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sanchez9457
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1}\int\limits_{0}^{1} xye^xe^y dydx\]

sanchez9457
 2 years ago
Best ResponseYou've already chosen the best response.0i just can't quite figure this one out! Anyone willing to help?

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{0}^{1}\int\limits_{0}^{1}xye^xe^ydydx=\int\limits_{0}^{1}ye^ydy \int\limits_{0}^{1}xe^xdx\] Because the "ye^y dy" part doesn't depend on x, it's like a constant so you can tak it out of the integral. Now you have a multiplication of two integrals that can be done by Integration By Parts... your typical xe^x integral :)

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1The answer will be 1 btw

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0\(\int y\cdot e^{y}\;dy = \int y\;d\left(e^{y}\right) = y\cdot e^{y}  \int e^{y}\;dy\)

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Do you need more help

sanchez9457
 2 years ago
Best ResponseYou've already chosen the best response.0@kirbykirby i really like that answer as you just answered another of my questions!! But i need some clarification if possible:

sanchez9457
 2 years ago
Best ResponseYou've already chosen the best response.0Okay so basically you just said that the double integral equals the multiple of the integral of each variable?

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, In this case you can do it because the term \[ye^ydy\]in the integral doesn't depend on x (recall you can just switch the dydx to dxdy and not worry about the bounds of integration since they are constants and the same for both integrals). So since ye^y dy doesn't depend on x, it acts like a constant (with respect to x), so you can just move it out of the integral

sanchez9457
 2 years ago
Best ResponseYou've already chosen the best response.0You @kirbykirby are a genius my good sir!
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