Answer
$$9$$
Work Step by Step
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:
$$= \int^3_0(-2y^2 + 6y)dy$$
$$= (-\frac{2}{3}y^3 + 3y^2)|^3_0$$
$$= [-\frac{2}{3}(3)^3 + 3(3)^2] - [-\frac{2}{3}(0)^3 + 3(0)^2]$$
$$= -18 + 27 + 0 + 0$$
$$= 9$$
