Evaluate $\int_a^b\frac{1}{ x^2}dx$ using limit of a sum definition 
From the definition of a definite integral as the limit of a sum,
  evaluate $$\int_a^b\frac{1}{ x^2}dx$$


Step 1
To simplify working with $x^2$, divide the interval $[a,b]$ using variable length intervals: $$[a,b]=\bigcup_{j=1}^n\bigg[a+\frac{\sqrt {j-1}(b-a)}{\sqrt n}, a+\frac{\sqrt j(b-a)}{\sqrt n}\bigg]$$ 
Step 2
Now rewriting the integral to remove the lower limit ($0\leq a\leq b$) : $$\int_a^b\frac1{x^2}dx=\int_0^b\frac{1}{x^2}dx-\int_0^a\frac{1}{x^2}dx$$
Step 3
Dividing $[0,b]$ as described in Step 1:$$[0,b]=\bigcup_{j=1}^n\bigg[\frac{\sqrt {j-1}(b)}{\sqrt n}, \frac{\sqrt j(b)}{\sqrt n}\bigg]$$ 
Step 4
Using limit of a sum definition on $\int_0^b\frac{1}{x^2}dx$ : 
(result can then be used for  $\int_0^a\frac{1}{x^2}dx$)
$$\begin{align}
\int_0^b\frac1{x^2} &= \lim_{n\to\infty}\sum_{j=1}^n\bigg(\frac{\sqrt j(b)}{\sqrt n}\bigg)^{-2}\times \bigg[\frac{\sqrt j(b)}{\sqrt n}-\frac{\sqrt {j-1}(b)}{\sqrt n} \bigg] \\
&=\frac1{b}\lim_{n\to\infty}\sqrt n \times \sum_{j=1}^n\frac1{j}\big[\sqrt j - \sqrt {j-1}\big]
\end{align}$$

After this point, I can not seem to find a nice form of $\sum_{j=1}^n\frac1{j}\big[\sqrt j - \sqrt {j-1}\big]$ to work with. Any help would be appreciated. 


We obtain for $0<a\leq  b$:
  \begin{align*}
\color{blue}{\int_a^b\frac{1}{x^2}\,dx}&=\lim_{n\to\infty}\sum_{j=1}^nf\left(a+j\frac{b-a}{n}\right)\frac{b-a}{n}\\
&=\lim_{n\to\infty}\sum_{j=1}^n\frac{1}{\left(a+j\frac{b-a}{n}\right)^2}\cdot\frac{b-a}{n}\\
&\,\,\color{blue}{=\lim_{n\to\infty}\frac{n}{b-a}\sum_{j=1}^n\frac{1}{\left(\frac{an}{b-a}+j\right)^2}}\tag{1}
\end{align*}

We  calcluate  the limit (1) by squeezing it with lower and upper bounds which can be easily calculated using telescoping.
We consider the inequality chain
\begin{align*}
\frac{1}{\left(\frac{an}{b-a}+j\right)\left(\frac{an}{b-a}+j+1\right)}
&\leq \frac{1}{\left(\frac{an}{b-a}+j\right)^2}\leq  \frac{1}{\left(\frac{an}{b-a}+j-1\right)\left(\frac{an}{b-a}+j\right)}\\
\frac{1}{\frac{an}{b-a}+j}-\frac{1}{\frac{an}{b-a}+j+1}
&\leq \frac{1}{\left(\frac{an}{b-a}+j\right)^2}\leq  \frac{1}{\frac{an}{b-a}+j-1}-\frac{1}{\frac{an}{b-a}+j}\tag{2}
\end{align*}
The left-most and  right-most   part admit  telescoping which  makes summation  and  taking  the limit an easy job.

We  start with the left-most  part of (2) and obtain
\begin{align*}
\color{blue}{\lim_{n\to\infty}}&\color{blue}{\frac{n}{b-a}\sum_{j=1}^n\left(\frac{1}{\frac{an}{b-a}+j}-\frac{1}{\frac{an}{b-a}+j+1}\right)}\\
&=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{1}{\frac{an}{b-a}+1}-\frac{1}{\frac{an}{b-a}+n+1}\right)\\
&=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{b-a}{an+b-a}-\frac{b-a}{bn+b-a}\right)\\
&=\lim_{n\to\infty}\left(\frac{n}{an+b-a}-\frac{n}{bn+b-a}\right)\\
&\,\,\color{blue}{=\frac{1}{a}-\frac{1}{b}}\tag{3}
\end{align*}
We continue with the right-most part of (2)
\begin{align*}
\color{blue}{\lim_{n\to\infty}}&\color{blue}{\frac{n}{b-a}\left(\frac{1}{\frac{an}{b-a}}-\frac{1}{\frac{an}{b-a}+n}\right)}\\
&=\lim_{n\to\infty}\frac{n}{b-a}\left(\frac{b-a}{an}-\frac{b-a}{bn}\right)\\
&\,\,\color{blue}{=\frac{1}{a}-\frac{1}{b}}\tag{4}
\end{align*}
We finally conclude since (1) is squeezed by (3) and (4)
\begin{align*}
\color{blue}{\int_a^b\frac{1}{x^2}\,dx=\frac{1}{a}-\frac{1}{b}}
\end{align*}

