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anonymous
 3 years ago
Int^ln2_0 int^ln5_0 (e^2xy) dx dy
anonymous
 3 years ago
Int^ln2_0 int^ln5_0 (e^2xy) dx dy

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\ln2} \int\limits_{0}^{\ln 5} e^{2xy} dx dy\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Using 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? :)

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge e^{y}_0^{\ln2} \quad \cdot\quad \frac{1}{2}e^{2x}_0^{\ln5}\]Something like this yes? :o

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i support to find dx then dy?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0_e ^ln 2* 1/2 e^2 ln 5?
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