Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.
The big idea in one line
Units 2 to 5 started with an amount and asked how fast it changes. Unit 6 turns this round. You know how fast something changes, and you want to know how much it changed.
The answer comes from area. If you graph the rate of change against time, the area between that graph and the time axis is the total change.
Starting simple: a constant rate
A hose fills a paddling pool at a steady 25 litres per minute for 8 minutes. The water added is
25 litres per minute × 8 minutes = 200 litres
Now draw the rate as a graph: a horizontal line at height 25 from t = 0 to t = 8. The region under it is a rectangle 8 wide and 25 tall. Its area is 8 × 25 = 200. The area of the rectangle is the water added.
Look at the units as well as the numbers. The height is measured in litres per minute and the width in minutes. Multiplying them gives
(litres/minute) × minutes = litres
which is exactly the unit of the quantity that is changing.
Rates that change in steps
Snow falls at 2 cm per hour for 3 hours, then at 0.5 cm per hour for the next 4 hours. Each step on the rate graph gives its own rectangle:
| Interval | Rate | Width | Area (change in depth) |
|---|---|---|---|
| 0 to 3 h | 2 cm/h | 3 h | 6 cm |
| 3 to 7 h | 0.5 cm/h | 4 h | 2 cm |
| Total | 8 cm |
Adding the rectangles adds the changes. The snow got 8 cm deeper over the 7 hours.
Rates that change smoothly
Most rates do not jump in steps; they vary all the time. You can still think of the region under the graph as made of very thin strips. On a thin enough strip the rate is almost constant, so the strip’s area is almost the change during that short time. Adding all the strips gives the total change.
Key fact. The area of the region between the graph of a rate of change and the horizontal axis, over an interval, gives the accumulated change in the quantity over that interval.
If the graph is made of straight segments and circular arcs, you can find that area exactly with geometry: rectangles, triangles, trapezoids and parts of circles. For curved graphs you will approximate the area with rectangles in Topic 6.2, and find it exactly later in the unit.
Units of an accumulated change
The rule is the same every time:
units of the area = units of the rate × units of the input variable
| Rate of change | Input | Area means | Units of area |
|---|---|---|---|
| Velocity, km/h | time, h | change in position | km |
| Power into a battery, kW | time, h | change in stored energy | kWh |
| People entering, people/min | time, min | change in number of people | people |
| Rate of change of temperature, °C/h | time, h | change in temperature | °C |
If your units come out as “per” something, you have multiplied wrongly or mixed up the rate with the quantity.
The sign of an accumulated change
A rate can be negative. Water can drain out, a battery can discharge and a drone can fly downwards.
- If the rate is positive on an interval, the graph is above the axis and the quantity increases. The accumulated change is positive.
- If the rate is negative on an interval, the graph is below the axis and the quantity decreases. The accumulated change is negative.
The region below the axis still has a positive geometric area. You attach the minus sign because the quantity is going down. So the net change over a long interval is
net change = (area above the axis) − (area below the axis)
Sometimes you want the total amount of change, ignoring direction. For motion this is the total distance travelled. Then you add the sizes of all the areas instead:
total distance = (area above) + (area below)
Worked example 1: a home battery
A home battery stores solar energy. Let P(t) be the rate, in kilowatts (kW), at which energy flows into the battery, where t is hours after 9:00. P is positive while the battery charges and negative while it supplies the house. The graph of P is made of straight segments joining (0, 0), (2, 3), (5, 3), (6, 0), (8, −2) and (10, −2). At t = 0 the battery holds 4 kWh.
(a) What are the units of the area between P and the t-axis? kilowatts × hours = kilowatt-hours (kWh), a unit of energy.
(b) Find the change in stored energy from t = 0 to t = 6. The region above the axis splits into three shapes:
- Triangle from t = 0 to t = 2: ½ × 2 × 3 = 3
- Rectangle from t = 2 to t = 5: 3 × 3 = 9
- Triangle from t = 5 to t = 6: ½ × 1 × 3 = 1.5
Total: 3 + 9 + 1.5 = 13.5 kWh. The graph is above the axis, so this is an increase. Between 9:00 and 15:00 the battery gained 13.5 kWh.
(c) Find the change from t = 6 to t = 10.
- Triangle from t = 6 to t = 8: ½ × 2 × 2 = 2
- Rectangle from t = 8 to t = 10: 2 × 2 = 4
The graph is below the axis, so the change is −(2 + 4) = −6 kWh. Between 15:00 and 19:00 the battery supplied 6 kWh to the house.
