$$\int\limits_0^\pi {xf\left( {\sin x} \right)dx} $$ is equal to

Let $$f:R \to R$$ be a differentiable function having $$f\left( 2 \right) = 6$$,
$$f'\left( 2 \right) = \left( {{1 \over {48}}} \right)$$. Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} $$ equals :

$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}}.... + {1 \over n}{{\sec }^2}1} \right]$$
equals

If $${I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } } $$ and $${I_4} = \int\limits_1^2 {{2^{{x^3}}}dx} $$ then