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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- Dec 5th 2008, 09:39 PMkhuezyHelp 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? - Dec 5th 2008, 09:51 PMmr fantastic
$\displaystyle \int_1^{\ln 8} \int_0^{\ln y} e^{x + y} \, dx \, dy = \int_1^{\ln 8} e^y \left( \int_0^{\ln y} e^{x} \, dx \right) \, dy$.

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

Finish things by using integration by parts.