If $f\left(x\right)=\frac{e^x}{1+e^x},l_1={\int }_{f(-a)}^{f\left(a\right)} xg\left\{x\right(1-x\left)\right\}dx$ .Then the value of $I_2/I_1$, is

Solution:

We have, $f\left(x\right)=\frac{e^x}{1+e^x}$ 

$\begin{array}{l}f(-x)=\frac{e^{-x}}{1+e^{-x}}=\frac{1}{1+e^x}\\ \therefore f\left(x\right)+f(-x)=\frac{e^x}{1+e^x}+\frac{1}{1+e^x}=1\\ \Rightarrow f\left(a\right)+f(-a)=1\end{array}$

Now, 

        $I_1={\int }_{f(-a)}^{f\left(a\right)} xg\left\{x\right(1-x\left)\right\}dx$                                   …(i)

Using ${\int }_a^b f(a+b-x)={\int }_a^b f\left(x\right)\text{ and }f(-a)+f\left(a\right)=1$, we get

$\begin{array}{l}{I}_1={\int }_{f(-a)}^{f\left(a\right)} (1-x)g\left\{\right(1-x\left)\right(1-(1-x)\left)\right\}dx\\ \Rightarrow I_1={\int }_{f(-a)}^{f\left(a\right)} (1-x)g\left\{x\right(1-x\left)\right\}dx\end{array}$

$\Rightarrow I_1={\int }_{f(-a)}^{f\left(a\right)} (1-x)g\left\{x\right(1-x\left)\right\}dx$                               ..(ii)

Adding (i) and (ii), we get

$\begin{array}{r}2I_1={\int }_{f(-a)}^{f\left(a\right)} g\left\{x\right(1-x\left)\right\}dx\\ \Rightarrow 2I_1=I_2\Rightarrow \frac{I_2}{I_1}=2\end{array}$