The area (in square units) enclosed between the curve $x^2-4y$ and the line $x = y$ is:

$\frac{8}{3}$

The correct answer is Option (4) → $\frac{8}{3}$

Given,

$x^2-4y=0$   ...(1)

$x=y$   ....(2)

from (1) and (2),

$x^2-4x=0$

$x(x-4)=0$

$x=0$ or $x=4$

So, the points of intersection are (0, 0) and (4, 4).

$A=\int\limits_0^4\left(x-\frac{x^2}{4}\right)dx$

$=\int\limits_0^4x\,dx-\int\limits_0^4\frac{x^2}{2}dx$

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