Question

Evaluate : $\stackrel{2}{\underset{-2}{\int }}\frac{x^2}{1+5^x}dx$.

Answer 1

Let
$I=\stackrel{2}{\underset{-2}{\int }}\frac{x^2}{1+5^x}dx$ .........(1)

By using property,

$\stackrel{b}{\underset{a}{\int }}f\left(x\right)dx=\stackrel{b}{\underset{a}{\int }}f(a+b-x)dx$, we will get

$I=\stackrel{2}{\underset{-2}{\int }}\frac{(-2+2-x{)}^2}{1+5^{(-2+2-x)}}dx$

$I=\stackrel{2}{\underset{-2}{\int }}\frac{x^2}{1+5^{-x}}dx=\stackrel{2}{\underset{-2}{\int }}\frac{x^2}{1+\frac{1}{5^x}}dx$

$I=\stackrel{2}{\underset{-2}{\int }}\frac{5^x.x^2}{5^x+1}dx$ .................(2)

Adding (1) and (2), we will get,
$2I=\stackrel{2}{\underset{-2}{\int }}[\frac{x^2}{1+5^x}+\frac{5^x.x^2}{1+5^x}]dx$

$2I=\stackrel{2}{\underset{-2}{\int }}\frac{x^2(1+5^x)dx}{(1+5^x)}$

$2I=\stackrel{2}{\underset{-2}{\int }}{x}^{2}dx=2\stackrel{2}{\underset{0}{\int }}{x}^{2}dx$

$[\because f(x)=x^2$ is an even function]

$\Rightarrow I=\frac{1}{2}\times 2{\left[\frac{x^3}{3}\right]}_0^2=\frac{1}{2}\times 2\times \frac{(2{)}^3}{3}$
Hence,
$I=\frac{8}{3}$