Let $f$  be a positive function,$I_1={\displaystyle \underset{1-k}{\overset{k}{\int }}xf\left(x\left(1-x\right)\right)dx}$  and $I_2={\displaystyle \underset{1-k}{\overset{k}{\int }}f\left(x\left(1-x\right)\right)dx}$ . Then $\frac{I_1}{I_2}=$

${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx=\frac{1}{2}}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx\to }$ 

$I_1=\underset{1-k}{\overset{k}{\int }}xf\left(x(1-x)\right)dx =\underset{1-k}{\overset{k}{\int }}(1-x)f\left((1-x)x\right)dx$

$\therefore 2I_1={\displaystyle \underset{k}{\overset{1-k}{\int }}\left(x+1-x\right)f\left(x\left(1-x\right)\right)dx}$

           $={\displaystyle \underset{k}{\overset{1-k}{\int }}f\left(x\left(1-x\right)\right)dx=I_2\to \frac{I_1}{I_2}=\frac{1}{2}}$