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sanchez9457

  • 2 years ago

Double Integral:

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  1. sanchez9457
    • 2 years ago
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    \[\int\limits_{0}^{1}\int\limits_{0}^{1} xye^xe^y dydx\]

  2. sanchez9457
    • 2 years ago
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    i just can't quite figure this one out! Anyone willing to help?

  3. kirbykirby
    • 2 years ago
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    \[\int\limits_{0}^{1}\int\limits_{0}^{1}xye^xe^ydydx=\int\limits_{0}^{1}ye^ydy \int\limits_{0}^{1}xe^xdx\] 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 :)

  4. kirbykirby
    • 2 years ago
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    The answer will be 1 btw

  5. tkhunny
    • 2 years ago
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    \(\int y\cdot e^{y}\;dy = \int y\;d\left(e^{y}\right) = y\cdot e^{y} - \int e^{y}\;dy\)

  6. kirbykirby
    • 2 years ago
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    Do you need more help

  7. sanchez9457
    • 2 years ago
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    @kirbykirby i really like that answer as you just answered another of my questions!! But i need some clarification if possible:

  8. sanchez9457
    • 2 years ago
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    Okay so basically you just said that the double integral equals the multiple of the integral of each variable?

  9. kirbykirby
    • 2 years ago
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    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

  10. sanchez9457
    • 2 years ago
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    AH!!!!!!!!!!

  11. sanchez9457
    • 2 years ago
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    You @kirbykirby are a genius my good sir!

  12. kirbykirby
    • 2 years ago
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    :) no problem!

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