Application of Integrals: Area Between Curves (Parabola & Line)

Substitute $y = 2x$ into $y^2 = 4x$: $$(2x)^2 = 4x \implies 4x^2 = 4x \implies 4x(x - 1) = 0 \implies x = 0 \text{ or } x = 1$$ Corresponding $y$-values: $y = 0$ when $x = 0$, and $y = 2$ when $x = 1$. Intersection points are $(0, 0)$ and $(1, 2)$.

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

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