Let$p\left(x\right)$ be a function defined on R such that $p\text{'}\left(x\right) = p\text{'}(1 – x)$, for all $x\in [0,1],p\left(0\right)=1$ and $p\left(1\right) = 41.$ Then ${\int }_0^1 p\left(x\right)dx$ equals

${\int }_0^1 p\left(x\right)dx={\int }_0^1 1\cdot p\left(x\right)dx=\left[xp\right(x){]}_0^1-I_1=p(1)-I_1$

where 

$\begin{array}{l}{I}_1={\int }_0^1 x{p}^{\mathrm{\prime }}\left(x\right)dx={\int }_0^1 (1-x)p^{\mathrm{\prime }}(1-x)dx\\ ={\int }_0^1 (1-x)p^{\mathrm{\prime }}\left(x\right)dx={\int }_0^1 p^{\mathrm{\prime }}\left(x\right)dx-I_1\end{array}$

$\Rightarrow 2I_1=\left[p\right(x){]}_0^1=p(1)-p(0)$

Thus, 

$\begin{array}{r}{\int }_0^1 p\left(x\right)dx=p\left(1\right)-\frac{1}{2}\left(p\right(1)-p(0\left)\right)\\ =\frac{1}{2}\left(p\right(1)+p(0\left)\right)=\frac{1}{2}(41+1)=21\end{array}$