The area (in sq. units) of the region bounded by the curve y = x2 and the line y = 4 is

$\frac{16}{3}$ $\frac{32}{3}$ 32 24

$\frac{32}{3}$

Symmetrical about y axis so  A = $2 × \int\limits_0^4 \sqrt{y} dy$ $=2 \times \frac{3}{3}\left[y^{3 / 2}\right]_0^4$ $=\frac{32}{3}$ sq. units Option: B