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?
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$\begingroup$Let $x_i=\dfrac{i}{n}$ then $x_1=\dfrac{1}{n}\to0$, $x_n=\dfrac{n}{n}\to1$and $\Delta x=\dfrac1n$ so $$\lim_{n\to\infty}\sum_{i=1}^n\dfrac{6}{9+4(\dfrac{i}{n})^2}\frac1n=\int_0^1\dfrac{6}{9+4x^2}dx$$
$\endgroup$ 1 $\begingroup$$$ \sum_{i=1}^{n} \frac{6n}{9n^2 + 4i^2} = \frac{1}{n} \sum_{i=1}^{n} \frac{6}{9 + 4(i/n)^2} \; \mathop{\longrightarrow} \limits_{n \to +\infty} \; \int_{0}^{1} \frac{6}{9 + 4t^2} \; dt. $$
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 $\displaystyle f \, : \, t \, \mapsto \, 6/(9 + 4t^2)$.
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