1、∫∫xsin(x+y)dxdy,D：0≤x≤π,0≤y≤二分之π2、∫∫(y+x²)dxdy,D是由y=x²和y²=x所围成的区域

1、∫∫xsin(x+y)dxdy,D：0≤x≤π,0≤y≤二分之π2、∫∫(y+x²)dxdy,D是由y=x²和y²=x所围成的区域
1、∫∫xsin(x+y)dxdy,D：0≤x≤π,0≤y≤二分之π
2、∫∫(y+x²)dxdy,D是由y=x²和y²=x所围成的区域

1、∫∫xsin(x+y)dxdy,D：0≤x≤π,0≤y≤二分之π2、∫∫(y+x²)dxdy,D是由y=x²和y²=x所围成的区域
利用二重积分的性质就可以解决了,根据性质要求这个二重积分的范围只需要求出被积函数在几分区域上的最值即可,也就是求X^2 Y^2在积分区域D内的最值.

∫∫_{D}xsin(x+y)dxdy=∫_{x:0->π}xdx∫_{y:0->π/2}sin(x+y)dy
=∫_{x:0->π}x[-cos(x+y)]|_{y:0->π/2}dx
=∫_{x:0->π}x[cos(x)-cos(x+π/2)]dx
=∫_{x:0->π/2}x[cos(x)-cos(x+π/2)]dx + ∫_{x:π/2->π}x[cos(x)-c...

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∫∫_{D}xsin(x+y)dxdy=∫_{x:0->π}xdx∫_{y:0->π/2}sin(x+y)dy
=∫_{x:0->π}x[-cos(x+y)]|_{y:0->π/2}dx
=∫_{x:0->π}x[cos(x)-cos(x+π/2)]dx
=∫_{x:0->π/2}x[cos(x)-cos(x+π/2)]dx + ∫_{x:π/2->π}x[cos(x)-cos(x+π/2)]dx
先看∫_{x:π/2->π}x[cos(x)-cos(x+π/2)]dx
设t=x-π/2, x:π/2->π, t:0->π/2. dt=dx.
∫_{x:π/2->π}x[cos(x)-cos(x+π/2)]dx = ∫_{t:0->π/2}(t+π/2)[cos(t+π/2)-cos(t+π)]dt
=∫_{t:0->π/2}(t+π/2)[cos(t+π/2)+cos(t)]dt
=∫_{x:0->π/2}(x+π/2)[cos(x+π/2)+cos(x)]dx
=∫_{x:0->π/2}x[cos(x+π/2)+cos(x)]dx + (π/2)∫_{x:0->π/2}[cos(x+π/2)+cos(x)]dx
∫∫_{D}xsin(x+y)dxdy=∫_{x:0->π/2}x[cos(x)-cos(x+π/2)]dx + ∫_{x:π/2->π}x[cos(x)-cos(x+π/2)]dx
=∫_{x:0->π/2}x[cos(x)-cos(x+π/2)]dx + ∫_{x:0->π/2}x[cos(x+π/2)+cos(x)]dx + (π/2)∫_{x:0->π/2}[cos(x+π/2)+cos(x)]dx
=2∫_{x:0->π/2}xcos(x)dx + (π/2)∫_{x:0->π/2}[cos(x+π/2)+cos(x)]dx
=2[xsin(x)]|_{x:0->π/2} - 2∫_{x:0->π/2}sin(x)dx + (π/2)[sin(x+π/2)+sin(x)]|_{x:0->(π/2)}
=2[π/2*sin(π/2)] + 2cos(x)|_{x:0->π/2} + (π/2)[sin(π)+sin(π/2) - sin(π/2) - sin(0)]
=π - 2 + (π/2)[0]
=π - 2
∫∫_{D}(y+x²)dxdy = ∫∫_{D}(y)dxdy + ∫∫_{D}(x²)dxdy
= ∫_{y:0->1}ydy∫_{x:y²->y^(1/2)}dx + ∫_{x:0->1}x²dx∫_{y:x²->x^(1/2)}dy
= ∫_{y:0->1}y[-y² + y^(1/2)]dy + ∫_{x:0->1}x²[-x² + x^(1/2)]dx
= ∫_{y:0->1}[-y^3 + y^(3/2)]dy + ∫_{x:0->1}[-x^4 + x^(5/2)]dx
= [-y^4/4 + (2/5)y^(5/2)]|_{y:0->1} + [-x^5/5 + (2/7)x^(7/2)]|_{x:0->1}
= [-1/4 + 2/5 ] + [-1/5 + 2/7]
= 2/7 + 1/5 - 1/4
= 2/7 - 1/20
= 33/140

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