Understanding Fermat's proof of the formula of integration for x^n

Question

This is the proof. Now I have two questions:-

1. Why did Fermat pick a lower limit of 0 here? Why not any other number? Is this formula provable with other numbers as the lower limit?

2. Where did 'r' come from?

Answer 1

1. An essential point of the proof is that all little rectangles are similar, and this for each fixed $r$. When you have some arbitrary point at the left end of the integration interval this does not longer hold. But when Fermat's proof is accomplished you can easily say that $$\int_a^b x^n\>dx=\int_0^b x^n\>dx-\int_0^a x^n\>dx={1\over n+1}(b^n-a^n)\ .$$
2. The $r$ is an artificial "working variable" invented by the prover. This $r$ was not present in the given problem. Note that when we split the interval $[0,a]$ into $N$ equal parts we also introduce such an artificial "working variable".

Answer 2

The proof as given in your link is not rigorous. The right approach is to let $0<a<b$ and partition $[a, b] $ via points $$x_k=ar^k, k=0, 1,2,\dots,n,r^n=b/a$$ The Riemann sum for $f(x) =x^m$ for this partition is $$\sum_{k=1}^{n}f(x_{k})(x_k-x_{k-1})$$ which is same as $$\sum_{k=1}^{n}a^{m+1}r^{km} (r^{k}-r^{k-1})=a^{m+1}(r-1)\sum_{k=1}^{n}r^{k(m+1)-1}$$ The above equals $$a^{m+1}(r-1)\cdot\frac{r^m(r^{n(m+1)}-1)}{r^{m+1}-1}$$ and the desired limit as $n\to\infty, r\to 1$ is $$\frac{a^{m+1}}{m+1}((b/a) ^{m+1}-1)=\frac{b^{m+1}-a^{m+1}}{m+1}$$ You can see that the variable $r$ is used only to simplify calculation and minimize typing effort. You can very well replace all instances of $r$ by $\sqrt[n] {b/a} $.

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The approach can't be used to deal with an interval which contains $0$ as it makes the definition of $r$ invalid. To deal with integrals over interval $[0,b]$ one can take limit of integral over $[a, b] $ as $a\to 0^+$.