Answer
$Area = 0$
Work Step by Step
$f(x) = cos^2~x~sin~x$
Under the graph on the interval $0 \leq x \leq 2\pi$, we can see that there is symmetry about the x-axis. That is, half the area is above the x-axis and half the area is below the x-axis. Therefore, the area is 0.
We can find the exact area $A$ under the graph:
$A = \int_{0}^{2\pi}cos^2~x~sin~x~dx$
$A = -\frac{1}{3}cos^3~x\Big\vert_{0}^{2\pi}$
$A = -\frac{1}{3}cos^3~(2\pi) - [-\frac{1}{3}cos^3~(0)]$
$A = -\frac{1}{3}(1)^3 +\frac{1}{3}(1)^3$
$A = 0$