So far, I have this for the n'th right Riemann sum: $\sum_{i=1}^{n} \left(a+\frac{(b-a) i}{n}\right)^3 \left(\frac{b-a}{n}\right)$
The first component is the height of the rectangle and the second is the width
Obviously I know this should eventually become $\frac{b^4}{4}-\frac{a^4}{4}$ since that's what the integration rules will tell us, but I'm not sure how the simplification can work.
What I'm doing right now is summations of $\sum_{i=1}^{n}1$, $i$, $i^2$ and $i^3$ and multiplying that by various $\frac{(b-a)i}{n}$ terms when taken to similar powers.
The strategy is that if I take the $\lim n \to \infty$ of this products, all those summation formulas would have $n$ taken to powers that would cancel when the $\frac{(b-a)i}{n}$ terms are taken to similar powers and I'd just be left with numbers.
What I have is $\lim_{n \to \infty} \left( \frac{b-a}{n}(a^3)(n)+ 3\left(\frac{b-a}{n}\right)^2\left(\frac{n(n+1)}{2}\right) + 3\left(\frac{b-a}{n}\right)^3\left(\frac{n(n+1)(2n+1)}{6}\right) + \left(\frac{b-a}{n}\right)^4 \left(\frac{n^2(n+1)^2}{4}\right) \right)$ and I'm hoping that this could simplify into the correct $\frac{b^4}{4}-\frac{a^4}{4}$
Still by using Riemann sums, is there a more elegant way of doing this? (Or am I making it too complicated)
First, we divide the area from a to b into n-equal subintervals. So, $\Delta x_i = \frac{(b-a)}{n}$ . From now on, let k = (b-a). So, as we know, the Riemann sum will be found by the summation $\sum_{i=1}^n f(x_i) \Delta x_i$, we need to find $x_i$. See $$x_0 = a$$ $$x_1 = a + \frac{k}{n}$$ $$x_2 = a + \frac{2k}{n}$$. So, one sees that the pattern illuminates $$x_i = a + \frac{ik}{n}$$ and that $$f(x_i) = (x_i)^3 = a^3 + \frac{3a^2ik}{n}\ + \frac{3ai^2k^2}{n^2}\ + \frac{i^3k^3}{n^3}.$$ Going back to our summation, we have $$\sum_{i=1}^n \frac{a^3k}{n}\ + \frac{3a^2ik^2}{n^2}\ + \frac{3ai^2k^3}{n^3}\ + \frac{i^3k^4}{n^4}.$$ This is equivalent to $$a^3k + \frac{3a^2k^2 (n+1)}{2n}\ + \frac{ak^3 (n+1)(2n+1)}{2n^2}\ + \frac{k^4(n+1)^2}{4n^2}$$ Now taking the limit as n approaches infinity, we have $$a^3k + \frac{3a^2k^2}{2}\ + ak^3 + \frac{k^4}{4}$$