$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{n + 1}}{{{n^2} + {1^2}}} + \frac{{n + 2}}{{{n^2} + {2^2}}} + \frac{{n + 3}}{{{n^2} + {3^2}}} + \dots + \frac{1}{n}} \right) = $

Given that $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x$. If $f: R \rightarrow R$ is defined by $f(x)=x^2+2$, then $\lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right)+\ldots+f(7)\right]=$

Evaluate the following definite integral as the limit of a sum:
$\int_{0}^{4} (x + e^{2x}) \, dx$

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n}\right]=$

For $a \in \mathbb{R}$ (the set of all real numbers),$a \neq -1$,if $\lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$,then $a$ is equal to:

The value of $\lim _{n}$ ${\rightarrow \infty} \left( \frac{\sqrt{n}}{\sqrt{n^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+4)^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+8)^{3}}}+\cdots +\frac{\sqrt{n}}{\sqrt{[n+4(n-1)]^{3}}} \right)$ is