1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Area under curve