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

$\frac{4}{3}$ sq units

The correct answer is Option (1) → $\frac{4}{3}$ sq units

We have to find the area enclosed by parabola $y^2 = x$ and the straight line $2y = x$.

Solving the above equations, we get

$∴\left( \frac{x}{2} \right)^2 = x$

$\Rightarrow x^2 = 4x \Rightarrow x(x - 4) = 0 \Rightarrow x = 0, 4$

When, $x = 4 \Rightarrow y = 2$ and when $x = 0 \Rightarrow y = 0$

So, the intersection points are $(0, 0)$ and $(4, 2)$.

$\text{Area enclosed by shaded region} = \int_{0}^{4} \left[ \sqrt{x} - \frac{x}{2} \right] dx$

$= \left[ \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{1}{2} \cdot \frac{x^2}{2} \right]_{0}^{4} = \left[ 2 \cdot \frac{x^{3/2}}{3} - \frac{x^2}{4} \right]_{0}^{4}$

$= \frac{2}{3} \cdot 4^{3/2} - \frac{16}{4} = \frac{2}{3} \cdot 8 - 4$

$= \frac{16}{3} - \frac{16}{4} = \frac{64 - 48}{12} = \frac{16}{12} = \frac{4}{3} \text{ sq. units}$