Chapter 9, Problem 56RE

a.

To determine

To find: The area of $R$ as an improper integral.

Answer to Problem 56RE

  $R={\displaystyle \underset{0}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx={\displaystyle \underset{0}{\overset{2}{\int }}x{e}^{\frac{-x}{2}}}dx+{\displaystyle \underset{2}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx$

Explanation of Solution

Given information:

  $R$ is the region bounded by $y=x{e}^{\frac{-x}{2}}$ and x-axis.

Formula used :

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx={\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)}dx+{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)}dx\text{ , }\forall c\in \left(a,b\right)$

Calculation:

  

Area of $R$ is shown in the adjacent figure.

Here, $y=x{e}^{\frac{-x}{2}}$ cuts the x-axis at $x=0$ and at $x=+\infty$

Therefore, $R={\displaystyle \underset{0}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx={\displaystyle \underset{0}{\overset{2}{\int }}x{e}^{\frac{-x}{2}}}dx+{\displaystyle \underset{2}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx$

The interval $\left[0,\infty \right)$ is divided into two parts as at $x=2$ the functional value is the maximum and in the first part the function is increasing and in the other part functional value is decreasing.

b.

To determine

To express: The integral of part (a) as a limit of a definite integral.

Answer to Problem 56RE

  $R=\underset{A\to \infty }{\lim }{\displaystyle \underset{0}{\overset{A}{\int }}x{e}^{\frac{-x}{2}}}dx$

Explanation of Solution

Given information:

  $R={\displaystyle \underset{0}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx={\displaystyle \underset{0}{\overset{2}{\int }}x{e}^{\frac{-x}{2}}}dx+{\displaystyle \underset{2}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx$

Concept Used:

Calculation:

  $R={\displaystyle \underset{0}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx=\underset{A\to \infty }{\lim }{\displaystyle \underset{0}{\overset{A}{\int }}x{e}^{\frac{-x}{2}}}dx$

c.

To determine

To find: The area of $R$

Answer to Problem 56RE

Required area $=$$2\left(1+\frac{1}{e}\right)$

Explanation of Solution

Given information:

  $R={\displaystyle \underset{0}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx$

Concept used :

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx={\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)}dx+{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)}dx\text{ , }\forall c\in \left(a,b\right)$

Calculation:

Required area $={\displaystyle \underset{0}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx$

  $={\displaystyle \underset{0}{\overset{2}{\int }}x{e}^{\frac{-x}{2}}}dx+{\displaystyle \underset{2}{\overset{\infty }{\int }}x{e}^{\frac{-x}{2}}}dx$

Now, $I_2={\displaystyle \underset{0}{\overset{2}{\int }}x{e}^{\frac{-x}{2}}}dx={\left[-2x\cdot e^{\frac{-x}{2}}\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}-2\cdot 1\cdot e^{\frac{-x}{2}}dx}\left[\text{applying By-parts rule}\right]$

  $\begin{array}{l}=-\frac{4}{e}+2{\displaystyle \underset{0}{\overset{2}{\int }}{e}^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx\\ =-\frac{4}{e}+{\left[-2e^{\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^2\\ =-\frac{4}{e}-\frac{2}{e}+2\\ =2-\frac{6}{e}\end{array}$

  $\begin{array}{c}{I}_2={\displaystyle \underset{2}{\overset{\infty }{\int }}x{e}^{-\frac{x}{2}}}dx\\ =\underset{A\to \infty }{\lim }{\displaystyle \underset{2}{\overset{A}{\int }}x{e}^{-\frac{x}{2}}}dx\end{array}$

  $\begin{array}{l}=\underset{A\to \infty }{\lim }\left\{{\left[-2x{e}^{-\frac{x}{2}}\right]}_2^A+2{\displaystyle \underset{2}{\overset{A}{\int }}{e}^{-\frac{x}{2}}}dx\right\}\\ =\underset{A\to \infty }{\lim }\left\{{\left[-2x{e}^{-\frac{x}{2}}\right]}_2^A-4{\left[e^{-\frac{x}{2}}\right]}_2^A\right\}\\ =-2\cdot \underset{A\to \infty }{\lim }\frac{A}{e^{\frac{A}{2}}}+\frac{A}{e}\cdot \underset{A\to \infty }{\lim }\frac{4}{e^{\frac{A}{2}}}+\frac{4}{e}\\ =-2\cdot \underset{A\to \infty }{\lim }\frac{2}{e^{\frac{A}{2}}}+\frac{8}{e}-0\\ =\frac{8}{e}\end{array}$

So, the required area = $I_1+I_2=2-\frac{6}{e}+\frac{8}{e}=2+\frac{2}{e}=2\left(1+\frac{1}{e}\right)$