Ta có $V=\pi \int\limits_{0}^{1}{{{\left[ 2(x-1){{e}^{x}} \right]}^{2}}}dx=4\pi \int\limits_{0}^{1}{({{x}^{2}}-2x+1){{e}^{2x}}dx}=4\pi {{I}_{1}}$

Đặt $\left\{ \begin{align} & u={{x}^{2}}-2x+1 \\ & dv={{e}^{2x}}dv \\ \end{align} \right.$ $\Rightarrow \left\{ \begin{align} & du=2x-2 \\ & v=\frac{{{e}^{2x}}}{2} \\ \end{align} \right.$ $\Rightarrow I=({{x}^{2}}-2x+1)\frac{{{e}^{2x}}}{2}\left| \begin{align} & 1 \\ & 0 \\ \end{align} \right.$ - $\int\limits_{0}^{1}{(x-1){{e}^{2x}}dx=-\frac{1}{2}-{{I}_{2}}}$

Đặt $\left\{ \begin{align} & {{u}_{1}}=x-1 \\ & d{{v}_{1}}={{e}^{2x}}dx \\ \end{align} \right.$ $\Rightarrow \left\{ \begin{align} & d{{u}_{1}}=dx \\ & {{v}_{1}}=\frac{{{e}^{2x}}}{2} \\ \end{align} \right.$ $\Rightarrow {{I}_{1}}=(x-1)\frac{{{e}^{2x}}}{2}\left| \begin{align} & 1 \\ & 0 \\ \end{align} \right.$ -$\frac{1}{2}\int\limits_{0}^{1}{{{e}^{2x}}dx}=\frac{1}{2}-\frac{{{e}^{2x}}}{4}\left| \begin{align} & 1 \\ & 0 \\ \end{align} \right.$ = $\frac{3}{4}-\frac{{{e}^{2}}}{4}$

Do vậy ${{I}_{1}}=\frac{{{e}^{2}}-5}{4}$ suy ra $V=\left( {{e}^{2}}-5 \right)\pi $