Question

If $p\left(x\right)$ be a function defined on $R$ such that $p\text{'}\left(x\right)=p\text{'}(1-x)$, $\forall x\in [0,1],p\left(0\right)=1$and $p\left(1\right)=41$. Then, ${\int }_0^{1}p\left(x\right)dx$

• A. $\sqrt{41}$
• B. $21$
• C. $41$
• D. $42$

Answer 1

The correct option is B

$21$

Explanation for the correct option.

Step 1: Find $p\left(x\right)$

Given that, $p^{\text{'}}\left(x\right)=p^{\text{'}}(1-x),x\in [0,1],p\left(0\right)=1,p\left(1\right)=41$.
Integrating both sides in $p^{\text{'}}\left(x\right)=p^{\text{'}}(1-x)$ we have:

$p\left(x\right)=-p(1-x)+c$
At $\mathrm{x}=0,\mathrm{p}\left(0\right)=-\mathrm{p}\left(1\right)+\mathrm{c}$
$\begin{array}{rcl}1& =& -41+c\\ \mathrm{c}& =& 42\end{array}$

Step 2: Integrate ${\int }_0^{1}p\left(x\right)dx$
$\mathrm{I}={\int }_0^1\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}...\left(1\right)$
$I={\int }_0^1\mathrm{p}(1-\mathrm{x})\mathrm{dx}..\left(2\right)$
Adding (1) and (2)
$\begin{array}{rcl}2\mathrm{I}& =& {\int }_0^1\mathrm{p}\left(\mathrm{x}\right)+\mathrm{p}(1-\mathrm{x})\mathrm{dx}\\ & =& 42{\int }_0^{1}dx\left[p\right(x)+p(1-x)=42]\\ 2\mathrm{I}& =& 42\\ I& =& 21\end{array}$

Hence, option B is correct.