Cho hàm số $y=f\left( x \right)$ liên tục và có đạo hàm trên...

Ta có: $5f\left( x \right)-7f\left( 1-x \right)=3\left( {{x}^{2}}-2x \right)$
Lần lượt chọn $x=0,x=1$, ta có hệ sau: $\left\{ \begin{aligned}
& 5f\left( 0 \right)-7f\left( 1 \right)=0 \\
& 5f\left( 1 \right)-7f\left( 0 \right)=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& f\left( 1 \right)=\dfrac{5}{8} \\
& f\left( 0 \right)=\dfrac{7}{8} \\
\end{aligned} \right.$
Tính $I=\int\limits_{0}^{1}{x.{f}'\left( x \right)}dx$
Đặt: $\left\{ \begin{aligned}
& u=x \\
& dv={f}'\left( x \right)dx \\
\end{aligned} \right. $ Chọn $ \left\{ \begin{aligned}
& du=dx \\
& v=f\left( x \right) \\
\end{aligned} \right..I=x.f\left( x \right)\left| _{0}^{1} \right.-\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{5}{8}-J$
Đặt $x=1-t\Rightarrow J=-\int\limits_{1}^{0}{f\left( 1-x \right)dt=\int\limits_{0}^{1}{f\left( 1-x \right)dx=K}}$
Suy ra $5J-7K=3\int\limits_{0}^{1}{\left( {{x}^{2}}-2x \right)dx=-2}$. Ta có $\left\{ \begin{aligned}
& J=K \\
& 5J-7K=-2 \\
\end{aligned} \right.\Leftrightarrow J=K=1$
Vậy $I=\dfrac{5}{8}-1=\dfrac{-3}{8}\Rightarrow \left\{ \begin{aligned}
& a=3 \\
& b=8 \\
\end{aligned} \right.\Rightarrow T=8a-3b=0$.