Question

Let p(x) be a function defined on R such that p’(x) = p’(1–x), for all x $ϵ$ [0, 1], p(0) = 1 and p(1) = 41. Then ${\int }_0^{1}p\left(x\right)dx$ equals

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

Answer 1

The correct option is A 21

$p^{\prime }\left(x\right)=p^{\prime }(1-x)$
$\Rightarrow p\left(x\right)=-p(1-x)+c$
$Putx=0$
$p\left(0\right)=-p\left(1\right)+c$
$\Rightarrow c=42$
$I={\int }_0^{1}p\left(x\right)dx$
$I={\int }_0^{1}p(1-x)dx$
$2I={\int }_0^1\left(p\right(x)+p(1-x\left)\right)dx={\int }_0^{1}cdx={\int }_0^{1}42dx$
$2I=42\Rightarrow I=21$