Worked examples
The question
Question: Water flows into a tank at r(t) = 4t + 5 litres per minute.
How much flows in from t = 0 to t = 4?
1 The net change
r(t) is a rate: how fast it flows in.
The total that flows in is the integral of the rate, from 0 to 4.
amount = ∫ from 0 to 4 of (4t + 5) dt
amount = ∫ from a to b of r(t) dt
The antiderivative: add 1 to each power, divide by the new power.
Now we have 2 t² + 5t.
= (2t2 + 5t) from 0 to 4
Put in 4, then take away the same with 0.
= (2(4)2 + 5(4)) − (2(0)2 + 5(0))
That is 52 minus 0: 52.
= 52 − 0 = 52
So 52 litres flow in.
amount = 52 litres
Careful!
r is the rate, not the amount.
The amount is the integral of the rate.
Careful: not r(4) − r(0) ✗
2 Check
Let's check with a calculator's integral: about 52.
Right!
calculator: ∫ from 0 to 4 of r(t) dt ≈ 52
Let's look back.
First: the net change.
Last: check.
Answer Careful: not r(4) − r(0) ✗
▶ Watch it solved
The question
Question: find the displacement and the total distance for v(t) = 2t − 4 from t = 0 to t = 5.
v(t) = 2t − 4
Displacement is where it ends up, compared with the start.
Distance counts every step, forward and back.
displacement: where it ends up
distance: how far it travels
1 The net change
The displacement is the integral of v.
Going back counts as negative.
displacement = ∫ from 0 to 5 of (2t − 4) dt
The antiderivative: add 1 to each power, divide by the new power.
Now we have t² − 4t.
= (t2 − 4t) from 0 to 5
Put in 5, then take away the same with 0.
= (52 − 4(5)) − (02 − 4(0))
That is 5 minus 0: 5.
= 5 − 0 = 5
So the displacement is 5 m.
displacement = 5 m
2 Where it turns around
It turns around where v changes sign.
First, where is v = 0?
2t − 4 = 0
Factor it.
2(t − 2) = 0
v is 0 at t = 2.
t = 2
Draw v.
From 0 to 2, v is negative: it moves back.
From 2 to 5, v is positive: it moves forward.
3 The total distance
From 0 to 2: the integral is −4, the same way as before.
∫ from 0 to 2 of (2t − 4) dt = −4
From 2 to 5: the integral is 9, the same way as before.
∫ from 2 to 5 of (2t − 4) dt = 9
Careful!
The integral of v is the displacement.
Going back cancels going forward.
For the distance, add each piece without its sign.
Careful: not distance = 5 ✗
Distance counts every piece as positive.
Add them: 13 m.
distance = |−4| + |9| = 13 m
4 Check
Let's check: the pieces with their signs add up to 5, the displacement.
Right!
−4 + 9 = 5
Let's look back.
First: the net change.
Then: where it turns around.
Then: the total distance.
Last: check.
Answer displacement = 5 m
▶ Watch it solved
The question
Question: find the displacement and the total distance for v(t) = t² − 4t + 3 from t = 0 to t = 4.
v(t) = t2 − 4t + 3
Displacement is where it ends up, compared with the start.
Distance counts every step, forward and back.
displacement: where it ends up
distance: how far it travels
1 The net change
The displacement is the integral of v.
Going back counts as negative.
displacement = ∫ from 0 to 4 of (t2 − 4t + 3) dt
The antiderivative: add 1 to each power, divide by the new power.
Now we have t³ over 3 − 2 t² + 3t.
= (t33 − 2t2 + 3t) from 0 to 4
Put in 4, then take away the same with 0.
= (433 − 2(4)2 + 3(4)) − (033 − 2(0)2 + 3(0))
That is 4/3 minus 0: 4/3.
= 43 − 0 = 43
So the displacement is 4/3 m.
displacement = 43 m
2 Where it turns around
It turns around where v changes sign.
First, where is v = 0?
t2 − 4t + 3 = 0
We need two numbers that multiply to 3 and add to −4: −1 and −3.
(t − 1)(t − 3) = 0
v is 0 at t = 1 and t = 3.
t = 1 or 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 4, v is positive: it moves forward.
3 The total distance
From 0 to 1: the integral is 4/3, the same way as before.
∫ from 0 to 1 of (t2 − 4t + 3) dt = 43
From 1 to 3: the integral is −4/3, the same way as before.
∫ from 1 to 3 of (t2 − 4t + 3) dt = −43
From 3 to 4: the integral is 4/3, the same way as before.
∫ from 3 to 4 of (t2 − 4t + 3) dt = 43
Careful!
The integral of v is the displacement.
Going back cancels going forward.
For the distance, add each piece without its sign.
Careful: not distance = 43 ✗
Distance counts every piece as positive.
Add them: 4 m.
distance = |43| + |−43| + |43| = 4 m
4 Check
Let's check: the pieces with their signs add up to 4/3, the displacement.
Right!
43 − 43 + 43 = 43
Let's look back.
First: the net change.
Then: where it turns around.
Then: the total distance.
Last: check.
Answer displacement = 43 m
▶ Watch it solved
