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I encounter some problem while evaluating this double integral. I have tried to use u-substitution and i keep using integration by parts but the equation becomes longer and longer. I found out that $$e^{-u^2}$$ is not an elementary function and if I want to integrate(indefinite integral) this, the result will be sqrt pi.(which is gaussian integral). inside the textbook I couldnt find any example similar to this. But I found examples online which most of them only dealing with one integral, this one is double integral. Could you point out on how can I evaluate this double integral? Thank you. $$\int_0^{1/2} \int_0^y e^{-(1-2x)^2} dxdy + \int_{1/2}^1\int_0^{1-y} e^{-(1-2x)^2}dxdy $$

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    $\begingroup$ Have you tried changing the order of integration? $\endgroup$
    – Mark Viola
    Apr 23, 2017 at 16:31

1 Answer 1

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We can change the order of integration (See Fubini's Theorem) to arrive at

$$\begin{align} \int_0^{1/2} \int_0^y e^{-(1-2x)^2} \,dx\,dy&=\int_0^{1/2} \int_x^{1/2} e^{-(1-2x)^2} \,dy\,dx\\\\ &=\int_0^{1/2}(1/2-x)e^{-(1-2x)^2}\,dx\\\\ &=\frac12\int_0^{1/2}(1-2x)e^{-(1-2x)^2}\,dx \end{align}$$

And similarly for the second integral

$$\begin{align} \int_{1/2}^1 \int_0^{1-y} e^{-(1-2x)^2} \,dx\,dy&=\int_0^{1/2} \int_{1/2}^{1-x} e^{-(1-2x)^2} \,dy\,dx\\\\ &=\int_0^{1/2}(1/2-x)e^{-(1-2x)^2}\,dx\\\\ &=\frac12\int_0^{1/2}(1-2x)e^{-(1-2x)^2}\,dx \end{align}$$

Can you finish?

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  • $\begingroup$ thank you for your reply.Yes! my answer is 1/4(e^-1 + 1) at the final. not sure if it is correct answer, coz i do not have answer sheet. I was about to ask you about my range is correct or not for the second part of the equation. sorry for late reply coz it took me some time to solve it. it was so kind of you, thank you so much for editing again the answer, thanks to your first part I get the same second part as yours!! am so glad. i spent so many hours on this before. thank you really!! $\endgroup$
    – cookick
    Apr 23, 2017 at 17:08
  • $\begingroup$ @cookick Well, you very welcome. And I'm pleased to hear that you have it now! -Mark ... P.S. The final answer is $\frac{1}{4}(1-e^{-1})$. $\endgroup$
    – Mark Viola
    Apr 23, 2017 at 17:10
  • $\begingroup$ yes i will up vote the answer once i accumulate enough reputation points ~ btw thanks for the final answer, I forgot the minus@@. Thank you! $\endgroup$
    – cookick
    Apr 23, 2017 at 17:22
  • $\begingroup$ You're welcome again! $\endgroup$
    – Mark Viola
    Apr 23, 2017 at 17:24

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