Question

Let $f$ be a positive function
Let $I_1={\int }_{1-k}^{k}xf\left\{x\right(1-x\left)\right\}dx$
$I_2={\int }_{1-k}^{k}f\left\{x\right(1-x\left)\right\}dx$
where $2k-1>0,$. If $I_2=p{I}_1$ then the value of $p$ is

Answer 1

$I_1={\int }_{1-k}^k(1-k+k-x)f\left\{\right(1-k+k-x\left)\right(1-(1-k+k-x)\left)\right\}dx$
$(\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x\left)dx\right)$
$={\int }_{1-k}^k(1-x)f\left(x\right(1-x\left)\right)dx$
$={\int }_{1-k}^{k}f\left\{x\right(1-x\left)\right\}dx-{\int }_{1-k}^{k}xf\left\{x\right(1-x\left)\right\}dx$
$I_1=I_2-I_1$
$\Rightarrow I_2=2I_1\Rightarrow p=2$