How do I find the upper and lower sum of the area of the graph by summation? Solve for the area $y=\sqrt{x}$ 



I have the following formulas for lower and upper sum.

$$LowerSum = \sum_{i=1} ^n f(m_i)\Delta x$$

$$Upper Sum = \sum_{i=1} ^n f(M_i)\Delta x$$

I know that the interval for $[a, b]$ is $[0, 1]$ and that there are $n$ rectangles. The change in x is $\Delta x = \frac{b-a}{n}$ so:

$$\Delta x = \frac{1}{n}$$

For the left endpoint, $f(m_i)$, the book has something like $f(m_i) = a + (i - 1)\frac{b-a}{n}$ so:

$$f(m_i) = \frac{i - 1}{n}$$

Now I have:

$$LowerSum = \sum_{i=1} ^n f\left(\frac{i - 1}{n}\right)\left(\frac{1}{n}\right)$$

Moving the change in x to outside and plug into the function:

$$\left(\frac{1}{n}\right)\sum_{i=1} ^n \sqrt{\frac{i - 1}{n}}$$

Now what do I do? The book doesn't have any example with a square root so in their example, they are left with a single $i$ or $i^2$ which there are summation formulas for those; what about a square root? Am I even doing it right?
Reman Sums can be applied in this case

For the function $f(x)=\sqrt x$:

$\lim\limits_{ n\to\infty} \sum\limits_{i=1}^{n} f(x_i)\Delta X$

let $x_i$ be the right end point, $x_i=\frac{i^2}{n^2}$

$\Delta x=\frac{1^2}{n^2}-\frac{(i-1)^2}{n^2}=\frac{2i-1}{n^2}$

Thus: $\lim\limits_{ n\to\infty} \sum\limits_{i=1}^n \sqrt{\frac{1^2}{n^2}}\frac{2i-1}{n^2}=\frac23$
Reman Sums can be applied in this case

For the function $f(x)=\sqrt x$:

$\lim\limits_{ n\to\infty} \sum\limits_{i=1}^{n} f(x_i)\Delta X$

let $x_i$ be the right end point, $x_i=\frac{i^2}{n^2}$

$\Delta x=\frac{1^2}{n^2}-\frac{(i-1)^2}{n^2}=\frac{2i-1}{n^2}$

Thus: $\lim\limits_{ n\to\infty} \sum\limits_{i=1}^n \sqrt{\frac{1^2}{n^2}}\frac{2i-1}{n^2}=\frac23$