yay, integrals
so since the 2nd equation is in terms of y, we differntiate with respect to y
so we see that they intersect at y=0 and y=3
so
y=x is on top so
[tex] \int\limits^3_0 {y-(y^2-2y)} \, dy [/tex]
[tex] \int\limits^3_0 {y-y^2+2y)} \, dy [/tex]
[tex] \int\limits^3_0 {y^2+3y} \, dy [/tex]
[tex][ \frac{1}{3}y^3+ \frac{3}{2}y^2]^3_0 [/tex]
9/2 is the area
