# How do I convert lim_(n->oo) sum_(i=0)^n (12n)/(16n^2 + 9i^2) into a definite integral?

## I'm new when it comes to Riemann sums. I also have trouble with finding $\Delta {x}_{i}$.

$\arctan \left(\frac{3}{4}\right)$
${\lim}_{n \to \infty} {\sum}_{i = 0}^{n} \frac{12 n}{16 {n}^{2} + 9 {i}^{2}} = {\lim}_{n \to \infty} {\sum}_{i = 0}^{n} \left(\frac{12}{16 + 9 {\left(\frac{i}{n}\right)}^{2}}\right) \frac{1}{n}$
and calling now $\xi = \frac{i}{n}$ and $d \xi = \frac{1}{n}$
${\lim}_{n \to \infty} {\sum}_{i = 0}^{n} \left(\frac{12}{16 + 9 {\left(\frac{i}{n}\right)}^{2}}\right) \frac{1}{n} = {\int}_{0}^{1} \frac{12}{16 + 9 {\xi}^{2}} d \xi = \arctan \left(\frac{3}{4}\right)$