The area (in square units) of the region bounded by the curve $x^2 = y$ and the straight line $y = 4$ in the first quadrant is equal to

$\frac{16}{3}$

The correct answer is Option (2) → $\frac{16}{3}$

Curve: $y=x^{2}$; line: $y=4$. In first quadrant intersection at $x^{2}=4\Rightarrow x=2$.

Area $= \displaystyle\int_{0}^{2}\big(4-x^{2}\big)\,dx$

$= \left[4x - \frac{x^{3}}{3}\right]_{0}^{2}$

$= \left(8 - \frac{8}{3}\right) - 0$

$= \frac{24-8}{3} = \frac{16}{3}$

The area is $\frac{16}{3}$ square units.