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?

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: $1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,\ldots$.
