The area lying between the curves y² = 4x and y = 2x is :

$\frac{1}{3}$

y² = 4x  ,    y = 2x

point of intersection

y = 2x        y² = 4x

⇒ (2x)² = 4x

⇒ x² = x

⇒ x = 1, 0

area = area under y² = 4x - area under y = 2x

so area = $\int\limits_0^1-2 x+2 \sqrt{x} d x$

$=\left[-x^2+\frac{4}{3} x^{3 / 2}\right]_0^1=\frac{4}{3}-1$

$= \frac{1}{3}$ sq. units