Question

求定积分的结果和过程

Answer 1

在 x∈[0,π/6) 时 sinx<1/2 ，有 1/2-sinx>0 ,|1/2-sinx|=1/2-sinx
在 x∈[π/6,π/2]时 sinx>=1/2,有 1/2-sinx<=0 ,|1/2-sinx|=sinx-1/2
所以
∫[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]
=π/12+√3/2-(0+1)+(-0-π/4)-(-√3/2-π/12)
=π/12-1+√3/2-π/4+√3/2+π/12
=√3-1-π/12

Answer 2

原式=∫(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)
=(π/12+√3/2)-(0+1)+(-0-π/4)-(-√3/2-π/12)
=π/12+√3/2-1-π/4+√3/2+π/12
=-π/12+√3-1