0
xe
−x/4 dx
(f)
∞
1
1
x p dx, where p is a positive real number
⊳
Improper Integrals Involving Unbounded Integrands
It is also possible for an integral to be improper due to the integrand being unbounded on
the interval of integration. For example, if we consider
1
0
1
√
x
dx,
we see that because f (x) =
1
√
x
has a vertical asymptote at x = 0, f is not continuous on
[0, 1], and the integral is attempting to represent the area of the unbounded region shown
at right in Figure 6.21.
Just as we did with improper integrals involving infinite limits, we address the problem
of the integrand being unbounded by replacing such an improper integral with a limit
of proper integrals. For example, to evaluate
1
0
1
√
x
dx, we replace 0 with a and let a
approach 0 from the right. Thus,
1
0
1
√
x
dx = lim
a→0 +
1
a
1
√
x
dx,