Giải bài 39 trang 175 SGK giải tích nâng cao 12

Thể tích khối tròn xoay cần tìm là

$V=\pi \int\limits_{0}^{1}{{{\left( x{{e}^{\frac{x}{2}}} \right)}^{2}}dx}=\pi \int\limits_{0}^{1}{{{x}^{2}}{{e}^{x}}dx}$

Đặt $\left\{ \begin{aligned} & {{x}^{2}}=u \\ & {{e}^{x}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & 2xdx=du \\ & v={{e}^{x}} \\ \end{aligned} \right.$

$\begin{aligned} \Rightarrow V&=\pi {{x}^{2}}{{e}^{x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.-2\pi \int\limits_{0}^{1}{x{{e}^{x}}dx} \\ & =\pi e-2\pi \int\limits_{0}^{1}{x{{e}^{x}}dx} \\ \end{aligned}$

Đặt $\left\{ \begin{aligned} & x=u \\ & {{e}^{x}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v={{e}^{x}} \\ \end{aligned} \right.$

$\begin{aligned} \Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}dx}&=x{{e}^{x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.-\int\limits_{0}^{1}{{{e}^{x}}dx} \\ & =e-{{e}^{x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right. \\ & =e-e+1=1 \\ \end{aligned}$

$\Rightarrow V=\pi e-2\pi =\pi \left( e-2 \right)$ (đvtt)