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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&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}fleft(frac{k}{n}right)\&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}sqrt{frac{k}{n}}\&=limlimits_{ntoinfty}frac{1}{n^{3/2}}sum^{n-1}_{k=0}sqrt{k}end{align}
I'm stuck here, what should I do? Thanks!

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

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

Hence,

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

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 $epsilon$.

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