double integration of e^(x+y)dx dy ranging from 0<x<lnx and 1<y<ln8

Hello!

It isn't a good idea, use the variable of integration in the limits of integration. I think you meant `0ltxltln(a)` for some constant `a` or `0ltxltln(y),` in the latter case integration by `x` is inner:

`int_1^(ln 8) (int_0^(ln a) e^(x+y) dx) dy.`

Consider the inner integral. It is simple because  `e^(x+y)=e^xe^y:`

`int_0^(ln a) e^(x+y) dx = e^yint_0^(ln a) e^(x) dx = e^y(a-1).`

This formula is true even if `a` is a function of `y.` If a is a constant, then the double integral is equal to

`(a-1) int_1^(ln 8) e^y dy = (a-1)(8-e).`

If `a=y` (in the case `0ltxltln y`), then the integral requires integration by parts. The indefinite integral is

`int e^y(y-1) dy = |u=y-1, du=dy, dv=e^y dy, v=e^y| =`

`= (y-1)e^y - int e^y dy =(y-2)e^y+C.`

And the definite integral is `8(ln8-2)+e approx 3.35.`

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