The area of the region enclosed between $y^2=4x$ and $x=1$ and $x=3$ in the first quadrant is :

$\frac{4}{3}(3\sqrt{3}-1)$

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

By symmetry

I = II

required area = I (first quardrant)

$=\int\limits_1^22\sqrt{x}dx$

$=\frac{4}{3}\left[x^{\frac{3}{2}}\right]_1^3$

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