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?
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?
$\displaystyle \int_1^{\ln 8} \int_0^{\ln y} e^{x + y} \, dx \, dy = \int_1^{\ln 8} e^y \left( \int_0^{\ln y} e^{x} \, dx \right) \, dy$.
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.