Let $p(x)$ be a function defined on $R$ such that $p'(x)=p'(1-x)$ for all $x \in[0,1], p(0)=1$ and $p(1)=41$. Then, $\int\limits_0^1 p(x) d x$ equals

21

We have,

$p'(x)=p'(1-x)$

i.e., $\frac{d}{d x} p(x)=\frac{d}{d(1-x)}\{p(1-x)\}$

$\Rightarrow p(x)=-p(1-x)+C$          .....(i)

Putting $x=0$ in (i), we get

$p(0)=-p(1)+C \Rightarrow C=42$           ......(ii)

Putting $C=42$ in (i), we get

$p(x)=-p(1-x)+42$

Now, $\int\limits_0^1 p(x) d x=\int\limits_0^1 p(1-x) d x$

$\Rightarrow \int\limits_0^1 p(x)=\int\limits_0^1\{42-p(x)\} d x$

$\Rightarrow 2 \int\limits_0^1 p(x) d x=42$

$\Rightarrow \int\limits_0^1 p(x) d x=21$