I=∫03∣x2−3x+2∣dx
=∫03∣(x−1)(x−2)∣dx
=∫01(x2−3x+2)dx−∫12(x2−3x+2)dx+∫23(x2−3x+2)dx
=[3x3−3(2x2)+2x]01−[3x3−3(2x2)+2x]12+[3x3−3(2x2)+2x]23
=(31−23+2)−[(38−6+4)−(31−23+2)]+[(9−227+6)−(38−6+4)]
=(32−3+4)−(316−12+8)+(9−227+6)
=20−6109=611
So,
12I=12×611=22