Laplace Transforms: transform f(t) = 2t + 6

f1player

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Feb 25, 2005
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Use the integral definition of the Laplace Transform to find the transform of the following function:

f(t) = 2t + 6

Now since i get the integral of a product i let u = 2t + 6 and dv/dt = e^(-st)

so du/dt = 2 and v = -1/s * e^(-st)

So i get the lim (as b >>> infinity) ((-2t + 6)/se^(st)) + 2/s lim (as b >>> infinity) integral (e^(-st) dt)

Is this right? Also what do i do with the upper and lower limits? do they stay as 0 and b or do i use the fact that u = 2t + 6 to change them?
 
It looks fine. Use the limits to evaluate the integral. The nature of Integration by Parts does not suggest changing the limits.
 
Yeah thanks for the reply but using the power of google, lol, i managed to find what i was looking for and did the question myself.
 
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