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Homework Help: Integration Problem

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm just having a bit of trouble with where to start on this integral.

    [tex]\int[1 - (r/0.11)]^1^/^5rdr[/tex]

    3. The attempt at a solution

    I've tried using integration by parts, "u" substitution and things like that but I dont seem to be getting anywhere with this. Any help would be greatly appreciated.
     
  2. jcsd
  3. Feb 28, 2010 #2

    Gib Z

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    Which substitutions did you try? Its obvious the stuff under the radical is giving you trouble, so why not try substituting it away.
     
  4. Feb 28, 2010 #3
    The first substitution that comes to mind is to let u = (1-(r/0.11)) so du = -1/0.11 dr. Rearranging for dr I get dr = -0.11du. If I sub this into the integral I still have an r in the equation.

    From here I rearranged the expression u = (1-(r/0.11)) for r to get r = (1-u)(0.11). If I sub this into the integral I get

    I = [tex]\int(u)^1^/^5(1-u)(0.11)(-0.11)du[/tex]
    = [tex]\int(u)^1^/^5(1-u)(-0.0121)du[/tex]

    It looks a bit better than before but I'm still stumped, hopefully the work I've done up to here is correct.
     
  5. Feb 28, 2010 #4

    Gib Z

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    It is correct. Now just take the constant outside, and expand the brackets.
     
  6. Feb 28, 2010 #5
    If I do that then the integral becomes

    I = -0.0121[tex]\int (u^1^/^5 - u^6^/^5)[/tex]
    = -(0.0121)[(5/6)u6/5 - (5/11)u11/5]

    So in order to get a solution I would also need to change the initial limits of integration using u = (1-(r/0.11)). If my initial limits of integration were 0 to 0.11 after subbing them into the equation for u my new limits of integration are from 1 to 0.
     
  7. Mar 1, 2010 #6

    Gib Z

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    Yes that is correct. If you want to follow convention you'll have to introduce a negative factor and swap the new limits of integration to ensure the upper limit of integration is larger than the lower limit of integration.
     
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