Finding the Laplace Transform of a Polynomial Function

[kidi3]

Homework Statement

Find the Laplace transform of the following function
t2 - 2t

The Attempt at a Solution

[itex]\[\begin{gathered}
f\left( t \right) = {t^2} - 2t \hfill \\
l\left( {f\left( t \right)} \right) = F\left( s \right) = \int_0^\infty {{e^{ - st}} \cdot \left( {{t^2} - 2t} \right)ds} \hfill \\
\Leftrightarrow \hfill \\
= \int_0^\infty {{e^{ - st}}{t^2} - 2t{e^{ - st}}ds} \hfill \\
= \int_0^\infty {{e^{ - st}}{t^2}} ds - \int_0^\infty {2t{e^{ - st}}} ds \hfill \\
= \int_0^\infty {{e^{ - st}}{t^2}} ds - \int_0^\infty {2t{e^{ - st}}} ds \hfill \\
{\text{Integration by parts}} \hfill \\
\int_0^\infty {{e^{ - st}}{t^2}} ds \hfill \\
\int_{}^{} {uv' = uv - \int_{}^{} {u'v} } \hfill \\
v'\left( t \right) = {e^{ - st}} \Rightarrow v = \frac{{ - 1}}{s}{e^{ - st}} \hfill \\
u\left( t \right) = {t^2} \Rightarrow u'\left( t \right) = 2t \hfill \\
\int_{}^{} {uv' = uv - \int_{}^{} {u'v} } \hfill \\
\int_0^\infty {{e^{ - st}}{t^2}} ds = \frac{{ - 1}}{s}{e^{ - st}} \cdot {t^2} - \int_{}^{} {2t \cdot } \frac{{ - 1}}{s}{e^{ - st}}ds = \frac{{ - 1}}{s}{e^{ - st}} \cdot {t^2} - \frac{{ - 2}}{s}\int_{}^{} {t \cdot } {e^{ - st}}ds \hfill \\
\end{gathered} \][/itex]am i doing it correctly?

 

[Karnage1993]

You're supposed to integrate with respect to ##t##, not ##s##! Switch the ##ds## with ##dt## and start over.

 

[kidi3]

Ahh.. It makes sense now.

 

[kidi3]

Could you tell me what i am doing wrong with this one..

Think i got it..