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Integral Problem

  1. Oct 11, 2003 #1
    Take the integral of:

    sin3x, with respect to x. [-a,a] - interval

    I end up getting 0/3:

    = [-cos(3x)/3]

    = [-cos(3a) - (- cos (-3a))]/3

    = [-cos (3a) + cos (3a)]/3 ---> cos(-3a) = cos 3a

    = 0/3 = 0

    I think I did something wrong, right?
     
  2. jcsd
  3. Oct 12, 2003 #2
    No you are correct , sine is an odd function and integrating it over any interval symmetric about the origin will give you 0.
     
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