## ranyai12 3 years ago 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

1. ranyai12

@Spacelimbus

2. ranyai12

@TuringTest

3. sami-21

@ranyai12 you there ?

4. ranyai12

yes i am

5. sami-21

ok

6. sami-21

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

7. ranyai12

sorry openstudy logged me out!

8. ranyai12

what values from the graph would we incorporate?

9. ranyai12

@sami-21 are you still there?

10. sami-21

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

11. Spacelimbus

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

12. ranyai12

lol yea i understand it kept on logging me out!

13. sami-21

it is piece wise function can be written as

14. ranyai12

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

15. sami-21

yes apply the limits

16. ranyai12

and the limits are the s right

17. sami-21

nopes!! they are for t :P

18. ranyai12

oh lol

19. ranyai12

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

20. sami-21

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

21. ranyai12

what do you mean first take 3 common

22. ranyai12

23. sami-21

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

24. sami-21

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

25. ranyai12

oh ok thank you

26. sami-21

you 're welcome :)

27. ranyai12

it ended up being wrong

28. ranyai12

wait nevermind i got it!

29. ranyai12

it was right! i plugged it in wrong

30. Spacelimbus

(-: !

31. sami-21

you must have missed minus sign with 3 :P

32. ranyai12

I did! lol