(1)2∫(0,∞)2x*e^(-2x)dx+3∫(0,∞)3x*e^(-3x)dx
首先求不定积分2∫2x*e^(-2x)dx+3∫3x*e^(-3x)dx
由分部积分法容易求得
2∫2x*e^(-2x)dx+3∫3x*e^(-3x)dx
=-(2x*e^(-2x)+3x*e^(-3x)+e^(-2x)+e^(-3x))+C
故
2∫(0,∞)2x*e^(-2x)dx+3∫(0,∞)3x*e^(-3x)dx
=lim(b--∞)[-(2b*e^(-2b)+3b*e^(-3b)+e^(-2b)+e^(-3b))+2]
=2
(2)∫(0,2)dx∫(0,2)(1/8)*xy*(x+y)dy
(i)∫(0,2)(1/8)*xy*(x+y)dy
=(1/8)*∫(0,2)(x^2*y+x*y^2)dy
=(1/8)*[∫(0,2)x^2*ydy+∫(0,2)x*y^2dy]
=(1/4)x^2+(1/3)x
(ii)∫(0,2)[(1/4)x^2+(1/3)x]dx
=(1/4)∫(0,2)x^2dx+(1/3)∫(0,2)xdx
=4/3
(iii)
∫(0,2)dx∫(0,2)(1/8)*xy*(x+y)dy=4/3
(3)∫(0,∞)(∫(0,∞)xye^(-x-y)dy)dx
(i)
∫xye^(-x-y)dy
=-∫xye^(-x-y)d(-x-y)
=-xe^(-x-y)(y+1)+C
故
∫(0,∞)xye^(-x-y)dy
=lim(b--∞)[-xe^(-x-b)(b+1)+xe^(-x)]
=xe^(-x)
(ii)
∫xe^(-x)dx
=-∫x^e(-x)d(-x)
=-∫xde^(-x)
=-e^(-x)(x+1)+C
故
∫(0,∞)xe^(-x)dx
=lim(b--∞)[-e^(-b)(b+1)+1]
=1
(iii)
∫(0,∞)(∫(0,∞)xye^(-x-y)dy)dx=1