sanchez9457
  • sanchez9457
Double Integral:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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sanchez9457
  • sanchez9457
\[\int\limits_{0}^{1}\int\limits_{0}^{1} xye^xe^y dydx\]
sanchez9457
  • sanchez9457
i just can't quite figure this one out! Anyone willing to help?
kirbykirby
  • kirbykirby
\[\int\limits_{0}^{1}\int\limits_{0}^{1}xye^xe^ydydx=\int\limits_{0}^{1}ye^ydy \int\limits_{0}^{1}xe^xdx\] 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 :)

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kirbykirby
  • kirbykirby
The answer will be 1 btw
tkhunny
  • tkhunny
\(\int y\cdot e^{y}\;dy = \int y\;d\left(e^{y}\right) = y\cdot e^{y} - \int e^{y}\;dy\)
kirbykirby
  • kirbykirby
Do you need more help
sanchez9457
  • sanchez9457
@kirbykirby i really like that answer as you just answered another of my questions!! But i need some clarification if possible:
sanchez9457
  • sanchez9457
Okay so basically you just said that the double integral equals the multiple of the integral of each variable?
kirbykirby
  • kirbykirby
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
sanchez9457
  • sanchez9457
AH!!!!!!!!!!
sanchez9457
  • sanchez9457
You @kirbykirby are a genius my good sir!
kirbykirby
  • kirbykirby
:) no problem!

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