What is the net area between #f(x) = x-4cos^2x # and the x-axis over #x in [0, 3pi ]#?

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Answer 1 · 2016-11-19T13:15:35

#=25.564# areal units, nearly

In #[0, 3pi]#, the periodic curve extends,

from #(0, -4) to (9.425, 5.425)#. #3pi=9.425 units, nearly.

The graph has been zoomed to include the part of the curve, under

reference.

The area #= int (x-4cos^2x)dx#, from #x = 0 to x = 3pi#

#= int (x-2(1+cos 2x)) dx#, for the limits

#= [x^2/2-2x-2(sin 2x)/2], between the limits

#=[9/2(pi)^2-6pi-0]#

#=25.564# areal units, nearly.

Note that there are two negative parts for the area, between x-axis

and the curve, that add up to the total
graph{x-4cos(x)cos(x) [-40, 40, -20, 20]}