Is it possible to prove that $\int_{a}^{b}x^{n}=\frac{b^{n+1}-a^{n+1}}{n+1}$ from the definition of the Riemann integral? In other words, is it possible to prove that $\int_{a}^{b}x^{n}=\frac{b^{n+1}-a^{n+1}}{n+1}$ without use of limits and derivatives? I have proved the special case for $n=0$.  
This is all I have so far, though, for the general case:

Since $x^{n}$ is continuous it suffices to show that
  $\underset{P}{\sup}L(f,P)=\frac{b^{n+1}-a^{n+1}}{n+1}$. If $n$ is odd
  then for a partition $P=\left \{ a=t_{0}<t_{1}<\cdots<t_{k-1}<t_k=b \right \}$ we have that $m_{i}=\inf\left \{t^{n}:t_{i-1}\leq t \leq t_{i}\right \}=t_{i-1}^{n}$ and
  $L(f,P)=\sum_{i=1}^{k}m_i(t_i-t_{i-1}).$

How can one continue from there?  
 A: 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.
A: So you want to use the definition of the Riemann integral via upper and lower Riemann sums to avoid talking about limits, right? This seems a little artificial, but here's one way. Assuming $b > a > 0$ and that $m$ is a positive integer, set $q = (\frac{b}{a})^{\frac{1}{m}}$. Let $P_m$ be the partition $a, aq, aq^2, ..., aq^m = b$. Then 
$ L(x^n,P_m) = \sum_{i=0}^{m-1} (aq^i)^n (aq^{i+1} - aq^i) = a^{n+1}(q-1) \sum_{i=0}^{m-1} (q^{n+1})^i = a^{n+1} (q - 1) \frac{q^{m(n+1)} - 1}{q^{n+1} - 1} $
Remembering that $q = (\frac{b}{a})^{\frac{1}{m}}$, we get
$ L(x^n,P_m)  = (b^{n+1} - a^{n+1}) \frac{q -1}{q^{n+1} - 1} = (b^{n+1} - a^{n+1}) \frac{1}{1 + \dotso + q^n }$
Now examine the second factor of the last expression. You can rephrase the following without explicitly talking about limits: because $q \rightarrow 1$ as $m \rightarrow \infty$, we may choose an $m$ for which $L(x^n,P_m)$ is arbitrarily close to $(b^{n+1} - a^{n+1}) \frac{1}{n+1}$, concluding that the sup of the lower Riemann sums is at least $(b^{n+1} - a^{n+1}) \frac{1}{n+1}$.
A similar argument should give you that the inf of the upper Riemann sums is at most  $(b^{n+1} - a^{n+1}) \frac{1}{n+1}$.
A: You might want to check the following proof I posted on a question. The method for solving the integral is pretty nice. Interpreting an integral under the Riemann Stieltjes form.
