请问一下这个题目怎么做?高二的数学题。
3个回答
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∫(0->π)|sinx-cosx|dx
=∫(0->π/4)|sinx-cosx|dx +∫(π/4->π )|sinx-cosx|dx
= ∫(0->π/4)(cosx-sinx)dx+∫(π/4->π)(sinx-cosx)dx
= ∫(0->π/4)cosxdx- ∫(0->π/4)sinxdx+∫(π/4->π)sinxdx-∫(π/4->π)cosxdx
=sinx|(0->π/4)+cosx|(0->π/4)-cosx|(π/4->π)-sinx|(π/4->π)
=(√2/2-0)+(√2/2-1)-(-1-√2/2)-(0-√2/2)
=√2/2+√2/2-1+√2/2+1+√2/2
=2√2
=∫(0->π/4)|sinx-cosx|dx +∫(π/4->π )|sinx-cosx|dx
= ∫(0->π/4)(cosx-sinx)dx+∫(π/4->π)(sinx-cosx)dx
= ∫(0->π/4)cosxdx- ∫(0->π/4)sinxdx+∫(π/4->π)sinxdx-∫(π/4->π)cosxdx
=sinx|(0->π/4)+cosx|(0->π/4)-cosx|(π/4->π)-sinx|(π/4->π)
=(√2/2-0)+(√2/2-1)-(-1-√2/2)-(0-√2/2)
=√2/2+√2/2-1+√2/2+1+√2/2
=2√2
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