The value of lim n → ∞ 1 n ∑ r = 0 2 n - 1 n 2 n 2 + 4 r 2 is:

Let

$I=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=0}^{2n-1}\left(\frac{n^2}{n^2+4r^2}\right)$

$I=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=0}^{2n-1}\left(\frac{n^2}{n^2+4r^2}\right)$

$I=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=0}^{2n-1}\left(\frac{1}{1+4{\left({\displaystyle \frac{r}{n}}\right)}^2}\right)$

Let $\frac{r}{n}=x$, lower limit $x=0$ when $r=0$ and upper limit $x=2$ when $r=2n-1$

$\Rightarrow I={\int }_0^2\frac{1}{1+4x^2}dx$

$\Rightarrow I=\frac{1}{4}{\int }_0^2\frac{1}{{\left({\displaystyle \frac{1}{2}}\right)}^2+x^2}dx$

$\Rightarrow I=\frac{1}{4}\times 2{\left[{\tan }^{-1}2x\right]}_0^2=\frac{1}{2}{\tan }^{-1}\left(4\right)$