Cho hàm số f(x) thỏa mãn x(f(2x)dx=1 và f(1)=-2 . Tích phân x^2f'(x)dx bằng

Đáp án D

Ta có $\underset{0}{\overset{1}{\int }}{x}^2.f^{\text{'}}\left(x\right)dx=\underset{0}{\overset{1}{\int }}{x}^{2}d\left[f\left(x\right)\right]=x^2.f\left(x\right)-\underset{0}{\overset{1}{\int }}f\left(x\right)d\left(x^2\right)$

$=f\left(1\right)-\underset{0}{\overset{1}{\int }}2x.f\left(x\right)dx=-2-2\underset{0}{\overset{1}{\int }}x.f\left(x\right)dx$

Xét $\underset{0}{\overset{\frac{1}{2}}{\int }}x.f\left(2x\right)dx=1$, đặt

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

$\Rightarrow \underset{0}{\overset{1}{\int }}x.f\left(x\right)dx=4\Rightarrow \underset{0}{\overset{1}{\int }}{x}^2.f^{\text{'}}\left(x\right)dx=-2-2.4=-10$. Chọn D.