Question

Let $f$ be a positive function.
Let $I_1=\underset{1-k}{\overset{k}{\int }}xf[x(1-x)]dx$,
$I_2=\underset{1-k}{\overset{k}{\int }}f[x(1-x)]dx$, where $2k-1>0$. Then $\frac{I_1}{I_2}$ is

Answer 1

Answer: 0.5

Given $f$ is a function and
$I_1=\underset{1-k}{\overset{k}{\int }}xf[x(1-x)]dx$
$I_2=\underset{1-k}{\overset{k}{\int }}f[x(1-x)]dx$
Now, $I_1=\underset{1-k}{\overset{k}{\int }}xf[x(1-x)]dx...(i)$
$\underset{1-k}{\overset{k}{\int }}(1-x)f[(1-x)x]dx...(ii)$
$[$ Using Property $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx]$
Adding (i) and (ii)
$2I_1=\underset{1-k}{\overset{k}{\int }}f[x(1-x)]dx=I_2$
$\therefore \frac{I_1}{I_2}=\frac{1}{2}=0.5$