Let p(x) be a function defined on R such that $\mathrm{p}\text{'}\left(\mathrm{x}\right)=\mathrm{p}\text{'}(1-\mathrm{x})$ for all $\mathrm{x}\text{\hspace{0.17em}∈\hspace{0.17em}\hspace{0.17em}}[0,1]$, $\mathrm{p}\left(0\right)=1$ and $\mathrm{p}\left(1\right)=41$ then ${\int }_0^1\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}=$

$\begin{array}{l}{\mathrm{p}}^{\mathrm{\prime }}\left(\mathrm{x}\right)={\mathrm{p}}^{\mathrm{\prime }}(1-\mathrm{x})\text{ integrating }\\ \mathrm{p}\left(\mathrm{x}\right)=-\mathrm{p}(1-\mathrm{x})+\mathrm{\lambda }\\ \mathrm{x}=0\Rightarrow \mathrm{p}\left(0\right)=-\mathrm{p}\left(1\right)+\mathrm{\lambda }\\ 1=-41+\mathrm{\lambda }\Rightarrow \mathrm{\lambda }=42\end{array}$

$$\begin{gathered} \left.\begin{array}{r}\begin{array}{l}\mathrm{I}=\underset{0}{\overset{1}{\int }}\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}....\left(1\right)\\ \mathrm{I}=\underset{0}{\overset{1}{\int }}\mathrm{p}(1-\mathrm{x})\mathrm{dx}...\left(2\right)\end{array}\\ \end{array}\right\} \\ Adding both \Rightarrow 2\mathrm{I}\underset{0}{\overset{1}{=\int }}(\mathrm{p}\left(\mathrm{x}\right)+\mathrm{p}(1-\mathrm{x}))\mathrm{dx} \end{gathered}$$

$=\underset{0}{\overset{1}{\int }}\left(42\right)\mathrm{dx}=42$