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

1 Answer
Apr 15, 2016

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.