Find the area of the region enclosed by the curves x^3-2x^2 and 2x^2-3x.

If you want the geometric area, then since the curves intersect at x=0,1,3 you need

∫[0,3] |(x^3-2x^2)-(2x^2-3x)| dx
= ∫[0,1] (x^3-2x^2)-(2x^2-3x) dx + ∫[1,3] (2x^2-3x)-(x^3-2x^2) dx = 37/12

Otherwise, if you want the algebraic area, that would be just
∫[0,3] ((x^3-2x^2)-(2x^2-3x)) dx = -9/4

They are equal at x = 0, x = 1, x = 3

so do your integral from 0 to 1 and from 1 to 3 and add the results
diff = x^3 - 2 x^2 -2 x^2 + 3 x = x^3 - 4 x^2 + 3 x
integral = x^4/4 - 4/3 x^3 + 3/2 x^2
at 1 that is 1/4 -4/3+ 3/2 = 3/12 - 16/12 + 18/12 = 5/12
at 0 it is 0
so from 0 to 1 the area is 5/12
now do from 1 to 3 and add the result