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Triangles and Definite Integrals

  1. Jul 4, 2012 #1
    I'm trying to figure out how to integrate a data set, without knowing the function. While doing this, I got to thinking about this:

    If the definite integral of a function can be represented by the area under that function, bound by the x axis, then shouldn't:

    [itex]\int_{a}^{b}2x\frac{\mathrm{d} }{\mathrm{d} x} = A = 1/2b*h[/itex]

    Where the integral is bound by a = 0 and b = 2, and the triangle's base is 2 and height is 2.

    but rather,

    [itex]\int_{0}^{2}2x\frac{\mathrm{d} }{\mathrm{d} x} = 4[/itex]


    [itex]A = 2[/itex]

    Where's the discrepancy?
  2. jcsd
  3. Jul 4, 2012 #2
    The triangle's height is not 2.
  4. Jul 4, 2012 #3
    Brain fart. I'm an idiot.
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