The area of the region bounded by $x^2=4y$ and $x=4y -2 $ (in square units ) is :

$\frac{9}{8}$

The correct answer is Option (4) → $\frac{9}{8}$

$x^2=4y$ and $x=4y -2 $

finding intersection

as $x^2=4y$

$x+2=4y$

$x^2=x+2$

$x=-1,2$

so area = $\int\limits_{-1}^2\frac{x+2}{4}-\frac{x^2}{4}dx$

$=\frac{1}{4}\int\limits_{-1}^2x+2-x^2dx=\frac{1}{4}\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^2$

$=\frac{9}{8}$ sq. units