Question
Let $p\left( x \right)$ be a function defined on $R$ such that $p'\left( x \right) = p'\left( {1 - x} \right),$ for all $x \in \left[ {0,1} \right]$ $p\left( 0 \right) = 1,p\left( 1 \right) = 41.$ then $\mathop \smallint \limits_0^1 p\left( x \right)dx = $

Answer

a
$p^{\prime}(x)=p^{\prime}(1-x)$

$\Rightarrow p(x)=-p(1-x)+c$

at $x=0$ $p(0)=-p(1)+c \quad $

$ \Rightarrow 42=c$

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

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

$I = \int\limits_0^1 p (x)dx = \int\limits_0^1 p (1 - x)dx$

$I = \int\limits_0^1 {\left( {42} \right)dx} \,\,\,\,\,\,\,\, \Rightarrow I = 21.$

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