Chapter 11, Problem 1RE

To determine

To find: The value of improper integral ${\displaystyle {\int }_0^{\infty }{e}^{-2x}}dx$, if it converges.

Answer to Problem 1RE

The value of ${\displaystyle {\int }_0^{\infty }{e}^{-2x}}dx$ is $\frac{1}{2}$.

Explanation of Solution

Given:

The improper integral is ${\displaystyle {\int }_0^{\infty }{e}^{-2x}}dx$.

Formula used:

Improper integrals:

“If f is continuous over the indicated interval and the limit exists, then

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

Calculation:

Compute the integral ${\displaystyle {\int }_0^{\infty }{e}^{-2x}}dx$ by using the formula for improper integral.

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }{e}^{-2x}}dx=\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^b{e}^{-2x}}dx\\ =\underset{b\to \infty }{\lim }{\left[-\frac{e^{-2x}}{2}\right]}_0^b\\ =-\frac{1}{2}\underset{b\to \infty }{\lim }\left(e^{-2b}-e^{-2\left(0\right)}\right)\\ =-\frac{1}{2}\left(\underset{b\to \infty }{\lim }\left(e^{-2b}\right)-\underset{b\to \infty }{\lim }\left(1\right)\right)\end{array}$

On further simplification,

$\begin{array}{c}{\displaystyle {\int }_0^{\infty }{e}^{-2x}}dx=-\frac{1}{2}\left(e^{-2\left(\infty \right)}-1\right)\\ =-\frac{1}{2}\left(0-1\right)\\ =\frac{1}{2}\end{array}$

That is, the limit exists. So, the improper integral converges.

Therefore, the value of ${\displaystyle {\int }_0^{\infty }{e}^{-2x}}dx$ is $\frac{1}{2}$.