When you are asked to find the area under the curve with a given equation, this is an application of integral calculus. The concept is that, any infinitesimal strip under the curve, when added together, equals the area. Thus, integrate the given equation with limits from 0 to 2.

A = [x³/3 - 4x]lim 0->2
A = (2³/3 - 4(2)] - (0³/3 - 4(0)]
A = 16/3 - 0
A = 16/3 ≈
5.3 sq. units