What is the average value of the function #h(x) = cos^4 x sin x# on the interval #[0,pi]#?

Question

Cameron Alexander · Accepted Answer

#2/(5pi)# The average value of #cos^4x sinx# on #[0,pi]# is #1/(pi-1) int_0^pi cos^4x sinx dx# Use #u = cosx# to integrate and get # - 1/pi [cosx/5]_0^pi = 2/(5pi)#

Aurora Alvarado · Answer

undefined To find the average value of the function $ h(x) = \cos^4(x) \sin(x) $ on the interval $[0, \pi]$, we first need to calculate the definite integral of $ h(x) $ over the interval $[0, \pi]$, and then divide the result by the length of the interval.

The definite integral of $ h(x) $ over the interval $[0, \pi]$ is given by:

$$
\int_{0}^{\pi} \cos^4(x) \sin(x) \, dx
$$

Using trigonometric identities, we can rewrite $ \cos^4(x) $ in terms of $ \sin(2x) $:

$$
\cos^4(x) = \left( \frac{1 + \cos(2x)}{2} \right)^2 = \frac{1}{4} (1 + 2\cos(2x) + \cos^2(2x))
$$

Then, we can rewrite $ \cos^2(2x) $ in terms of $ \sin(4x) $:

$$
\cos^2(2x) = \frac{1 + \cos(4x)}{2} = \frac{1}{2} + \frac{1}{2}\cos(4x)
$$

Substituting these expressions back into $ h(x) $, we get:

$$
h(x) = \frac{1}{4} \left( \sin(x) + 2\sin(x)\cos(2x) + \sin(x)\cos(4x) \right)
$$

Now, we can integrate $ h(x) $ over the interval $[0, \pi]$:

$$
\begin{aligned}
\int_{0}^{\pi} h(x) \, dx &= \frac{1}{4} \left( \int_{0}^{\pi} \sin(x) \, dx + 2\int_{0}^{\pi} \sin(x)\cos(2x) \, dx + \int_{0}^{\pi} \sin(x)\cos(4x) \, dx \right) \
&= \frac{1}{4} \left( -\cos(x) \Big|_{0}^{\pi} + 2\int_{0}^{\pi} \sin(x)\cos(2x) \, dx + 0 \right) \
&= \frac{1}{4} \left( -(-1) - 0 + 0 \right) \
&= \frac{1}{4} (1) \
&= \frac{1}{4}
\end{aligned}
$$

Therefore, the average value of the function $ h(x) $ on the interval $[0, \pi]$ is $ \frac{1}{4} $.