Calculate, using direct integration, the value of e^2t

given that s > -2.

and...

e^t = 1/s-1

for s > 1

Any help on how to do this direct integration would be great guys, thank you

Printable View

- Aug 10th 2009, 02:56 PMtone999Direct Integration
Calculate, using direct integration, the value of e^2t

given that s > -2.

and...

e^t = 1/s-1

for s > 1

Any help on how to do this direct integration would be great guys, thank you - Aug 10th 2009, 11:51 PMmatheagle
s,t? Are you asking about LaPlace transforms?

- Aug 11th 2009, 03:26 AMtone999
yes. sorry, i didnt put the laplace sign at the front

- Aug 11th 2009, 03:36 AMmr fantastic
By definition: $\displaystyle LT[f(t)] = \int_0^{+\infty} e^{-st} f(t) \, dt$.

Therefore: $\displaystyle LT\left[ e^{2t} \right] = \int_0^{+\infty} e^{-st} e^{2t} \, dt = \int_0^{+\infty} e^{-t(s-2)} \, dt $.

It's expected you can calculate an improper integral like this one.