Integration by Parts [SOLVED]

SkyEye

Ugh. I don't even remember if this was in Calc I or II.

Anyway, I'm integrating exp(-Dt)dt. If I set u=exp(-D), v=exp(t), I get exp(-Dt). If I set u=1, x=exp(-Dt), I get -1/D*exp(-Dt). Those are not the same. Why aren't they the same?

Oh, good luck, ceres!

ceres

It's Calc I. I have it now. The final is next week. Please kill me.

Usagi

so you're integrating e^(-Dt) dt or e^(-Dt) dD?

also, this might help

SkyEye

Oh, sorry. The differential term is dt. I'm pretty sure the correct answer is the second one but I can't figure out why the first one doesn't work.

Usagi

ok, so you're integrating e^(-Dt) dt where D is a constant

why are you integrating by parts?

ecco the dolphin

Hey, just wondering - if:

u = exp(-D)
v = exp(t)

Then

uv = exp(-D) * exp(t) = exp(-D + t) =/= exp(-Dt)

?

Clipse

SkyEye wrote: »

Ugh. I don't even remember if this was in Calc I or II.

Anyway, I'm integrating exp(-Dt)dt. If I set u=exp(-D), v=exp(t), I get exp(-Dt). If I set u=1, x=exp(-Dt), I get -1/D*exp(-Dt). Those are not the same. Why aren't they the same?

The product of exp(-D) and exp(t) is exp(-D + t).

EDIT: Oops, eeccccoo beat me to it. Guess I should actually read the other replies before making my own :P

SkyEye

Goddamn, I knew this would make me look stupid somehow.

ecco the dolphin

SkyEye wrote: »

Goddamn, I knew this would make me look stupid somehow.

Nah, don't fuss.

You're not the first person to have made this mistake.

Besides, now that you've made the mistake publicly, you'll remember when it comes to your exam!