A bird flies for 4 seconds with velocity $v(t)=t-2$. What total distance does it cover from $t=0$ to $t=4$?

What you have to watch for with “distance”

Distance is always nonnegative, so when velocity changes sign you cannot just integrate $v(t)$ directly. Instead, find where $v(t)=0$ and add up the areas using speed $|v(t)|.

Find when the bird changes direction

Set the velocity to zero: $$t-2=0 \Rightarrow t=2.$$ So the bird moves in the negative direction from $t=0$ to $t=2$, then in the positive direction from $t=2$ to $t=4$.

Compute total distance using $\int |v(t)|\, dt$

On $[0,2]$, $v(t)<0$ so $|v(t)|=2-t$: $$\text{Distance}_1=\int_0^2 (2-t)\, dt=\left[2t-\frac{t^2}{2}\right]_0^2=4-2=2.$$

On $[2,4]$, $v(t)>0$ so $|v(t)|=t-2$: $$\text{Distance}_2=\int_2^4 (t-2)\, dt=\left[\frac{t^2}{2}-2t\right]_2^4=0-(-2)=2.$$

Add them: $$\text{Total distance}=2+2=4.$$

Quick check (why displacement is different)

If you integrate velocity without absolute value, $$\int_0^4 (t-2)\, dt=0,$$ which means the net displacement is $0$, not that the bird did not move.