The area of the region enclosed between the curves $4 x^2=y$ and $y=4$ is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{16}{3}$ sq. units
$4 x^2=y$, $y=4$
at $y=4$, $4 x^2=4$
$x=±1$
area required = $4×2-\int\limits_{-1}^14x^2dx$
$=4×2-2\int\limits_0^14x^2dx$
$=8-8\left[\frac{x^3}{3}\right]_0^1$
$=\frac{16}{3}$ sq. units