If the area of the region bounded by the curves  $y^2-2y=-x,x+y=0$ is A, then 8A is equal to ____

Given curves are
 $y^2-2y=-x\mathrm{.............}\left(1\right)$
And y = -x .........(2)
From (1) & (2)  $\Rightarrow y^2-2y=y$
 $\Rightarrow y^2-3y=0\Rightarrow y=0,y=3$
Point of intersections are A (0, 0)  B (-3, 3)
Required area  $A=\underset{0}{\overset{3}{\int }}[2y-y^2-(y\left)\right]dy$

$=\underset{0}{\overset{3}{\int }}{(3y-y)}^{2}dy={(\frac{3y^2}{2}-\frac{y^3}{3})}_0^3=\frac{9}{2}sq.units$