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

Question

Evelyn Agard · Accepted Answer

The average value is #0#.

This can be seen in two ways: in length and in width. I'll outline the quick fix. #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#. Consequently, #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.

Ezekiel Alexander · Answer

undefined To find the average value of the function $ F(t) = \sin^3(t) \cos^3(t) $ on the interval $[- \pi, \pi]$, we need to compute the definite integral of the function over that interval and then divide by the length of the interval. The formula for the average value of a function $ f(x) $ on the interval $[a, b]$ is given by:

$$ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx $$

So, for $ F(t) = \sin^3(t) \cos^3(t) $ on the interval $[- \pi, \pi]$, we have:

$$ \text{Average value} = \frac{1}{\pi - (-\pi)} \int_{-\pi}^{\pi} \sin^3(t) \cos^3(t) \, dt $$

$$ = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \sin^3(t) \cos^3(t) \, dt $$

$$ = \frac{1}{2 \pi} \left[ \frac{1}{6} \sin^6(t) \right]_{-\pi}^{\pi} $$

$$ = \frac{1}{12 \pi} \left( \sin^6(\pi) - \sin^6(-\pi) \right) $$

$$ = \frac{1}{12 \pi} \left( 0 - 0 \right) $$

$$ = 0 $$

Therefore, the average value of the function $ F(t) = \sin^3(t) \cos^3(t) $ on the interval $[- \pi, \pi]$ is $0$.