[Math] Riemann Sum of $\sqrt{x}$ with uniform partition

Question

How to find the Riemann sums ( Lower and Upper) of $\sqrt{x}$ using only the uniform partition in $[0,1]$??

It's really hard to find some expression to calculate $\displaystyle\sum_{i=1}^{n}\sqrt{i}$

Answer 1

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

$$\int_0^1\sqrt{x} \, dx = \lim_{n \to \infty}\frac1{n^{3/2}}\sum_{k=1}^n\sqrt{k} = \frac{2}{3}.$$