The area of the region bounded by $x^2=4y$ and $x=4y -2 $ (in square units ) is :
Answer & explanation
Correct answer: option 4
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