# 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.

• You need to start by writing the sums you mention. After that, you’ll look at how to compute those. Apr 30 at 8:16
• There isn't a unique choice of Riemann sum, but let's take the case where each strip of area is of equal width. Can you see how to evaluate $\lim_{n\to\infty}\frac1n(a-1)\sum_{k=0}^{n-1}\sqrt{1+\frac{k}{n}(a-1)}$ (without, of course, rewriting it as an integral)?
– J.G.
Apr 30 at 8:34
• @J.G.: sorry, I can't - I only started learning how to convert definite integrals to Riemann sums recently, so I'm only a beginner on that Apr 30 at 8:37
• I suggest applying the partition $a^{k/n},$ where $k=0,1,\ldots, n.$ Apr 30 at 9:11
• – J.G.
Apr 30 at 9:48

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.

• $a^{1/n}-1=(a^{1/(2n)}-1)(a^{1/(2n)}+1)$ and similarly decompose the denominator. Then Hospital is not needed. May 1 at 12:11

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*}