Question

let $f$ be a positive function.
If $I_1={\int }_{1-k}^{k}xf\left[x\right(1-x\left)\right]dx$ and
$I_2={\int }_{1-k}^{k}f\left[x\right(1-x\left)\right]dx$, where $2k-1>0$.
Then, $\frac{I_1}{I_2}$ is

• A. $2$
• B. $k$
• C. $1/2$
• D. $1$

Answer 1

The correct option is C $1/2$

$\begin{array}{cc}{I}_1={\int }_{1-K}^{K}xf\left[x\left(1-x\right)\right]dx& \left[\begin{array}{c}usingproperty\\ {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\end{array}\right]\\ a=1-K,b=K& \\ a-b-K=1-K+k-x=1-x& \\ I_1={\int }_{1-k}^K\left(1-x\right)f\left[\left(1-x\right)x\right]dx\Rightarrow {\int }_{1-K}^{K}f[x\left(1-x\right)dx-{\int }_{1-K}^{K}f\left[x\left(1-x\right)\right]dx& \\ I_1+I_2=I_2& \\ 2I_1=I_2& \\ \frac{I_1}{I_2}=\frac{1}{2}& \end{array}$