Cho hàm số y = f(x) có đạo hàm liên tục trên R

Xét tích phân $I=\underset{0}{\overset{1}{\int }}xf\left(x\right)dx.$

Đặt $\left\{\begin{array}{l}u=f\left(x\right)\\ dv=xdx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=f\text{'}\left(x\right)dx\\ v=\frac{x^2}{2}\end{array}\right.$

$\Rightarrow I=\frac{x^2}{2}f\left(x\right)\left|\begin{array}{l}1\\ 0\end{array}\right.-\frac{1}{2}\underset{0}{\overset{1}{\int }}{x}^{2}f\text{'}\left(x\right)dx$

     $=\frac{1}{2}f\left(1\right)-\frac{1}{2}\underset{0}{\overset{1}{\int }}x\left(e^x-1\right)dx=-\frac{1}{2}J$

Ta có $J=\underset{0}{\overset{1}{\int }}x\left(e^x-1\right)dx=\underset{0}{\overset{1}{\int }}x{e}^{x}dx-\underset{0}{\overset{1}{\int }}xdx=\underset{0}{\overset{1}{\int }}x{e}^{x}dx-\frac{1}{2}.$

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

$\Rightarrow \underset{0}{\overset{1}{\int }}x{e}^{x}dx=x{e}^x\left|\begin{array}{l}1\\ 0\end{array}\right.-\underset{0}{\overset{1}{\int }}{e}^{x}dx=e-\left(e-1\right)=1.$

Vậy $I=-\frac{1}{2}\left(1-\frac{1}{2}\right)=-\frac{1}{4}.$

Chọn B.