∫[0,1] dx∫[-x^2,1] f(x,y)dy+∫[1,e] dx∫[lnx,1] f(x,y)dy交换积分次序∫[0,1] dy∫[0,1] f(x,y)dx=∫[0,1] x| [0,1]dy= ∫[0,1] dy=∫y| [0,1]=1?

来源：学生作业帮助网编辑：作业帮时间：2024/05/12 13:40:41

∫[0,1] dx∫[-x^2,1] f(x,y)dy+∫[1,e] dx∫[lnx,1] f(x,y)dy交换积分次序∫[0,1] dy∫[0,1] f(x,y)dx=∫[0,1] x| [0,1]dy= ∫[0,1] dy=∫y| [0,1]=1?
∫[0,1] dx∫[-x^2,1] f(x,y)dy+∫[1,e] dx∫[lnx,1] f(x,y)dy交换积分次序
∫[0,1] dy∫[0,1] f(x,y)dx=∫[0,1] x| [0,1]dy= ∫[0,1] dy=∫y| [0,1]=1?

∫[0,1] dx∫[-x^2,1] f(x,y)dy+∫[1,e] dx∫[lnx,1] f(x,y)dy交换积分次序∫[0,1] dy∫[0,1] f(x,y)dx=∫[0,1] x| [0,1]dy= ∫[0,1] dy=∫y| [0,1]=1?
∫[0,1] dx∫[-x^2,1] f(x,y)dy
=∫[-1,0] dy∫[(-y)^(1/2),1] f(x,y)dx+∫[0,1] dy∫[0,1] f(x,y)dx
∫[1,e] dx∫[lnx,1] f(x,y)dy
=∫[0,1] dy∫[1,e^y] f(x,y)dx
∫[0,1] dx∫[-x^2,1] f(x,y)dy+∫[1,e] dx∫[lnx,1] f(x,y)dy
=∫[-1,0] dy∫[(-y)^(1/2),1] f(x,y)dx+∫[0,1] dy∫[0,1] f(x,y)dx (我这答案是对的,除非你输错了,围成的部分包括x轴的下面还有上面部分)
+
∫[0,1] dy∫[1,e^y] f(x,y)dx
补充的有问题!
除非f(x,y)=1.

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