Question

What is the average value of the function `F(t)=Sin^3(t)cos^3(t)` on the interval `[-pi, pi]`?

Answer 1

The average value is `0`.

Explanation:

There is a long way and a short way to see this.

I'll show the short way.

`F(t) = sin^3(t)cos^3(t)` is an odd function.

Therefore, for any `a` in the domain, `int_-a^a F(t) dt = 0`

Consequently, the average value on `[-pi,pi]` is

`1/(pi-(-pi)) int_-pi^pi F(t) dt = 0/(2pi) = 0`

Theorem

If `f` is an odd function that is integrable on `[-a,a]`, then `int_-a^a f(x) dx = 0`

Proof:

Let `f` be an odd function, integrable on `[-a,a]`.

`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`

Substitute `u=-x` so `x=-u`, dx = -du`and when`x=-a`,`u = a#. The integral above becomes:

`= -int_0^a f(-u) * (-du) = int_0^a f(-u) du`

But `f` is odd, so `f(-u) = -f(u)` and so we have:

`int_-a^0 f(x) dx = int_0^a -f(u) du = -int_0^a f(x) dx`.

Therefore,

`int_-a^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`

`= 0` `" "` `" "` Q.E.D.