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第1个回答 2021-12-09
原式 = ∫<-1, 2>ydy∫<y^2-1, y+1>3x^2dx
= ∫<-1, 2>ydy[x^3]<y^2-1, y+1>
= ∫<-1, 2>y[(y^2-1)^3-(y+1)^3]dy
= ∫<-1, 2>y(y^6-3y^4-y^3-3y-2)dy
= ∫<-1, 2>(y^7-3y^5-y^4-3y^2-2y)dy
= [y^8/8-(1/2)y^6-y^5/5-y^3-y^2]<-1, 2>
= 32-32-32/5-8-4-1/8+1/2-1/5-1+1 = -18.225本回答被提问者采纳
= ∫<-1, 2>ydy[x^3]<y^2-1, y+1>
= ∫<-1, 2>y[(y^2-1)^3-(y+1)^3]dy
= ∫<-1, 2>y(y^6-3y^4-y^3-3y-2)dy
= ∫<-1, 2>(y^7-3y^5-y^4-3y^2-2y)dy
= [y^8/8-(1/2)y^6-y^5/5-y^3-y^2]<-1, 2>
= 32-32-32/5-8-4-1/8+1/2-1/5-1+1 = -18.225本回答被提问者采纳
第2个回答 2021-12-10
原式=∫<-1,2>y[(y+1)^3-(y^2-1)^3]dy
=[(y/4)(y+1)^4-（1/20）(y+1)^5-(1/8)(y^2-1)^4]|<-1,2>
=81/2-243/20-81/8
=(81/40)(20-6-5)
=729/40.
=[(y/4)(y+1)^4-（1/20）(y+1)^5-(1/8)(y^2-1)^4]|<-1,2>
=81/2-243/20-81/8
=(81/40)(20-6-5)
=729/40.
第3个回答 2021-12-12
此题已经是最佳次序积分。先对 x 积分后得：
∫{-1,2} [(y+1)^3 - (y^2-1)^3] y dy
= 729/40
∫{-1,2} [(y+1)^3 - (y^2-1)^3] y dy
= 729/40