How to demonstrate that `int_(-a)^af(x)dx=0` for `f`-odd?

By definition:

`f` is an odd function `iff f(-x)=-f(x)`

Using this definition then we note that if `f` is odd then

`f(-x)=-f(x) => f(x) = -f(-x)`

Using a property of integrals:

`int_a^b \ f(x) \ dx = int_a^c \ f(x) \ dx + int_c^b \ f(x) \ dx`

we can write the integral we seek to evaluate as

`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`

For the first integral we can perform simply substitution, let:

`u=-x => (du)/dx = -1` and `u=-x`

And we can change the limits of integration: When:

`{ (x=-a), (x=0) :} => { (u=a), (u=0) :}`

So the integral can be rewritten as:

`int_(-a)^(a) \ f(x) \ dx = int_(a)^(0) \ -f(u) \ (-1)du + int_(0)^(a) \ f(x) \ dx`
`" " = int_(a)^(0) \ f(u) \ du + int_(0)^(a) \ f(x) \ dx`

And using another property of integrals (corollary of of the earlier property)

`int_a^b \ f(x) \ dx = - int_b^a \ f(x) \ dx`

We can write

`int_(-a)^(a) \ f(x) \ dx = -int_(0)^(a) \ f(u) \ du + int_(0)^(a) \ f(x) \ dx`

And finally noting that definite integral are independent of the integration variable that is :

`int_a^b \ f(x) \ dx = int_a^b \ f(u) \ du`

we can write:

`int_(-a)^(a) \ f(x) \ dx = -int_(0)^(a) \ f(x) \ dx + int_(0)^(a) \ f(x) \ dx`
`" " = 0 \ \ \ ` QED