(d) How much energy is stored at t = 10? Start with the amount at t = 0 and add each change:
4 + 13.5 + (−6) = 11.5 kWh
Check. The net change over the whole day is 13.5 − 6 = 7.5 kWh, and 4 + 7.5 = 11.5. Notice also that the stored energy is greatest at t = 6: before that the rate is positive, after it the rate is negative. At t = 6 the battery held 4 + 13.5 = 17.5 kWh.
Worked example 2: a drone’s height
A drone moves straight up and down. Its vertical velocity is v(t) metres per second, t seconds after a timer starts, for 0 ≤ t ≤ 12. Positive v means rising. The graph of v is:
- a semicircle above the axis from t = 0 to t = 4, with centre (2, 0) and radius 2;
- a straight segment from (4, 0) to (6, −3);
- the horizontal segment v = −3 from t = 6 to t = 10;
- a straight segment from (10, −3) to (12, 0).
At t = 0 the drone is 30 m above the ground.
(a) Find the drone’s change in height from t = 0 to t = 12.
- Above the axis. A semicircle of radius 2 has area ½ × π × 2² = 2π. Units: (m/s) × s = m. The drone rises 2π ≈ 6.283 m.
- Below the axis. Two triangles, each ½ × 2 × 3 = 3, and a rectangle 4 × 3 = 12. Total 3 + 12 + 3 = 18. The drone falls 18 m.
- Net change = 2π − 18 ≈ −11.717 m. The drone ends about 11.7 m lower than it started.
(b) Find the total distance the drone travels. Add the sizes: 2π + 18 ≈ 24.283 m.
(c) How high is the drone at t = 12? 30 + 2π − 18 = 12 + 2π ≈ 18.283 m above the ground.
Check. The highest point is at t = 4, where v changes from positive to negative: 30 + 2π ≈ 36.283 m. From there it falls 18 m, to 18.283 m. The two routes agree.
Why equal scales matter. “Semicircle” describes the shape on the graph, so its radius is 2 units on each axis. On an exam graph, trust the description and the labelled points, not your ruler.
Worked example 3: writing an interpretation
Sometimes you are given an area and asked what it means.
Question. R(t) is the rate, in passengers per minute, at which passengers enter a railway station, where t is minutes after 7:00. The area of the region between the graph of R and the t-axis from t = 10 to t = 30 is 2,450. Interpret this value.
Answer. Between 7:10 and 7:30, 2,450 passengers entered the station.
A full interpretation has four parts:
- What accumulated: the number of passengers who entered.
- The interval, in the context’s own terms: 7:10 to 7:30.
- The size, and the direction if the rate can be negative.
- The units: passengers, because (passengers/minute) × minutes = passengers.
Two wrong answers to avoid. “There are 2,450 passengers in the station at 7:30” confuses the change with the amount: some may have boarded trains, and some were there before 7:10. “Passengers enter at 2,450 per minute” confuses the area with the rate. The average rate over those 20 minutes is 2,450 ÷ 20 = 122.5 passengers per minute.
Common misconceptions
- “The area is the amount at the end.” The area is the change. Add it to the starting amount.
- “Area below the axis adds on.” If the rate is negative, the quantity decreases. Subtract that area for the net change.
- “Net change and total distance are the same.” They are equal only if the rate never changes sign. Otherwise total distance is larger.
- Units copied from the rate. The area under a graph in litres per minute is in litres, not litres per minute.
- Reading the height as the answer. The height of the rate graph at one time is a rate at that instant. The area over an interval is a change.
- Ignoring the scale. A gap of 1 on the t-axis might be 10 minutes. Use the axis values, not the number of grid squares.
- Forgetting the interval in an interpretation. “The battery gained 13.5 kWh” is incomplete. Say between which times.
Where this leads
In Topic 6.2 you will estimate areas under curved rate graphs with rectangles and trapezoids, when geometry alone is not enough. Topic 6.3 names the exact area a definite integral. Later topics connect it to antiderivatives, and Unit 8 builds on the distance and net change ideas here. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.
Try the practice questions now, then use the revision notes and the checklist to consolidate.
Quick questions
Is this page for Calculus AB or Calculus BC?
Both. Topic 6.1 is common content, so the same page serves AB and BC students.
Is the area the amount of the quantity, or the change in it?
The change. To get the amount at the end, add the accumulated change to the amount you started with.
Can an area be negative?
A geometric area is never negative. But when the rate graph is below the axis, the quantity is decreasing, so that area counts as a negative change. Keep the size and the sign separate in your head.