题目

求定积分:∫（上标是(π/2),下标是0）|sinx-cosx|dx=

提问时间：2020-10-09

答案

∫（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