Question

Can someone help me differentiate \[\int_0^1e^{-st}tdt + \int_1^{\infty} e^{-st}(2-t)dt\]

Answer 1

My answer differs from the book unfortunately -_-

Answer 2

looks lagrangian

Answer 3

or is it laplace? i mix those guys up alot

Answer 4

Laplace transform

Answer 5

But like all you gotta do is integrate

Answer 6

\[\int_0^1e^{-st}t~dt + \int_1^{\infty} e^{-st}(2-t)~dt\]

well, a by parts on the poly usually sheds some light if youve forgotten the tables, and i tend to forget the tables

e^-st
t e^-st/ -s
-1 e^-st/ (-s)^2
0

\[\int_0^1e^{-st}t~dt=-\frac{t}{s}e^{-st}-\frac{1}{(-s)^2}e^{-st}\]
at t=0 and t=1

Answer 7

\[\int_1^{\infty} e^{-st}(2-t)~dt\]

(2-t) same run down
1 on this side
0

\[\int_1^{\infty} e^{-st}(2-t)~dt=e^{-st}(-\frac{2-t}{s}+\frac{1}{(-s)^2})\]

Answer 8

Ok thanks. Not sure what tables u r referring to though

Answer 9

laplace tranform tables, they present them after they make you sweat this stuff out.

Answer 10

ohhhhhhh Thats what u mean. Ya cant use those I gotta do it by hand :(

Answer 11

\[-e^{-s}(\frac{1}{s}+\frac{1}{(-s)^2})+e^{0}(\frac{0}{s}+\frac{1}{(-s)^2})\]plus
\[\cancel{e^{-\infty}(-\frac{2-(\infty)}{s}+\frac{1}{(-s)^2})}^0-e^{-s}(-\frac{2-1}{s}+\frac{1}{(-s)^2})\]

Answer 12

Yaaaa thats the answer in the book not sure where I went wrong
THANKS :)

Answer 13

youre welcome; the error tends to be in the algebra with these things :)

Answer 14

Yaaaa and I can never find my mistakes. Thanks :)