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

Question

Christopher Allers · Accepted Answer

#2/3# areal units.

Net area = int f(x) dx, between the limits #0 and 3pi# #=int cos^2x sin x dx#, between the limits #0 and 3pi# #=-int cos^2x d(cos x)#, between the x-limits #0 and 3pi# #=-[(cos x)^3]#, between the limits #0 and 3pi# #=-((cos(3pi))^3-(cos 0)^3)# #=-((-1)^3-1)# #=2/3# Note that f is periodic with period #2pi# and, Interestingly, the net periodic area ( here up to #x = 2pi# ) =#2/3-2/3=0#..

Luke Alvarez · Answer

undefined To find the net area between $ f(x) = \cos^2(x)\sin(x) $ and the x-axis over $ x $ in $ [0, 3\pi] $, you need to integrate $ f(x) $ from $ x = 0 $ to $ x = 3\pi $ and then take the absolute value of the result. This is because the function $ f(x) $ dips below the x-axis within the interval, which would otherwise result in negative area. So:

$$ \text{Net area} = \left| \int_{0}^{3\pi} \cos^2(x)\sin(x) \, dx \right| $$