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

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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} ).

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