Without the loss of generality, let just deal with integrals with a lower bound of $a = 0$ (noting that in general we have that $\int_a^b = \int_0^b - \int_0^a$). So, we must show that $\int_0^b x^n = \frac{b^{n+1}}{n+1}$
Let $\mathcal{P}_m = \{t_0,t_1,\dots,t_m\} = \{0, b\cdot\frac{1}{m}, b\cdot\frac{2}{m}, \dots, b\cdot\frac{m-1}{m},b\}$ (when computing Riemann integrals from scratch this should be your go-to partition). Observe that $t_{i} - t_{i-1} = b\cdot\frac{1}{m}$ and that $M_{i} = \big(b \cdot \frac{i}{m}\big)^n = (\frac{b}{m})^n \cdot i^n$ (let's deal with upper sums). Accordingly,
$U(\mathcal{P}_m, x^n) = \displaystyle\sum_{i=1}^m M_i(t_i - t_{i-1}) = \displaystyle\sum_{i=1}^m (\frac{b}{m})^n \cdot i^n \frac{b}{m} = (\frac{b}{m})^{n+1}\displaystyle\sum_{i=1}^m i^n$
Let us pause. We need deal with the sum $\displaystyle\sum_{i=1}^m i^n$ somehow. I will refer you to here. Now,
$ U(\mathcal{P}_m, x^n) = (\frac{b}{m})^{n+1}\displaystyle\sum_{i=1}^m i^n = (\frac{b}{m})^{n+1}\bigg(\frac{1}{n+1}\displaystyle\sum_{k=0}^n (-1)^k \binom{n+1}{k}B_k m^{n+1-k} \bigg) $
which simplifies to
$ \frac{b^{n+1}}{n+1} \bigg(B_0 + \text{terms involving negative powers of m} \bigg)$
Noting that $B_0 = 1$, we have the limit $\displaystyle\lim_{m \to \infty} U(\mathcal{P}_m, x^n) = \frac{b^{n+1}}{n+1}$. I hope that you would agree that we are done now.