设f(x)=sinx,x∈[0,π/2) f(x)= 1,x∈[π/2,2].则f(x)dx在0到2上的积分为
人气:161 ℃ 时间:2020-05-31 04:42:11
解答
∫上2下0;f(x)dx
=∫上2下π/2;f(x)dx+∫上π/2下0;f(x)dx
=∫上2下π/2;1dx+∫上π/2下0;sinxdx
=x│上2下π/2-cosx│上π/2下0
=(2-π/2)-(0-1)
=3-π/2
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