Chapter 7.3, Problem 46E

To determine

To find: The volume of the solid.

Answer to Problem 46E

The volume of the solid is 0.594.

Explanation of Solution

Given:

  $A\left(x\right)=x{e}^{-x},0\le x\le 2$ .

Concept used:

The volume of a solid of known integrable cross-section area $A\left(x\right)$ from $x=a$ to $x=b$ is the integral of $A$ from $a$ to $b$ ,

  $V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}$

Calculation:

The volume of the solid is,

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

Now consider,

  ${\displaystyle \int x{e}^{-x}dx}$

Let $u=x$ and $dv=e^{-x}dx$ .

Now by differentiating $u=x$ with respect to $x$ , it follows that

  $\begin{array}{l}\frac{du}{dx}=1\\ du=dx\end{array}$

Now by integrating $dv=e^{-x}dx$ , it follows that

  $\begin{array}{c}v={\displaystyle \int e^{-x}dx}\\ =-e^{-x}\end{array}$

Therefore,

  $\begin{array}{c}{\displaystyle \int x{e}^{-x}dx}=x\left(-e^{-x}\right)-{\displaystyle \int \left(-e^{-x}\right)dx}\\ =-x{e}^{-x}-e^{-x}+C\end{array}$

Therefore, the volume is

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{2}{\int }}x{e}^{-x}dx}={\left[-x{e}^{-x}-e^{-x}\right]}_0^2\\ =\left[-2e^{-2}-e^{-2}\right]-\left[-0\cdot e^0-e^0\right]\\ =-2e^{-2}-e^{-2}+1\\ \approx 0.594\end{array}$

Hence, the volume of the solid is 0.594.