Math Help - integrate e^(2+y) 0<= y <= ln2

1. integrate e^(2+y) 0<= y <= ln2

how to integrate e^(2+y) dy where 0<= y <= ln2?

Thanks

2. integrate e^(2+y) dy 0<= y <= ln2

Note: the real problem is integrating e^(x + y) dydx where 0<= y <= ln2 and 0 <= x <= ln3. My solution manual says that the inner integral is e^x. I just can't see how you get e^x from integral e^(x + y) dy = e^x with the given limits (0 <= y <= ln2). I replaced x with 2 since x is constant in the inner integral on my original post.

3. $e^{2+y} = e^2 e^y$.

So you have...

$\int_0^{\ln(2)} e^{2+y} dy = e^2 \int_0^{\ln(2)} e^y dy = e^2 \left [e^y \right]_0^{\ln(2)} dy$

$= e^2 (e^{\ln(2)} - e^0) = e^2 (2 - 1) = e^2$.

4. integrate e^(2+y) dy 0<= y <= ln2

Thanks. I get it now. The mistake I was making was this:

e^(2 + y) = e^2 + e^y instead of e^2e^y. Now it works out.

Many thanks.