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Shortcut to the area under a curve?

  1. May 13, 2009 #1
    I'm not up to calculus yet, but I was playing with functions and I had this idea that I wanted to ask about.

    Is there an easy way to get the area under a curve given these restrictions:

    1. The curve is described by second degree polynomial
    2. The area we are interested in is the complete range between the zero's on the x-axis

    In other words, what is the area under the curve between the zero's of a(x-b)(x-c)

    It seems to be that this should be narrow enough to be calculable without doing series and integrals and whatnot (calculus, I don't know the exact approach yet), but I cant seem to find anything that fits.

    So, is there a nice solution to this?
     
  2. jcsd
  3. May 13, 2009 #2

    HallsofIvy

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  4. May 13, 2009 #3
    If you had a way to find the area under the specific curve 1-x2 then you could find the area under any other parabola by shifting and rescaling vertically and horizontally. You can do this conversion algebraically by completing the square and a change of variables.

    Of course, the challenge remains to find the area under 1-x2.
     
  5. May 13, 2009 #4
    Maze: That's some real nice insight, I didn't even consider that. You are 100% correct of course, no need to worry about a, b, and c but just focus on 1-x^2. Thanks.

    From what you are saying I gather that there is no known exact solution to the area under 1-x^2?

    HallsofIvy I've read some popular recounts of Archimedes' approach and yes that is segmentation and summation. My thought was that perhaps there are certain classes of curves that are easier to get the area of, like second degree polynomials with integer zeroes, for example.

    k
     
  6. May 13, 2009 #5
    Uh, no, there is a known solution to the area under 1-x^2. Calculus makes it pretty easy to find.

    I think the method of exhaustion, previously mentioned, states that this area - bounded by 1 - x^2 and the x-axis - equals 2/3 of the area of the smallest rectangle that covers the region... or is it 1/3... anyway, you might want to look into this.
     
  7. May 13, 2009 #6

    CRGreathouse

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    You have f(x) = a(x-b)(x-c). Choose b <= c.

    The area under the curve between b and c is
    [tex]a(b-c)^3/6[/tex]
     
  8. May 14, 2009 #7
    CRGreathouse: That's exactly the sort of thing i was trying to "construct". Can you show how that was derived or would that require calculus?

    k
     
  9. May 14, 2009 #8

    HallsofIvy

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    Yes, CRGreathouse used calculus to produce a general formula that could then be used as a "shortcut" for specific cases!
     
  10. May 14, 2009 #9
    HallsofIvy ok, thank you. I'll just leave this until I get to calculus I guess.

    k
     
  11. May 29, 2009 #10
    I just read an explanation of Archimedes' method, and wanted to say "Wow!"

    It is not really segmentation/summation at all, the idea that what is true for parts must be true for the sum is just a tiny part of the proof. I don't want to spoil how he did it but rather

    recommend that everyone look into how he did it.

    http://www.math.ubc.ca/~cass/archimedes/parabola.html

    Basically using nothing but primary school geometry.

    k
     
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