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

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 :

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

$$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{e^{{r \over n}}}} $$ is