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

Given curves are 
${\mathrm{y}}^2-2\mathrm{y}=-\mathrm{x}$
And $y=-x$
From (1) & (2) $\Rightarrow {\mathrm{y}}^2-2\mathrm{y}=\mathrm{y}$
$\Rightarrow {\mathrm{y}}^2-3\mathrm{y}=0\Rightarrow \mathrm{y}=0,\mathrm{y}=3$
Point of intersections are A (0, 0) B (-3, 3)
Required area $\mathrm{A}={\int }_0^3 \left[2\mathrm{y}-{\mathrm{y}}^2-\left(\mathrm{y}\right)\right]\mathrm{dy}$

$={\int }_0^3 (3\mathrm{y}-\mathrm{y}{)}^2\mathrm{dy}={\left(\frac{3{\mathrm{y}}^2}{2}-\frac{{\mathrm{y}}^3}{3}\right)}_0^3=\frac{9}{2}\text{ sq.units }$