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

Answer 1

Ezra Adams

The answer is #=3/8pi=1.178u^2#

We start, by simplifying

#sin2x=2sinxcosx#

and

#cos2x=1-2sin^2x#

#sin^2x=(1-cos2x)/2#

#sin^2 2x=1/2(1-cos4x)#

Therefore,

#f(x)=cos^2xsin^2x#

#=1/4sin^2 2x#

#=1/4*1/2*(1-cos4x)#

#=1/8(1-cos4x)#

The area is

#=1/8int_0^(3pi)(1-cos4x)dx#

#=1/8[x-1/4sin4x]_0^(3pi)#

#=1/8(3pi-1/4sin12pi-0-0)#

#=3/8pi=1.178#

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Answer 2

Elijah Abram

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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Answer from HIX Tutor

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.