$\lim _{n}$ ${\rightarrow \infty} \frac{1}{n} \left[ \frac{1}{n} \sin ^{-1} \frac{1}{n} + \frac{2}{n} \sin ^{-1} \frac{2}{n} + \dots + \frac{n}{n} \sin ^{-1} \frac{n}{n} \right] =$

$\lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2n+1}}+\frac{n}{(n+2) \sqrt{2(2n+2)}}+\frac{n}{(n+3) \sqrt{3(2n+3)}}+\ldots n \text{ terms}\right]=\int_0^1 f(x) d x$,then $f(x)=$

$\mathop {Lim}\limits_{n \to \infty } \frac{\pi }{{6n}}\left[ {{{\sec }^2}\left( {\frac{\pi }{{6n}}} \right) + {{\sec }^2}\left( {2 \cdot \frac{\pi }{{6n}}} \right) + \dots + {{\sec }^2}\left( {(n - 1)\frac{\pi }{{6n}}} \right) + \frac{4}{3}} \right]$ has the value equal to

$\int_{0}^{1} a^k x^k dx =$

If $U_{n}=\left(1+\frac{1^{2}}{n^{2}}\right)^{1}\left(1+\frac{2^{2}}{n^{2}}\right)^{2} \ldots\left(1+\frac{n^{2}}{n^{2}}\right)^{n}$,then $\lim _{n \rightarrow \infty}\left(U_{n}\right)^{\frac{-4}{n^{2}}}$ is equal to :

$\mathop {Lim}\limits_{n \to \infty } \,\,\sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}{x^2}}}} $,$x > 0$ is equal to