CPM Homework Help : CALC3RD Problem 6-167

Determine what makes each of the integrals below “improper.” Use a limit to rewrite the integral, then evaluate. If the integral diverges, say so.

Hint:

Improper Integrals

There are two types of improper integrals:

Type I: The interval of integration is approaches infinity or negative infinity.

Type II: The integrand becomes infinitely large within the closed interval.

Examples of Type I:	(a) $\int _ { 1 } ^ { \infty } \frac { 1 } { x } d x$	(b) $\int _ { - \infty } ^ { 1 } \frac { 1 } { 1 + x ^ { 2 } } d x$
Examples of Type II:	(c) $\int _ { 0 } ^ { 5 } \operatorname { ln } ( x ) d x$	(d) $\int _ { 0 } ^ { 2 } \frac { 1 } { x ^ { 2 } } d x$	(e) $\int _ { - 1 } ^ { 1 } \frac { 1 } { x ^ { 2 } } d x$

In order to evaluate an improper integrals, we must use limits, as shown below.

(a) $\int _ { 1 } ^ { \infty } \frac { 1 } { x } d x = \lim\limits _ { b \rightarrow \infty } \int _ { 1 } ^ { b } \frac { 1 } { x } d x$ (d) $\int _ { 1 } ^ { \infty } \frac { 1 } { x } d x = \lim\limits _ { b \rightarrow \infty } \int _ { 1 } ^ { b } \frac { 1 } { x } d x$

When the infinitely large value is not at an endpoint of the interval (such as (e)), we must separate the original integral into the sum of two integrals, each written as a limit:

(e) $\int _ { - 1 } ^ { 1 } \frac { 1 } { x ^ { 2 } } d x = \lim\limits _ { b \rightarrow 0 ^ { - } } \int _ { - 1 } ^ { b } \frac { 1 } { x ^ { 2 } } d x + \lim\limits _ { b \rightarrow 0 ^ { + } } \int _ { b } ^ { 1 } \frac { 1 } { x ^ { 2 } } d x$

If the limit exists, we say that the integral converges, if not, the integral diverges.

In the examples, (b), (c), (d), and (e) converge while (a) diverges.

$\int _ { 0 } ^ { \infty } e ^ { - x } d x$
Hint (a):
Does the graph of the integrand, $y = e^{-x}$ , have any asymptotes on the bounded domain, $[0, ∞)$ ?
Steps (a):
$\lim \limits_{a\to \infty }\int_{0}^{a}e^{-x}dx=\lim \limits_{a\to \infty }-e^{-x}\Big|_0^a=\underline{\ \ \ \ \ \ \ }$
Answer (a):
The answer should be very pleasing.
$\int _ { - 3 } ^ { 2 } \frac { 1 } { 2 x + 6 } d x$
Hint (b):
Does the graph of the integrand, $y= \frac{1}{2x-6}$ have any asymptotes on the bounded domain, $[−3, 2]$ ?
$\int _ { 1 } ^ { \infty } \frac { \operatorname { ln } ( x ) } { x } d x$
Hint (c):
Before you rewrite as a limit, use $U$ -substitution on the integrand and the bounds. Let $U = \ln(x)$ .