Demonstrations - Calculating the area under a curve - integration - functions of one variable Please, I wish to help with this issue. I tried with the principle of mathematical induction but could not.
Consider the following:
Let the function $$f:[a,b]\subset\mathbb{R}\to \mathbb{R},$$
The partition of the interval $ [a, b] $:
$$a={x_0} < {x_1} < {x_2} < ... < {x_{n - 1}} < {x_n} = b,$$
subinterval length:
$$\Delta x = \frac{{b - a}}{n},$$
the sum below:
$$A({R_n}) = \sum\limits_{i = 0}^{n - 1} {f({x_i})\Delta x} $$
and the upper sum
$$A({S_n}) = \sum\limits_{i = 1}^n {f({x_i})\Delta x}.$$
Well, now I want to resolve this problem:
Let $A_a^b(x^m)$ be the area under the curve $f(x) = x^m$ over the closed interval $[a, b]$. 
Prove that $$A_a^b(x^m)=\frac{b^{m+1}}{m+1}-\frac{a^{m+1}}{m+1}$$ whenever $m \geqslant 0$.

The wording of the problem seems to me that he has been given the partition to work with, he can not use another one. To OP: Below I give a lemma that is crucial for solving the problem with the particular choice of partition of the interval given in the question. I leave setting up the Riemann sums, learning Big-Oh notation and filling in any details for steps you do not see immediately to you.
Proof by Induction that $ \sum_{k=1}^n k^{p-1} = \frac{n^p}{p} + O\left( n^{p-1} \right) $ where $ p \in \mathbb{N} $:
The base case $p=1$ is easily verified to be true. We make an induction hypothesis that there exists $ m \in \mathbb{N} $ such that the above equality holds for $ p = 1,2,3, \cdots m $. From the binomial theorem we have $$ (k+1)^{m+1} - k^{m+1} = \sum_{t=1}^{m+1} \binom{m+1}{t} k^{m+1-t}. $$ Letting $ k = 1,2,3, \cdots n $ and summing gives $ (n+1)^{m+1} - 1 = (m+1) \sum_{k=1}^n k^m + O(n^m) $, where the lower order terms were $ O(n^m) $ by the induction hypothesis. Thus, $ \sum_{k=1}^n k^m = \frac{n^{m+1}}{m+1} + O(n^m) $ which completes the inductive step.
From the above result, $$ \frac{ \sum_{k=1}^n k^{p} }{n^{p+1}} = \frac{1}{p+1} + O\left( \frac{1}{n}\right) \to \frac{1}{p+1} $$  as  $ n\to \infty .$
