Answer
$6\sqrt{3}$
Work Step by Step
Over the interval $x\displaystyle \in[-\frac{\pi}{3},\frac{\pi}{3}]$,
the graph of $f(x)=8\cos x$ is above the graph of $g(x)=\sec^{2}x$.
$A=\displaystyle \int_{a}^{b}(y_{above}-y_{below})dx=\int_{-\pi/3}^{\pi/3}[8\cos x-\sec^{2}x]dx$.
$=\left[8\sin x -\tan x \right]_{-\pi/3}^{\pi/3}$
$=8(\displaystyle \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2})-(\sqrt{3}+\sqrt{3})$
$=6\sqrt{3}$
