The Student Room Group

Integrate e^-5

How is it e^-5/-1? Shouldn't it be simply e^-5 cause the differential of -5 is 0? Can't get my head around this
1/4 e^-4
Original post by lilypear
1/4 e^-4

Wat. e=exponential function, not x.
Original post by JacobAlevels
Wat. e=exponential function, not x.


Wat
Original post by lilypear
Wat

Well you can't integrate e like you'd integrate x..
Original post by JacobAlevels
How is it e^-5/-1? Shouldn't it be simply e^-5 cause the differential of -5 is 0? Can't get my head around this


What is the question:confused: is it the integral of e^-5x if so then thats -(e^-5x)/5 +C.
Oh sorry, it's actually e^-s, not e^-5. Sorry lol
Original post by JacobAlevels
Oh sorry, it's actually e^-s, not e^-5. Sorry lol


It's very important when asking questions here to double check to make sure you have typed the question exactly as asked if you want useful answers.

The integral of e^-s with respect to s is simply -e^-s.
Original post by JacobAlevels
Oh sorry, it's actually e^-s, not e^-5. Sorry lol


Next time tell us what you are integrating with respect to and also try to read the question properly!
Original post by lilypear
1/4 e^-4


That's so wrong it's frightening.
Original post by Mr M
That's so wrong it's frightening.


Please forgive me :P I wasn't paying attention xD
Reply 11
Original post by JacobAlevels
Oh sorry, it's actually e^-s, not e^-5. Sorry lol


So what are you integrating with respect to?

If you're integrating w.r.t.s then the answer will be es+c-e^{-s} + c exactly as if you were integrating exe^{-x} w.r.t.x.

(You really need to be a bit more precise when asking things like this, otherwise people go chasing round in all directions trying to integrate irrelevant functions! :smile: )
Original post by JacobAlevels
...


I=es ds=s=xex dx= \displaystyle \begin{aligned} I & = \int e^{-s} \text{ d}s \\ & \overset{-s = x}= - \int e^{x} \text{ d}x \\ & = \ \dots \end{aligned}

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