Step by step
The question
Question: find the displacement and the total distance for v(t) = t² − 4t + 3 from t = 0 to t = 5.
Displacement is where it ends up, compared with the start.
Distance counts every step, forward and back.
1 The net change
The displacement is the integral of v: in a tiny time it moves about v times that time.
Going back counts as negative.
The antiderivative: add 1 to each power, divide by the new power.
Now we have t³ over 3 − 2 t² + 3t.
Put in 5, then take away the same with 0.
That is 20/3 minus 0: 20/3.
So the displacement is 20/3 ft.
2 Where it turns around
It turns around where v changes sign.
First, where is v = 0?
To factor t² − 4t + 3, the rule: we need two numbers that multiply to 3 and add to −4.
They are −1 and −3: they multiply to 3 and add to −4.
So we write it like this.
v is 0 at t = 1 and t = 3.
Draw v.
From 0 to 1, v is positive: it moves forward.
From 1 to 3, v is negative: it moves back.
From 3 to 5, v is positive: it moves forward.
3 The total distance
Call the antiderivative F.
Work it out at 1 and 3 too.
From 0 to 1: the same antiderivative at 1, minus at 0.
4/3 minus 0 is 4/3.
From 1 to 3: the same antiderivative at 3, minus at 1.
0 minus 4/3 is −4/3.
From 3 to 5: the same antiderivative at 5, minus at 3.
20/3 minus 0 is 20/3.
Careful!
The integral of v is the displacement.
Going back cancels going forward.
For the distance, add each piece without its sign.
Distance counts every piece as positive.
Add them: 28/3 ft.
4 Check
Let's check: the pieces with their signs add up to 20/3, the displacement.
Right!
Let's look back.
First: the net change.
Then: where it turns around.
Then: the total distance.
Last: check.