求定积分∫(上限为π/2.下限为0)|1/2-sin x| dx
问题描述：
求定积分∫(上限为π/2.下限为0)|1/2-sin x| dx
答
把区间分为(0,π/6),(π/6,π/2)
∫(0,π/2)|(1/2)-sinx| dx
=∫(0,π/6)[(1/2)-sinx]dx+∫(π/6,π/2)[sinx-(1/2)]dx
=[(x/2)+cosx]|(0,π/6)+[-cosx-(x/2)]|(π/6,π/2)
=(√3)-1-π/12