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 :

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

The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,} $$ is

The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx $$ is