Question

\[\int\limits_{1}^{\infty}e^{-x} dx\] anyone know how do you evaluate this integal?

Answer 1

yes

Answer 2

\[-e ^{-x}\] evaluated from \[[1,\infty)\]

Answer 3

Skim through your textbook for 'Improper integral'.

Answer 4

So you have \[-e ^{-1}- (-e ^{-\infty}) = e ^{-\infty}-\frac{ 1 }{ e }\]

Answer 5

oh, thank you. My answer was zero :(.

Answer 6

Nope

Answer 7

What did I do wrong.......

Answer 8

@klllerbee thinks the answer is 0

Answer 9

ohhh.. I see. \[e ^{-\infty}= \frac{ 1 }{ e ^{\infty} }= 0\]

Answer 10

so the answer is zero?

Answer 11

nope

Answer 12

so now we have \[0-\frac{ 1 }{ e } = -\frac{ 1 }{ e }\]

Answer 13

so -1/e = 0 ?

Answer 14

nope, not 0

Answer 15

It won't be 0 because you've already integrated and input your limits of integration. the only way \[-\frac{ 1 }{ e } =0\] is if we were still taking the limit. the answer is \[-\frac{ 1 }{ e }\]

Answer 16

1/e

Answer 17

so 1/e = -1/e?

Answer 18

wai... 0-(-1/e) ..... yes.

Answer 19

No i forgot my negative sign. -.- @Zarkon is correct. the answer is (1/e)

Answer 20

Thanks @Zarkon

Answer 21

Oh cool. Zarkon is an android, so he can do it all in his head.

Answer 22

That's not true,maybe he's just seen the problem before, or worked it out himself.

Answer 23

But I can see his picture, he's data

Answer 24

TY both ;D

Answer 25

No problem :)

Answer 26

you should probably also work on notation too

\[\int\limits_{1}^{\infty}e^{-x} dx=\lim_{t\to\infty}\int\limits_{1}^{t}e^{-x} dx\]

Answer 27

Awww yiisss.. I hated that substitution.

Answer 28

My notation?