Answer
$\displaystyle{V=\frac{4\pi}{e}}$
Work Step by Step
$\displaystyle{y=e^x\qquad y=e^{-x}}\\
\displaystyle{e^x=e^{-x}}\\
\displaystyle{\ln e^x=\ln e^{-x}}\\
\displaystyle{x\ln e=-x\ln e}\\
\displaystyle{x=-x}\\
\displaystyle{2x=0}\\
\displaystyle{x=0}\\$
$\displaystyle{V=\int_0^1 (2\pi x)\left(e^x-e^{-x}\right)\ dx}\\
\displaystyle{V=2\pi\int_0^1 x\left(e^x-e^{-x}\right)\ dx}\\$
$\displaystyle \left[\begin{array}{ll} u=x & dv=e^x-e^{-x} \\ & \\ du=1 & v=e^x+e^{-x} \end{array}\right]$ Integration by parts
$\displaystyle{V=2\pi \left(\left[x\left(e^x+e^{-x}\right)\right]_0^1-\int_0^1e^x+e^{-x}\ dx\right)}\\
\displaystyle{V=2\pi \left(e+\frac{1}{e}-[e^x-e^{-x}]_0^1\right)}\\
\displaystyle{V=2\pi \left(e+\frac{1}{e}-e+\frac{1}{e}\right)}\\
\displaystyle{V=\frac{4\pi}{e}}$
