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Find the area between two curves - Help finding the limits of integration

  1. Apr 30, 2010 #1
    1. The problem statement, all variables and given/known data

    The curves are:

    [tex] f(x)= x^{2/3} [/tex]

    and

    [tex] g(x)=x^{3/2} [/tex]

    2. Relevant equations

    I am assuming that:

    [tex] x^{2/3} = x^{3/2} [/tex]

    is going to give me the limits of integration but I don't know how to solve for x on this equation.

    Could also put it this way:

    [tex] \sqrt[3]{x^{2}} = \sqrt{x^{3}} [/tex]

    but that doesn't help me much..

    3. The attempt at a solution

    no real attempt because I don't know what to do.
     
  2. jcsd
  3. Apr 30, 2010 #2
    I don't know how to do with algebra but, here's a hint:

    What two numbers can be raised to any power and still equal themselves?
     
  4. Apr 30, 2010 #3
    I would say, 0 and 1, right? because -1 will change signs if raised to an even power so that doesn't fit. If that's right what was the clue that that would be the answer?
     
  5. Apr 30, 2010 #4
    Your reasoning is correct. Just subtract the two terms so the difference is equal to 0 and factor out x2/3. Then find the roots.
     
  6. Apr 30, 2010 #5
    @VeeEight

    I think you read the problem wrong because I cant factor out the x^(2/3) as the lines are x^3/2 and x^2/3, not the same, or else I could easily solve for x
     
  7. Apr 30, 2010 #6
    Perhaps this would be useful.

    xm*xn=xm+n
     
  8. Apr 30, 2010 #7
    If you have the equation x2/3 - x3/2 = 0, write this as x2/3(something - 1)
     
  9. Apr 30, 2010 #8
    @VeeEight

    So with:

    [tex] x^{2/3}(x^{5/6}-1) = 0 [/tex]

    can it be solved for x like so:

    [tex] \sqrt[3]{x^{2}} = 0 \rightarrow x = 0 [/tex]

    and

    [tex] \sqrt[6]{x^{5}}-1 = 0 [/tex]

    [tex] \sqrt[6]{x^{5}}=1 \rightarrow x =1 [/tex]

    so the limits will be 0 to 1? (I know those are the right limits because when the integral is evaluated from 0 to 1 the answer checks out in the back of the book).
     
  10. Apr 30, 2010 #9
    Yes, that is the correct approach.
     
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