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Help with Double Integral?

  1. Mar 14, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    Double integral of y*e^(x^4-1)
    with bounds
    0=<y=<1
    y^(2/3)=<x=<1


    2. Relevant equations


    3. The attempt at a solution

    Well, the first key thing to recognize is that we need the correct order for the bounds to compute this double integral.
    So I switch it from x=y^(2/3) and x=1 TO y=x^(3/2) and y=1
    and x=0 to x=1 becomes the x boundaries.

    So now I integrate with respects to the y boundary first as that is the only way to solve this problem.
    I get y^2/2*e^(x^4-1) from y=1 to y=x^(3/2)

    This then becomes the integral from x=0 to x=1 of 1/2(e^(x^4-1)-x^3*e^(x^4-1))dx

    And I am clueless on how to solve this. I've been trying to do u-substitution for a while now knowing that letting u = x^4-1 and du=x^3 I can work with something.

    Did I do something wrong in my previous steps?
     
  2. jcsd
  3. Mar 14, 2015 #2

    SammyS

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    Graph the region over which the integration takes place.

    If ##\displaystyle\ x > y^{2/3}\,,\ ## then ##\displaystyle\ y < x^{3/2}\ ## .

    So your limits of integration are not correct for integral with the order of integration reversed.
     
  4. Mar 14, 2015 #3

    RJLiberator

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    I believe I did, where it is 'close' to a x^2 parabola and the region is beneath x^(3/2) from x=0 to x=1 and from y=0 to y=x^3/2

    No?
     
  5. Mar 14, 2015 #4

    SammyS

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    Right.

    y goes from 0 to x3/2..

    What is ##\displaystyle\ \int_0^{x^{3/2}} y\, dy\ ## ?
     
  6. Mar 14, 2015 #5

    RJLiberator

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    Well, that is just y^2/2 evaluated at the bounds so
    x^(3) is what it is.

    However, ah........
    So you are saying my bounds for y in the initial setup were messed up.

    I had them going from y=x^(3/2) to y=1 and you are claiming that they go from y=0 to y=x^(3/2)

    Hm. Let me check to see if this makes sense. I'm not sure how this makes sense.
    Original bounds have y=0 to y=1. And x=y^(2/3) to x=1.
    Switching them results in x=0 to x=1 and y=x^(3/2) to y=1, no?
     
    Last edited: Mar 14, 2015
  7. Mar 14, 2015 #6

    RUber

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    You should integrate from y=0. Then that -1 will go away.
     
  8. Mar 14, 2015 #7

    RJLiberator

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    Why can I integrate from y=0 to y=x^(3/2) instead of y=x^(3/2) to y=1.

    How can I do this?
     
  9. Mar 14, 2015 #8

    SammyS

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    Look at the following.

    Let me repeat, graph the region of integration.
     
  10. Mar 14, 2015 #9

    RJLiberator

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    BINGO.
    That makes sense.
    In my original interpretation I was taking the incorrect region. This means I need to use y=0 to y=x^(3/2) as the correct region. BINGO. Now let's see how I handle this.
     
  11. Mar 14, 2015 #10

    RJLiberator

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    My answer becomes 1/8-1/8e
    and I am fairly confident that this is correct thanks to you guys showing me my wrong interpretation of the region. Thank you.
     
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