Question

double integral of e^(x+y)/(x-y) dx dy
Evaluate the integral where R is the trapezoidal region with corners at (1,0), (2,0), (0,2), (0,-1)

Answer 1

\[\int\limits_{?}^{?}\int\limits_{?}^{?}e ^{(x+y)/(x-y)}dxdy\]

Answer 2

Are you sure you typed the corner points right?

Answer 3

(1,0),
(2,0), (0,-2), (0,-1)

Answer 4

sorry about that.

Answer 5

You have to divide your integral into two. One from x=0 to x=1 and one from x=1 to x=2

Answer 6

i don't follow...2 double integrals??

Answer 7

\[ \int_0^1\int_{x-2}^{x-1} f(x,y) dy dx +
\int_1^2\int_{x-2}^{0} f(x,y) dy dx
\]

Answer 8

i don't really understand why you did that but i can kinda see where you are coming from. i don't really understand your bounds

Answer 9

Wrtie the equations of the two parallel lines of the trapeziods and tell me what they are?

Answer 10

on the first part of the integration why does the x only go from 0 to 1

Answer 11

Answer my question above first. It will help you understand

Answer 12

y=x-2 and y=x-1?

Answer 13

actually i don't really understand the equations that you got. i would really apprecciate if you could explain it to me

Answer 14

oh nevermind...got it

Answer 15

after thinking about it, it would be lot easier to compute your integral after a change of variables.
u= x - y and v = x

Answer 16

that doesn't make sense. then it would be v+y/ u??

Answer 17

You will have one itntegral to worry about
\[ 1\le u \le 2\\
0\le v \le u

\]
In the uv plane. Do not forget the jacobian determinant.
Your integral will be
\[
\frac{3 \left(e^2-1\right)}{4
e}
\]

Answer 18

Yes it does make sense
x+y = 2 v -u and x-y =u

Answer 19

Your integral becomes
\[
\int _1^2\int _0^ue^{\frac{2
v}{u}-1}dvdu
\]

Answer 20

The inner integral
\[
\int_0^u e^{\frac{2 v}{u}-1}
\, dv=\frac{\left(e^2-1\right) u}{2
e}
\]

Answer 21

Integrate what you got above
\[
\frac{\left(e^2-1\right)
\int_1^2 u \, du}{2 e}=\frac{3 \left(e^2-1\right)}{4
e}
\]