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Homework Help: Line integral, problems with substitution (probably!)

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data

    I am trying to solve a line integral (bear with me, I am new to calculus!) and my basic skills of integration seem to fail me. I am sure the mistake is quite obvious, as I keep getting the wrong answer, 2, when it should be ~2.69

    2. Relevant equations
    [tex]\int_C 4x^3dS[/tex]

    C is the curve given by [tex]x=t, y=t^3-1[/tex] and [tex]0 \leq t \leq 1[/tex]

    3. The attempt at a solution
    [tex]\int_C 4x^3 dS = \int_0^1 4t^3 \sqrt{1^2+(3t^2)^2} dt = \int_0^1 4t^3 \sqrt{9t^4+1} dt[/tex]
    I attempt substitution in order to solve the integral:
    [tex]u=9t^4+1 \Rightarrow \frac{1}{36}du=t^3 dt[/tex].
    The limits are now [tex]9 \cdot 0^4=0[/tex] and [tex]9 \cdot 1^4=9[/tex], so by substitution we have:
    [tex]\frac{4}{36} \int_0^9 \sqrt{u} \, du = \frac{1}{9} \left[\frac{2}{3} u^{\frac{3}{2}}\right]_0^9 = \frac{2}{27} \left(9^{\frac{3}{2}} - 0\right)[/tex]

    Which equals.... Two! Well, there's a mistake in there somewhere, so, no it doesn't. Hopefully a non-mathematician will have mercy on me so that the punishment isn't too hard. o:)

    -- Sarah
    Last edited: Mar 26, 2009
  2. jcsd
  3. Mar 26, 2009 #2
    Nevermind, I found the problem. I did not correctly calculate the new limits, they should be 10 and 1 respectively. :biggrin:

    -- Sarah
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