# What is the net area between #f(x) = cos^2xsin^2x # and the x-axis over #x in [0, 3pi ]#?

The answer is

We start, by simplifying

and

Therefore,

The area is

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To find the net area between ( f(x) = \cos^2(x) \sin^2(x) ) and the x-axis over ( x ) in the interval ( [0, 3\pi] ), we need to compute the definite integral of the absolute value of ( f(x) ) over the given interval.

[ \int_{0}^{3\pi} |f(x)| , dx ]

Using the trigonometric identity ( \sin^2(x) = 1 - \cos^2(x) ), we can rewrite the function ( f(x) ) as:

[ f(x) = \cos^2(x)(1 - \cos^2(x)) ]

Then,

[ |f(x)| = \cos^2(x)(1 - \cos^2(x)) ]

[ \int_{0}^{3\pi} |f(x)| , dx = \int_{0}^{3\pi} \cos^2(x)(1 - \cos^2(x)) , dx ]

Now, you can compute this integral to find the net area.

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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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