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Andresfon12

  • 3 years ago

Int^ln2_0 int^ln5_0 (e^2x-y) dx dy

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  1. Andresfon12
    • 3 years ago
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    \[\int\limits_{0}^{\ln2} \int\limits_{0}^{\ln 5} e^{2x-y} dx dy\]

  2. zepdrix
    • 3 years ago
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    Using a law of exponents, we can write it like this, \[\large \int\limits\limits_{0}^{\ln2} \int\limits\limits_{0}^{\ln 5} e^{2x}\cdot e^{-y} \;dx dy\] And from there , since we don't have any x or y in our limits, we can separate the integrals if we want!\[\large \int\limits\limits\limits_{0}^{\ln2} e^{-y}\;dy \quad \cdot \quad \int\limits\limits\limits_{0}^{\ln 5} e^{2x} \;dx dy\]Can you solve it from here? :)

  3. zepdrix
    • 3 years ago
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    \[\large \int\limits\limits\limits\limits_{0}^{\ln2} e^{-y}\;dy \quad \cdot \quad \int\limits\limits\limits\limits_{0}^{\ln 5} e^{2x} \;dx\]Woops, that last term shouldn't have a dy on it :) my bad

  4. Andresfon12
    • 3 years ago
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    2e^2x-y dx | ln 5_0

  5. zepdrix
    • 3 years ago
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    \[\huge -e^{-y}|_0^{\ln2} \quad \cdot\quad \frac{1}{2}e^{2x}|_0^{\ln5}\]Something like this yes? :o

  6. Andresfon12
    • 3 years ago
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    i support to find dx then dy?

  7. Andresfon12
    • 3 years ago
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    _e ^-ln 2* 1/2 e^2 ln 5?

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