# Direct Integration

• August 10th 2009, 02:56 PM
tone999
Direct 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
• August 10th 2009, 11:51 PM
matheagle
• August 11th 2009, 03:26 AM
tone999
yes. sorry, i didnt put the laplace sign at the front
• August 11th 2009, 03:36 AM
mr fantastic
Quote:

Originally Posted by tone999
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

By definition: $LT[f(t)] = \int_0^{+\infty} e^{-st} f(t) \, dt$.

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

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