Chapter 8, Problem 56EP

i.

To determine

To calculate: To solve the area of $R$ as an improper integral.

Answer to Problem 56EP

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

Explanation of Solution

Given information: The given curve as follows:

The area of $R$ as an improper integral. The Infinite region $R$ in the first quadrant under the curve $y=x{e}^{-\frac{x}{2}}$ .

Calculation:

Since , the area can be calculated as:

  $Area={\displaystyle {\int }_a^{b}f\left(x\right)dx}$

The area of $R$ as an improper integral.

The area in an improper integral form can be represented as follows:

  $\text{Area}={\displaystyle {\int }_0^{\infty }x{e}^{-\frac{x}{2}}dx}$ .

ii.

To determine

To calculate: To solve the integral as a limit of a definite integral.

Answer to Problem 56EP

The Area of an improper integral is $\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^{b}x{e}^{-\frac{x}{2}}dx}$ .

Explanation of Solution

Given information: The given curve as follows:

The area of $R$ as an improper integral. The Infinite region $R$ in the first quadrant under the curve $y=x{e}^{-\frac{x}{2}}$ .

Calculation:

Since , the area can be calculated as:

  $Area={\displaystyle {\int }_a^{b}f\left(x\right)dx}$

The area of $R$ as an improper integral.

The integral as a limit of a definite integral can be represented as follows:

  $\text{Area}=\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^{b}x{e}^{-\frac{x}{2}}dx}$ .

i.

To determine

To calculate: To solve the area of $R$ .

Answer to Problem 56EP

The Area of an improper integral is $4$ .

Explanation of Solution

Given information: The given curve as follows:

The area of $R$ as an improper integral. The Infinite region $R$ in the first quadrant under the curve $y=x{e}^{-\frac{x}{2}}$ .

Calculation:

Since , the area can be calculated as:

  $Area={\displaystyle {\int }_a^{b}f\left(x\right)dx}$

The area of $R$ as an improper integral.

Here, we can solve as:

  $\begin{array}{c}\text{Area}=\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^{b}x{e}^{-\frac{x}{2}}dx}\\ $ =\underset{b\to \infty }{\lim }{[-2\left(x+2\right)e^{-\frac{x}{2}}]}_0^b$=\underset{b\to \infty }{\lim }[-2\left(b+2\right)e^{-\frac{b}{2}}+2\left(2\right)e^0]\\ =\underset{b\to \infty }{\lim }[-2\left(b+2\right)e^{-\frac{b}{2}}+4]\\ =4-2\underset{b\to \infty }{\lim }[\frac{-2\left(b+2\right)}{e^{\frac{b}{2}}}]\\ =4-2\underset{b\to \infty }{\lim }[\frac{-2}{\frac{-1}{2}{e}^{\frac{b}{2}}}]\\ =4-2\left(0\right)\\ =4\end{array}$