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Total change from a rate: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

“Use the definite integral of a rate of change to find the total or net change in a quantity over an interval”

8 min read

Quick answer

If f′(t)f'(t) is the rate of change of a quantity, then ∫abf′(t) dt=f(b)−f(a)\int_a^b f'(t)\,dt = f(b)-f(a) is the net change over [a,b][a,b]. Adding the initial value recovers the quantity itself: f(b)=f(a)+∫abf′(t) dtf(b)=f(a)+\int_a^b f'(t)\,dt.

Jump to a section
  1. What this dot point is asking
  2. The accumulation principle
  3. Recovering the quantity itself
  4. Reading the rate graph
  5. Net versus total accumulation
  6. Exam-style questions

Syllabus placement. In the SCSA Mathematics Methods ATAR Year 12 syllabus, integrals are Unit 3 (Topic 3.2), not Unit 4. This page sits in our Unit 4 folder for historical reasons; the content is Unit 3.

What this dot point is asking

SCSA Unit 4 applies the definite integral to accumulation: recovering total change from a rate. This is the Fundamental Theorem of Calculus read in context, and this dot point asks you to interpret and compute it in real situations such as flow, growth and consumption. It is examined in both sections.

The accumulation principle

The Fundamental Theorem of Calculus says that integrating a rate of change gives the net change in the original quantity.

Net change from a rate

If R(t)=f′(t)R(t)=f'(t) is the rate of change of ff, then

∫abR(t) dt=f(b)−f(a),\int_a^b R(t)\,dt = f(b)-f(a),

the net change in ff from t=at=a to t=bt=b.

The units confirm the interpretation: a rate in litres per minute integrated over minutes gives litres. A positive integral means a net increase; a negative integral means a net decrease.

This is the same idea as the kinematics result that integrating velocity gives displacement, generalised to any quantity and its rate. A flow rate integrated over time gives a volume; a marginal cost integrated over output gives a total cost; a rate of population change integrated over time gives the change in population. In each case the definite integral of the rate, read with its units, is the net change in the underlying quantity, and the Fundamental Theorem of Calculus is what guarantees the equality. Recognising a quantity described "per unit time" or "per unit output" as a rate to be integrated is the key modelling step these questions test.

Net change in volume from a flow rate

Water flowing into a tank

Water flows into a tank at R(t)=20−2tR(t)=20-2t litres per minute for 0≤t≤100\le t\le 10. Find the net change in volume over the first 88 minutes.

Integrate the rate over [0,8][0,8].

∫08(20−2t) dt=[20t−t2]08=(160−64)−0=96.\int_0^8 (20-2t)\,dt = \left[20t-t^{2}\right]_0^8 = (160-64) - 0 = 96.

The net change is 9696 litres added over the first 88 minutes. Markers reward integrating the rate, applying the limits, and stating the units.

Recovering the quantity itself

If the initial value of the quantity is known, add it to the accumulated change to find the quantity at a later time.

Recovering a total from a rate and initial value

Population from a growth rate

A colony grows at R(t)=300e0.1tR(t)=300e^{0.1t} organisms per day, starting at P(0)=5000P(0)=5000. Find the population after 1010 days.

Compute the accumulated change.

∫010300e0.1t dt=300[10.1e0.1t]010=3000(e1−e0)=3000(2.71828−1).\int_0^{10} 300e^{0.1t}\,dt = 300\left[\frac{1}{0.1}e^{0.1t}\right]_0^{10} = 3000\left(e^{1}-e^{0}\right) = 3000(2.71828-1).

≈3000×1.71828≈5155.\approx 3000\times 1.71828 \approx 5155.

Add the initial population.

P(10)=5000+5155=10155 organisms (to the nearest whole).P(10) = 5000 + 5155 = 10155 \text{ organisms (to the nearest whole)}.

Markers reward integrating the rate, the 10.1\tfrac{1}{0.1} factor, and adding the initial value.

Reading the rate graph

Many SCSA questions give the rate as a graph rather than a formula and ask for the total change as an area. Because ∫abR(t) dt\int_a^b R(t)\,dt is the signed area under the rate curve, the net change is the area above the time axis minus the area below it. A region where the rate is positive adds to the quantity; a region where it is negative subtracts. When the rate graph is made of straight lines, the area can be found with triangle and trapezium formulas instead of integration, which is a quick calculator-free route. Always check whether the rate dips below the axis, since that signals a period when the quantity is decreasing.

Net versus total accumulation

When the rate changes sign, the definite integral gives the net change, with inflow and outflow cancelling. If a question asks for the total amount that flowed in (ignoring outflow), split the integral where the rate is zero and add the absolute values of the positive and negative parts, exactly as for distance travelled in kinematics.

Common errors
Forgetting the initial value
The integral gives only the change; add the starting amount to find the quantity itself.
Confusing net change with total flow
When the rate changes sign, the integral nets inflow against outflow; split it if the total is required.
Dropping units
A net-change answer should carry the units of the original quantity.
Integrating the quantity instead of the rate
You integrate the rate R(t)R(t), which is the derivative of the quantity, to recover the change.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks
Calculator-assumed. Water leaks from a tank at a rate R(t)=8e−0.2tR(t)=8e^{-0.2t} litres per hour. (a) Find the total volume that leaks out during the first 55 hours. (b) If the tank initially holds 6060 litres, how much remains after 55 hours?
Show worked answer →

A net-change-from-a-rate question with an exponential rate.

(a) Total leaked =∫058e−0.2t dt=8[1−0.2e−0.2t]05=−40(e−1−1)=40(1−e−1)≈40×0.632=25.3=\displaystyle\int_{0}^{5}8e^{-0.2t}\,dt=8\left[\dfrac{1}{-0.2}e^{-0.2t}\right]_{0}^{5}=-40\bigl(e^{-1}-1\bigr)=40(1-e^{-1})\approx 40\times 0.632=25.3 litres.

(b) Remaining =60−25.3=34.7=60-25.3=34.7 litres.

Markers reward integrating the rate, the 1−0.2\dfrac{1}{-0.2} factor, and subtracting the loss from the initial volume.

WACE 2024-style5 marks
Calculator-free. The rate of change of a quantity is R(t)=4t−t2R(t)=4t-t^{2} units per second on 0≤t≤40\le t\le 4. (a) Find the net change over [0,4][0,4]. (b) State at what time the quantity is greatest and justify.
Show worked answer →

A net-change calculation with a turning point.

(a) Net change =∫04(4t−t2) dt=[2t2−t33]04=32−643=323=\displaystyle\int_{0}^{4}(4t-t^{2})\,dt=\left[2t^{2}-\dfrac{t^{3}}{3}\right]_{0}^{4}=32-\dfrac{64}{3}=\dfrac{32}{3} units.

(b) The quantity increases while R(t)>0R(t)>0 and decreases while R(t)<0R(t)<0. R(t)=t(4−t)=0R(t)=t(4-t)=0 at t=0t=0 and t=4t=4, and R(t)>0R(t)>0 throughout 0<t<40<t<4, so the quantity rises across the whole interval and is greatest at t=4t=4.

Markers reward the definite integral and a justification using the sign of the rate.

Practise this

Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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