Let $f$ be a positive function. Let

$I_1=\int\limits_{1-k}^k x f\left\{x(1-x\} d x, I_2=\int\limits_{1-k}^k f\{x(1-x\} d x\right.$

where $2 k-1>0$. Then, $\frac{I_1}{I_2}$ is

1/2

Using $\int\limits_0^a f(x) d x=\int\limits_0^a f(a-x) d x$, we have

$I_1 =\int\limits_{1-k}^k(1-x) f\{(1-x)(1-(1-x))\} d x$

$\Rightarrow I_1 =\int\limits_{1-k}^k(1-x) f\{x(1-x)\} d x$

$\Rightarrow \quad I_1=I_2-I_1 \Rightarrow 2 I_1=I_2 \Rightarrow \frac{I_1}{I_2}=\frac{1}{2}$