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Math Help - Double integral (iterated integral)

  1. #1
    MHF Contributor arbolis's Avatar
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    Double integral (iterated integral)

    I must calculate the following integral in the 2 senses ( dxdy and dydx).
    \int _0^2 \int _1^{e^x} dydx.
    My attempt : \int _0^2 \int_1^{e^x} dydx=\int _0^2 y \big | _1^{e^x}dx=e^2-3.

    I'm not sure how to proceed when it comes to change the order of integration. Here's what I did : y goes from 1 to e^2.
    x goes from 0 to 2.
    So I have the double integral \int_{1}^{e^2} \int_{0}^{2} dxdy = \int_{1}^{e^2} 2dy=2(e^2-1) \neq e^2-3. And I think they must be equal, so I made an error.
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  2. #2
    Super Member Random Variable's Avatar
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     \int^{e^{2}}_{1} \int^{2}_{ln(y)} \ dx \ dy
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Random Variable View Post
     \int^{e^{2}}_{1} \int^{2}_{ln(y)} \ dx \ dy
    Oh... nice. Let me think on this!
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  4. #4
    Super Member Random Variable's Avatar
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    You have to draw the region to figure out the limits.
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  5. #5
    Senior Member Spec's Avatar
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    1\leq y \leq e^x \implies x \textcolor{green}{\geq} \ln y (notice the part in green; that tells you it's the lower limit)

    So we have \ln y \leq x \leq 2
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  6. #6
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    Because e^2 is a constant, the limits

    \int_1^{e^2}\int_0^2

    would define a rectangle, not a curvilinear region as in

    \int_0^2\int_1^{e^x}.
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