The question reads:
Evaluate the area of the region bounded by y=sec^2x and y = 4 between x = -pi/3 and x = pi/3.
Thank You!
$\displaystyle \int_{-\pi/3}^{\pi/3}4-\sec^2(x)dx=4x-\tan(x) \bigg|_{-\pi/3}^{\pi/3}=4\left( \frac{\pi}{3}\right) -\tan\left( \frac{\pi}{3}\right)-4\left( \frac{-\pi}{3}\right)+\tan\left( \frac{-\pi}{3}\right)= $
$\displaystyle 8\left( \frac{\pi}{3}\right)-2\tan\left( \frac{\pi}{3}\right) \approx 4.91$