Cho hàm số $f\left( x \right)$ liên tục trên $\left[ 0;1 \right].$...

HD: Ta có: $I=\int\limits_{0}^{1}{\left[ x.{f}'\left( 1-x \right)-f\left( x \right) \right]}dx=\int\limits_{0}^{1}{x.{f}'\left( 1-x \right)dx}-\int\limits_{0}^{1}{f\left( x \right)dx}$
Đặt $t=1-x\Rightarrow dt=-dx.$ Đổi cận $\left| \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=0 \\
\end{aligned} \right.,$ ta có:
$\int\limits_{0}^{1}{x.{f}'\left( 1-x \right)dx}=\int\limits_{1}^{0}{\left( 1-t \right){f}'\left( t \right)\left( -dt \right)}=\int\limits_{0}^{1}{\left( 1-x \right){f}'\left( x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=1-x \\
& dv={f}'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-dx \\
& v=f\left( x \right) \\
\end{aligned} \right. $ ta có $ \int\limits_{0}^{1}{\left( 1-x \right){f}'\left( x \right)dx}=\left( 1-x \right)f\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.+\int\limits_{0}^{1}{f\left( x \right)dx}=-f\left( 0 \right)+\int\limits_{0}^{1}{f\left( x \right)dx}$
Suy ra $I=-f\left( 0 \right)\Rightarrow f\left( 0 \right)=-\dfrac{1}{2}.$ Chọn C.