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Homework Help: Change of Variable issue with Integration

  1. May 24, 2006 #1
    I have the following equation:

    [tex]
    I = \frac{a}{2} \int_{-a/2}^{a/2} x sin^{2}\left[ \pi (\frac{x}{a}-\frac{1}{2})\right], dx
    [/tex]

    I have set

    [tex]
    y= \frac{x}{a}-\frac{1}{2}
    [/tex]

    and

    [tex]
    dy = dx/a
    [/tex]

    When I substitute the two latter equations into the first equation I should get this:

    [tex]
    I = {a} \int_{-1}^{0} (2y+1) sin^{2}\left[ \pi y\right], dy
    [/tex]

    For some reason I get this instead:

    [tex]
    I = \frac{a}{2} \int_{-1}^{0} (y+\frac{1}{2}) sin^{2}\left[ \pi y\right], dy
    [/tex]

    I'm off by a factor of four. What am I doing wrong?
     
    Last edited: May 25, 2006
  2. jcsd
  3. May 24, 2006 #2
    It looks like you substitued y = x/a + 1/2 in the integral, and used x/a - 1/2 to determine the new limits of integration.

    [tex]
    I = \frac{a}{2} \int_{-a/2}^{a/2} x sin^{2}\left[ \pi (\frac{x}{a}+\frac{1}{2})\right], dx
    [/tex]

    Or should have this been

    [tex]
    I = \frac{a}{2} \int_{-a/2}^{a/2} x sin^{2}\left[ \pi (\frac{x}{a}-\frac{1}{2})\right], dx
    [/tex]
     
  4. May 25, 2006 #3
    The given equation should have been

    [tex]
    I = \frac{a}{2} \int_{-a/2}^{a/2} x sin^{2}\left[ \pi (\frac{x}{a}-\frac{1}{2})\right] dx
    [/tex]

    I have corrected this in the initial post.

    Unfortunately I'm still stuck same answer (and this answer does not match the one in my book).
     
  5. May 25, 2006 #4
    If [tex]y= \frac{x}{a}-\frac{1}{2}[/tex] then [tex]x = a(y + \frac{1}{2})[/tex]

    and

    [tex]dx = ady [/tex]

    So we get for the integral

    [tex] I = \frac{a}{2} \int_{-1}^{0} a(y + \frac{1}{2}) sin^{2}\left[ \pi y \right] ady [/tex]

    marlon
     
  6. May 25, 2006 #5
    Well, this book does have some errors in it.
     
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