Confused about Applying Fubini's Theorem to a Double Integral?

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Discussion Overview

The discussion revolves around the application of Fubini's theorem to a specific double integral involving an integrable function. Participants explore the transformation of the integration domain and seek clarification on the reasoning behind the change of limits in the integral.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about applying Fubini's theorem to the integral \(\int_0^a\int_x^a \frac{1}{t}|f(t)|dtdx\) and seeks an explanation for the transformation to \(\int_0^a \int_0^t \frac{1}{t}|f(t)|dxdt\).
  • Another participant simplifies the function to \(\frac{1}{t}|f(t)| = t\) and attempts to understand the change in the integration area, asking for clarification on the new limits.
  • A third participant suggests visualizing the integration region in the x-t plane, indicating that the region is bounded by specific curves and explaining how the limits change based on the fixed value of \(t\).
  • A later reply reveals that the original misunderstanding stemmed from a misinterpretation of the integration area, which was actually the upper triangle of the square [0,1]^2 rather than the entire square.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial confusion regarding the application of Fubini's theorem, but there is a shared understanding of the correct integration area once clarified.

Contextual Notes

The discussion highlights the importance of visualizing integration regions and understanding the conditions under which Fubini's theorem applies, but does not resolve all uncertainties regarding the transformation of limits.

Bernoulli
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Hi, I usually don't have any problems with Fubini's theorem, but there is something I just can't figure out. Let f be integrable, and a some positive constant. How do i apply the theorem to this integral:
[tex]\int_0^a\int_x^a \frac{1}{t}|f(t)|dtdx[/tex]
Really; I know the answer is
[tex]\int_0^a \int_0^t \frac{1}{t}|f(t)|dxdt[/tex]
but I just don't get it. To me this is not obvious (should it be?). Can someone explain this to me?
 
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I guess I have made it more difficult then it really is. I just want to know why the second integration domain turns out like that. For simplicity put [tex]\frac{1}{t}|f(t)| = t[/tex] for example (and [tex]t\in (0,a)[/tex]). Then the integration area becomes
[tex]\int_0^a\int_x^a t dtdx = \int_?^?\int_?^?t dxdt[/tex]
Why?
 
Draw a picture in the x-t plane. Your region is bounded by the curves t=a, x=0 and x=t. For a fixed t, x ranges from 0 to t. t itself can range from 0 to a.
 
Oh, man... I must be tired :) The original problem was not posted like this. I did not realize that the 2D integration area was infact the upper triangle of the square [0,1]^2. I thought I was dealing with the hole square... Stupid me :)

Then of course it is easy.
Thanks
 

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