Question

Let $f$ be a positive function
Let $I_1={\int }_{1-k}^{k}xf\left\{x\right(1-x\left)\right\}dxand{I}_2={\int }_{1-k}^{k}f\left\{x\right(1-x\left)\right\}dx$
where $2k-1>0,$ then $\frac{I_1}{I_2}$ is

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

Answer 1

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

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