Evaluate the following as a limit of sum:
$$\int^b_a{\frac{1}{\sqrt{x}}}dx$$
We can write it as
$$\lim_{h \to 0;h>0} h\left(\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{a+h}} + \frac{1}{\sqrt{a+2h}}+…\frac{1}{\sqrt{a+(n-1)h}}\right)$$ where $h=\dfrac{b-a}{n}$
I'm not sure how to proceed further.
I also tried to expand everything binomially. It didn't lead to anything.
Best Answer
Note that with $a > 0$,
$$\underbrace{h\sum_{k=0}^{n-1}\frac{2}{\sqrt{a + kh} + \sqrt{a + (k+1)h}}}_{=2\sum_{k=0}^{n-1}\left(\sqrt{a + (k+1)h} - \sqrt{a + kh}\right)}\leqslant h\sum_{k=0}^{n-1}\frac{1}{\sqrt{a + kh}} \\\leqslant \frac{h}{\sqrt{a}}+ \underbrace{ \sum_{k=1}^{n-1}\frac{2}{\sqrt{a + kh} + \sqrt{a + (k-1)h}}}_{=2\sum_{k=1}^{n-1}\left(\sqrt{a + kh} - \sqrt{a + (k-1)h}\right)}$$
See if you can finish yourself by evaluating the telescoping sums on the LHS and RHS, and then applying the squeeze theorem in taking limits as $h \to 0$.