Integral of Power

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Theorem

$\ds \forall n \in \R_{\ne -1}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$


Conventional Proof

From the Fundamental Theorem of Calculus:

$(1): \quad \ds \int_0^b x^n \rd x = \bigintlimits {\map F x} 0 b = \map F b - \map F 0$

where $\map F x$ is a primitive of $x^n$.


By Primitive of Power, $\dfrac {x^{n + 1} } {n + 1}$ is a primitive of $x^n$.


Then:

\(\ds \int_0^b x^n \rd x\) \(=\) \(\ds \intlimits {\frac {x^{n + 1} } {n + 1} } 0 b\) substituting $\dfrac {x^{n + 1} } {n + 1}$ for $F$ in $(1)$
\(\ds \) \(=\) \(\ds \frac {b^{n + 1} } {n + 1} - \frac {0^{n + 1} } {n + 1}\)
\(\ds \) \(=\) \(\ds \frac {b^{n + 1} } {n + 1}\)

$\blacksquare$


Fermat's Proof

This proof is is valid only for positive rational numbers, that is, it proves that:

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


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 Darboux sum $\map U S$ over $S$ (starting from the right):

\(\ds \map U S\) \(=\) \(\ds b^n \paren {b - r b} + \paren {r b}^n \paren {r b - r^2 b} + \paren {r^2 b}^n \paren {r^2 b - r^3 b} + \cdots\)
\(\ds \) \(=\) \(\ds b^{n + 1} \paren {1 - r} + b^{n + 1} r^{n + 1} \paren {1 - r} + b^{n + 1} r^{2 n + 2} \paren {1 - r} + \cdots\)
\(\ds \) \(=\) \(\ds b^{n + 1} \paren {1 - r} \paren {1 + r^{n + 1} + r^{\paren {n + 1}^2} + \cdots}\)
\(\ds \) \(=\) \(\ds \frac {b^{n + 1} \paren {1 - r} } {1 - r^{n + 1} }\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds \frac {b^{n + 1} } {1 + r + r^2 + \cdots + r^n}\)

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

Thus:

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

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:

\(\ds \frac {1 - r} {1 - r^{n + 1} }\) \(=\) \(\ds \frac {1 - s^q} {1 - \paren {s^q}^{p / q + 1} }\)
\(\ds \) \(=\) \(\ds \frac {1 - s^q} {1 - s^{p + q} }\)
\(\ds \) \(=\) \(\ds \frac {\paren {1 - s^q} / \paren {1 - s} } {\paren {1 - s^{p + q} } / \paren {1 - s} }\)
\(\ds \) \(=\) \(\ds \frac {1 + s + s^2 + \cdots + s^{q - 1} }{1 + s + s^2 + \cdots + s^{p + q - 1} }\)

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

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

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

$\blacksquare$


Historical Note

The conventional proof of Integral of Power of course holds for all real $n \ne -1$, not just where $n$ is a strictly positive rational.

However, the real point of this page is Fermat's proof, which demonstrates how integration was achieved before the full machinery of calculus had been thoroughly constructed.

Bonaventura Francesco Cavalieri had previously made progress with this problem, proving it for integral $1 \le n \le 9$ but the algebra for the proof of each power was more difficult than the previous one, and he found $10$ too much hard work. The clear beauty of Fermat's approach was that it works for all $n$, rational as well as integral.


Sources

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