Step 1: Sketch the two pieces of $f(t)$.
Between $t=0$ and $t=2$, $f(t)$ sits at a constant height of $1$, so this part traces a flat rectangle. Between $t=2$ and $t=3$, $f(t)=-t+3$ is a straight line that starts at $f(2)=1$ and ends at $f(3)=0$, so this part traces a downward sloping segment reaching zero exactly at $t=3$. Outside $[0,3]$ the function is zero, so it adds nothing to the area.
Step 2: Read the area under the flat part as a rectangle.
This piece is a rectangle of width $2-0=2$ and height $1$, so its area is
\[ \text{Area}_1 = 2\times1 = 2 \]
Step 3: Read the area under the sloped part as a triangle.
The sloped piece goes from the point $(2,1)$ down to $(3,0)$, forming a right triangle with base $3-2=1$ along the $t$-axis and height $1$ (the value at $t=2$). Its area is
\[ \text{Area}_2 = \frac{1}{2}\times1\times1 = 0.5 \]
Step 4: Add the two areas.
The integral over all $t$ is just the sum of the areas of these two shapes, since $f$ is zero everywhere else:
\[ \int_{-\infty}^{\infty} f(\tau)\,d\tau = 2+0.5 = 2.5 \]
\[ \boxed{2.5} \]
