Cho hàm số y = f(x) có đạo hàm liên tục trên R và thỏa mãn $f(-2)=1;{\int }_1^{2}f(2x-4)dx=1.$ Tính $I={\int }_{-2}^{0}xf\text{'}\left(x\right)dx.$  

+) $1=\underset{1}{\overset{2}{\int }}f(2x-4)dx=\frac{1}{2}\underset{1}{\overset{2}{\int }}f(2x-4)d(2x-4)=\frac{1}{2}\underset{-2}{\overset{0}{\int }}f\left(t\right)dt=\frac{1}{2}\underset{-2}{\overset{0}{\int }}f\left(x\right)dx\Rightarrow \underset{-2}{\overset{0}{\int }}f\left(x\right)dx=2$ 

+) $I=\underset{-2}{\overset{0}{\int }}xf\text{'}\left(x\right)dx.$ 

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

$\Rightarrow I=xf\left(x\right)\left|{}_{{}_{-2}}^{{}^0}\right.-\underset{-2}{\overset{0}{\int }}f\left(x\right)dx=2f(-2)-\underset{-2}{\overset{0}{\int }}f\left(x\right)dx=2.1-2=0.$

Chọn B.