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.
Best Answer
Since no one has answered yet:
Since $\sqrt{x}$ is integrable on $[1,a]$, we know that the Riemann sums corresponding to a sequence of partitions $P_n$ will converge to $\int_{1}^{a} \sqrt{x} \,dx$ if the maximum width of the partitions converges to zero as $n \to \infty$.
Therefore, we are free to take a sequence of partitions $P_n$ such that the corresponding Riemann sums are easy to calculate. To find such a partition is a matter of looking at the integrand and simply trying; I use the partitions $$ P_n = \{ a^{k/n} \,\,| \,\, k=0,1,\ldots, n \}, $$ as suggested by Ryszard Szwarc in the comments. Then if we evaluate $\sqrt{x}$ in the starting point of each interval, the $n$'th Riemann sum is $$ \sum_{k=0}^{n-1} a^{k/2n} \left( a^{(k+1)/n}-a^{k/n}\right) \\= (a^{1/n}-1) \sum_{k=0}^{n-1} a^{\frac{3k}{2n}} \\=(a^{1/n}-1) \frac{1-a^{3/2}}{1-a^{3/{2n}}} \\=\frac{a^{3/2+1/n}-a^{3/2}-a^{1/n}+1}{a^{3/{2n}}-1}. $$ Calculating the limit $n\to \infty$, say with l'Hopital's rule, gives the result.