Integral of $1 / \sqrt x$ using Limits Actually the problem here is to find out the INTEGRAL of $\frac{1}{\sqrt x}$ using the limit definition. I am very well able to solve the question using POWER rule but that is not allowed in the question.
$$b-a=nh$$
where $h$ is very small.
Then
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}{\frac{h}{\sqrt x} + \frac{h}{\sqrt{x+h}}+...+\frac{h}{\sqrt{x+(n-1)h}}}$$
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}({\frac{1}{\sqrt x} + \frac{1}{\sqrt{x+h}}+...+\frac{1}{\sqrt{x+(n-1)h}}})h$$
Every problem starts from this very point as from here I am unable to cancel out the $h$
1
Tried to remove the sqrt from the denominator
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}({\frac{\sqrt x}{x} + \frac{\sqrt{x+h}}{x+h}+...+\frac{\sqrt{x+(n-1)h}}{x+(n-1)h}})h$$
After which it seems much more complicated
2
Taking the $h$ in the numerator to the denominator
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}{\frac{1}{\sqrt{\frac{x}{h^2}}} + \frac{1}{\sqrt{\frac{x+h}{h^2}}}+...+\frac{1}{\sqrt{\frac{x+(n-1)h}{h^2}}}}$$
After some cancelling
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}{\frac{1}{\sqrt{\frac{x}{h^2}}} + \frac{1}{\sqrt{\frac{x}{h^2}+\frac{1}{h}}}+...+\frac{1}{\sqrt{\frac{x}{h^2}+\frac{(n-1)}{h}}}}$$
Which left me in a frenzy
3
Lastly i tried to rake common first
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}{\frac{h}{\sqrt x} + \frac{h}{\sqrt{x+h}}+...+\frac{h}{\sqrt{x+(n-1)h}}}$$
Taking $\frac{h}{\sqrt x}$ common
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}\frac{h}{\sqrt x}({1 + \frac{1}{\sqrt{1+\frac{h}{x}}}+...+\frac{1}{\sqrt{1+\frac{(n-1)h}{x}}}})$$
I was not able to remove $h$ from this method even
 A: If the interval is positive and avoids $0$, the function is continuous, and thus integrable. Since it is integrable, any refinements of partitions will approach the correct answer. We then would like to set up a partition $a,aq,aq^2,\cdots$ where $q=(b/a)^{1/N}$ for an order $N$ partition. Note that Riemann’s integral does not require partitions to be equally spaced; he only requires that in refinement, the largest spacing (the “mesh”) goes to zero, which this geometric partition does. Also note that this geometric partition runs from $a\to b$ which is of course also necessary! Your sum with equal spacing would also work, but... it is extremely difficult to evaluate, whereas this one is easier as we shall soon see.
Note also that this method will integrate $x^a$ for any rational power of $a\neq-1$; do this as an exercise, maybe!
Our partial sums are:
$$\begin{align}\sum_{n=0}^{N-1}(aq^n)^{-1/2}(aq^{n+1}-aq^n)&=\sum_{n=0}^{N-1}(aq^n)^{-1/2}aq^n(q-1)\\&=\sum_{n=0}^{N-1}a^{1/2}q^{n/2}(q-1)\\&=a^{1/2}(q-1)\sum_{n=0}^{N-1}q^{n/2}\end{align}$$
Hopefully you are familiar with the geometric series:
$$a^{1/2}(q-1)\sum_{n=0}^{N-1}q^{n/2}=a^{1/2}(q-1)\frac{q^{N/2}-1}{q^{1/2}-1}=a^{1/2}(q-1)\frac{(b/a)^{1/2}-1}{q^{1/2}-1}$$
Which resolves to:
$$a^{1/2}((b/a)^{1/2}-1)\frac{q-1}{q^{1/2}-1}=(b^{1/2}-a^{1/2})\cdot\frac{q-1}{q^{1/2}-1}$$
I have for the last fraction (difference of two squares):
$$\frac{q-1}{q^{1/2}-1}=q^{1/2}+1\to2,\,N\to\infty$$
Since $\lim_{N\to\infty}(b/a)^{1/N}=1$, assuming $(b/a)\gt0$ which we have done. If you want to generalise this, put the power (as a rational) as $r/s$, and put $\tau=q^{1/s}$; you'll get two geometric series, and using a similar argument you'll get the right answer. To generalise this to irrational powers $\neq-1$, you use the continuity of exponentiation and the mean value theorem for integration to approximate it closer and closer with integrals of rational power; you arrive at the power law for integration at the end, and together with the fundamental theorem of calculus you arrive at a formal derivation of the power law of differentiation for all non-zero real powers! I found this personally to be a very instructive from-first-principles derivation: this is all due to Courant's Differential and Integral Calculus, a very old but pretty good textbook.
The final answer is then:
$$\int_a^b\frac{1}{\sqrt{x}}\,\mathrm{d}x=\lim_{N\to\infty}\sum_{n=0}^{N-1}\frac{1}{\sqrt{aq^n}}(aq^{n+1}-aq^n)=(b^{1/2}-a^{1/2})\cdot\lim_{N\to\infty}\frac{q-1}{q^{1/2}-1}=2(b^{1/2}-a^{1/2})$$
As required.
A: I'm posting another answer based on fixed-width partitions and a "harmonic mean approximation" for fun.  I think this answer can hardly be generalized to other rational powers of $x$.  The image and the arguments were copied from my former messages on Discord.

