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(Apologies for asking another question based on Julian Havil's "The Irrationals")

On page 86 of Havil's "The Irrationals" the author outlines how John Wallis approximated integration by applying a variant of the "Cavalieri Principle" (that two solids are of equal volume if the areas of their cross sections are everywhere equal) as applied to two dimensions.

He outlines this in this fashion - for a "positive value continuous function $f(x)$ defined on the positive real axis with the maximum value of $M_N$ on the interval $[0,N]$":

$$\int_0^1 f(x)\,dx = \lim_{N\rightarrow \infty} \frac{\sum^N_{r=0} f(r)}{\sum^N_{r=0}M_N} = \lim_{N \rightarrow \infty}\frac{\sum^N_{r=0} f(r)}{M_N(N+1)}$$

And writes "In words, add up the lengths of the vertical strips from the $x$-axis to the curve and then divide this by the same order approximation of the area of the enclosing rectangle."

What has totally flummoxed me is the idea that the area under the curve of a function between 0 and 1 is related to its value at infinity. I can see that we can approximate the value of the integral by summing the values of the function and then dividing by the interval - but not the sum presented here: could someone explain the reasoning here, please?

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    $\begingroup$ No problem, your discomfort is reasonable, the formula is just wrong. Could be fixed, by a scaling/descaling procedure, but the details are different. $\endgroup$
    – André Nicolas
    Commented Aug 5, 2013 at 17:37
  • $\begingroup$ When I read that my first reaction was that I must have copied it down wrong - but I have rechecked and that is how it is printed in the book :( $\endgroup$
    – adrianmcmenamin
    Commented Aug 5, 2013 at 21:48
  • $\begingroup$ I managed to check also, the Google preview carried it. $\endgroup$
    – André Nicolas
    Commented Aug 5, 2013 at 21:53
  • $\begingroup$ I have spotted a few misprints in the book, but the suggestion here is of an outright error in methodology? $\endgroup$
    – adrianmcmenamin
    Commented Aug 5, 2013 at 21:56
  • $\begingroup$ It can be fixed, we want a Riemann-type sum. $\endgroup$
    – André Nicolas
    Commented Aug 5, 2013 at 22:00

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This page - http://www.math.wpi.edu/IQP/BVCalcHist/calc1.html - gives an excellent explanation of Cavalieri's and Wallis's thinking.

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I think the following link might help in answering your question...

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.196.3331&rep=rep1&type=pdf

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  • $\begingroup$ Not really. It doesn't explain the Willis-Cavalieri technique and the reference it gives to it is merely an abstract for a presentation :( $\endgroup$
    – adrianmcmenamin
    Commented Aug 24, 2013 at 12:24
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I had exactly the same reaction while reading Havil's book today. It lead me here and to the articles referenced above. And got me thinking...

I think the key point is that this is not a general method. The area ratio approach only works if the function is self-similar in the sense that it has the same characteristic shape over the interval 0 to A independent of A. Clearly simple power functions have this property.

That is the sense in which (quoting the question) "...the area under the curve of a function between 0 and 1 is related to its value at infinity".

So having found the precise ratio of areas (for any A), you then apply it to the unit interval of the integral between 0 and 1 (i.e. when A is 1). If the coefficient of the power in the function is unity then the area of the rectangle (actually a square now) is 1, and so the ratio becomes numerically equal to area under the curve (just the units have changed).

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