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

  • Thread starter tainted
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  • #1
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Homework Statement



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]

Homework Equations


If f is odd, then
##f(-x) = -f(x)##
##u=-x##
Our TA told us to set u equal to -x.

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!
 

Answers and Replies

  • #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.
 
  • #3
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##\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!
 
  • #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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