Thanks a lot for your help, but I have come here to tell you guys that I've just figured out my mistake. I got the areas of the rectangles wrong. The sum of the areas would be,
[math]A_r = (a - ar)a^n + (ar - ar^2)(ar)^n + (ar^2 - ar^3)(ar^2)^n + \cdots[/math]
[math]A_r = a^{n+1}(1 - r) \left(1 + r^{n+1} + r^{2(n+1)} + \cdots \right)[/math]
[math]A_r = \frac{a^{n+1}(1 - r)}{1 - r^{n+1}}[/math]
DrRocket, yes, it is a crude approximation, which is why Fermat reasoned that the width of each rectangle must be made small. And for this, r must be close to 1.
[math]A_r = \frac{a^{n+1}(1 - r)}{1 - r^{n+1}}[/math]
[math]A_r = \frac{a^{n+1}(1 - r)}{(1 - r)(1 + r + r^2 + \cdots + r^n)}[/math]
[math]A_r = \frac{a^{n+1}}{1 + r + r^2 + \cdots + r^n}[/math]
As we let [imath]r \rightarrow 1[/imath], each term is the denominator tends to 1, resulting in,
[math]A = \frac{a^{n+1}}{n + 1}[/math]
And that is the integration formula [math]\int_{0}^{a} x^n dx = \frac{a^{n+1}}{n + 1}[/math]
.