Let $0 < a < b$. Use Riemann sums to compute
$$ \int_{a}^{b}x^{-2}\ \mathrm{d}x\ . $$
So far, I have gotten to the step where
\begin{align} \int_{a}^{b} x^{-2}\ \mathrm{d}x & = \lim_{n \to \infty}\ \sum_{i = 1}^{n} \left[\left(a + \frac {b - a} {n} \ i\right)^{-2}\left(\frac {b - a} {n}\right)\right] \\[5 mm] & = \lim_{n\to\infty}\left[\frac{b - a}{n}\ \sum_{i = 1}^{n}\left(a + \frac {b - a} {n}\ i\right)^{-2}\ \right]. \end{align}
I assume my equation is correct, but I am not sure how to evaluate the sum.
I also know how to compute
$$ \int_{a}^{b}x^{2} \ \mathrm{d}x\ , $$
but I am not sure how to carry forward the idea (if it is even relevant) to this problem (if it is even relevant).
Riemann sums were just covered in my module and I am still trying to get used to it. I would appreciate some help/guidance on this!.
With a uniform partition, let $X_k = a + \frac{b-a}{n}k$. The Riemann sum is $R_n = \frac{b-a}{n}\sum_{k=1}^n X_k^{-2}$ and it can be squeezed between sums whose limits are easily evaluated, viz.
$$\tag{*}\frac{b-a}{n}\sum_{k=1}^n X_k^{-1}X_{k+1}^{-1}\leqslant R_n \leqslant \frac{b-a}{n}\sum_{k=1}^n X_{k-1}^{-1}X_k^{-1}$$
Note that
$$\frac{b-a}{n}\sum_{k=1}^n X_k^{-1}X_{k+1}^{-1} = \sum_{k=1}^n (X_k^{-1}-X_{k+1}^{-1}) = X_1^{-1} - X_{n+1}^{-1} \\ = \left(a+ \frac{b-a}{n} \right)^{-1}- \left(a+ (b-a)\frac{n+1}{n} \right)^{-1} \\ \underset{n \to \infty} \to(a^{-1} - b^{-1})$$
Similarly, we can show that the sum on the RHS of (*) also converges to $a^{-1} - b^{-1}$, and by the squeeze theorem
$$\lim_{n \to \infty}R_n = a^{-1} - b^{-1} = \int_a^b \frac{dx}{x^2}$$