Question

Hello Guys,
can someone help me to understand why the integral of e^-x is -e^-x and vice versa. Thanks for your help.

Answer 1

\[\int\limits_{}^{}e^{-x} dx\]
let u=-x => du=-dx
so we have
\[\int\limits_{}^{}e^{u} (-du) =- \int\limits_{}^{}e^u du=- e^u+C=-e^{-x}+C\]

Answer 2

If you want proof put e^-x itslef as t x=-ln t dx=-(1/t)dt so t and t cancel you have just inegral of -dt which is -t=-e^-x

Answer 3

its hard for me to read that

Answer 4

Thank you very much. I was confuse between e^x and e^-x. I know that integral of e^x is e^x, and was thinking if integral of e^-x is the same as e^-x. But i know now that if it is not e^x then i must do substitution. I hope my comments are correct.

Answer 5

So you look like you made substitution

Answer 6

\[x=-\ln(t) \]

Answer 7

\[dx=-\frac{1}{t} dt\]

Answer 8

@Myininaya
i didn't do the substitution. i just wrote what its in the book. I didn't understand why the author got that answer. But your answer made it clear to me

Answer 9

@myininaya are u telling me? yeah i did make that substitution, if you wanted proof that integral of e^x is e^x only i'm not too good with Writing those equations visually...

Answer 10

\[\int\limits_{}^{}e^{-(-\ln(t))} \frac{-1}{t} dt=-\int\limits_{}^{}e^{\ln(t)} \frac{1}{t} dt=- \int\limits_{}^{}t \cdot \frac{1}{t} dt=- \int\limits_{}^{} 1 dt =- t+c\]

Answer 11

so if \[x=-\ln(t) => -x=\ln(t)=>e^{-x}=e^{\ln(t)}=>e^{-x}=t\]

Answer 12

so both ways work

Answer 13

\[-e^{-x}+c\]

Answer 14

gj i like the first way that had above better though

Answer 15

err... whatever... \m/

Answer 16

shank i wasn't saying you were wrong
i was just having trouble reading it
i wanted to see it clearly sorry

Answer 17

I never said you said i was wrong :) \m/

Answer 18

ok :)

Answer 19

you don't have to make substitutions once you get use to what these answers look like

Answer 20

Yeah of course i just thought he/she wanted the proof :D

Answer 21

right she has two proofs now
your proof and my proof

Answer 22

she/he*