Converting riemann sum to definite integral Another Riemann sum I'm struggling to convert to a definite integral... $\lim_{n\to \infty}$ $\sum_{i=1}^n$ $\frac{6n}{9n^2+4i^2}$. Any ideas as to what my $x_i$ should be in this case?

$$ \sum_{i=1}^{n} \frac{6n}{9n^2 + 4i^2} = \frac{1}{n} \sum_{i=1}^{n} \frac{6}{9 + 4(i/n)^2} \; \mathop{\longrightarrow} \limits_{n \to +\infty} \; \int_{0}^{1} \frac{6}{9 + 4t^2} \; dt. $$

This follows from the following result. 
If $f$ is a continuous real-valued function defined on $[a,b]$ ($a < b$) then:
$$ \lim \limits_{n \to +\infty} \frac{b-a}{n} \sum_{k=1}^{n} f\Big( a + k\frac{b-a}{n} \Big) = \int_{a}^{b} f(t) \; dt. $$
Take $a = 0$, $b = 1$ and $\displaystyle f \, : \, t \, \mapsto \, 6/(9 + 4t^2)$. 
