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?
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?