It does not matter what the function is. The general formula are
$$
I \approx \frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{1}{k}\right)$$
using the left side of rectangles.
$$
I \approx \frac{1}{n}\sum_{k=1}^{n} f\left(\frac{1}{k}\right)$$
using the right side of rectangles.
You can average the two to get Trapezoidal integration.
For your problem
$$
f\left(\frac{1}{k}\right) = e^{\frac{-1}{k^2}}
$$
Added in response to OP's question
Here are calculations for $N=4$
$$
\frac{ e^{-(1/4)^2}+ e^{-(2/4)^2}+ e^{-(3/4)^2}+ e^{-(4/4)^2}}{4}
\\
\frac{ 0.9394+ 0.7788+ 0.5698+ 0.3679}{4}
= 0.664$$
For $N=6$
$$\frac{ e^{-(1/6)^2}+ e^{-(2/6)^2}+ e^{-(3/6)^2}+ e^{-(4/6)^2}+ e^{-(5/6)^2}+ e^
{-(6/6)^2}}{6}\\
\frac{0.972604+ 0.894839+ 0.778801+ 0.64118+ 0.499352+ 0.367879}{6}
= 0.692443
$$
For $N=8$:
$$
\frac{e^{-(1/8)^2}+ e^{-(2/8)^2}+ e^{-(3/8)^2}+ e^{-(4/8)^2}+ e^{-(5/8)^2}+ e^
{-(6/8)^2}+ e^{-(7/8)^2}+ e^{-(8/8)^2}}{8}
\\
=\frac{0.9845+ 0.9394+ 0.8688+ 0.7788+ 0.6766+ 0.5698+ 0.465+ 0.3679}{8}
= 0.7064
$$
If you look carefully, half the numbers for $N=8$ are already calculated for $N=4$. So if you double $N$, you can save half the calculations.
If you know Richardson Extrapolation, you can estimate the limit $N\rightarrow \infty$ easily.
If you are just trying to show that the integral exists, then since $f(x) = \sqrt{x}$ is increasing, the upper and lower sums for a uniform partition $P = (0, 1/n, 2/n, \ldots, 1)$ are
$$U(P,f) = \frac1{n}\sum_{k=1}^n\sqrt{k/n}= \frac1{n^{3/2}}\sum_{k=1}^n\sqrt{k},\\L(P,f) = \frac1{n}\sum_{k=1}^n\sqrt{(k-1)/n}= \frac1{n^{3/2}}\sum_{k=1}^n\sqrt{k-1.}$$
Hence,
$$U(P,f)-L(P,f) = \frac1{n^{3/2}}\sum_{k=1}^n[\sqrt{k}-\sqrt{k-1}|.$$
Since the sum is telescoping, we have for $n > 1/\epsilon$,
$$U(P,f)-L(P,f) = \frac{\sqrt{n}}{n^{3/2}}= \frac1{n}< \epsilon.$$
Therefore, $f$ is integrable by the Riemann criterion -- since for any $\epsilon >0$ there exists a partition for which the difference between upper and lower sums is less than $\epsilon$.
If you want to calculate the integral as a limit, then using the upper sum we have
$$\int_0^1\sqrt{x} \, dx = \lim_{n \to \infty}\frac1{n^{3/2}}\sum_{k=1}^n\sqrt{k}.$$
Using the binomial expansion $(k > 0),$
$$(k-1)^{3/2} = k^{3/2} - \frac{3}{2}\sqrt{k} + O(1/\sqrt{k}),$$
and
$$\sqrt{k} = \frac{2}{3}\left[k^{3/2} - (k-1)^{3/2}\right] + O(1/\sqrt{k}).$$
Hence,
$$\frac1{n^{3/2}}\sum_{k=1}^n\sqrt{k}=\frac{2}{3n^{3/2}}\sum_{k=1}^n[k^{3/2}-(k-1)^{3/2} + O(1/\sqrt{k})] \\= \frac{2}{3}+ O(1/n),$$
and
$$\int_0^1\sqrt{x} \, dx = \lim_{n \to \infty}\frac1{n^{3/2}}\sum_{k=1}^n\sqrt{k} = \frac{2}{3}.$$
Best Answer
If the function is integrable then it really does not matter which points you pick as long as the $\delta x $ goes to zero as $n$ goes to infinity.
Not every function is integrable so one way to prove that a given function is not integrable is to show that there are two Riemann sums which tend to different limits.
That is why we learn different methods.