Step by step
The question
Question: find the displacement and the total distance for v(t) = t² − 1 from t = 0 to t = 3.
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.
Going back counts as negative.
The antiderivative: add 1 to each power, divide by the new power.
Now we have t³ over 3 − t.
Put in 3, then take away the same with 0.
That is 6 minus 0: 6.
So the displacement is 6 ft.
2 Where it turns around
It turns around where v changes sign.
First, where is v = 0?
x² − 1 is a difference of two squares: x² minus 1 squared.
So it is (x − 1)(x + 1).
v is 0 at t = 1.
Draw v.
From 0 to 1, v is negative: it moves back.
From 1 to 3, v is positive: it moves forward.
3 The total distance
From 0 to 1: the integral is −2/3, the same way as before.
From 1 to 3: the integral is 20/3, the same way as before.
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: 22/3 ft.
4 Check
Let's check: the pieces with their signs add up to 6, the displacement.
Right!
Let's look back.
First: the net change.
Then: where it turns around.
Then: the total distance.
Last: check.