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 $m \geqslant 0$.