Question

二重积分的计算 ∫dx∫(3/(2x^4)(y^3)) dy x的积分上限是无穷,下限1 y的积分上限是x,下限是1/x

解题过程能写下吗，谢谢

Answer 1

∫dx∫(3/(2x^4)(y^3)) dy
= -1/2∫(1,+∞)(3/(2x^4)(y^（-2）)|(1/x,x)) dx

= -3/4∫(1,+∞)(1/(x^4)*(1/x²-x²) dx
=-3/4∫(1,+∞)(1/x^6-1/x^2) dx
=-3/4∫(1,+∞)(1/x^6-1/x^2) dx
=-3/4 [-1/5 x^(-5)|(1,+∞)+1/x|(1,+∞)]
=-3/4(0+1/5-1)
=3/5

Answer 2

x| 0->无穷
y| 1/x-> x

∫dx∫(3/(2x^4)(y^3)) dy
=∫ 3/(2x^4) [1/(-2y^2)](1/x,x) dx
= ∫ 3/(-4x^4) ( 1/x^2 - x^2) dx
= (-3/4) ∫ (1/x^6-1/x^2)dx
= (-3/4) [ 1/(-5x^5)- 1/(-x) ] (1,无穷)
=(-3/4)[ 1/5- 1]
=(-3/4) ( -4/5)
= 3/5

Answer 3

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