Answer
The volume of the solid formed by revolving the function $f(x)=e^{-x} $ about the x-axis from $x=-2$ to $x=1$, is
$$
\begin{aligned}
V &=\int_{a}^{b} \pi [f(x)]^{2}dx\\
&= \int_{-2}^{1} \pi [e^{-x} ]^{2}dx \\
&=\frac{\pi\left(e^{4}-e^{-2}\right)}{2} \\
& \approx 85.55 .
\end{aligned}
$$
Work Step by Step
$$
\left( f(x)=e^{-x}, \quad y =0, \quad x=-2, \quad x=1 \right)
$$
If $f(x) $ is nonnegative and $R$ is the region between $f(x) $ and the x-axis from $x=-2$ to $x=1$, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\int_{a}^{b} \pi [f(x)]^{2}dx
$$
So,the volume of the solid formed by revolving the function $f(x)=e^{-x} $ about the x-axis from $x=-2$ to $x=1$, is
$$
\begin{aligned}
V &= \int_{-2}^{1} \pi [e^{-x} ]^{2}dx \\
&=\int_{-2}^{1} \pi e^{-2x} dx\\
&=\pi \int_{-2}^{1} e^{-2 x} d x \\
&=\left.\frac{\pi e^{-2 x}}{-2}\right|_{-2} ^{1} \\
&=\frac{\pi e^{-2}}{-2}+\frac{\pi e^{4}}{2} \\
&=\frac{\pi\left(e^{4}-e^{-2}\right)}{2} \\ & \approx 85.55
\end{aligned}
$$
