Question

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

Answer 1

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

Answer 2

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? :)

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

i support to find dx then dy?

Answer 7

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