计算二重积分∫∫e^(x+y)dxdy

数学积分问题2

化为累次积分 ∫∫D(x^2-y^2)dxdy =∫(0,π)dx∫(0,sinx)(x^2-y^2)dy(积分限是0到π和0到sinx) =∫(0,π)(x^2y-y^2/3)|(0,sinx)dx =∫(0,π)[x^2sinx-(sinx)^3]dx =∫(0,π)x^2sinxdx-∫(0,π)(sinx)^3dx ∫x^2sinxdx=-x^2cosx-∫-2xcosxdx =-x^2cosx+∫2xcosxdx=-x^2cosx+2xsinx-∫2sinxdx =-x^2cosx+2xsinx+2cosx+C, ∫(0,π)x^2sinxdx=π^2-4, ∫(sinx)^3d...全部

  化为累次积分 ∫∫D(x^2-y^2)dxdy =∫(0,π)dx∫(0,sinx)(x^2-y^2)dy(积分限是0到π和0到sinx) =∫(0,π)(x^2y-y^2/3)|(0,sinx)dx =∫(0,π)[x^2sinx-(sinx)^3]dx =∫(0,π)x^2sinxdx-∫(0,π)(sinx)^3dx ∫x^2sinxdx=-x^2cosx-∫-2xcosxdx =-x^2cosx+∫2xcosxdx=-x^2cosx+2xsinx-∫2sinxdx =-x^2cosx+2xsinx+2cosx+C, ∫(0,π)x^2sinxdx=π^2-4, ∫(sinx)^3dx=∫sinx*(sinx)^2dx =∫[1-(cosx)^2]d(-cosx)=∫[(cosx)^2-1]d(cosx) =(cosx)^3/3-cosx+C ∫(0,π)(sinx)^3dx=4/3, 原积分=π^2-4-4/3=π^2-16/3。
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