Question

求定积分:∫（上标是(π/2)，下标是0）|sinx-cosx|dx=

Answer 1

∫（0->π/2）|sinx-cosx|dx
=∫（0->π/4)|sinx-cosx|dx +∫（π/4->π/2)|sinx-cosx|dx
= ∫（0->π/4)(cosx-sinx)dx+∫（π/4->π/2)(sinx-cosx)dx
= ∫（0->π/4)cosxdx- ∫（0->π/4)sinxdx+∫（π/4->π/2)sinxdx-∫（π/4->π/2)cosxdx
=sinx|（0->π/4)+cosx|（0->π/4)-cosx|（π/4->π/2)-sinx|（π/4->π/2)
=(√2/2-0)+(√2/2-1)-(0-√2/2)-(1-√2/2)
=√2/2+√2/2-1+√2/2-1+√2/2
=2√2-2

Answer 2

0≤x≤π/4时,cosx≥sinx,π/4≤x≤π/2时,cosx≤sinx
∫(π/2,0) |sinx-cosx| dx
=∫(π/4,0) (cosx-sinx)dx-∫(π/2,π/4) (cosx-sinx)dx
=∫(π/4,0) cosx dx-∫(π/4,0) sinx dx-∫(π/2,π/4) cosx dx+∫(π/2,π/4) sinx dx
=sinx|(π/4,0) -cosx|(π/4,0)-sinx|(π/2,π/4)-cosx|(π/2,π/4)