Question

How to prove this integral...

Answer 1

\[2/e ^{5}\le \int\limits_{}^{}\int\limits_{D}^{}e ^{-(x ^{2}+y ^{2})}dxdy \le 2\]

Answer 2

:/ good luck :)

Answer 3

without any bounds on the "D" stuff you might wanna dx it first and keep the +C

Answer 4

ohh....D = [0,1]x[0,2]

Answer 5

\[2/e ^{5}\le \int\limits_{0}^{1}\int\limits_{0}^{2}e ^{-(x ^{2}+y ^{2})}dxdy \le 2\]

Answer 6

yes

Answer 7

i dont know hot to start it...i cant substitute, no per partes.

Answer 8

focus in the dx first and consider y a constant

\[\int e^{-x^2-C_y}\ dx\]

Answer 9

the -x^2 has issues since its not from an elementary function

Answer 10

we might have to do a laplace transform on it

Answer 11

can i use function gama?

Answer 12

maybe, i aint familiar with the gamma enough to say

Answer 13

i odnt like this at all:(

Answer 14

can we do integration by parts?
\[\int e^{-x^2}\ e^{-C_y}\ dx\]

Answer 15

i dont think that ibp would be useful ....

Answer 16

if i dont know hot to solve this integral, can i say at least that it is between this two values because......i dont know what

Answer 17

laplace transform is about the only notion I have that might be useful

Answer 18

you cant prove something by simple stating it as fact :)

Answer 19

do you maybe know how would this e^-(...) look like(graph)

Answer 20

not without the wolfram

Answer 21

seems to resemble a cbrt function

Answer 22

repeated integration

\[\int\limits_{0}^{1}\int\limits_{0}^{2}\epsilon ^{-(x ^{2}+y ^{2})}dxdy = \int\limits_{0}^{1}(\int\limits_{0}^{2}\epsilon ^{-(x ^{2}+y ^{2})}dx)dy\]

Answer 23

and how can i integrate this e^-(x^2+y^2)?

Answer 24

Wolfraam output in file

Answer 25

:)....but how to get there, by hand, not wolfram:)

Answer 26

Wolfraam gives an erf function, so we have a start

Answer 27

\[\int\limits_{0}^{2}e ^{-(x^2+y^2)}dx = \int\limits_{0}^{2}e ^{-y^2} e^{-x^2}dx \]