Question

Let $f\left(x\right)$ and $g\left(x\right)$ be two functions satisfying $f\left(x^2\right)+g(4-x)=4x^3$ and $g(4-x)+g\left(x\right)=0,$ then the value of $\underset{-4}{\overset{4}{\int }}f\left(x^2\right)dx=$

Answer 1

$I=\underset{-4}{\overset{4}{\int }}f\left(x^2\right)dx=2\underset{0}{\overset{4}{\int }}f\left(x^2\right)dx\cdots \left(i\right)$
$(\because f(x^2\left)\text{is even function}\right)$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$I=2\underset{0}{\overset{4}{\int }}f(4-x{)}^{2}dx\cdots (ii)$
Adding $\left(i\right)$ and $\left(ii\right):$
$\Rightarrow 2I=2\underset{0}{\overset{4}{\int }}\left[f\right(x^2)+f(4-x{)}^2]dx\cdots (iii)$
Given, $f\left(x^2\right)+g(4-x)=4x^3\cdots \left(iv\right)$
Replacing $x\to (4-x):$
$\Rightarrow f\left(\right(4-x{)}^2)+g(x)=4(4-x{)}^3\cdots \left(v\right)$
Adding $\left(iv\right)$ and $\left(v\right):$
$\Rightarrow f\left(x^2\right)+f\left(\right(4-x{)}^2)+g(x)+g(4-x)=4[x^3+(4-x{)}^3]\Rightarrow f\left(x^2\right)+f\left(\right(4-x{)}^2)=4[x^3+(4-x{)}^3]\cdots \left(vi\right)$
From $\left(iii\right)$ & $\left(vi\right)\Rightarrow I=4\underset{0}{\overset{4}{\int }}[x^3+(4-x{)}^3]dx=512$