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Double integrals

  1. Jun 12, 2007 #1
    I'm having some trouble with this particular question.

    ∫∫x dA bound by y = 4x^3 - x^4 and y = 3 - 4x + 4x^2.

    All I can think to do is equate the two equations to find where they intercept to give the bounds for the double integral giving 0 = x^4 - 4x^3 + 4x^2 - 4x + 3. But I don't know where to go from here.

    Any help would be appreciated.
     
  2. jcsd
  3. Jun 12, 2007 #2
    That's pretty nasty.
    Maybe you could try Newton's method? Take an initial guess, e.g. x=1...
    Actually, it looks like x=1 works.. (1 - 4 + 4 - 4 + 3 = -3 + 4 - 4 + 3 = 0)
    So since we have that, divide through by (x-1) to get the other roots.
    Have fun. :D
     
  4. Jun 12, 2007 #3

    NateTG

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    The only possible rational roots are:
    [tex]\pm 1, \pm 3[/tex]
    so you could start by checking whether those are intersections.
     
  5. Jun 12, 2007 #4
    What method did you use to find those roots?
     
  6. Jun 12, 2007 #5

    NateTG

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  7. Jun 12, 2007 #6
    Cheers mate. Can't say I've ever heard of the rational zero theorem.
     
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