Question

I'm somehow mixing up when solving the integral from 0 to infinity of kte^(-st)dt aka laplace transform

Answer 1

i see u still didnt figure out this one , so k is constant leave it out of the integral u'll just multiply with it at the end , then for te^(-st)dt use partial integration with u=t and dv=e^(-st)

Answer 2

lol, glad your here, so far i have k[(-1/s)te^)-st) and then i can't seem to get the second part where they somehow get an -1/s^S

Answer 3

i understad to concept, its the "algebra part that gets em

Answer 4

is that what you wrote \[k[-\frac{ 1 }{ s }te^{-st}]\]

Answer 5

yes, i got the first, the second I'm not getting, isn't dv=e^-st which make v=-e^-st?

Answer 6

there is the mistake integral of e^(-st) is \[-\frac{ 1 }{ s }e^{-st}\]

Answer 7

can you show me where (in general) the fractions is coming from, i understand the integral of e^x=xe^x, and integral of e^-x=-e^-x, so wheres the fraction coming from

Answer 8

and don't you derive from dv to v?

Answer 9

ok so if you have dv in order to get v you need to integrate so we would have
\[dv=e^{-st}\]
if we integrate both side we get :
\[\int\limits_{}^{}{dv}=\int\limits_{}^{}{e^{-st}dt}\]
which is
\[v=\int\limits_{}^{}{e^{-st}dt}\]

Answer 10

now to solve that integral we have u substitution u=-st => du=-sdt => -du/s=dt

Answer 11

aahhh....

Answer 12

so the integral would be
\[-\frac{ 1 }{ s }\int\limits_{}^{}{e^{u}du}\]

Answer 13

now i belive you can solve that integral :P

Answer 14

so i have k[-1/s t e^-st - 1/s e ^-st] from o to infinity which then k[0-0+1/s+1/s] => k(1/s^2) => k/s^2m thank you

Answer 15

0.0 i got no idea what you did there how you get from 1/s+1/s to i/s^2

Answer 16

after integrating, to find the limit you sub t for infinity and 0 and subtract them, since e^infinity is 0 the first part of the equation is 0, then e^0=1 which is multiplied by the fraction (which is the same) and the positives are coming from the subtraction with the negatives making them positive

Answer 17

yeah that what u just wrote sounds correct just that e^infinity=infinity but e^-infinity = 0 , but u have e^-infinity so yeah its 0

Answer 18

...lol thanx for your help, i have a couple more problems which I'm sure ill post, hope you'll be standing by

Answer 19

if i am here i'll try help ^^