Lời giải:Gọi V là thể tích cần tìm:
\(V = \pi \int\limits_0^2 {{{\left( {x{e^x}} \right)}^2}dx = \pi \int\limits_0^2 {{x^2}{e^{2x}}dx} } \)
Đặt: \(\left\{ \begin{array}{l}
u = {x^2}\\
dv = {e^{2x}}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = 2xdx\\
v = \frac{1}{2}{e^{2x}}
\end{array} \right.\)
\(V = \left. {\frac{{\pi {x^2}{e^{2x}}}}{2}} \right|_0^2 - \pi \int\limits_0^2 {x{e^{2x}}dx} = 2\pi {e^4} - \pi \int\limits_0^2 {x{e^{2x}}dx} \)
Đặt: \(\left\{ \begin{array}{l}
u = x\\
dv = {e^{2x}}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \frac{1}{2}{e^{2x}}
\end{array} \right.\)
\(V = 2\pi {e^4} - \pi \int\limits_0^2 {x{e^{2x}}dx = 2\pi {e^4} - \left[ {\left. {\frac{{\pi x{e^{2x}}}}{2}} \right|_0^2 - \frac{\pi }{2}\int\limits_0^2 {{e^{2x}}dx} } \right]} \)
\( = 2\pi {e^4} - \left[ {\pi {e^4} - \left. {\frac{\pi }{4}{e^{2x}}} \right|_0^2} \right] = 2\pi {e^4} - \pi {e^4} + \frac{\pi }{4}\left( {{e^4} - 1} \right)\)
\( = \frac{\pi }{4}\left( {5{e^4} - 1} \right)\)
