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## anonymous 5 years ago Double integration... with x+y dy dx where y goes from x to infinity and x goes from 0 to 1/2... anybody up for this???

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1. anonymous

$P(X \le1/2)=(I think)\int\limits_{0}^{1/2} \int\limits_{x}^{\infty} (x+y)dy dx$

2. anonymous

$\int\limits_{0}^{1/2}\int\limits_{0}^{x}(x+y)dydx$ is your integral. Since all they gave you was the probability x is at least 1/2 or smaller, the probability has to be between 0 and 1, so your dy limit will start at zero and go as far as x. That is just like takin the limit as x goes to infinity on the upper limit. But it does have to be at least zero by order of the definition of a probability. So you just solve that and you've got it. I got 1/64 +1/24, which is some small decimal and consistant with the definition. I'm sure you can do the integration just fine.

3. anonymous

Thank You!!! the infinities were making me dizzy....

4. anonymous

No problem, I had to spend some time looking it up, and I almost confused myself.

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