# Thread: integrating e^(x+y)

1. ## integrating e^(x+y)

i am working on some problems involving triple integrals and a lot of them have e^(x+y) but i just don't get how to integrate this with respect to x or y.

can somebody please explain me how this works?

Thanks

2. Candy

If you integrate e^(x+1) I'm sure you'd say e^(x+1) and you'd be right

When intergrating e^(x+y) wrt x treat y as a constant

When integratin e^(x+y) wrt y treat as x as a constant

In either case you get e ^(x+y)

3. Originally Posted by Calculus26
Candy

If you integrate e^(x+1) I'm sure you'd say e^(x+1) and you'd be right

When intergrating e^(x+y) wrt x treat y as a constant

When integratin e^(x+y) wrt y treat as x as a constant

In either case you get e ^(x+y)
You need to bring the 1/constant out too because of integration by parts (or whatever the opposite of the chain rule is)

If x is constant and with respect to y:

$\int e^{x+y}dy = e^{x+y} + C$

If y is constant and with respect to x:

$\int e^{x+y}dx = e^{x+y} + C$

4. e^ipi

your formula is for integrating e^(xy) not e^(x+y)

5. Now, what if i want to integrate ∫∫∫ ze^(x+y) dV to find the mass of the rectangular box determined by 0≤x≤1, 0≤y≤1 and 0≤z≤5.

i tried to solve the integral in the following order..
∫from 0 to 1 ∫from 0 to 1 ∫from 0 to 5 (ze^(x+y))dzdydx
i use the method Calculus 26 described and i got 58.38 but it's wrong!
can you tell mewhat i did wrong, is my set up right?

Thank you

6. See attachment

7. now i get it, thanks

8. $\iint e^{x+y}dxdy=\iint e^x e^y dxdy = \left(\int e^x dx\right)\left(\int e^y dy\right)$