Let p(x) be a function defined on R such that p'(x) = p'(1-x), for all x [0,1], p(0)=1 and…

Question

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

Accepted Answer

$$p'\left( x \right) = p'\left( {1 - x} \right)$$

$$ \Rightarrow p\left( x \right) =  - p\left( {1 - x} \right) + c$$

at  $x=0$

$$p\left( 0 \right) =  - p\left( 1 \right) + c \Rightarrow 42 = c$$

Now, $$p\left( x \right) =  - p\left( {1 - x} \right) + 42$$

$$ \Rightarrow p\left( x \right) + p\left( {1 - x} \right) = 42$$

$$ \Rightarrow I = \int\limits_0^1 {p\left( x \right)dx\,\,\,\,\,\,\,\,\,\,...\left( i \right)} $$

$$ \Rightarrow I = \int\limits_0^1 {p\left( {1 - x} \right)} dx\,\,\,\,\,\,...\left( {ii} \right)$$

on adding $(i)$ and $(ii),$

$$\,\,\,\,\,\,\,\,\,\,\,$$ $2I = \int\limits_0^1 {\left( {42} \right)dx \Rightarrow I = 21} $

Solutions · 1