Calculate the integral by Riemann $\int_{0}^{1}\sqrt{x}\,dx$ Calculate the integral by Riemann $\displaystyle \int _{0}^{1}\sqrt{x} \, dx$
$$\displaystyle a_{k} =0+k\cdot \frac{1}{n} =\frac{k}{n}$$
$$\displaystyle \Delta x=\frac{1-0}{n} =\frac{1}{n}$$
\begin{align*}
\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\sqrt{\frac{k}{n}}\right) \cdotp \frac{1}{n} & = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{\sqrt{k}}{\sqrt{n}}\right) \cdotp \frac{1}{n} \\[1ex]
& = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{1}{n\sqrt{n}} \cdotp \sqrt{k}\right) \\[1ex]
& = \lim_{n\rightarrow \infty }\frac{1}{n \sqrt{n}} \cdotp \sum_{k=1}^{n}\left(\sqrt{k}\right)
\end{align*}
I am stuck on the sum of $\sqrt k$ maybe have another way to solve this question?
 A: As suggested in the linked question and other answers posted here, you need not use equally-spaced subintervals in the partition of $[0,1]$. Instead, consider the sequence of intervals
$$\left\{\left[\left(\frac{i-1}n\right)^2, \left(\frac in\right)^2\right] \,\bigg| \, 1 \le i \le n \right\}_{n\in\Bbb N}$$
each with length $\frac{i^2}{n^2} - \frac{(i-1)^2}{n^2} = \frac{2i - 1}{n^2}$.

Then the (right-endpoint) Riemann sum is
$$\sum_{i=1}^n \sqrt{\left(\frac in\right)^2} \frac{2i-1}{n^2} = \sum_{i=1}^n \frac{2i^2-i}{n^3} = \frac{(n+1)(4n-1)}{6n^2}$$
As $n\to\infty$, the sum converges to the definite integral, and we have
$$\int_0^1 \sqrt x \, dx = \lim_{n\to\infty} \frac{(n+1)(4n-1)}{6n^2} = \frac23$$
which agrees with the known antiderivative result,
$$\int_0^1 x^{\frac12} \, dx = \frac23 x^{\frac32}\bigg|_0^1 = \frac23$$
A: Maybe this will help you (i hope this is true).
First let's make a change of variable: $t=\sqrt{x}$ so we get $\int_{0}^{1}\sqrt{x}dx=\int_{0}^{1}2t^2dt$
Now lets compute: $\int_{0}^{1}2t^2dt$ using Riemman integrable.
Thus by doing exactly as you did we get: $lim_{n \to \infty } \frac{2}{n^3} \sum_{k=1}^{n}k^2$
And we know that the sum of squares of n natural numbers can be calculated using the formula $[n(n+1)(2n+1)] / 6$. So: $lim_{n \to \infty } \frac{2}{n^3} [n(n+1)(2n+1)] / 6 = 2/3$
This is correct because: $\int_{0}^{1}\sqrt{x}dx = 2/3$
So we did exactly what it was asked:

Calculate the integral by Riemann $\int_{0}^{1}\sqrt{x}dx$

But by calculating integral by Riemman on a simplest function.
A: Here is a direct way to look at this if you would like to stick with a Riemann sum that uses equally spaced subintervals. The strategy is, in small doses, to replace the irrational $\sqrt{k}$ values with nearby integers.
$$\begin{align}
\lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k}
&=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k}
\end{align}$$
Already this might give you pause. But for one thing, as $n\to\infty$, so does $n^2$. The sum on the right is like the sum on the left but between one value of $n$ and the next, you add
$$\frac{1}{n^2\sqrt{n^2}}\left(\sqrt{(n-1)^2+1}+\sqrt{(n-1)^2+2}+\cdots+\sqrt{(n-1)^2+(2n-1)}\right)$$
That is, you add $2n-1$ terms that are each less than or equal to $n$, and the result is divided by $n^3$. So this tail that you add with each step, all by itself, converges to $0$. So the two sums have the same limit. Starting again:
$$\begin{align}
\lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k}
&=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k}\\
&=\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n^2}\right)\\
&\leq\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{9}+\sqrt{9}+\cdots+\sqrt{n^2}\right)\\
\end{align}$$
Here we have one $\sqrt{1}$s, then three $\sqrt{4}$s, then five $\sqrt{9}$s, and so on.
$$\begin{align}
&=\lim_{n\to\infty}\frac{1}{n^3}\left(1+3(2)+5(3)+\cdots+(2n-1)n\right)\\
&=\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n(2k-1)k\\
&=\lim_{n\to\infty}\frac{1}{n^3}\left(2\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}2\right)\\
&=\lim_{n\to\infty}\left(\frac13(1+1/n)(2+1/n)-\frac{(1+1/n)}{2n}\right)\\
&=\frac23
\end{align}$$
So we have an upper bound and the limit you are after is at most $\frac23$. (Of course we know independently that the limit actually is $\frac23$.)
Do you see how to modify this to get a lower bound? Hint: $1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,\ldots$.
A: In the spirit of Reimann sums, you do not need to make the sequence $a_k$ arithmetic, you can use any sequence that covers the interval.
In this case, since you need to take the square root, it will make things simpler if you make $a_k$ a quadratic sequence...
$$a_k = 0 + \frac {k^2 }{n^2}$$
Then $\sqrt{a_k}$ simplifies to $\frac kn$ , but your $\Delta x$ will now depend of $k$...
$$ \Delta x_k = a_{k+1}-a_k = \frac {2k+1}{n^2}$$
So
$$\int_0^1 \sqrt x dx = \lim_{n \to \infty} \sum_{k=1}^n a_k \Delta x_k $$
$$= \lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k(2k+1)$$
$$= 2\lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k^2 + \lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k$$
Now you can use the formulas...
$$ \sum_{k=1}^n k^2 = \frac n6 (2n+1)(n+1) \text{ and } \sum_{k=1}^n k=\frac n2(n+1)   $$
to obtain the required result
A: If the answer upper isn't satisfying here an oher.
Choose a partition as follow: $x_0=0 , x_1 = 1/n^2 ,x_2=(2^2)/n^2 ,...,x_n=n^2/n^2=1$ so $x_k-x_{k-1}=\Delta x_k=\frac{2k-1}{n^2}$.
$\sum_{k=1}^{n}\frac{2k-1}{n^2}\frac{k}{n}=\sum_{k=1}^{n}\frac{2k^2-k}{n^3}$
And we know a formula for each of those sums.
A: I thought it might be instructive to present a way forward that uses creative telescoping and simple estimates.  To that end we proceed.

Note that we can write
$$\begin{align}
n^{3/2}-1&=\sum_{k=1}^n \left((k+1)^{3/2}-k^{3/2}\right)\\\\
&=\frac32 \sum_{k=1}^n \left(\sqrt{k} +O(k^{-1/2})\right)\tag1
\end{align}$$
Rearranging $(1)$ reveals
$$\frac1{n^{3/2}}\sum_{k=1}^n \sqrt k=\frac23 \left(1-n^{-3/2}\right)+\frac1{n^{3/2}}\underbrace{\sum_{k=1}^n O(k^{-1/2})}_{\le O(n)}$$
Letting $n\to \infty$ yields the coveted result
$$\lim_{n\to\infty}\frac1{n^{3/2}}\sum_{k=1}^n \sqrt k=\frac23$$
