∫[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?
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![∫[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?](/uploads/image/z/11686313-65-3.jpg?t=%E2%88%AB%5B0%2C1%5D+dx%E2%88%AB%5B-x%5E2%2C1%5D+f%28x%2Cy%29dy%2B%E2%88%AB%5B1%2Ce%5D+dx%E2%88%AB%5Blnx%2C1%5D+f%28x%2Cy%29dy%E4%BA%A4%E6%8D%A2%E7%A7%AF%E5%88%86%E6%AC%A1%E5%BA%8F%E2%88%AB%5B0%2C1%5D+dy%E2%88%AB%5B0%2C1%5D+f%28x%2Cy%29dx%3D%E2%88%AB%5B0%2C1%5D+x%7C+%5B0%2C1%5Ddy%3D+%E2%88%AB%5B0%2C1%5D+dy%3D%E2%88%ABy%7C+%5B0%2C1%5D%3D1%3F)
∫[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.