Giả sử hàm f có đạo hàm cấp hai trên R thỏa

Đáp án C

Từ giả thiết $f\left(1-x\right)+x^2{f}^{\text{'}\text{'}}\left(x\right)=2x\Rightarrow f\left(1\right)=0$  (thay x=0  )

$\Rightarrow \underset{0}{\overset{1}{\int }}{x}^2{f}^{\text{'}\text{'}}\left(x\right)dx=\underset{0}{\overset{1}{\int }}2xdx-\underset{0}{\overset{1}{\int }}f\left(1-x\right)dx$

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

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

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

Suy ra .$1-2I=1+I\Rightarrow I=0$