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?

Let $x_i=\dfrac{i}{n}$ then $x_1=\dfrac{1}{n}\to0$, $x_n=\dfrac{n}{n}\to1$and $\Delta x=\dfrac1n$ so
$$\lim_{n\to\infty}\sum_{i=1}^n\dfrac{6}{9+4(\dfrac{i}{n})^2}\frac1n=\int_0^1\dfrac{6}{9+4x^2}dx$$
