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

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

The upper branch of \(y^2 = x\) is \(y = \sqrt{x}\) and the lower branch is \(y = -\sqrt{x}\). From the figure, the shaded region has two parts:

    from x = 0 to x = 2, between the two branches (above and below the x-axis);
from x = 2 to x = 3, between the upper branch and the x-axis only.

First part:

    \( \displaystyle A_1 = \int_{0}^{2} \left(\sqrt{x} - (-\sqrt{x})\right) dx = 2\int_{0}^{2} x^{1/2}\,dx \)

    \( \displaystyle = 2\left[\frac{2}{3}x^{3/2}\right]_{0}^{2} = \frac{4}{3} \times 2\sqrt{2} = \frac{8\sqrt{2}}{3} \)

Second part:

    \( \displaystyle A_2 = \int_{2}^{3} \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_{2}^{3} \)

    \( \displaystyle = \frac{2}{3}\left(3\sqrt{3} - 2\sqrt{2}\right) = 2\sqrt{3} - \frac{4\sqrt{2}}{3} \)

Total area:

    \( \displaystyle A_1 + A_2 = \frac{8\sqrt{2}}{3} + 2\sqrt{3} - \frac{4\sqrt{2}}{3} = \frac{4\sqrt{2}}{3} + 2\sqrt{3} \)