Laplace Transform of a Heaviside function

Question

Find the Laplace transform.

$$g(t)= (t-1) u_1(t) - 2(t-2) u_2(t) + (t-3) u_3(t)$$

I understand that the $\mathcal{L}\{u_c(t) f(t-c)\} = e^{-cs}*F(s)$

Finding $F(s)$ is the hard part for me. My professor has used, for example, $$f(t-2)=t^2$$ let $$s = t-2$$ $$t= s+2$$ $$f(s) = (s+2)^2$$ therefore $f(t) = (t+2)^2$

But then he said that $f(t-2) = 1$ therefore $f(t) = 1$. But why/how?

By the previous logic if you let $s = t-2$ then $t= s+2$, and $f(s) = s+2$, so $f(t) = t+2$ not $1$.

I'm having a tough time figuring this out.

Answer 1

Using the notation $u_{c}(t) = H(t-c)$, where $H(t)$ is the Heaviside step function, then $$g(t)= (t-1) u_1(t) - 2(t-2) u_2(t) + (t-3) u_3(t)$$ is evaluated as follows.

Let $g(s) = \mathcal{L}\{g(t)\}$ such that: \begin{align} g(s) &= \int_{0}^{\infty} e^{-s t} \left[ (t-1) \, H(t-1) - 2 \, (t-2) \, H(t-2) + (t-3) \, H(t-3) \right] \, dt \\ &= \int_{1}^{\infty} (t-1) \, e^{-st} \, dt - 2 \, \int_{2}^{\infty} e^{-s t} (t-2) \, dt + \int_{3}^{\infty} e^{-s t} (t -3) \, dt \\ &= \int_{0}^{\infty} e^{-s(x+1)} \, x \, dx - 2 \, \int_{0}^{\infty} e^{-s(x+2)} \, x \, dx + \int_{0}^{\infty} e^{-s(x+3)} \, x \, dx \\ &= \left( e^{-s} - 2 \, e^{-2 s} + e^{-3s} \right) \, \int_{0}^{\infty} e^{-s x} \, x \, dx \\ &= \frac{e^{-s} \, (1-e^{-s})^{2}}{s^{2}} \end{align}

Answer 2

Your function is $f(t)=t$ but not $f(t)=1$.
You need to use the formula $$\mathcal{L}\{u_c(t) f(t-c)\} = e^{-cs}*F(s)$$ 3 times with $c=1,2, 3$ to get the answer.

The answer derived by Leucippus looks correct.