Question

Let f be a given function such that f(x)>0, ∀x \in D_f. 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 $2k-1>0$
then $\frac{I_1}{I_2}$ is equal to

• A. $1$
• B. $k$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{1}{2}$

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