Recall that
[tex]\Theta(t)=\begin{cases}0&\text{for }t<0\\1&\text{for }t\ge0\end{cases}[/tex]
We're looking for an equivalent form of
[tex]f(t)=\begin{cases}0&\text{for }t<0\\t^2&\text{for }t\in[0,1]\\t&\text{for }t>1\end{cases}[/tex]
in terms of [tex]\Theta(t)[/tex]. For starters, we have
[tex]t^2\Theta(t)=\begin{cases}0&\text{for }t<0\\t^2&\text{for }t\ge0\end{cases}[/tex]
which takes care of the negative domain, but we have "excess function" for [tex]t>1[/tex]. To get rid of it, note that
[tex]t^2\Theta(t-1)=\begin{cases}0&\text{for }t<1\\t^2&\text{for }t\ge1\end{cases}[/tex]
so we can subtract this to get
[tex]t^2\bigg(\Theta(t)-\Theta(t-1)\bigg)=\begin{cases}0&\text{for }t<0\text{ and }t>1\\t^2&\text{for }t\in[0,1]\end{cases}[/tex]
Finally, we just add another shifted Heaviside function scaled by [tex]t[/tex]:
[tex]f(t)\equiv t^2\bigg(\Theta(t)-\Theta(t-1)\bigg)+t\Theta(t-1)[/tex]
which can also be expressed as
[tex]f(t)\equiv t^2\Theta(t)+(t-t^2)\Theta(t-1)[/tex]
