Prove that $f(x)=\sqrt{x}$ is Riemann integrable over $[0,1]$

Question

Prove that $f(x)=\sqrt{x}$ is Riemann integrable over $[0,1]$

MY WORK

Consider the Riemann sum

\begin{align}\int^{1}_{0}f(x)dx&=\lim\limits_{n\to\infty}\frac{1}{n}\sum^{n-1}_{k=0}f\left(\frac{k}{n}\right)\\&=\lim\limits_{n\to\infty}\frac{1}{n}\sum^{n-1}_{k=0}\sqrt{\frac{k}{n}}\\&=\lim\limits_{n\to\infty}\frac{1}{n^{3/2}}\sum^{n-1}_{k=0}\sqrt{k}\end{align} I'm stuck here, what should I do? Thanks!

Answer 1

There are a number of possible approaches.

If you are trying to show that the integral exists with Riemann (Darboux) sums, then since $f(x) = \sqrt{x}$ is increasing, the upper and lower sums for a uniform partition $P = (0, 1/n, 2/n, \ldots, 1)$ are

$$U(P,f) = \frac1{n}\sum_{k=1}^n\sqrt{k/n}= \frac1{n^{3/2}}\sum_{k=1}^n\sqrt{k},\\L(P,f) = \frac1{n}\sum_{k=1}^n\sqrt{(k-1)/n}= \frac1{n^{3/2}}\sum_{k=1}^n\sqrt{k-1}$$

Hence,

$$U(P,f)-L(P,f) = \frac1{n^{3/2}}\sum_{k=1}^n\left(\sqrt{k}-\sqrt{k-1}\right)$$

Since the sum is telescoping, we have for $n > 1/\epsilon$,

$$U(P,f)-L(P,f) = \frac{\sqrt{n}}{n^{3/2}}= \frac1{n}< \epsilon$$

Therefore, $f$ is integrable by the Riemann criterion -- since for any $\epsilon >0$ there exists a partition for which the difference between upper and lower sums is less than $\epsilon$.

Answer 2

Recall that a continuous function $f:[a,b] \to \mathbb{R}$ is integrable on the interval $[a,b]$.