This is shorter and more specific version of certain questions about a rather simple quadrature method. The answers I got were great but not what I asked. The terms in the title for $y=x^p$ look strange to us, but might not have to some 17th century mathematicians. They considered this question:

Find the area $A_p(a,b)$ of the region under $x^p$ for $a \le x \le b.$

We would write this with an integral sign and use the Fundamental theorem of Calculus but they did not have those tools (as is the case with modern day Calculus students for a lecture or two). Even without these tools they (and modern day Calculus students) show for $p=0,1,2,3$ that $$A_p(a,b)=\frac{b^{p+1}-a^{p+1}}{b-a}.$$ It is easy to get as a corollary the cases $p=1/2,1/3$ and the conjecture is and was obvious: The equation should hold for "all" $a,b,p$ except when it doesn't (make sense). There is much further supporting evidence, but that is not a proof. Useful identities for $p=1,2,3$ are $S_2(n)=\sum_1^nk^2=\frac{n(n+1)(2n+1)}{6}$ along with $S_1(n)=\frac{n(n+1)}{2}$ and $S_3(n)=S_1(n)^2.$ The desired method is a familiar kind of geometric/algebraic argument using the area under step functions over $[a,b].$

To be specific (in a way favorable to my approach): A partition $P$ of $[a,b]$ is a finite sequence $a=x_0 \lt x_1 \lt \cdots \lt x_n=b$. The mesh of $P$ is $\max(x_{i}-x_{i-1}).$ (There is rarely a reason to have unequal length sub-intervals, but Fermat gave one.) We use the resulting sub-intervals, in two ways, as the bases of an assemblage of rectangles with heights determined by the endpoints. Since $x^p$ is monotonic, one is covered by the region and the other covers it. So the two areas provide a lower and an upper bound.

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$$ \sum_1^nx_{i-1}^p(x_i-x_{i-1}) \lt A_p(a,b) \lt \sum_1^nx_{i}^p(x_i-x_{i-1}).$$ If we manage to compute or bound these bounds and show that, when the mesh goes to zero, they have a common limit (the one we expect), we are done. The actual bounds we compute are of value only for the interesting, but secondary, topic of speed of convergence. And anyway, if $m(P) \lt \epsilon$, then the difference between the two bounds is less than $(b^p-(b-\epsilon)^p)(b-a) \lt p \epsilon b^p (b-a)$ which converges to zero. (For $p \lt 0,$ use $a-(a+\epsilon)^p$.)

Brief historical digression: This seems like masochism but many Calculus classes have students do it (at least for $p=2$) before learning the Fundamental Theorem of Calculus. The method normally used involves equal length sub-intervals and depends on the formulas above for $S_p(n)$ . Cavelieri gave a similar treatment for integral $0 \le p \le 9$ with the main difficulty being finding the formula for $S_p(n)=\sum_{k=0}^nn^p.$ Pascal showed $S_p(n)=\frac{n^{p+1}}{p+1}+\frac{n^p}{2}+O(n^{p-1})$ which does allow arbitrary positive integer $p.$ Fermat had an ingenious technique using finite geometric progressions which applies to arbitrary (rational) values of $p$ (see the previous question or this.) The finite method is valid for $a \gt 0$ , I ignore a variation for $a=0$ with infinite partitions and geometric progressions.

My specific questions involve another method described immediately below. They are

1) Have you seen this before?

2) Why ( if you care to speculate) do you think it was not used in the 17th century? Were the "rules" that one must arrive at the answer pretending not to know it?

The method :

I propose to instead assign to each sub-interval $[u,v]$ the height $h(u,v)=\frac{v^{p+1}-u^{p+1}}{(p+1)(v-u)}$ and "compute" $\sum_1^nh(x_{i-1},x_i)(x_i-x_{i-1})$ which immediately collapses to, of course, $\frac{b^{p+1}-a^{p+1}}{p+1}.$

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Establishing that this has any relevance requires showing that the height $h(u,v)$ is between $u^p$ and $v^p$. This is easy in practice if one simplifies. If you simplify first, then the whole thing can look like magic until you see what was done.

So for $p=-3$, obviously $$\frac{1}{v^3} = \frac{1}{2v^3}+\frac{1}{2v^3} \lt h(u,v)=\frac{1}{2v^2u}+\frac{1}{2vu^2} \lt \frac{1}{u^3}.$$ OK, so what? Why not use the average or the geometric mean or $\left(\frac{u+v}{2}\right)^{-3}$? Well, check that $(v-u)h(u,v)=\frac{1}{2u^2}-\frac{1}{2v^2}$, so $\sum_1^nh(x_{i-1},x_i)(x_i-x_{i-1})$ collapses to $\frac{1}{2a^2}-\frac{1}{2b^2}$.

I'll frame the situation as follows: The two historic methods above use specific partitions and careful elementary but somewhat involved calculations of two bounding sums which converge to the conjectured final answer. My method (which I am fully confident is not new) starts at the correct answer and calmly stays there, with no further computations, no matter what partition is specified. I don't immediately see that it applies to any other functions, but $x^p$ does have a certain primary importance.

  • $\begingroup$ 1. I have not seen this before. (But I have not read all those XVII century papers). 2. I suspect that you found this argument because you knew the answer in advance:-) And Fermat probably did not. But I agree that this is very elegant. $\endgroup$ – Alexandre Eremenko Dec 6 '12 at 13:02
  • $\begingroup$ I don't understand, what's the purpose of a method that only works if you know the answer beforehand? $\endgroup$ – Ketil Tveiten Dec 6 '12 at 13:14
  • $\begingroup$ @Ketil: The purpose is that this is how one often finds the answer. You "guess" an answer (by intuition or after spotting a pattern), go through the computation, and see if it works. If it does you're done. If not, maybe you can figure out a correction and get closer to the final solution. I said something similar in an answer to http//mathoverflow.net/questions/115032/…. $\endgroup$ – Rodrigo A. Pérez Dec 6 '12 at 14:56
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    $\begingroup$ Isn't your proof exactly following the standard proof of the fundamental theorem of calculus with an explicit version of the mean value theorem? $\endgroup$ – Franz Lemmermeyer Dec 6 '12 at 15:28
  • $\begingroup$ @Alexandre: I think that Fermat definitely knew the answer in advance. @Franz: I got Google hits for "converse of Mean Value Theorem for integrals" but not pertinent ones. And yes, given $\int\frac{1}{1+x^2}dx=\arctan(x)$ , the MVTfI shows $\frac{1}{1+b^2} \lt \frac{\arctan{b}-\arctan{a}}{b-a} \lt \frac{1}{1+a^2}$ when $0 \le a \lt b.$ A geometric derivation (is there one?) of this fact would provide by "my" method an independent derivation of certain areas (definite integrals). I stuck to monotonic functions to keep things simple. $\endgroup$ – Aaron Meyerowitz Dec 6 '12 at 18:18

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