Answer
$\frac{5}{2}$
Work Step by Step
Step 1. Using the figure given in the Exercise, find the intersections between the functions. We can get point $(0,1)$ for the y-intercept of the red curve, point $(1,2$ between the red and green curves, and point $(2,1)$ between the green and blue curves.
Step 2. The enclosed area can be written as the sum of two integrals with respect to $y$:
$A=\int_{0}^1 (2\sqrt y-0)dy + \int_1^2 (3-y-(y-1)^2)dy =\frac{4}{3}y^{3/2}|_{0}^1+3y|_{1}^2 - \frac{1}{2}y^2|_{1}^2 - \frac{1}{3}(y-1)^3|_{1}^2=\frac{4}{3}+(6-3) -(2-\frac{1}{2}) - \frac{1}{3}=\frac{5}{2}$
