∫∣x-2∣dx(区间为1-3)

分段
1<x<2
|x-2|=2-x
2<x<3,|x-2|=x-2
所以原式=∫(1-2)(2-x)dx+∫(2-3)(x-2)dx
=(2x-x^2/2) (1-2) +(x^2/2-2x) (2-3)
=1/2+1/2
=1

当x属于(-3,1)
x-2<0

积分:(1,-3)|x-2|dx
=积分:(1,-3)(2-x)dx
=(2x-1/2x^2)|(1,-3)
=2-1/2+6+9/2
=12

4