Explanation

(b)
Let $I = \lim_{n \rightarrow \infty}{\frac{1}{n}\sum_{r = 1}^{2n}\frac{r}{\sqrt{n^{2} + r^{2}}}}$
$= \lim_{n \rightarrow \infty}\frac{1^{\ }}{n}\sum_{r = 1}^{2n}\frac{r}{n\sqrt{1 + (r/n)^{2}}}$
$= \lim_{n \rightarrow \infty}\frac{1^{\ }}{n}\sum_{r = 1}^{2n}\frac{r/n}{\sqrt{1 + (r/n\ )^{2}}}$
Put $\frac{r}{n} = x,\ \frac{1}{n}dx,\ \lim_{n \rightarrow \infty}\ \sum_{r = 1}^{2n}{= \int_{0}^{2}\ }$
$\therefore I = \int_{0}^{2}{\frac{x}{\sqrt{1 + x^{2}}}dx = \lbrack\sqrt{1 + x^{2}}\left. \ \ \right\rbrack_{0}^{2}} = \sqrt{5} - 1$