Answer
$\displaystyle{V=\pi\left(1-\frac{1}{e}\right)}$
Work Step by Step
$\displaystyle{V=\int_{0}^{1}(2\pi x)\left(e^{-x^2}\right)\ dx}\\
\displaystyle{V=2\pi\int_{0}^{1}xe^{-x^2}\ dx}$
$\displaystyle \left[\begin{array}{ll} u=x^2 & =) \\ & \\ \frac{du}{dx}=2x & \frac{du}{2x}=dx \end{array}\right]$ Integration by substitution
$\displaystyle{V=2\pi\int_{0}^{1}xe^{-u}\ \frac{du}{2x}}\\
\displaystyle{V=\pi\int_{0}^{1}e^{-u}\ du}\\
\displaystyle{V=\pi\left[-e^{-u}\right]_{0}^{1}}\\
\displaystyle{V=\pi\left(\left(-e^{-1}\right)-\left(-e^{0}\right)\right)}\\
\displaystyle{V=\pi\left(1-\frac{1}{e}\right)}$
