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

问题描述:

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

∫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