第1个回答 2019-06-03
∫<0, 2π>|cosx|dx
= ∫<0, π/2>cosxdx + ∫<π/2, 3π/2> -cosxdx + ∫<π/2, 2π>cosxdx
= [sinx]<0, π/2> - [sinx]<π/2, 3π/2> + [sinx]<3π/2, 2π>
= 1 + 2 + 1 = 4
∫<-1, 1>(x^3+x+1)dx = ∫<-1, 1>(x^3+x)dx + ∫<-1, 1>dx
= 0 + 2 = 2
这类题目不胜枚举。例如:
∫<-π, π>|sinx|dx ,
∫<-π/2, π/2>[|sinx|+>|cosx|]dx
∫<-π, 0>√[2(1+cos2x)]dx
∫<0, π>√[2(1-cos2x)]dx
∫<-1, 1>[xcos2xe^(-x^2)+5]dx
∫<-2, 2>[ln(x+√(x^2+1)]+(3/4)x^2]dx
∫<-π/2, π/2>[cosx+e^(-x^2)sinxln(1+x^2)]dx
∫<-1, 1>[3x^2+xsin(1+x^4)]dx本回答被网友采纳