Step by step
The question
Question: Sand flows into a pile at r(t) = 9 - t^2 cubic metres per hour.
How much flows in from t = 0 to t = 3?
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 3.
amount = ∫ from 0 to 3 of (9 − t2) 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 9t − t³ over 3.
= (9t − t33) from 0 to 3
Put in 3, then take away the same with 0.
= (9(3) − 333) − (9(0) − 033)
That is 18 minus 0: 18.
= 18 − 0 = 18
So 18 cubic metres flow in.
amount = 18 cubic metres
Careful!
r is the rate, not the amount.
The amount is the integral of the rate.
Careful: not r(3) − r(0) ✗
2 Check
Let's check with a calculator's integral: about 18.
Right!
calculator: ∫ from 0 to 3 of r(t) dt ≈ 18
Let's look back.
First: the net change.
Last: check.
Answer
Careful: not r(3) − r(0) ✗