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


My progress so far:
$\begin{align}
\int_a^b\frac{1}{\sqrt x}dx & =\lim_{n\to\infty}h\times\bigg(f(a)+f(a+h)+\cdots+f\big(a+(n-1)h\big)\bigg) \\
& = \lim_{n\to\infty}h\times\bigg(\frac1{\sqrt a}+\frac1{\sqrt {a+h}} +\cdots+\frac1{\sqrt {a+(n-1)h}}\bigg)
\end{align}$

After this point, I cant't think of a way to combine the reciprocals of square roots into a neat form. Usually a convenient summation formula is applicable but how would I proceed in this case?
 A: We assume $0\leq a\leq b$. In order to cope with the square root function $\frac{1}{\sqrt{x}}$ it is convenient to use variable length intervals with length $j^2\frac{b-a}{n^2}$. When taking square roots we can factor out $j$  and summation is expected to become simpler. The corresponding Riemann sum is
\begin{align*}
\int_{a}^{b}\frac{1}{\sqrt{x}}\,dx=\lim_{n\to \infty}\sum_{j=1}^n\frac{1}{\sqrt{\color{blue}{a}+ j^2\frac{b-a}{n^2}}}\cdot\left(j^2\frac{b-a}{n^2}-(j-1)^2\frac{b-a}{n^2}\right)\tag{1}
\end{align*}
But it's still not easy due to the constant $\color{blue}{a}$. To overcome this difficulty we write the integral as difference of two improper integrals
\begin{align*}
\int_{a}^{b}\frac{1}{\sqrt{x}}\,dx=\int_{0}^{b}\frac{1}{\sqrt{x}}\,dx-\int_{0}^{a}\frac{1}{\sqrt{x}}\,dx\tag{2}
\end{align*}
If the limit of the Riemann sums of the improper integrals exists we have found the wanted Riemann integral and we are done.

We subdivide the interval $[0,b]$ in 
  \begin{align*}
[0,b]=\bigcup_{j=1}^n\left[(j-1)^2\frac{b}{n^2},j^2\frac{b}{n^2}\right]
\end{align*}
  with  length
  \begin{align*}
j^2\frac{b}{n^2}-(j-1)^2\frac{b}{n^2}=\frac{(2j-1)b}{n^2}\qquad\qquad1\leq  j\leq n
\end{align*}
  and  obtain the Riemann sum
  \begin{align*}
\color{blue}{\int_{0}^b\frac{1}{\sqrt{x}}\,dx}&=\lim_{n\to\infty}\sum_{j=1}^n\frac{1}{\sqrt{\frac{j^2b}{n^2}}}\cdot\frac{(2j-1)b}{n^2}\tag{3}\\
&=\sqrt{b}\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^n\left(2-\frac{1}{j}\right)\tag{4}\\
&=\sqrt{b}\lim_{n\to\infty}\frac{1}{n}\left(2n-H_n\right)\tag{5}\\
&\,\,\color{blue}{=2\sqrt{b}}\tag{6}
\end{align*}
We finally  conclude  from (2) and (6)
  \begin{align*}
\color{blue}{\int_{a}^b\frac{1}{\sqrt{x}}\,dx=2\left(\sqrt{b}-\sqrt{a}\right)}
\end{align*}

Comment:


*

*In (3) we have the  now convenient representation (1) evaluated at  $a=0$.

*In (4) we   do   some  simplifications and     factor  out    terms not dependent on $n$ resp.   $j$.

*In (5) we note the Harmonic numbers $H_n=\sum_{j=1}^n \frac{1}{j}$ grow asymptotically with  $\log n$, so that
\begin{align*}
\lim_{n\to\infty}\frac{H_n}{n}=\lim_{n\to\infty}\frac{\log n}{n}=0.
\end{align*}
