Shown below is the graph of parabola $y^2=x$, the area (in sq. units) of the shaded region is:

$\frac{4\sqrt{2}}{3}+2\sqrt{3}$

The correct answer is Option (2) → $\frac{4\sqrt{2}}{3}+2\sqrt{3}$

Shaded region = area inside parabola $y^2=x$ from $x=0$ to $x=2$ (both above and below $x$-axis) plus area under upper branch from $x=2$ to $x=3$.

Area$_1$ (for $0\le x\le2$): $ \displaystyle \int_{0}^{2} \big(\sqrt{x}-(-\sqrt{x})\big)\,dx = 2\int_{0}^{2} x^{1/2}\,dx = 2\left[\frac{2}{3}x^{3/2}\right]_0^2 = \frac{4}{3}\cdot 2^{3/2} = \frac{8\sqrt{2}}{3}$.

Area$_2$ (for $2\le x\le3$, above $x$-axis): $ \displaystyle \int_{2}^{3} \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}\big(3^{3/2}-2^{3/2}\big) = \frac{2}{3}\big(3\sqrt{3}-2\sqrt{2}\big)$.

Total area $=$ Area$_1$ + Area$_2$

$\displaystyle = \frac{8\sqrt{2}}{3} + \frac{2}{3}\big(3\sqrt{3}-2\sqrt{2}\big) = \frac{2}{3}\big(2\sqrt{2}+3\sqrt{3}\big) = \frac{4\sqrt{2}+6\sqrt{3}}{3}$.

Required area = $\displaystyle \frac{4\sqrt{2}+6\sqrt{3}}{3}$