Chapter 8.4, Problem 17E

To determine

To evaluate: The improper integral or state that it diverges.

Answer to Problem 17E

The improper integral is $\frac{3e^{-2}}{4}$ , the given improper integral converges.

Explanation of Solution

Given information: ${\displaystyle \underset{1}{\overset{\infty }{\int }}x{e}^{-2x}dx}$

Concept used: Consider that $f\left(x\right)$ is continuous on $\left[a,\infty \right)$ , then ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)dx=\underset{b\to \infty }{\lim }}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

Calculation:

Let $I={\displaystyle \underset{1}{\overset{\infty }{\int }}x{e}^{-2x}dx}$ . In order to express the improper integral as a limit definite integral, consider a real number $b$ such that ,

  $\begin{array}{c}I={\displaystyle \underset{1}{\overset{\infty }{\int }}x{e}^{-2x}dx}\\ =\underset{b\to \infty }{\lim }{\displaystyle \underset{1}{\overset{b}{\int }}x{e}^{-2x}dx}\end{array}$

Consider that $u=x$ such that $du=dx$ . Also, $dv=e^{-2x}dx$ and $v=-\frac{e^{-2x}}{2}$ .

So, by the integration by parts formula, the integral $I$ can be calculated as shown below.

  $\begin{array}{c}{\displaystyle \underset{1}{\overset{b}{\int }}x{e}^{-2x}dx}={\left[-\frac{x{e}^{-2x}}{2}\right]}_1^b-{\displaystyle \underset{1}{\overset{b}{\int }}\left(-\frac{e^{-2x}}{2}\right)dx}\\ ={\left[-\frac{x{e}^{-2x}}{2}\right]}_1^b+{\left[-\frac{e^{-2x}}{4}\right]}_1^b\\ =\left[-\frac{1}{e^{2b}}\left(\frac{b}{2}+\frac{1}{4}\right)+\frac{3e^{-2}}{4}\right]\end{array}$

Now, determine the value of $\underset{b\to \infty }{\lim }{\displaystyle \underset{1}{\overset{b}{\int }}x{e}^{-2x}dx}$ .

  $\begin{array}{c}\underset{b\to \infty }{\lim }{\displaystyle \underset{1}{\overset{b}{\int }}x{e}^{-2x}dx}=\underset{b\to \infty }{\lim }\left[-\frac{1}{e^{2b}}\left(\frac{b}{2}+\frac{1}{4}\right)+\frac{3e^{-2}}{4}\right]\\ =\frac{3e^{-2}}{4}\end{array}$

Since the given improper integral has some value, it converges.