Answer
$$ 2\pi (e^2+1)$$
Work Step by Step
Since
\begin{align*}
V&=\int_{a}^{b}(2 \pi r) h d x\\
&=2 \pi \int_{0}^{2} x e^{x} d x
\end{align*}
Let
\begin{align*}
u&=x\ \ \ \ \ \ \ \ \ \ \ \ dv= e^xdx\\
du&=dx\ \ \ \ \ \ \ \ \ \ \ v=e^x
\end{align*}
Then
\begin{align*}
V&=\int_{a}^{b}(2 \pi r) h d x\\
&=2 \pi \int_{0}^{2} x e^{x} d x\\
&=2\pi \left( xe^x \bigg|_{0}^{2} - \int_{0}^{2} e^{x} d x\right)\\
&= 2\pi \left( xe^x -e^x\bigg|_{0}^{2} \right)\\
&= 2\pi (e^2+1)
\end{align*}
