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Calculus AB · Unit 6 · 6.1
AP Calculus · Integration and Accumulation of Change
Exploring Accumulations of Change
The area under a rate graph is the amount that accumulated — and area below the axis takes some back.
key idea$\text{Area under a RATE graph}=\text{the accumulated AMOUNT}$
What it says
Accumulation of change
If a graph shows a rate, then the area beneath it over an interval is the total amount that accumulated. Rate times time gives amount — and area is exactly a product of a height (the rate) and a width (the time).
Signed area
Area below the horizontal axis counts as negative. A negative rate means the quantity is decreasing, so that stretch subtracts from the total. This is why accumulation gives net change rather than a running total of activity.
| If the graph shows | then the area gives | Units |
|---|
| velocity (m/s) against time (s) | displacement in metres | m/s $\times$ s $=$ m |
| flow rate (L/min) against time (min) | volume in litres | L/min $\times$ min $=$ L |
| marginal cost ($/item) against items | total cost in dollars | $/item $\times$ items $=$ $ |
| population growth rate (people/yr) against years | change in population | people/yr $\times$ yr |
| power (watts) against time (s) | energy in joules | W $\times$ s $=$ J |
Multiply the two axis units together and the answer's units appear. That product is also a reliable check that you have read the graph correctly.
Reading the notation
Rate graphA graph whose height is a rate of change. Its area carries meaning; its height does not directly give the amount.
Accumulated changeThe net amount added over an interval — the signed area.
Net changeFinal amount minus initial amount. Equal to the signed area, not to the total area.
Total amountRequires a starting value: initial amount plus accumulated change. A rate graph alone can never supply it.
Signed areaArea above the axis counts positive, below counts negative.
height versus area
- A rate graph's height is a speed of change, not an amount.
- Its area is the amount — and that is what accumulates.
- Below the axis subtracts. Net change can be zero with a great deal happening.
- To get a total, you must be told the starting value.
Worked examples
EX 1
Area as distance
constant then changingA car's velocity is $60$ km/h for the first $2$ hours, then $80$ km/h for the next hour. Find the distance travelled using areas.
1
Draw the rate graph
A step graph: height $60$ over $[0,2]$, height $80$ over $[2,3]$.
2
First rectangle
Height $60$, width $2$, so area $=120$.
3
Second rectangle
Height $80$, width $1$, so area $=80$.
5
Attach units
km/h $\times$ h $=$ km, so the car travelled $200$ km.
EX 2
Signed area and net change
a rate that goes negativeWater flows into a tank at $5$ L/min for $4$ minutes, then out at $3$ L/min for $6$ minutes. Find the net change.
1
Read the signs
Inflow is a positive rate; outflow is negative: $-3$ L/min.
2
First region
$5\times4=+20$ litres, above the axis.
3
Second region
$-3\times6=-18$ litres, below the axis.
4
Add the signed areas
$20+(-18)=+2$ litres.
5
Interpret
The tank has 2 litres more than it started with. Note that $38$ litres of water moved in total — net change and total activity are different questions.
EX 3
Net change is not the total
starting value neededA tank's inflow rate over $[0,10]$ minutes encloses a signed area of $+45$ litres. How much water is in the tank at $t=10$?
1
Identify what the area gives
The change in volume over the interval: $+45$ litres.
2
Ask what is missing
The volume at $t=0$. The area says nothing about it.
3
State the relationship
$V(10)=V(0)+45$.
4
Answer honestly
Unanswerable as posed. If $V(0)=12$, then $V(10)=57$ litres.
5
The general principle
Accumulation gives change. Converting change into a total always requires one known value.
EX 4
Zero net change
symmetric rateA particle's velocity is $+4$ m/s for $3$ seconds, then $-4$ m/s for $3$ seconds. Find the displacement and the distance travelled.
1
First region
$4\times3=+12$ metres.
2
Second region
$-4\times3=-12$ metres.
3
Displacement is the signed sum
$12-12=0$ metres — the particle is back where it started.
4
Distance uses magnitudes
$\lvert12\rvert+\lvert-12\rvert=24$ metres.
5
The distinction
Displacement $0$, distance $24$. “Nothing accumulated” and “nothing happened” are entirely different claims.
Boundary cases
Height is a rate, area is an amount
A tall rate graph means change is fast, not that the quantity is large. Read the area for the amount.
Below the axis subtracts
A negative rate removes from the accumulated total. Signed area is what makes net change meaningful.
Accumulation gives change, not the total
You need one known value to convert. A rate graph alone cannot tell you how much is there.
Net change and total activity differ
Example 4 accumulates zero while the particle travels 24 metres. Ask which the question wants.