Find integral of $\sqrt{x}$ using Riemann sum definition Let $a > 1$ be a real number. Evaluate the definite integral
\begin{equation}
\int_{1}^{a} \sqrt{x} \,dx
\end{equation}
from the Riemann sum definition.
My approach I know a Riemann sum consists of a sigma notation with a width and function. However, I am confused and not sure where to start. Any hints/answers are appreciated. Thanks.

Using the hints and tips provided, I successfully proved the integral from the Riemann sum definition:
The function $f(x)=\sqrt{x}$ is continuous on $[1,a]$, hence integrable on $[1,a]$. For every positive integer $n$, we consider the left Riemann sum of $f$ with respect to the partition $[1,a^{1/n},a^{2/n},a^{1/n}...a^{n/n}]$ of $[1,a]$ into $n$ subintervals. Then,
\begin{align*} 
\int_{1}^{a} \sqrt{x} \,dx  &=  \lim_{n\to\infty} \sum_{k=0}^{n-1} \sqrt{a^{k/n}}(a^{\frac{k+1}{n}}-a^{k/n}) \\ 
&= \lim_{n\to\infty} (a^{\frac{1}{n}}-1) \sum_{k=0}^{n-1} (a^{\frac{3}{2n}})^k \\
&= \lim_{n\to\infty} (a^{\frac{1}{n}}-1) \frac{(a^{\frac{3}{2n}})^n-1}{a^{\frac{3}{2n}}-1} \\
&= (a^{\frac{3}{2}}-1) \lim_{n\to\infty} \frac{a^{\frac{1}{n}}-1}{a^{\frac{3}{2n}}-1} \\
&= (a^{\frac{3}{2}}-1) \lim_{n\to\infty} \frac{(a^{\frac{1}{2n}}-1)(a^{\frac{1}{2n}}+1)}{(a^{\frac{1}{2n}}-1)(a^{\frac{1}{n}}+a^{\frac{1}{2n}}+1)} \\
&= (a^{\frac{3}{2}}-1) \lim_{n\to\infty} \frac{a^{\frac{1}{2n}}+1}{a^{\frac{1}{n}}+a^{\frac{1}{2n}}+1} \\
&= (a^{\frac{3}{2}}-1) \frac{a^0+1}{a^0+a^0+1} \\
&= \frac{2}{3}(a^{\frac{3}{2}}-1)
\end{align*}
