How to compute integrals using Riemann sums

Question

Compute the integral using Riemann sums $$\int_0^s x^2 \,dx$$ Find the sum $ U_n $ of all rectangles below the function $y = x^3$
Find the sum $O_n$ of all rectangles above the function $y = x^3$
Take the limits to show that $$\lim \limits_{x \to ∞}U_n = \lim \limits_{x \to ∞}O_n = \frac{s^3}3$$

I am vaguely familiar with Riemann sums, however, given this problem I really am not sure how to approach it.
I know that $\Delta$x = $\frac{s}n$ and I know that $x_i$ = $\frac{si}n$.
The next thing I believe I am supposed to do is compute $f(x_i) = (\frac{si}n)^2$ and then use that to compute $$\lim \limits_{x \to ∞} \sum_{i=0}^n [\frac{s^2i^2}{n^2}]·\frac{s}n $$ At this point it feels like things start getting convoluted and I am not sure how I would use this to find the sum of the rectangles above and below the function $y=x^3$.

Guidance to solving this question would be much appreciated.

Answer 1

Factor out what can be factored: $$\sum_{i=1}^n \frac{s^2i^2}{n^2}·\frac{s}n=\frac{s^3}{n^3}\sum_{i=1}^{n}i^2=\frac{s^3}{n^3}\frac{n(n+1)(2n+1)}6$$ for the upper Riemann sum, and almost the same for the lower one ($i=0$ to $n-1$).

Answer 2

Hint :

$$\lim_{n \to \infty} \frac{\sum_{i=0}^{n} i^2}{n^3} = \lim_{n \to \infty}\frac{n(n+1)(2n+1)}{6n^3} = \frac{1}{3} $$

Answer 3

If we have function where $f'>0$ in the domain such as this, if we take the area of the rectangle in a given area $x\in[a,b]$ then it is clear that $(b-a)f(b)>A$ and $(b-a)f(a)<A$ so we end up with a limit we can squeeze: $$(b-a)f(a)<A<(b-a)f(b)$$ now to cover the whole domain we are interested in just take the sum