Question

Double integral of dxdy/(3+x-y)^2

Answer 1

\[\int\limits_{1}^{2}\int\limits_{0}^{x+1} \frac{ dxdy }{ (3+x-y)^2 }\]

Answer 2

let (3-y) be some constant value wrt.x

Answer 3

\[\int \frac{1}{(k+x)^2}~dx\]

Answer 4

∫1(k+x)2 dx

Answer 5

you sure your limits are correct? dxdy should amount to a limit of x from a to f(y)

Answer 6

they gave me the domain

Answer 7

what i don't understand is this : \[\int\limits_{0}^{x+1} \frac{ dy }{ (3+x-y)^2 }= \frac{ 1 }{ 3+x-y } from 0 \to x+1\]

Answer 8

I think you are correct

Answer 9

y from 0 to x+1

Answer 10

Here 3+x assumed to be constant

Answer 11

is there a formula for this? i don't have it in my book..

Answer 12

they gave you the domain, and you deduced the limits?

and came up with\[\Large\int\limits_{1}^{2}\int\limits_{0}^{\color{red}{x+1}} \frac{ 1}{ (3+x-y)^2 }~\color{red}{dx}~dy\]

Answer 13

yes
D: x=1 x=2, y=x+1

Answer 14

if we solve this as is ... then for dy we have all x parts and its just y

Answer 15

the integrals are reversed

Answer 16

first dy and after dx...