Find the area lying between the curve y^2 = 4x and y = 2x.

The correct answer is Option (1) → $\frac{1}{3}$

The area lying between the curve $y^2 = 4x$ and $y = 2x$ is represented by the shaded area $OBAO$ as shown in the figure.

The points of intersection of the curves are $O(0, 0)$ and $A(1, 2)$.

We draw $AC$ perpendicular to the $X$-axis such that the coordinate of $C$ is $(1, 0)$.

$\text{Area of } OBAO = \text{Area of } OCABO - \text{Area of } \Delta OCA$

$A = \int_{0}^{1} 2\sqrt{x} \, dx - \int_{0}^{1} 2x \, dx$

$= 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} - 2 \left[ \frac{x^2}{2} \right]_{0}^{1}$

$= \left[ \frac{4}{3} - 1 \right]$

$= \left| \frac{1}{3} \right|$

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