The area of the region bounded by the curve $y^2=x$ and the lines $x=1, x=4 $ is :

$\frac{7}{3}sq.units$ $\frac{28}{3}sq.units$ $\frac{14}{3}sq.units$ $\frac{35}{3}sq.units $

$\frac{28}{3}sq.units$

The correct answer is Option (2) → $\frac{28}{3}sq.units$ By symmetry Area (I) = Area (II) ⇒ Area = $2×\int\limits_1^4\sqrt{x}dx$ $=\left[2×\frac{2}{3}x^{\frac{3}{2}}\right]_1^4$ $=\frac{4}{3}(8-1)=\frac{28}{3}sq.units$