Question

If $f(x)=\frac{e^{x}}{1+e^{x}}, I_{1}=\int\limits_{f(-a)}^{f(a)} x g(x(1-x)) d x$ and $I_{2}=\int\limits_{f(-a)}^{f(a)} g(x(1-x)) d x$, then find the value of $\frac{I_{2}}{I_{1}}$.

Answer 1

$f(a)+f(-a)=1$
$I _{1}=\int\limits_{ f (- a )}^{ f ( a )} xg ( x (1- x )) dx$
$I _{1}=\int\limits_{ f (- a )}^{ f ( a )}(1- x ) g ((1- x ) x ) dx$
$\Rightarrow I _{1}+ I _{1}=\int\limits_{ f (- a )}^{ f ( a )} g ( x (1- x )) dx$
$\Rightarrow 2 I _{1}= I _{2}$
$\Rightarrow \frac{ I _{2}}{ I _{1}}=2$