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 $\int_0^1 p(x) \, dx$ equals

Given $p'(x)=p'(1-x)$ for $x\in[0,1]$, define $f(x)=p(x)-p(1-x)$. Then $$f'(x)=p'(x)+p'(1-x).$$ Using the given condition, $$f'(x)=2p'(x).$$ A simpler symmetric idea is to differentiate $p(x)+p(1-x)$: $$\frac{d}{dx}\bigl(p(x)+p(1-x)\bigr)=p'(x)-p'(1-x)=0.$$ So $p(x)+p(1-x)$ is constant on $[0,1]$. Evaluate it at $x=0$: $$p(0)+p(1)=1+41=42.$$ Hence for every $x\in[0,1]$, $$p(x)+p(1-x)=42.$$ Integrate from $0$ to $1$: $$\int_0^1 p(x)\,dx+\int_0^1 p(1-x)\,dx=\int_0^1 42\,dx.$$ By the substitution $u=1-x$, the two integrals on the left are equal, so $$2\int_0^1 p(x)\,dx=42.$$ Therefore, $$\int_0^1 p(x)\,dx=21.$$ This matches option $\mathrm{(C)}$.