Question

\[\int\limits_{2}^{\infty}te^{-st}dt\]where "s" is some constant

Answer 1

Looks like Laplace transform.

Answer 2

Hmm is this coming from like.. part of a Laplace Transform of a piecewise?

Answer 3

Yes sir it is

Answer 4

oh cool c:

Answer 5

Use integration by parts.

Answer 6

didn't u get it last time ?

Answer 7

I'll post the original question

Answer 8

\(\large u=t\), \( \large dv = e^{-st}dt\)

Answer 9

\(\\ \huge \color{red}{\int e^x[f(x)+f’(x)]dx=e^xf(x)+c}
\\\)

Answer 10

here f(t) =t+1

Answer 11

Using the definition to determine the Laplace transform if the given function
f(t)= {0, 0<t<2
{t, 2<t

Answer 12

So you were able to figure out the first piece alright? Just having trouble with the other piece since it's by parts I guess ?:D

Answer 13

@hartnn told me that @gohangoku58 I'm having trouble seeing how it applies

Answer 14

\(\\ \huge \color{red}{\int e^{-st}[(t+1)-1]dx=e^{-st}(t+1)/(-s)+c}
\\\)

Answer 15

ya, its been a couple years since I had to use parts. I'd like to see it done both methods shown above.

Answer 16

*dt

Answer 17

@gohangoku58 ahh, thanks. I forgot about subtracting 1. But if f(t) = t + 1 f'(t) = 1

Answer 18

ohh... then it will be t-1

Answer 19

f(t) = t-1

Answer 20

sorry :\

Answer 21

This would be the set up by parts, just in case you wanted to see it :O hopefully I didn't skip too many steps there.