Quick Integration Question w/ e

[Weather Freak]

Quick Integration Question w/ "e"

Hey folks,

On my calc homework tonight, I have to solve the integral $$\int (1/e^z) dz$$ as part of my differential equation. Anywho, I thought the answer to that was $$ln (e^z) = z$$ but my TI-89 claims that it's really $$-e^-z$$.

Could someone please tell me which of us is correct, and more importantly, why? I was under the impression that the antiderivative of e was simply itself if there was no inner function (and since z is just a variable, I don't think there is).

Thank you!

 

[Zurtex]

I'm a little confused by your question but:

$$\int \frac{1}{e^z}dz = \int e^{-z} dz = -e^{-z} + \mathcal{C}$$

 

[Weather Freak]

I'm a little confused by your question but:

$$\int \frac{1}{e^z}dz = \int e^{-z} dz = -e^{-z} + \mathcal{C}$$

Oh, wow. I didn't even think of it like that, thank's a lot! Yeah, the question does seem odd now that I know the answer.

 

[what]

Your calc. was right this time, your answer was incorrect. You should remember that you can differentiate to check and $$D_x(ln(e^z))$$ is just $$1$$.