Q. The area (in sq. units) of the region $A = \left\{\left(x,y\right) : \frac{y^{2}}{2} \le x \le y+4\right\} $ is :
Solution:
$y^{2} = 2x $
$ x-y-4 = 0 $
$ \left(x-4\right)^{2} = 2x $
$ x^{2} +16-8x-2x=0 $
$ x^{2} -10x+16=0 $
$ x=8,2 $
$ y =4,-2 $
$ A = \int^{4}_{-2} \left(y+4 - \frac{y^{2}}{2}\right)dy $
$ = \frac{y^{2}}{2} \bigg|^4_{-2} + 4 y \bigg|^4_{-2} - \frac{y^3}{6} \bigg|^4_{-2} $
$ = ( 8 - 2) + 4 (6) - \frac{1}{6} (64 + 8)$
$ 6+ 24 - 12 = 18 $
