<s>[INST] I need to compute the integral $\int _0^s\:x^2$ using Riemann sums. I need to compute the integral $\int _0^s\:x^2$  using Riemann sums.
a) Consider the function $y=x^2$  on the interval [0, s] and divide it into n equal intervals. Each of these subintervals has length $\frac{s}{n}$.
b) Find the sum $U_n$ of all rectangles below the function $y=x^2$
$$U_n=\frac{s}{n}\left(0^3+\left(\frac{s}{n}\right)^3+\left(2\frac{s}{n}\right)^3+...+\left(n-1\right)\left(\frac{s}{n}\right)^3\right)= \frac{s^3}{n^3}\sum _{k=0}^{n-1}\:k^3$$
c) Now find the sum $O_n$ of all rectangles above the function $y=x^2$
$$O_n=\frac{s^3}{n^3}\left(1^3+2^3+3^3+...+n^3\right)=\frac{s^3}{n^3}\sum _{t=1}^nk^3$$
d) Take the limits to show that:
$$\lim _{n\to \infty }\left(U_n\right)=\lim \:_{n\to \:\infty \:}\left(O_n\right)$$
So My answer is:
$$U_n=\frac{s^3}{n^3}\sum _{k=0}^{n-1}\:k^3=\frac{s^3}{n^3}((\frac{n\left(n+1\right)}{6})-n^3$$
Therefore $U_n=\frac{s^3}{n^3}=\lim _{n\to \infty }\left(\frac{s^3}{3}+\frac{s^3}{2n}-\frac{s^3}{n}\right)=\frac{s^3}{3}$
Is it correct? I'd appreciate some guide and feedback if I am wrong.
 [/INST] First a comment. If you assume that $x \mapsto x^2$ is Riemann integrable, you don't need to compute both upper and lower Riemann sums and verify that those are equal. This would be required to prove the Riemann integrability.
Now regarding the computation itself. You have
$$U_n=\frac{s}{n}\left(0^2+\left(\frac{s}{n}\right)^2+\left(2\frac{s}{n}\right)^2+...+\left(n-1\right)\left(\frac{s}{n}\right)^2\right)= \frac{s^3}{n^3}\sum_{k=0}^{n-1} k^2.$$
The power is $2$ and not $3$ as you're integrating $x \mapsto x^2$.
Using
$$\sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}$$ and taking the limit you get the right answer $$\int_0^s x^2 \ dx = \frac{s^3}{3}.$$
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