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Thread: John Wallis Integration question

  1. #1
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    John Wallis Integration question

    I have just been reading about John Wallis ( 17th Century English Mathematician) on wikipedia.
    https://en.wikipedia.org/wiki/John_Wallis

    Can anyone help with this:

    the area enclosed between the curve y = xm, the axis of x, and any ordinate x = h, and he proved that the ratio of this area to that of the parallelogram on the same base and of the same height is 1/(m + 1),
    Which parallelogram are we talking about here?
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  2. #2
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    Re: John Wallis Integration question

    Quote Originally Posted by rodders View Post
    I have just been reading about John Wallis ( 17th Century English Mathematician) on wikipedia.
    https://en.wikipedia.org/wiki/John_Wallis

    Can anyone help with this:

    the area enclosed between the curve y = xm, the axis of x, and any ordinate x = h, and he proved that the ratio of this area to that of the parallelogram on the same base and of the same height is 1/(m + 1),
    Which parallelogram are we talking about here?
    It looks to me like the word "rectangle" rather than "parallelogram" would be better here (of course, a rectangle [b]is[\b] a type of parallelogram). That is, the rectangle with vertices at (0, 0), (h, 0), ((h, h^m), and (0, h^m). That rectangle has length a and height a^m so area a^{m+1}. What is the are or the region bounded by the x-axis, y= x^m, and x= h?\
    Thanks from rodders
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  3. #3
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    Re: John Wallis Integration question

    That makes sense but presumably only between x=0 and x=h , rather than x=a to x=h ? Doesn't feel like much of a remarkable result??
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