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A question about Fermat's method of calculating areas under curves

  1. Nov 7, 2011 #1
    I am currently reading the book "e: The Story of a Number" by Eli Maor. And I got stuck at something. In chapter 7 of the book, the author described the method Fermat used to calculate areas under curves of the form [itex]y = x^n[/itex], where n is a positive integer. I am quoting the relevant bit here (sorry, I can't show the figure, but from the description, you can easily receate it):

    Now I can't get to the final formula. The areas of each rectangle I found are [itex]a^{n+1}, (ar)^{n+1}, (ar^{2})^{n+1},[/itex] and so on. Their sum,

    [tex]A_{r} = a^{n+1} + (ar)^{n+1} + (ar^{2})^{n+1} + \cdots[/tex]
    [tex]= a^{n+1}\left(1 + r^{n+1} + r^{2(n+1)} + \cdots \right)[/tex]
    [tex]= \frac{a^{n+1}}{1 - r^{n+1}}[/tex]

    Where am I getting wrong?
    Last edited: Nov 7, 2011
  2. jcsd
  3. Nov 7, 2011 #2


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    You are using the ordinates to get the heights of the rectangles. I think the formula is based on using the averages of the adjacent ordinates to get the rectangle heights,
  4. Nov 8, 2011 #3
    Thanks, but that was not it. I've just figured out my mistake. I got the areas of the rectangles wrong. The sum of the areas would be,

    [tex]A_r = (a - ar)a^n + (ar - ar^2)(ar)^n + (ar^2 - ar^3)(ar^2)^n + \cdots[/tex]
    [tex]A_r = a^{n+1}(1 - r) \left(1 + r^{n+1} + r^{2(n+1)} + \cdots \right)[/tex]
    [tex]A_r = \frac{a^{n+1}(1 - r)}{1 - r^{n+1}}[/tex]
  5. Nov 8, 2011 #4

    Stephen Tashi

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    I don't understand how you got those areas. Does the base of the first rectangle have length = a or does it have length = (a - ar)?
  6. Nov 8, 2011 #5
    Yes, that's the mistake I made. I posted the correct calculation in this post: https://www.physicsforums.com/showpost.php?p=3604692&postcount=3"

    The bases are (a - ar), (ar - ar2), (ar2 - ar3), and so on.
    Last edited by a moderator: Apr 26, 2017
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