What is the area of the region between the parabola $y^{2} = 4x$ and the line $y = 2x$?

Explanation

1. 01

   $4x^{2} = 4x$ gives $x = 0, 1$; between them $y = 2\sqrt{x}$ lies above $y = 2x$.

2. 02

   $\int_0^{1} (2\sqrt{x} - 2x)\,dx = \frac{4}{3} - 1 = \frac{1}{3}$ square units.