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Proving integrand of odd function from -a to a is 0

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose f is a continuous function defined on an interval [-a, a]. Show that if f is odd, then
    [itex] [-a,a]\int f(x)\,dx = 0 [/itex]
    2. Relevant equations
    If f is odd, then
    ##f(-x) = -f(x)##
    ##u=-x##
    Our TA told us to set u equal to -x.
    3. The attempt at a solution
    ## u = -x ##
    ## -du = dx ##

    ##-[a,-a]\int f(x)\,dx##
    ##[a,-a]\int -f(x)\,dx##
    definition of odd function
    ##[a,-a]\int f(-x)\,dx##

    ##-\int f(u)\,du##
    ##-F(-x)|[a,-a]##
    ##F(-x)|[-a,a]##
    ##F(-a) - F(-(-a))##
    ##F(-a) - F(a)##
    ## -2F(a) ##
    Obviously I went wrong in my proof somewhere or I got distracted and did useless steps.
    Thanks in advance!
     
  2. jcsd
  3. Sep 16, 2012 #2

    LCKurtz

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    Write ##\int_{-a}^a f(x)\ dx = \int_{-a}^0 f(x)\ dx +\int_{0}^a f(x)\ dx ## and make that substitution on the first integral only.
     
  4. Sep 17, 2012 #3
    ##\int_{-a}^a f(x)\ dx = \int_{-a}^0 f(x)\ dx +\int_{0}^a f(x)\ dx##
    ## \int_{-a}^0 f(x)\ dx ##
    ##-\int_{0}^{-a} f(x)\ dx ##
    ##\int_{0}^{-a} -f(x)\ dx ##
    ##\int_{0}^{-a} f(-x)\ dx ##
    ##\int f(u)\,dx ##
    ## -F(u) = -F(-x) ##
    ## -F(-x)|_{0}^{-a} ##
    ## -[F(-(-a)) - F(0)] ##
    ## [F(0) - F(a)] ##

    ##\int_{0}^a f(x)\ dx##
    ## F(x)|_{0}^{a} ##
    ##[F(a) - F(0)]##


    ##F(0) - F(a) + F(a) - F(0) = 0##

    Thanks man!
     
  5. Sep 17, 2012 #4

    LCKurtz

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    It is much less convoluted to do it like this with ##\int_{-a}^0 f(x)\, dx##. Let ##u = -x,\, du=-dx## giving ##\int_a^0 f(-u)(-du)=-\int_0^a f(u)\, du##, which cancels out the ##\int_0^a f(x)\, dx##.
     
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