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求文档: 求曲线y=sinx,y=cosx与直线x=-π/4,x=π/4所围成图形的面积
人气:475 ℃ 时间:2019-10-25 10:06:33
解答
当x∈[-π/4,π/4]时,有cosx>sinx
∴A=∫(cosx-sinx)dx 积分限为[-π/4,π/4]
=sinx+cosx
=[sin(π/4)+cos(π/4)]-[sin(-π/4)+cos(-π/4)]
=[√2/2+√2/2]-[-√2/2+√2/2]
=√2
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