<s>[INST] Prove $\int \limits_0^b x^3 = \frac{b^4}{4} $ by considering partitions $[0, b]$ in $n$ equal subinvtervals. I was given this question as an exercise in real analysis class. Here is what I came up with. Any help is appreciated!

Prove $\int \limits_0^b x^3$ = $\frac{b^4}{4} $ by considering partitions [0, b] in $n$ equal subinvtervals.

Consider $f$ on the interval $[0, b]$ where $b > 0$. For a partition $ P = \{0=t_0 < t_1 < ...< t_n = b\}$ we have:
$U(f, P) = $$\sum_{k=1}^n  t^3_k (t_k - t_1) $
If we choose $t_k = \frac{kb}{n} $ then we can say 
$U(f, P) = $$\sum_{k=1}^n  \frac{k^3b^3}{n^3}\ . \frac{b}{n}$
=>  $U(f, P) = \frac{b^4}{n^4} \sum_{k=1}^n  k^3$
=>  $U(f, P) = \frac{b^4}{n^4}\ . \frac{n^2(n + 1)}4$
so $L(f) \ge \frac{b^4}{n^4}$ . Therefore $f(x) = x^3$ is integrable on $[0, b]$ and $\int \limits_0^b x^3$ = $\frac{b^4}{4} $.
 [/INST] You can consider the Riemann sums. For instance, the left Riemann sum

$$ \sum_{i=0}^{n-1} f(x_i) \Delta x_i = \sum_{i=0}^{n-1} \left(\frac{b}{n}i\right)^3\frac{b}{n} = \frac{b^4}{n^4}\sum_{i=0}^{n-1} i^3 = \frac{b^4}{n^4}\left({n}^{4}/4 - {n}^{3}/2+ {n}^{2}/4\right)\longrightarrow_{n\to \infty} \frac{b^4}{4} $$

To find the last sum, see here. 
</s>