This isn't too hard to puzzle out but it will help you to graph the 3 functions involved.
On (-∞,1] ∪ [2,∞) y = 1
On [1,3/2] y = -2x + 3
and
on [3/2,2] y = 2x-3
The integral will, therefore, need to be defined on the 4 intervals defined above.
For example on (-∞,1} the integral is x.
On [1,3/2] the integral is x2- 3x.
If you need to define the interval across the division points, you need to add the values for the integral.

Paul M.
12/07/18
Jeb S.
how did you get the inverval [1,3/2], im only getting (-∞,1], [2,∞), and (1,2)12/07/18