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Homework Help: Integration problem

  1. May 21, 2007 #1
    Sorry this is the first time using this forum.. so I'm not too good with writing out the equations

    1. The problem statement, all variables and given/known data
    Find the area of the region between the curves f(x) = 3x^2 and g(x) = sqrt(x/3) for 0 <= x <= 1.

    2. Relevant equations

    3. The attempt at a solution
    So.. since g(x) is greater than f(x) from 0 to 1/3, and less from 1/3 to 1.

    I set up the integrals

    [int. from 0 to 1/3 of g(x) - int. from 0 to 1/3 of f(x)] + [int. from 1/3 to 1 of f(x) - int. from 1/3 to 1 of g(x)]

    Is that set up correctly?

    Next is where my problem is... :confused:
    integral of 3x^2 is x^3, but how do I integrate sqrt(x/3) ?

    I used the power rule and got [2(x/3)^3/2]/3, but that doesn't seem right, please help me out with this.

  2. jcsd
  3. May 21, 2007 #2


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    Staff Emeritus
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    Gold Member

    Your method seems fine but your integration of the square root is slightly out. Try integrating the following and see why you are slightly out.

    [tex] \frac{1}{\sqrt{3}}\times x^{\frac{1}{2}}[/tex]
    Last edited: May 21, 2007
  4. May 21, 2007 #3


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    Yeah, just factor out the 1/sqrt[3] and treat the sqrt[x] on its own.
  5. May 22, 2007 #4
    So would I end up with 1/sqrt(3) * int. (x^3/2) / (3/2) ?
  6. May 22, 2007 #5


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    Homework Helper

    Yes, of course, but without the integral sign thingy. :)
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