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$\begingroup$

UPDATE: Apparently a bounded region with 3 sides can expand to a bounded region with 4 sides after a transformation. I was not aware of this. Though it seems like a reasonable thing to happen.

I've been working on this problem for about 6 hours.

$$ I = \int^1_0dx\int^x_0\frac{(x+y)e^\frac{x+y}{x-y}}{x^2}dy$$

with the change of variables:

$ u = y/x$

$ v = x+y$

I calculated the jacobian determinant to be:

$|J| = \frac{v}{1+u^2}$

alot of stuff cancels nicely since,

$x^2 = \frac{v^2}{(1+u)^2}$

then I just becomes a simple integral over the new region Q

$$ I = \int\int_Qe^vdvdu$$

To get this region I noted that the integral region in x,y is a triangle which is the intersection of the following lines:

$x=1$

$y=0$

$x=y$

which correspond to the triangular region Q of u-v plane given by the intersection of:

$ u-v = -1$

$ v = 0 $

$ u = 1 $

However pluging these boundaries into the integral gives:

$$ I = \int^2_0e^vdv\int^1_{v-1}du$$

This clearly evaluates to $I = e^2 -3$ whereas the correct answer from the back of the textbook is $ I = e^2-e-1$

this is kind of drivin' me nuts, would appreciate any insight as to what i might be doing wrong.

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    is my region Q too big or something?2012-08-04
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    Note that you can get displayed equations using double dollar signs instead of single dollar signs. It makes fractions and integrals easier to read (and also centres the equations).2012-08-04
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    let me try that2012-08-04
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    By the way, the integration region is not the *intersection* of three lines but the region *bounded* by those three lines. If you wanted to describe it as an intersection, it would have to be the intersection of three half-planes bounded by those three lines.2012-08-04
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    i'm having issue finding the bounded region in uv2012-08-04

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