Integral of Power/Fermat's Proof

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Theorem

$\displaystyle \forall n \in \Q_{>0}: \int_0^b x^n \mathrm d x = \frac {b^{n + 1} } {n + 1}$


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 $\left[{0 \,.\,.\, b}\right]$ defined as:

$S = \left\{{0, \ldots, r^2 b, r b, b}\right\}$

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


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

\(\displaystyle U \left({S}\right)\) \(=\) \(\displaystyle b^n \left({b - r b}\right) + \left({r b}\right)^n \left({r b - r^2 b}\right) + \left({r^2 b}\right)^n \left({r^2 b - r^3 b}\right) + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle b^{n + 1} \left({1 - r}\right) + b^{n + 1} r^{n + 1} \left({1 - r}\right) + b^{n + 1} r^{2 n + 2} \left({1 - r}\right) + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle b^{n + 1} \left({1 - r}\right) \left({1 + r^{n + 1} + r^{\left({n + 1}\right)^2} + \cdots}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {b^{n + 1} \left({1 - r}\right)}{1 - r^{n + 1} }\) Sum of Geometric Progression
\(\displaystyle \) \(=\) \(\displaystyle \frac {b^{n + 1} }{1 + r + r^2 + \cdots + r^n}\)

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

Thus:

$\displaystyle \lim_{r \mathop \to 1} S = \frac {b^{n + 1} } {n + 1}$

That is:

$\displaystyle \int_0^b x^n \mathrm d x = \frac {b^{n + 1} } {n + 1}$

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:

\(\displaystyle \frac {1 - r} {1 - r^{n + 1} }\) \(=\) \(\displaystyle \frac {1 - s^q} {1 - \left({s^q}\right)^{p / q + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 - s^q} {1 - s^{p + q} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({1 - s^q}\right) / \left({1 - s}\right)}{\left({1 - s^{p + q} }\right) / \left({1 - s}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 + s + s^2 + \cdots + s^{q - 1} }{1 + s + s^2 + \cdots + s^{p + q - 1} }\)

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

\(\displaystyle \frac {1 - r} {1 - r^{n + 1} }\) \(\to\) \(\displaystyle \frac q {p + q}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {p / q + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {n + 1}\)

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

$\blacksquare$


Historical Note

The integral of a power for rational power was used by Fermat, and predated the work done by Newton and Leibniz by a considerable period.


Sources