∫∫D(x^2 /y^2)dxdy,其中D是由xy=2,y=1+x^2,x=2围成的区域∫∫D(x^2 /y^2)dxdy=∫[1,2]x^2dx ∫[2/x,1+x^2]1/y^2 dy=∫[1,2]x^2dx (-1/y) [2/x,1+x^2]=∫[1,2]{x^3/2-x^2/(1+x^2)}dx=∫[1,2]{x^3/2-1+1/(1+x^2)}dx=1/8*x^4-x+arctanx [1,2]=7/8-
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![∫∫D(x^2 /y^2)dxdy,其中D是由xy=2,y=1+x^2,x=2围成的区域∫∫D(x^2 /y^2)dxdy=∫[1,2]x^2dx ∫[2/x,1+x^2]1/y^2 dy=∫[1,2]x^2dx (-1/y) [2/x,1+x^2]=∫[1,2]{x^3/2-x^2/(1+x^2)}dx=∫[1,2]{x^3/2-1+1/(1+x^2)}dx=1/8*x^4-x+arctanx [1,2]=7/8-](/uploads/image/z/7268986-10-6.jpg?t=%E2%88%AB%E2%88%ABD%EF%BC%88x%5E2+%2Fy%5E2%29dxdy%2C%E5%85%B6%E4%B8%ADD%E6%98%AF%E7%94%B1xy%3D2%2Cy%3D1%2Bx%5E2%2Cx%3D2%E5%9B%B4%E6%88%90%E7%9A%84%E5%8C%BA%E5%9F%9F%E2%88%AB%E2%88%ABD%EF%BC%88x%5E2+%2Fy%5E2%29dxdy%3D%E2%88%AB%5B1%2C2%5Dx%5E2dx+%E2%88%AB%5B2%2Fx%2C1%2Bx%5E2%5D1%2Fy%5E2+dy%3D%E2%88%AB%5B1%2C2%5Dx%5E2dx+%28-1%2Fy%29+%5B2%2Fx%2C1%2Bx%5E2%5D%3D%E2%88%AB%5B1%2C2%5D%7Bx%5E3%2F2-x%5E2%2F%281%2Bx%5E2%29%7Ddx%3D%E2%88%AB%5B1%2C2%5D%7Bx%5E3%2F2-1%2B1%2F%281%2Bx%5E2%29%7Ddx%3D1%2F8%2Ax%5E4-x%2Barctanx+%5B1%2C2%5D%3D7%2F8-)
∫∫D(x^2 /y^2)dxdy,其中D是由xy=2,y=1+x^2,x=2围成的区域∫∫D(x^2 /y^2)dxdy=∫[1,2]x^2dx ∫[2/x,1+x^2]1/y^2 dy=∫[1,2]x^2dx (-1/y) [2/x,1+x^2]=∫[1,2]{x^3/2-x^2/(1+x^2)}dx=∫[1,2]{x^3/2-1+1/(1+x^2)}dx=1/8*x^4-x+arctanx [1,2]=7/8-
∫∫D(x^2 /y^2)dxdy,其中D是由xy=2,y=1+x^2,x=2围成的区域
∫∫D(x^2 /y^2)dxdy
=∫[1,2]x^2dx ∫[2/x,1+x^2]1/y^2 dy
=∫[1,2]x^2dx (-1/y) [2/x,1+x^2]
=∫[1,2]{x^3/2-x^2/(1+x^2)}dx
=∫[1,2]{x^3/2-1+1/(1+x^2)}dx
=1/8*x^4-x+arctanx [1,2]
=7/8-π/4+arctan2
arctan1=π/4,为什么有负号?倒数第二步到最后一步不懂
∫∫D(x^2 /y^2)dxdy,其中D是由xy=2,y=1+x^2,x=2围成的区域∫∫D(x^2 /y^2)dxdy=∫[1,2]x^2dx ∫[2/x,1+x^2]1/y^2 dy=∫[1,2]x^2dx (-1/y) [2/x,1+x^2]=∫[1,2]{x^3/2-x^2/(1+x^2)}dx=∫[1,2]{x^3/2-1+1/(1+x^2)}dx=1/8*x^4-x+arctanx [1,2]=7/8-
1/8*x^4-x+arctanx [1,2]
=[(1/8)*2^4-2+arctan2]-[(1/8)*1^4-1+arctan1]
=arctan2-(-7/8+π/4)
=arctan2+7/8-π/4