Giả sử hàm $f$ có đạo hàm cấp 2 trên $\mathbb{R}$ thỏa mãn...

Đặt $\left\{ \begin{aligned}
& u={f}'\left( x \right) \
& dv=xdx \

\end{aligned} \right.\Leftrightarrow \left\{ $$\begin{array}{rl}& du={f^{\prime }}^{\prime }\left(x\right)dx\\ & v={\displaystyle \frac{x^2}{2}}\end{array}$$ \right.\Rightarrow 2I=\left. {{x}^{2}}{f}'\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{{{x}^{2}}{{f}'}'\left( x \right)dx}$$\Leftrightarrow 2I=1-\int\limits_{0}^{1}{{{x}^{2}}{{f}'}'\left( x \right)dx}$ suy ra

Ta có $f(1-x)+x^2{f^{\prime }}^{\prime }\left(x\right)=2x\iff \underset{0}{\overset{1}{\int }}f(1-x)dx+\underset{0}{\overset{1}{\int }}{x}^2{f^{\prime }}^{\prime }\left(x\right)dx=\underset{0}{\overset{1}{\int }}2xdx$
$\iff \underset{0}{\overset{1}{\int }}f\left(x\right)dx+1-2I={x^2|}_0^1=1\Rightarrow$ (1)

Đặt $\{\begin{array}{rl}& u=f\left(x\right)\\ & dv=dx\end{array}\iff \{\begin{array}{rl}& du=f^{\prime }\left(x\right)dx\\ & v=x\end{array}\Rightarrow$ (2)
Từ (1), (2) suy ra $2I=1-I\to I={\displaystyle \frac{1}{3}}$.