How do you find the volume of the solid generated by revolving the region bounded by the graphs y=e^-x, y=0, x=0, x=1, about the x axis?

$V = \frac{\pi \left({e}^{2} - 1\right)}{2 {e}^{2}}$

Explanation:

At any distance x, they coordinate happens to be the radius of the revolved element
$y = {e}^{-} x$
$r = {e}^{-} x$
Circular area generated by revolving around x axis is
$A = \pi {r}^{2}$
volume of the elementary solid of thickness dx is
$\mathrm{dV} = A \mathrm{dx}$
The lower limit is given to be x=0
The upper limit is given to be x=1

$\mathrm{dV} = A \mathrm{dx}$

$\mathrm{dV} = \pi {r}^{2} \mathrm{dx}$
${r}^{2} = {\left({e}^{-} x\right)}^{2}$
${r}^{2} = {e}^{-} \left(2 x\right)$

$\mathrm{dV} = \pi {e}^{-} \left(2 x\right) \mathrm{dx}$
Integrating between x=0 and x=1

${\int}_{0}^{1} \mathrm{dV} = {\int}_{0}^{1} \pi {e}^{-} \left(2 x\right) \mathrm{dx}$

$V = - \frac{\pi}{2} | {e}^{- 2 x} {|}_{0}^{1}$

$V = - \frac{\pi}{2} \left({e}^{- 2 \times 1} - {e}^{- 2 \times 0}\right)$
$V = - \frac{\pi}{2} \left({e}^{- 2} - 1\right)$

$V = - \frac{\pi}{2} \left(\frac{1}{e} ^ 2 - 1\right)$

$V = \frac{\pi}{2} \left(1 - \frac{1}{e} ^ 2\right)$

$V = \frac{\pi}{2} \frac{{e}^{2} - 1}{e} ^ 2$

$V = \frac{\pi \left({e}^{2} - 1\right)}{2 {e}^{2}}$