Im given a graph of $f(t)$ and i need to find the Laplace transform of $f(t)$. From looking at the graph i have
$$f(t) =
\begin{cases}
t, & \text{$0 \le t \le 1 $} \\
0, & \text{$1 \lt t \lt \infty$} \\
\end{cases}$$
The previous problems ive worked all involved the unit step function, so assuming that's the right direction with this one, i setup $f(t)$ as
$$f(t)=tu(t)-tu(t-1)$$
Then
$$\begin{align}
F(s)&=\mathcal{L}\{tu(t)\}+\mathcal{L}\{tu(t-1)\} \\
& = \int_0^1e^{-st}tdt-\int_1^\infty e^{-st}tdt \\
& = \left(\frac{-1}{s}-\frac{1}{s^2}\right)e^{-s}-\int_1^\infty e^{-st}tdt \\
\end{align}$$
I dont know how to go any further, or if the setup is right?. The answer the book has is $$F(s)=\frac{1-e^{-s}-se^{-s}}{s^2}$$
Best Answer
I hope that this gets you going in the right direction:
$$\begin{align} F(s)&= \int_0^1e^{-st}tdt \\ & = \left.\left(\frac{-t}{s}-\frac{1}{s^2}\right)e^{-st}\right|_0^1 \\ & = \left(\frac{-1}{s}-\frac{1}{s^2}\right)e^{-s} -\left(0-\frac{1}{s^2}\right)\\ \end{align}$$
You should be able to get to the book answer from there. I did the integration by looking it up in an integral table here: