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Archimedes' rule that area of a parabola is 4/3 times an inscribed triangle

  1. Apr 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Let ABC be a piece of a parabola. The point B is chosen such that the tangent to the parabola at B is parallel to the line AC. Archimedes proved the Area of parabola inside ABC is 4/3 times the triangle ABC. Prove this using calculus


    2. Relevant equations
    The integral from a to c (Where pt A=(a,a^2), pt B=(b,b^2), and pt C=(c,c^2)) of (x^2)-(x+c)


    3. The attempt at a solution
    I have been trying to figure this out for hours. I know you can assume the parabola to be in the standard form y=x^2 without loss of generality because the problem relies on ratios of areas. And from this, I thought you could assume the segment AC has standard form y=x+c. Then, since we are looking for area, we could use the method of area between two curves, which is why I feel like the integral from a to c of (x+c)-(x^2) would be right. But when I calculate this, it should end up looking something like 4/3(base*height*1/2) and it definitely doesnt look that way. Any help would be greatly appreciated!
     
  2. jcsd
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