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

1. Sep 16, 2012

### tainted

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
$[-a,a]\int f(x)\,dx = 0$
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.

2. Sep 16, 2012

### LCKurtz

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. Sep 17, 2012

### tainted

$\int_{-a}^a f(x)\ dx = \int_{-a}^0 f(x)\ dx +\int_{0}^a f(x)\ dx$
$\int_{-a}^0 f(x)\ dx$
$-\int_{0}^{-a} f(x)\ dx$
$\int_{0}^{-a} -f(x)\ dx$
$\int_{0}^{-a} f(-x)\ dx$
$\int f(u)\,dx$
$-F(u) = -F(-x)$
$-F(-x)|_{0}^{-a}$
$-[F(-(-a)) - F(0)]$
$[F(0) - F(a)]$

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

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

Thanks man!

4. Sep 17, 2012

### LCKurtz

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 $\int_0^a f(x)\, dx$.