Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
$y = e^x, y = 0, x = -1, x = 1$ rotated by x-axis.
So I don't know how to draw the graph but...this is the integral I get:
$$ \pi \int_{-1}^1 e^{2x} \, dx$$
$$ = \pi \left [ \frac{e^{2x}}{2} \right ]_{-1}^1$$
$$ = \frac{e^2}{2} - \frac{e^{-2}}{2}$$
But wolfram alpha says it is this. What is this even?
The function $\sinh$ is defined by$$\sinh x = \frac{e^x-e^{-x}}{2}$$ so Wolfram Alpha is giving you the answer $\frac{e^2-e^{-2}}{2}$ which is the answer you got.
Note that you dropped the $\pi$ factor on the last line so the answer is actually $\pi \sinh 2$.