带绝对值的定积分怎么求－－－－－讨论,把绝对号去掉.
∫(0,2)︱x-1︱dx=∫(0,1)︱x-1︱dx+∫(1,2)︱x-1︱dx=∫(0,1)(-x+1)dx+∫(1,2)(x-1)dx
=[(-1/2)x^2+x](0,1)+[(1/2x^2-x](1,2)=1/2-(-1/2)=1