Question

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

Answer 1

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