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Expand integrand

  1. Dec 8, 2009 #1
    Is this an expand integrand?
    [tex]\int 9x[/tex] [tex]^{2}[/tex] [tex]/[/tex] (3 - x)[tex]^{4}[/tex]

    I set u = ( 3 - x)
    du = -1dx

    and so if i treat x = 3 - u , i might get this integral

    [tex]\int[/tex] 9(3-u)[tex]^{2}[/tex] (u)[tex]^{4}[/tex]

    the answer is
    (3[tex]/[/tex]x - 1) [tex]^{-3}[/tex] + c
    but i can't get it...

    Originally, from the book, it gave a simple example like this

    [tex]\int[/tex] [tex]x[/tex] (2-x)[tex]^{1/2}[/tex]

    negative [tex]\int[/tex] [tex](2-u) [/tex] u[tex]^{1/2}[/tex]

    it sets
    u = 2 - x
    du = -dx
    and x = 2-u

    I just don't get what EXPANDED INTEGRAND is really doing...
    Last edited: Dec 8, 2009
  2. jcsd
  3. Dec 8, 2009 #2


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    Science Advisor
    Homework Helper

    Hi jwxie! :smile:

    (have an integral: ∫ and try using the X2 tag just above the Reply box :wink:)

    "expand the integrand" simply means multiply it out, so that you get a/u4 + b/u3 + c/u2, and then integrate that

    and after you've done that, you should get a polynomial plus a constant of integration … then you can subtract a multiple of u3/u3 from the constant, and complete the cube :wink:
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