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Double integral - reversing order

  1. Apr 4, 2010 #1
    1. The problem statement, all variables and given/known data

    \displaystyle\int^1_0 \int^{e^x}_{1}dydx

    2. Relevant equations

    3. The attempt at a solution
    the above integral i can do with no problem, but changing the order of integration give me a totally different answer and need to know if i am doing it correct

    First off
    \displaystyle\int^1_0 \int^{e^x}_{1}dydx = e^1 - 2

    To reverse the order of integration i get:
    \displaystyle\int^{e^1}_1 \int^{ln(y)}_{0}dxdy
    which gives me 1 which is wrong

    Before i post how i went about my solution I want to know if i am doing my limit right?
    Last edited: Apr 4, 2010
  2. jcsd
  3. Apr 4, 2010 #2


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    Did you draw a graph of the integration region? x doesn't go from 0 to ln(y).
  4. Apr 4, 2010 #3
    it goes from 0 to ln(e^1) which is 0 to 1

    considering y=e^x then x = ln(y)

    but im guessing my understanding is wrong
    Last edited: Apr 4, 2010
  5. Apr 4, 2010 #4
    Ok, these are two questions alike. Can you please tell me if i am completely misunderstanding the region of integration.
    I need to show by reversing the order that i still get the same answer.


    EDIT: ok, another quick question. For Question 4 in the image here, is my region of integration on the wrong side of the line???
    Last edited by a moderator: Apr 25, 2017
  6. Apr 4, 2010 #5


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    Your graph is on the correct side of the line. But look at it and imagine integrating dx. Isn't ln(y) the LOWER bound for x?
  7. Apr 4, 2010 #6
    that seemed so hard to wrap my head around at the time but is so simple now
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