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I am stuck with this problem:

$$\int \int_Q e^{-x-y}dA$$ where Q is the first quadrant of the XY plane.

I then rewrite it as $\int_0^ \infty \int_0 ^\infty e^{-x-y}dxdy$. So far so good. If we start with the inner integral we get $\int_0 ^\infty e^{-x-y}dx = -[e^{u}]_0^{-\infty}=-1$. Then when we get back to the double integral, we now got $\int_0^ \infty-1dy=[-y]_0^\infty=-\infty$. However the correct answer is 1. What am I doing wrong? I have watched multiple similar problems and I understand the concepts overall but this particular one is different.

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  • $\begingroup$ You should probably put limits at the end $\endgroup$
    – tryst with freedom
    Oct 5, 2020 at 9:19

3 Answers 3

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You have $$\int_{0}^{\infty}e^{-x-y}dx=\int_{0}^{\infty}e^{-x}e^{-y}dx$$ $$=e^{-y}\big[-e^{-x}\big]_{x=0}^{x\rightarrow\infty}=e^{-y}[-0+1]=e^{-y}.$$

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  • $\begingroup$ Ohhh, I put my u=-x-y when doing u substitution. I see why I got the wrong answer, however how could I have prevented this from happening? Is it even possible to do u substitution in my case? $\endgroup$
    – J. Doe
    Oct 5, 2020 at 9:25
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    $\begingroup$ If $u=-x-y$ then at $x=0$ we have $u=-y$, so both integration limits change. Thus we have $$\int_{-y}^{-\infty}-e^{u}du=[-e^{u}]_{u=-y}^{u\rightarrow -\infty}=[-0+e^{-y}]=e^{-y}.$$ $\endgroup$
    – Alessio K
    Oct 5, 2020 at 9:30
  • $\begingroup$ ohh I see, I really need to pay better attention to the small details. Thank you for your help. This was invaluable :) $\endgroup$
    – J. Doe
    Oct 5, 2020 at 9:32
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Hint:$$\int\limits_0^ \infty \int\limits_0 ^\infty e^{-x-y}dxdy=\left(\int\limits_0 ^\infty e^{-x}dx\right)^2$$

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  • $\begingroup$ How does that work? $\endgroup$
    – J. Doe
    Oct 5, 2020 at 9:23
  • $\begingroup$ Simply separate variables. Or you mean something other? $\endgroup$
    – zkutch
    Oct 5, 2020 at 9:25
  • $\begingroup$ I mean, the whole Y-part disappears and turn into a "^2". When can I do this? Why did this work? $\endgroup$
    – J. Doe
    Oct 5, 2020 at 9:27
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    $\begingroup$ Generally, when $f(x,y)=f_1(x)\cdot f_2(y)$. One variable becomes parameter and you can take appropriate factor out of inner integral and calculating goes separately. $\endgroup$
    – zkutch
    Oct 5, 2020 at 9:32
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$$\int_0^{\infty}\int_0^{\infty}e^{-x-y}dxdy=\int_0^{\infty}e^{-x}dx\times\int_0^{\infty}e^{-y}dy=1\times 1$$

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  • $\begingroup$ How do I know when I can separate the double integral like this? $\endgroup$
    – J. Doe
    Oct 5, 2020 at 9:26
  • $\begingroup$ @J. Doe : simply observe that $e^{-x-y}=e^{-x}\cdot e^{-y}$ $\endgroup$
    – tommik
    Oct 5, 2020 at 9:27
  • $\begingroup$ And then there was no need for my u-substitution? $\endgroup$
    – J. Doe
    Oct 5, 2020 at 9:28
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    $\begingroup$ @J.Doe : exactly... no need $\endgroup$
    – tommik
    Oct 5, 2020 at 9:29

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