Chapter 9, Problem 46RE

To determine

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

Answer to Problem 46RE

The answer is $-\frac{1}{9}$ .

Explanation of Solution

Given information :

The given improper integral is ${\displaystyle {\int }_{-\infty }^{0}x{e}^{3x}}dx$ .

Calculation :

Since, if $f\left(x\right)$ is continuous on $\left(-\infty ,b\right]$ . Then,

  ${\displaystyle {\int }_{-\infty }^{b}f\left(x\right)}dx=\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{b}f\left(x\right)}dx$ .

Therefore

  $\begin{array}{l}{\displaystyle {\int }_{-\infty }^{0}x{e}^{3x}}dx=\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{0}x{e}^{3x}}dx\\ {\displaystyle {\int }_{-\infty }^{0}x{e}^{3x}}dx=\underset{a\to -\infty }{\lim }{\left[\frac{1}{3}x{e}^{3x}\right]}_a^0-\frac{1}{3}\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^0{e}^{3x}}dx\\ {\displaystyle {\int }_{-\infty }^{0}x{e}^{3x}}dx=\underset{a\to -\infty }{\lim }{\left[\frac{1}{3}x{e}^{3x}-\frac{1}{9}{e}^{3x}\right]}_a^0\\ {\displaystyle {\int }_{-\infty }^{0}x{e}^{3x}}dx=\underset{a\to -\infty }{\lim }\left[\left(\frac{1}{3}0e^0-\frac{1}{9}{e}^0\right)-\left(\frac{1}{3}a{e}^{3a}-\frac{1}{9}{e}^{3a}\right)\right]\\ {\displaystyle {\int }_{-\infty }^{0}x{e}^{3x}}dx=\left[\left(0-\frac{1}{9}\right)-\left(0-0\right)\right]\\ {\displaystyle {\int }_{-\infty }^{0}x{e}^{3x}}dx=-\frac{1}{9}\end{array}$

Hence,

The answer is $-\frac{1}{9}$ .