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Simple or double integration?

  1. Mar 21, 2007 #1
    If I want to compute the area of the region bounded by the graphs y=x and y=x^2-2x, can I simply compute the integral of (x-(x^2-2x)) dx from x=0 to x=3, or do I have to use double integration?

    In any case, why?
  2. jcsd
  3. Mar 21, 2007 #2
    For two curves, f(x) and g(x) the area of the element between x and x+dx is simply |f(x)-g(x)|dx. The magnitude sign is there only if you don't want the area to go negative when the curves cross.

    So the total area = Int |f(x)-g(x)|dx
  4. Mar 21, 2007 #3
    "Ordinary" integration will do.
  5. Mar 21, 2007 #4


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    Staff Emeritus
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    I'm a bit confused by the question. Since you know about "double integration", you must have been studying calculus for some time- and should have learned to find the "area between two curves" quite a while ago!

    Here, y= x is above y= x2- 2x for x between 0 and 3 so the area is given by
    [tex]\int_0^3 (x- (x^2- 2x))dx= \int_0^3 (3x- x^2)dx[/tex]

    Of course, you could use "double integration" with '1' as integrand:
    [tex]\int_{x=0}^3 \int_{y= x^2- 2x}^x dy dx[/tex]
    but after the first integration, it reduces to the single integral form.
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