Integral of Power/Fermat's Proof

Theorem

 * $\ds \forall n \in \Q_{>0}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$

Proof
First let $n$ be a positive integer.

Take a real number $r \in \R$ such that $0 < r < 1$ but reasonably close to $1$.

Consider a subdivision $S$ of the closed interval $\closedint 0 b$ defined as:
 * $S = \set {0, \ldots, r^2 b, r b, b}$

that is, by taking as the points of subdivision successive powers of $r$.

Now we take the upper sum $\map U S$ over $S$ (starting from the right):

Now we let $r \to 1$ and see that each of the terms on the bottom also approach $1$.

Thus:
 * $\ds \lim_{r \mathop \to 1} S = \frac {b^{n + 1} } {n + 1}$

That is:
 * $\ds \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$

for every positive integer $n$.

Now assume $n = \dfrac p q$ be a strictly positive rational number.

We set $s = r^{1/q}$ and proceed:

As $r \to 1$ we have $s \to 1$ and so that last expression shows:

So the expression for the main result still holds for rational $n$.