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

##### 1 Answer

We start by checking for x-intercepts within the given interval.

#f(x) = cosxsinx -> f(x) = 1/2sin2x#

If we set

#0 = 1/2sin2x#

#0 = sin2x#

#2x = 2pi or 2x = pi#

#x = pi or x = pi/2#

Since trig functions repeat, and the period of this function is

graph{sinxcosx [-4.5, 15.5, -5.16, 4.84]}

This means that there are

Our expression for area will be

#A = 3(int_0^(pi/2) 1/2sin2x dx- int_(pi/2)^pi 1/2sin2x dx)#

#A = 3([-cos2x]_ 0^(pi/2) - [-cos2x]_(pi/2)^pi)#

#A = 3(-cospi - cos0 - (-cos2pi - cospi))#

#A = 3(1 - 1 + 1 - 1)#

#A = 0#

A geometric view of this gives sense to this result: whenever a positive area occurs, it is cancelled out by the negative area, giving a final result of

Hopefully this helps!