Evaluate the integral from -1 to 2: |x-x^2|?

Question

i got 3/2 as answer but apparently its not correct. Thanks!

Answer 1

∫|x-x^2| dx, from -1 to 2
= -∫x-x^2 dx, from -1 to 0 + ∫x-x^2 dx, from 0 to 1 - ∫x-x^2 dx, from 1 to 2
= 1 - (1/2)(2^2-1) + (1/3)(2^3-1)
= 11/6

Answer 2

integral from x is (x^2)/2 and from x^2 is (x^3)/3 so solving the problem ure result will be: -5/6

Answer 3

sign of x(1-x) is --------0+++++++1 ---------
so divide the integral
Int(-1,0) x^2-x dx + Int (0,1)x-x^2 dx +Int(1,2) x^2-x dx
= -1/3+1/2 +1/2-1/3 +8/3-2 -1/3+1/2 = -1+3/2+2/3 = 7/6