概率论题目,划线的部分是怎样算出六分之一的?

2025-04-23 00:36:48
推荐回答(1个)
回答1:

∫(0->∞) k e^[-(2x+3y) ] dx
=k e^(-3y) [ -(1/2) e^(-2x) ] |(0->∞)
=(1/2) k e^(-3y)
∫(0->∞) [∫(0->∞) k e^[-(2x+3y) ] dx ] dy
=∫(0->∞) (1/2) k e^(-3y) dy
=-(1/6)k [ e^(-3y)] |((0->∞)
=(1/6)k