If $$\int\limits_0^\pi {xf\left( {\sin x} \right)dx = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx,} } $$ then $$A$$ is

If $$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx} $$
and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${{{I_2}} \over {{I_1}}}$$ is

The value of the integral $$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $$ is

$$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}}$$ - $$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}$$