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Area of a Parabola

January 12, 2012

In my unending quest to find  out seemingly simple bits of maths that I didn’t know, one of my year 10 students found this:



So three questions:

1.       Is this true?

2.       Is this useful?

3.       Is it better/easier than integrating?

If it is true, I think it falls into the “wow that’s neat” category.


From → Uncategorized

  1. kjrose permalink

    Yep, it’s totally truestart withf(x) = (b/2 – x)(b/2 +x)*c = (b^2*c/4 – c*x^2)h = b^2*c/4 => c = h*4/b^2sof(x) = h-h*4*x^2/b^2solve the integral from 0 to b/2 and multiply by 2 and you gethb*2/3pretty sexy actually.

  2. kjrose permalink

    I think appolonius may have a geometric proof of this as well.

  3. daveinstpaul permalink

    Yes. It’s even true if the parabolic segment is not symmetric. This is (more or less) how Archimedes described the quadrature of the parabola. have used it occasionally, and it can be easier than integrating.

  4. pat ballew permalink

    dave, As daveinstpaul pointed out, Archimedes did this, and he did it with perhaps the first use of summing an infinite geometric series in math history.

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