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 $m \geqslant 0$.

HINT:
Instead of setting up the successive $x_i$ in arithmetic progression, set them up in geometric progression. (I will not say explicitly how - I think it will become apparent).
Then, carry out the sums. Just do them. They will be geometric series, easily calculated, and the answer will appear.
If you are still stuck, then post your progress and where you are stuck.
