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
Answer & explanation
Correct answer: option 2
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.