Question

Let $f\left(x\right)$ be a positive function. 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

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

Answer 1

Answer: (C)

$\frac{1}{2}$

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