<s>[INST] 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
 [/INST] 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
$$\left|S - 2\left(\sqrt{\mathstrut b} - \sqrt{\mathstrut a}\right)\right| < \left(\frac{b-a}{\sqrt2 a} + 2 \right) h^{1/2}.$$
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