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sanchez9457Best ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{1}\int\limits_{0}^{1} xye^xe^y dydx\]
 one year ago

sanchez9457Best ResponseYou've already chosen the best response.0
i just can't quite figure this one out! Anyone willing to help?
 one year ago

kirbykirbyBest 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 :)
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
The answer will be 1 btw
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
\(\int y\cdot e^{y}\;dy = \int y\;d\left(e^{y}\right) = y\cdot e^{y}  \int e^{y}\;dy\)
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Do you need more help
 one year ago

sanchez9457Best 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:
 one year ago

sanchez9457Best ResponseYou've already chosen the best response.0
Okay so basically you just said that the double integral equals the multiple of the integral of each variable?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Yes, 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
 one year ago

sanchez9457Best ResponseYou've already chosen the best response.0
You @kirbykirby are a genius my good sir!
 one year ago
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