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Andresfon12 Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{\ln2} \int\limits_{0}^{\ln 5} e^{2xy} dx dy\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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? :)
 one year ago

zepdrix Group TitleBest 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
 one year ago

Andresfon12 Group TitleBest ResponseYou've already chosen the best response.0
2e^2xy dx  ln 5_0
 one year ago

zepdrix Group TitleBest 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
 one year ago

Andresfon12 Group TitleBest ResponseYou've already chosen the best response.0
i support to find dx then dy?
 one year ago

Andresfon12 Group TitleBest ResponseYou've already chosen the best response.0
_e ^ln 2* 1/2 e^2 ln 5?
 one year ago
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