Chapter 9.4, Problem 20E

To determine

To find: the value of improper integral.

Answer to Problem 20E

The value of integral is $\infty$ and integral is diverge.

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_0^{\infty }\left(x+1\right)e^{-x}dx}$ .

Calculation: by using definition of improper integrals with infinite integration limits, the improper integral can be written as:

  ${\displaystyle {\int }_0^{\infty }\left(x+1\right)e^{-x}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\left(x+1\right)e^{-x}dx}$

To find the value of integral, first integrate and then simplify,

  $\begin{array}{l}{\displaystyle {\int }_0^{\infty }\left(x+1\right)e^{-x}dx}=\underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\left(x+1\right)e^{-x}dx}\\ {\displaystyle {\int }_0^{\infty }\left(x+1\right)e^{-x}dx}=\underset{a\to \infty }{\lim }{\left[-\left(x+2\right)e^{-x}\right]}_0^a\text{ [}{\displaystyle \int uvdx=u{\displaystyle \int vdx}-{\displaystyle \int \frac{du}{dx}}{\displaystyle \int vdxdx}\text{ and simplify}]}\\ {\displaystyle {\int }_0^{\infty }\left(x+1\right)e^{-x}dx}=\underset{a\to \infty }{\lim }\left[-\left(a+2\right)e^{-a}+\left(0+2\right)e^{-0}\right]\text{ [simplify]}\\ {\displaystyle {\int }_0^{\infty }\left(x+1\right)e^{-x}dx}=\infty +2\text{ [simplify]}\\ {\displaystyle {\int }_0^{\infty }\left(x+1\right)e^{-x}dx}=\infty \end{array}$

Thus, the value is $\infty$ .