We would make a little trick of approximating $1/\sqrt{x}$ by the harmonic mean of the integrand evaluated at midpoints of neighbouring partitions
$$\frac{1}{\sqrt{x}} \approx \frac{2}{\sqrt{x - h/2} + \sqrt{x + h/2}}.\label{hma}\tag{$\Large\star$}$$
Using the difference of squares identity, it's easy to see that
$$\frac{2}{\sqrt{x - h/2} + \sqrt{x + h/2}} = \frac{2 (\sqrt{x + h/2} - \sqrt{x - h/2})}{h},$$
so the right Riemann sum can be approximated by
$$\begin{aligned} S &= \sum_{k = 1}^n \frac{1}{\sqrt{x_k}} \cdot h \\
&\approx \sum_{k = 1}^n \frac2h \cdot \left(\sqrt{x_k + h/2} - \sqrt{x_k - h/2} \right) \cdot h \\
&= 2 \left(\sqrt{b+h/2} - \sqrt{a+h/2} \right)
\end{aligned}$$
Here we have $n = (b-a)/h$ partitions with partition points $x_k = a + kh$ for $k \in \{1, \dots, n\}$.  Using the following elementary identity, I'm going to find the order of the error in terms of mesh $h$.

Exercise: Show that for all $a,b\ge0$, $\sqrt{\mathstrut a+b} \le \sqrt{\mathstrut a} + \sqrt{\mathstrut b}.$
Hence, show that for all $a>0$ and $b \in [0,a]$, $\sqrt{\mathstrut a-b} \le \sqrt{\mathstrut a} - \sqrt{\mathstrut b}.$

These two basic inequalities enable us to establish upper and lower bounds for the denominator in our "harmonic mean approximation" \eqref{hma}
$$2\sqrt{\mathstrut x} - \sqrt{\mathstrut h/2} < \sqrt{\mathstrut x - h/2} + \sqrt{\mathstrut x + h/2} < 2 \sqrt{\mathstrut x} + \sqrt{\mathstrut h/2}.$$
Take reciprocal, multiply by 2, then minus the integrand $1/\sqrt{x}$ to observe that the actual error in our "harmonic mean approximation" \eqref{hma} can be bounded by other another fraction.
Calculate
$$\frac{2}{2\sqrt{x} \pm \sqrt{h/2}} - \frac{1}{\sqrt{x}} = \frac{\mp\sqrt{h/2}}{(2\sqrt{x} \pm \sqrt{h/2}) \sqrt{x}}$$
to get the bound
$$\left|\frac{2}{2\sqrt{x} \pm \sqrt{h/2}} - \frac{1}{\sqrt{x}}\right| = \frac{\sqrt{h/2}}{\sqrt{a} \cdot \sqrt{a}} < \frac{1}{\sqrt2 a} \, h^{1/2}$$
for $x\in[a,b]$.
Use the triangle inequality $|a+b|\le|a|+|b|$ to take summation out of the absolute sign.
$$
\begin{aligned}
& \left|\underbrace{\sum_{k=1}^n\frac{2}{\sqrt{x_k - h/2} + \sqrt{x_k + h/2}} \cdot h }_{2(\sqrt{b+h/2} - \sqrt{a+h/2})} - \underbrace{\sum_{k=1}^n\frac{1}{\sqrt{x_k}}  \cdot h}_{S} \right| \\
&\le \sum_{k=1}^n \underbrace{\left| \frac{2}{\sqrt{x_k - h/2} + \sqrt{x_k + h/2}} - \frac{1}{\sqrt{x_k}} \right|}_{< h^{1/2}/(\sqrt2 a)} \cdot h \\
&< nh \, \frac{h^{1/2}}{\sqrt{2} a} \\
&= \frac{b-a}{\sqrt2 a} \, h^{1/2}
\end{aligned}
\label{err1}\tag{main error}
$$
It remains to take away the $h/2$ in $\sqrt{a+h/2}$ and $\sqrt{b+h/2}$.
\begin{gather*}
\sqrt{\mathstrut b} < \sqrt{\mathstrut b + h/2} < \sqrt{\mathstrut b} + \sqrt{\mathstrut h/2} \\
-\sqrt{\mathstrut a} - \sqrt{\mathstrut h/2} < -\sqrt{\mathstrut a + h/2} < -\sqrt{\mathstrut a}
\end{gather*}
Add these two inequalites to see that
$$\left|\left(\sqrt{\mathstrut b + h/2}-\sqrt{\mathstrut a + h/2}\right)-\left(\sqrt{\mathstrut b}-\sqrt{\mathstrut a}\right)\right|< \sqrt{\mathstrut h/2}.\label{err2}\tag{minor error}$$
Use \ref{err1}, \ref{err2} and the triangle inequality to see that
$$\left|S - 2\left(\sqrt{\mathstrut b} - \sqrt{\mathstrut a}\right)\right| < \left(\frac{b-a}{\sqrt2 a} + 2 \right) h^{1/2}.$$
