I was working on a question from Chapter 13 of Spivak's Calculus (Pg 266). It asked us to calculate $\int_{0}^{b}x^{3}$ using the Darboux sums. I was able to do this successfully, but then I got to thinking myself and out of curiosity asked how would I go about calculating $\int_{0}^{b}x^{3}$ using the Riemann Sum ?, since both forms are to produce an equivalent answer.
So in using the idea for the upper and lower Darboux sums I arrive at a situation where after taking partitons $P_{n} = \{t_{0}, \dots, t_{n}\}$ of $n$ equal subintervals, thus implying that $|t_{i} -t_{i-1}| = \frac{b}{n}$ we have
$$L(f,P) = \sum_{i = 1}^{n}\bigg(\frac{(i-1)b}{n}\bigg)^{3}\frac{b}{n} \\ U(f,P) = \sum_{i = 1}^{n}\bigg(\frac{ib}{n}\bigg)^{3}\frac{b}{n}$$
After some algebra (specifically using the formula for $\sum_{i=1}^{n}i^{3}$ which is:
$$\frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4}$$ and concluding that $x^{3}$ is integrable, I arrive at $\frac{b^{4}}{4}$ as the solution.
So now form this I was thinking I could perhaps adapt this idea to using Riemann sums, so I would have a scenario I envision whereby:
$$\sum_{i = 1}^{n}f(x_{i})\frac{b}{n} \\ = \sum_{i = 1}^{n}x_{i}^{3}\frac{b}{n} $$
With the Riemann integral, even with using equal sized subintervals, I still have to choose the $x_{i}$ I want to use in each interval $(t_{i} - t_{i-1})$ to create in this case my tagged partition. So I don't have the luxury of just using $m_{i}$ and $M_{i}$ which are my minimum and maximum on each subinterval respectively. Using those values allowed for the use of the technique above, but I'm struggling to see how it can be translated to this situation.