Application of Integrals: Area Bounded Between Parabola and Line

Substitute $y = 2x$ into $y^2 = 4x$: $(2x)^2 = 4x \implies 4x^2 = 4x \implies 4x(x - 1) = 0$ Thus $x = 0$ or $x = 1$. The points of intersection are $(0, 0)$ and $(1, 2)$.

For $x \in [0, 1]$, the curve $y = 2\sqrt{x}$ lies above the line $y = 2x$: $\text{Area} = \int_{0}^{1} (2\sqrt{x} - 2x) \, dx$

$\text{Area} = 2 \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{1} = 2 \left( \frac{2}{3}(1) - \frac{1}{2}(1) \right) = 2 \left( \frac{4 - 3}{6} \right) = 2 \left( \frac{1}{6} \right) = \frac{1}{3} \text{ sq. units}$