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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.

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  • $\begingroup$ You need to start by writing the sums you mention. After that, you’ll look at how to compute those. $\endgroup$
    – mathcounterexamples.net
    Apr 30 at 8:16
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    $\begingroup$ 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)? $\endgroup$
    – J.G.
    Apr 30 at 8:34
  • $\begingroup$ @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 $\endgroup$
    – ordinary
    Apr 30 at 8:37
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    $\begingroup$ I suggest applying the partition $a^{k/n},$ where $k=0,1,\ldots, n.$ $\endgroup$
    – Ryszard Szwarc
    Apr 30 at 9:11
  • $\begingroup$ Relevant. $\endgroup$
    – J.G.
    Apr 30 at 9:48

2 Answers 2

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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.

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

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