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Area under curve

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the area bound by y= x - 2[tex]\sqrt{x}[/tex] and y=0

    3. The attempt at a solution

    X intercept = 4
    Y intercept = 0

    Using verticle rectangles..

    dA = ((Upper curve)-(Lower curve)) dx
    [tex]\int[/tex](dA = [tex]\int[/tex](0-(x-2[tex]\sqrt{x}[/tex])) dx

    A= -(1/2X^2)-((2x^(3/2))/(2/3)) Evaluated from 0 to 4

    A= 16

    I understand this is an extremely easy question but I just cannot see where I have gone wrong with it. Is there a problem with my signs since the area I am finding is below the x-axis? Sorry if this is a bit messy, any help is appecricated, thanks.
  2. jcsd
  3. Oct 22, 2008 #2
    You should be calculating [tex]\int[/tex]2sqrt(x) - x from 0 to 4.
  4. Oct 23, 2008 #3
    That was easy. Thanks.
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