## Calculus: Early Transcendentals 8th Edition

$$9$$
Area = $\int^d_c f(y) - g(y) dy$ For this question, $f(y) = 2y-y^2$ and $g(x) = y^2 - 4y$ The limits of integration are the y-coords where the two graphs intercept at $(-3,3)$ and $(0,0)$. Therefore, $c = 0$ and $d = 3$ $$\int^d_c f(y) - g(y) dy$$ $$= \int^3_0(2y-y^2) - (y^2 - 4y) dy$$ Solve the definite integral, plug in the limits of integration, and simplify to get the answer: $$= \int^3_0(-2y^2 + 6y)dy$$ $$= (-\frac{2}{3}y^3 + 3y^2)|^3_0$$ $$= [-\frac{2}{3}(3)^3 + 3(3)^2] - [-\frac{2}{3}(0)^3 + 3(0)^2]$$ $$= -18 + 27 + 0 + 0$$ $$= 9$$