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

The average value is

This can be seen in two ways: in length and in width.

I'll outline the quick fix.

Theorem

Proof:

Consequently,

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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).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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