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 $\displaystyle\sum_{i=1}^n \sqrt {i}$
 A: 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,
$$\lim_{n \to \infty}\frac1{n \sqrt{n}}\sum_{i=1}^n\sqrt{i} = \frac{2}{3}.$$
A: Let $f$ be bounded on closed interval $[a, b]$. Let $P = \{x_{0} = a, x_{1}, x_{2}, \ldots, x_{n} = b\}$ be a partition of $[a, b]$ then the upper Darboux sum for $f$ over $P$ is given by $$U(P, f) = \sum_{i = 1}^{n}M_{i}(x_{i} - x_{i - 1})\tag{1}$$ and the lower Darboux sum of $f$ over $P$ is given by $$L(P, f) = \sum_{i = 1}^{n}m_{i}(x_{i} - x_{i - 1})\tag{2}$$ where $$M_{i} = \sup\, \{f(x) \mid x \in [x_{i - 1}, x_{i}]\},\, m_{i} = \inf\,\{f(x)\mid x \in [x_{i - 1}, x_{i}]\}\tag{3}$$ Now it is much easier to calculate the area under $f(x) = x^{p}$ for general $p > 0$ on interval $[a, b]$ via limit of Darboux sums. And then we can put $p = 1/2, a = 0, b = 1$ to get the answer to the current question.
For this question it is better to assume that $0 < a < b$ let the partition $P$ of $[a, b]$ be given by $x_{i} = ar^{i}$ when $b = x_{n} = ar^{n}$ and as $n \to \infty$ we have $r \to 1$. Thus the points of partition are in geometric progression (instead of the usual points $x_{i} = a + ih, b = x_{n} = a + nh$ in arithmetic progression). Then since $f(x) = x^{p}$ is increasing we have $m_{i} = x_{i - 1}^{p}, M_{i} = x_{i}^{p}$ and therefore
\begin{align}
U(P, f) &= \sum_{i = 1}^{n}x_{i}^{p}(x_{i} - x_{i - 1})\notag\\
&= \sum_{i = 1}^{n}a^{p}r^{ip}(ar^{i} - ar^{i - 1})\notag\\
&= a^{p + 1}(r - 1)\sum_{i = 1}^{n}r^{ip + i - 1}\notag\\
&= a^{p + 1}(r - 1)r^{p}\frac{r^{(p + 1)n} - 1}{r^{p + 1} - 1}\notag\\
&= a^{p + 1}r^{p}(r^{(p + 1)n} - 1)\frac{r - 1}{r^{p + 1} - 1}\notag\\
&= a^{p + 1}r^{p}((b/a)^{p + 1} - 1)\frac{r - 1}{r^{p + 1} - 1}\notag\\
\end{align}
This tends to $$a^{p + 1}((b/a)^{p + 1} - 1)\frac{1}{p + 1} = \frac{b^{p + 1} - a^{p + 1}}{p + 1}$$ as $r \to 1$. It can be proved in similar manner that $L(P, f)$ also tends to same limit. Hence we have $$\int_{a}^{b}x^{p}\,dx = \frac{b^{p + 1} - a^{p + 1}}{p + 1}\tag{4}$$ Note that in the above we have assumed $p > 0, b > a > 0$. The same result holds if $-1 < p \leq 0$ (only values of $m_{i}$ and $M_{i}$ are changed). Also it is possible to use the result in this question to show that the formula $(4)$ is valid for $0 \leq a < b$ also.
Now putting $p = 1/2, a = 0, b = 1$ in formula $(4)$ we get the desired area under graph of $f(x) = \sqrt{x}$ between $x = 0$ and $x = 1$ as $$\int_{0}^{1}\sqrt{x}\,dx = \frac{1^{3/2} - 0^{3/2}}{3/2} = \frac{2}{3}$$
