Please check my work. Did I calculate the following Laplace Transform correctly? Our function is $f(t)=t$ where $(1<t<4)$ and $f(t)=0$ everywhere else.
My solution:
First express function with the unit step function. While doing this notice that said function is windowed between $t=1$ and $t=4$. Therefore we have...
$$f(t)=t\cdot u(t-1)-t\cdot u(t-4)$$
For simplicity we shall break up our function into two parts (one for each term)...
$$f_1(t)=t\cdot u(t-1)$$
$$f_2(t)=t\cdot u(t-4)$$
where...
$$f(t)=f_1(t)-f_2(t)$$
The following corollary from the second shifting theorem (t-shift) shall be applied separately to each term...
$$\mathcal{L}\{g(t)u(t-a)\}=e^{-as}\mathcal{L}\{g(t+a)\}$$
where...
$$f(t)=g(t+a)$$
with $g_1(t)=t$ and $a=1$
$$g_1(t+1)=t+1$$
$$\mathcal{L_1}\{g_1(t+1)\}=e^{-s}\cdot \mathcal{L}\{t+1\}$$
$$F_1(s)=e^{-s}\cdot [\frac{1}{s}+\frac{1}{s^2}]$$
with $g_2(t)=t$ and $a=4$
$$g_2(t+4)=t+4$$
$$\mathcal{L_2}\{g_2(t+4)\}=e^{-4s}\cdot \mathcal{L}\{t+4\}$$
$$F_2(s)=e^{-4s}\cdot [\frac{4}{s}+\frac{1}{s^2}]$$
hence...
$$F(s)=F_1(s)-F_2(s)$$
$$F(s)=e^{-s}\cdot [\frac{1}{s}+\frac{1}{s^2}]-e^{-4s}\cdot [\frac{4}{s}+\frac{1}{s^2}]$$
Your solution is correct. However, I'd like to show you another, and in my opinion simpler method to solve such problems. Note that the derivative of the given function is
$$f'(t)=\delta(t-1)+u(t-1)-u(t-4)-4\delta(t-4)\tag{1}$$
The Laplace transform of (1) is very simple:
$$\mathcal{L}\{f'(t)\}=e^{-s}+\frac{e^{-s}}{s}-\frac{e^{-4s}}{s}-4e^{-4s}= e^{-s}\left(1+\frac{1}{s}\right)-e^{-4s}\left(4+\frac{1}{s}\right)\tag{2}$$
Using
$$\mathcal{L}\{f'(t)\}=sF(s)-f(0)$$
you get with $f(0)=0$ the same expression for $F(s)$ that you derived.