Can someone please help me!!!
Use the definition of the Laplace Transform to find the transform of the function shown in the link
https://instruct.math.lsa.umich.edu/webwork2_course_files/ma216-u12/tmp/gif/rilayian-103-sethomework6prob13image1.png

- anonymous

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- anonymous

- anonymous

- anonymous

@ranyai12 you there ?

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## More answers

- anonymous

yes i am

- anonymous

ok

- anonymous

you have given piecewise function .first the definition of Laplace is
\[[\Large F(s)=\int\limits\limits_{0}^{\infty}e^{-st}f(t)dt\]\]

- anonymous

sorry openstudy logged me out!

- anonymous

what values from the graph would we incorporate?

- anonymous

@sami-21 are you still there?

- anonymous

yes i am here . openstudy crashed me :( i was typing everything went off :(
let me try again.

- anonymous

OpenStudy is a bit of a pain today, I was opening this question about 20 minutes ago, and by now I am here hehe.

- anonymous

lol yea i understand it kept on logging me out!

- anonymous

it is piece wise function can be written as

##### 1 Attachment

- anonymous

do you get the integration to be (-3e^(-st))/s

- anonymous

yes apply the limits

- anonymous

and the limits are the s right

- anonymous

nopes!! they are for t :P

- anonymous

oh lol

- anonymous

ok so the answer would be (3e^(-5s)(3e^s-1))/s

- anonymous

you did a mistake .
first take 3 common also the second one should be e^(-2s)

- anonymous

what do you mean first take 3 common

- anonymous

so the answer is 3(e^5s/s-e^2s/s)

- anonymous

\[\Large \frac{-3}{s}|e^{-st}|_{2}^{5}=\frac{-3}{s}[e^{-5s}-e^{-2s}]\]

- anonymous

@ranyai12 yes it is just you missed minus sign with the s in the exponent.

- anonymous

oh ok thank you

- anonymous

you 're welcome :)

- anonymous

it ended up being wrong

- anonymous

wait nevermind i got it!

- anonymous

it was right! i plugged it in wrong

- anonymous

(-: !

- anonymous

you must have missed minus sign with 3 :P

- anonymous

I did! lol

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