The area of the region bounded by the parabola $y^2 = x$ and the straight line $2y = x$ is

4/3 sq. units

The correct answer is Option (3) → 4/3 sq. units

Given curves: $y^2 = x$ and $x = 2y$

Substitute $x = 2y$ into $y^2 = x$:

$y^2 = 2y \Rightarrow y^2 - 2y = 0 \Rightarrow y(y - 2) = 0$

Points of intersection: $y = 0$ and $y = 2$

Area between curves = $\int_{0}^{2} (2y - y^2) \, dy$

$= \int_{0}^{2} 2y \, dy - \int_{0}^{2} y^2 \, dy$

$= \left[ y^2 \right]_0^2 - \left[ \frac{y^3}{3} \right]_0^2$

$= 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3}$