含绝对值的函数的定积分,

解
∵x-3≥0时,/x-3/=x-3
∴x≥3
∵x-3≤0时,/x-3/=-(x-3)=3-x
∴0≤x≤3
∴∫(4.0)|x-3|dx
=∫(0,3)(3-x)dx+∫(3,4)(x-3)dx
=3x-1/2x²|(0,3)+1/2x²-3x|(3,4)
=(9-9/2)-(0-0)+(8-12)-(9/2-9)
=(9-4)+(-9/2-9/2+9)=5