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Math Help - Help with double integral of exponential

  1. #1
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    Help with double integral of exponential

    I=
    [integral from 1 to ln(8)] [integral from 0 to ln(y)] e^(x+y) dxdy

    solving for the dx (inside) I get
    (can I break e^(x+y) to (e^x)(e^y))
    (e^y)(e^ln(y)) - e^(y)
    (e^y)(e^ln(y) -1)

    outer integral
    e^((ln(y) -1 + y)) dy

    Am I doing this wrong?
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  2. #2
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    Quote Originally Posted by khuezy View Post
    I=
    [integral from 1 to ln(8)] [integral from 0 to ln(y)] e^(x+y) dxdy

    solving for the dx (inside) I get
    (can I break e^(x+y) to (e^x)(e^y))
    (e^y)(e^ln(y)) - e^(y)
    (e^y)(e^ln(y) -1)

    outer integral
    e^((ln(y) -1 + y)) dy

    Am I doing this wrong?
    \int_1^{\ln 8} \int_0^{\ln y} e^{x + y} \, dx \, dy = \int_1^{\ln 8} e^y \left( \int_0^{\ln y} e^{x} \, dx \right) \, dy.

    Note that  \int_0^{\ln y} e^{x} \, dx = \left[ e^x\right]_0^{\ln y} = e^{\ln y} - e^0 = y - 1.

    Finish things by using integration by parts.
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