Question

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

Answer 1

Let $I={\int }_{-2}^2\frac{x^2}{1+5^x}$

we know that ${\int }_a^{b}f\left(x\right)={\int }_a^{b}f(a+b-x)$

Therefore $I={\int }_{-2}^2\frac{x^2}{1+5^x}$ and $I={\int }_{-2}^2\frac{{(-x)}^2}{1+5^{-x}}={\int }_{-2}^2\frac{x^2{5}^x}{1+5^x}$

Add above two , we get $2I={\int }_{-2}^2\frac{x^2{5}^x}{1+5^x}+{\int }_{-2}^2\frac{x^2}{1+5^x}={\int }_{-2}^2{x}^2=\frac{{\left(2\right)}^3}{3}-\frac{{(-2)}^3}{3}=\frac{16}{3}$

Therefore $I=\frac{8}{3}$