## Andresfon12 2 years ago Int^ln2_0 int^ln5_0 (e^2x-y) dx dy

1. Andresfon12

$\int\limits_{0}^{\ln2} \int\limits_{0}^{\ln 5} e^{2x-y} dx dy$

2. zepdrix

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

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

4. Andresfon12

2e^2x-y dx | ln 5_0

5. zepdrix

$\huge -e^{-y}|_0^{\ln2} \quad \cdot\quad \frac{1}{2}e^{2x}|_0^{\ln5}$Something like this yes? :o

6. Andresfon12

i support to find dx then dy?

7. Andresfon12

_e ^-ln 2* 1/2 e^2 ln 5?