Use Darboux sums to calculate the area of $\sqrt x$ in $[0,1]$ I know that $$\Delta x=\frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n}$$
So $$\overline{D}=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^n\sup f(x_i)\Delta x=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^nf(x_i)\Delta x=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^n \frac{1}{n}\sqrt {\frac{i}{n}}=\displaystyle\lim_{n\to\infty}\frac{1}{n\sqrt n}\displaystyle\sum_{i=1}^n \sqrt {i} $$
I don't know how to calculate the last part $\displaystyle\sum_{i=1}^n \sqrt {i}$

It is possible to compute the area (or integral) directly as the limit of the upper Darboux sum.
As you showed
$$\overline{D} = \lim_{n\to\infty}\frac{1}{n\sqrt n} \sum_{i=1}^n \sqrt {i}.$$
We can use the binomial expansion $(1-x)^{\alpha} =1 - \alpha x + O(x^2).$
For $i > 1,$ we have
$$\begin{align}(i-1)^{3/2} &=  i^{3/2}(1 - i^{-1})^{3/2} \\ &= i^{3/2}\left[ 1  - \frac{3}{2}i^{-1} + O(i^{-2})\right] \\ &= i^{3/2} - \frac{3}{2}\sqrt{i} + O(1/\sqrt{i}).\end{align}$$
Hence,
$$\sqrt{i} = \frac{2}{3}\left[i^{3/2} - (i-1)^{3/2}\right] + O(1/\sqrt{i}),$$
and,
$$\begin{align}\frac1{n\sqrt{n}}\sum_{i=1}^n\sqrt{i} &=\frac{1}{n \sqrt{n}}\left(1 + \frac{2}{3}\sum_{i=2}^n[i^{3/2}-(i-1)^{3/2} + O(1/\sqrt{i})]\right) \\ &= \frac{1}{n \sqrt{n}}\left(1 + \frac{2}{3}n^{3/2}- \frac{2}{3} + O(n)\right) \\ &= \frac{2}{3} +\frac{1}{3n \sqrt{n}} + O(1/\sqrt{n}). \end{align}$$
Thus,
$$\lim_{n \to \infty}\frac1{n \sqrt{n}}\sum_{i=1}^n\sqrt{i} = \frac{2}{3}.$$
