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(D) The given curves are $y^2-2y=-x$ and $x+y=0$.From the second equation,$x=-y$.Substituting this into the first equation: $y^2-2y=-(-y) \Rightarrow

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$36$

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(D) The given curves are $y^2-2y=-x$ and $x+y=0$.From the second equation,$x=-y$.Substituting this into the first equation: $y^2-2y=-(-y) \Rightarrow y^2-2y=y \Rightarrow y^2-3y=0$.Thus,$y(y-3)=0$,which gives $y=0$ and $y=3$.When $y=0$,$x=0$. When $y=3$,$x=-3$.The area $A$ is given by the integral of the difference between the curves with respect to $y$:$A = \int_{0}^{3} (x_{	ext{right}} - x_{	ext{left}}) dy = \int_{0}^{3} (-y^2+2y - (-y)) dy = \int_{0}^{3} (-y^2+3y) dy$.Evaluating the integral:$A = \left[ -\frac{y^3}{3} + \frac{3y^2}{2} ight]_{0}^{3} = \left( -\frac{27}{3} + \frac{3(9)}{2} ight) - 0 = -9 + 13.5 = 4.5 = \frac{9}{2}$.Therefore,$8A = 8 	imes \frac{9}{2} = 36$.

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

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