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求定积分:∫(上标是(π/2),下标是0)|sinx-cosx|dx=
人气:232 ℃ 时间:2019-12-08 19:23:04
解答
∫(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
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