When definite integrals are amenable to exact valuation, it is typically the case that the more expedient approach involves an anti-derivative rather than the limit of a Riemann sum. Often computation of the limit may be straightforward or even trivial, but somewhat tedious, as is the case for integrals of $f: x \mapsto x$ or $f: x \mapsto x^2$.
On the other hand, integrals with simple integrands and easily recognized anti-derivatives such as $f: x\mapsto x^{-2}$ are more challenging with regard to the limit of Riemann sum -- and in that sense the Riemann sum may be "interesting."
To make this more explicit, consider computing the integral
$$\int_a^b x^{-2} \, dx = \lim_{n \to \infty}S_n$$
where
$$S_n =\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-2}.$$
We have
$$\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k+1)\right)^{-1} \leqslant S_n \\ \leqslant \frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k-1)\right)^{-1},$$
and decomposing into partial fractions,
$$\sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}k\right)^{-1}-\left(a + \frac{b-a}{n}(k+1)\right)^{-1}\right\} \leqslant S_n \\\leqslant \sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}(k-1)\right)^{-1}-\left(a + \frac{b-a}{n}k\right)^{-1}\right\}. $$
Since the sums are telescoping, we have
$$\left(a + \frac{b-a}{n}\right)^{-1}-\left(a + \frac{b-a}{n}(n+1)\right)^{-1} \leqslant S_n \leqslant a^{-1} - b^{-1}.$$
By the squeeze theorem, we get the value of the integral as
$$\lim_{n \to \infty}S_n = a^{-1} - b^{-1}.$$
An example where I found the Riemann sum an interesting and, perhaps, most expedient approach is:
Bronstein Integral 21.42
Best Answer
The correct Riemann sum is
$$S_n = \frac{15}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}[(i/n)^2+(j/n)^2]=\frac{15}{n^4}\sum_{i=1}^{n}\sum_{j=1}^{n}(i^2+j^2)=\frac{15}{n^4}\frac{2n^2(n+1)(2n+1)}{6}.$$
Note that
$$\sum_{i=1}^{n}\sum_{j=1}^{n}(i^2+j^2)=\sum_{i=1}^{n}\left[ni^2+\frac{n(n+1)(2n+1)}{6}\right]=\frac{2n^2(n+1)(2n+1)}{6}.$$
Then
$$\lim_{n \rightarrow \infty}S_n = 5\lim_{n \rightarrow \infty}(1+1/n)(2+1/n)=10$$