Let $p(x)$ be a function defined on $R$ such that $\lim_{x \rightarrow \infty}{\frac{f(3x)}{f(x)} = 1,\ \ \ }p^{'}(x) = p^{'}(1 - x),$ for all $x\epsilon\lbrack 0,1\rbrack,\ p(0) = 1\ and\ p(1) = 41.Then,\ \int_{0}^{1}{p(x)dx\ }$equals

Question

Dr. Abubacker · Accepted Answer

(b)
We have,
$p^{'}(x) = p^{'}(1 - x),\forall x \in \lbrack 0,1brack,p(0) = 1,p(1) = 41$
$\Rightarrow \ p(x) = - p(1 - x) + C$
At $x = 0,p(0) = - p(1) + C$
$\Longrightarrow \ \ 1 = - 41 + C$
$\Rightarrow \ C = 42$
$	herefore p(x) + p(1 - x) = 42$
Now, $I = \int_{0}^{1}{p(x)dx = \int_{0}^{1}{p(1 - x)dx}}$
$\Rightarrow \ 2I = \int_{0}^{1}{\left( p(x) + p(1 - x) ight)dx = \int_{4}^{1}42dx = 42}$
$\Rightarrow \ I = 21$

Q. Let $p(x)$ be a function defined on $R$ such that $\lim_{x \rightarrow \infty}{\frac{f(3x)}{f(x)} = 1,\ \ \ }p^{'}(x) = p^{'}(1 - x),$ for all $x\epsilon\lbrack 0,1\rbrack,\ p(0) = 1\ and\ p(1) = 41.Then,\ \int_{0}^{1}{p(x)dx\ }$equals

Explanation

Reviewed by Dr. Abubacker

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