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I'm trying to set up a double integral on $e^{-(x+y)}$ and the range listed is: $0 < x < y < \infty.$

I'm interpreting this as $0 < x < y$ for $x$'s range and $x < y < \infty$ as $y$'s range. I've put those bounds on my two integrals and proceeded with the mission. However, $y$ is still lingering in my result. (The homework is already turned in and I'm sure I've done it wrong; now I just want to know how to do it properly.)

My result of the double integral was: $-((e^{-2y})/2) + (1/2)$. I suspect that how I set up the double integral was the source of the problem. How should it have looked?

The integral I calculated was $$\int_0^y\int_x^\infty e^{-(x+y)}dydx$$

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  • $\begingroup$ Hi, Welcome to MSE! Here we use MathJax to typeset mathematics, and avoid linking to images. Please also expand on what you mean by range in this context $\endgroup$ – K.Power Jun 6 '18 at 23:40
  • $\begingroup$ I've fixed the part about the range. I'm attempting to figure our MathJax... $\endgroup$ – AvdotyaC Jun 6 '18 at 23:45
  • $\begingroup$ I have edited your post, so don't worry about the MathJax for now, but try learn from my edit and the link for any future posts :) $\endgroup$ – K.Power Jun 6 '18 at 23:54
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If I'm interpreting your question correctly, I believe what they're asking you to do is to calculate the definite integral of the function $f(x,y)=e^{-(x+y)}$ over the region bounded below by the $y=x$ line, and to the "left" by $x=0$. In other words calculate the integral $$\int_{0}^\infty\int_x^\infty e^{-(x+y)}dy dx.$$

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  • $\begingroup$ Doing the integral gives me a much nicer, cleaner answer to work with for the subsequent steps. That makes a lot of sense $\endgroup$ – AvdotyaC Jun 7 '18 at 0:04
  • $\begingroup$ I'm glad. If they give the domain of integration in such an equality it is always a good idea to first look at the Cartesian plane and try figure out exactly what area of it satisfies the inequality. Setting up the integral bounds is a lot easier once you can actually draw the domain of integration $\endgroup$ – K.Power Jun 7 '18 at 0:09
  • $\begingroup$ Thanks for pointing me on the right path. $\endgroup$ – AvdotyaC Jun 7 '18 at 0:19
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$$\int^{y}_{0}\int^{\infty}_{x}e^{-(x+y)}dydx$$

First separate the integral and solve $$\int^{\infty}_{x}e^{-(x+y)}dy$$ After calculating you will get $$\int^{\infty}_{x}e^{-(x+y)}dy=e^{-2x}$$ and $$\int^{y}_{0}e^{-2x}dx=-\frac12e^{-2y}+\frac12$$

$$\int^{y}_{0}\int^{\infty}_{x}e^{-(x+y)}dydx=-\frac12e^{-2y}+\frac12$$

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