#absu = {(u,"if",u >= 0),(-u,"if",u < 0) :}#
#abs(2x-3) = {(2x-3,"if",2x-3 >= 0),(-(2x-3),"if",2x-3 < 0) :}#
Solving the inequalities, we have
#abs(2x-3) = {(2x-3,"if",x >= 3/2),(-(2x-3),"if",x < 3/2) :}#
Because #3/2# is in the interval over which we are integrating, we must split the integral.
#int_0^3 abs(2x-3) dx = int_0^(3/2) -(2x-3) dx + int_(3/2)^3 (2x-3) dx#
# = - int_0^(3/2)(2x-3) dx + int_(3/2)^3 (2x-3) dx#
# = -(-9/4)+9/4 = 9/2#
Note
Once you've seen why, your grader may permit you to use:
#int_0^3 abs(2x-3) dx = abs(int_0^(3/2) (2x-3) dx) +abs( int_(3/2)^3 (2x-3) dx)#
Essentially, integrate between zeros of the argument and make any negative answer positive. Then add up the answers.
So, for example
#int_0^5 abs(x^2-4x+3) dx#
The zeros of #x^2-4x+3# are #1# and #3#, so we'll integrate each of
#int_0^1 (x^2-4x+3) dx#, #int_1^3 (x^2-4x+3) dx#, and #int_3^5 (x^2-4x+3) dx#
Change any negative answer to positives, then add the three positive numbers.
To write it correctly we write
#int_0^5 abs(x^2-4x+3) dx = abs(int_0^1 (x^2-4x+3) dx) + abs(int_1^3 (x^2-4x+3) dx) + abs(int_3^5 (x^2-4x+3) dx)#
(The middle one will be negative.)
