Loav · VCE Mathematical Methods · Units 3 & 4
M34-INT-03 · Integration · Current study design
1. A rate becomes a change, then a quantity
If Q(t) is a changing quantity and Q′(t) = r(t), then ∫ab r(t) dt = Q(b) − Q(a). Integrating a rate gives the change in the quantity, not automatically its final amount. To recover the final amount, add the starting amount: Q(b) = Q(a) + ∫ab r(t) dt. This distinction is central to interpreting application questions.
Use units to decide what the integral can mean. Integrating litres per minute with respect to minutes gives litres. Integrating metres per second with respect to seconds gives metres. Integrating temperature in degrees Celsius over hours instead gives degree-hours, so a further division by elapsed time is required if the question asks for average temperature. An integral symbol does not erase the units of either axis.
The sign convention must be explicit. A positive net flow may mean fluid entering a tank, with negative values representing net outflow. A positive velocity may mean motion in the chosen positive direction. An expression described as a non-negative loss rate has a different convention: its integral measures the amount lost, which must be subtracted from the initial quantity. Read the definition before deciding whether to add or subtract.
2. Accumulation can itself be a function
Define A(t) = ∫at r(u) du for continuous r. The letter u is a dummy variable; t is the moving upper endpoint. This notation avoids using t simultaneously as the integration variable and the endpoint. A(a) = 0, because no interval has accumulated yet. If an initial amount Q(a) is given, then Q(t) = Q(a) + A(t).
The fundamental theorem gives A′(t) = r(t). A positive rate makes A increase; a negative rate makes A decrease; a zero rate gives a stationary point of A, though not necessarily a maximum or minimum. To classify a candidate, inspect the sign of r on both sides. A change from positive to negative gives a local maximum of accumulated quantity.
For an absolute maximum or minimum on a closed time interval, check endpoints as well as stationary candidates. A model may begin with its largest quantity and decline throughout, so there need not be an interior turning point. Keep the model’s time domain in view. Algebraic roots outside it do not represent valid events, and an extrapolated negative volume can signal that the physical model no longer applies.
Worked example 1: A tank with changing net flow
A tank contains 50 L at t = 0. Its net inflow is r(t) = 12 − 2t L min−1 for 0 ≤ t ≤ 8, where positive values mean net entry. The change over the first eight minutes is ∫08(12 − 2t) dt = [12t − t2]08 = 32 L. The final amount is therefore 82 L.
The amount function is V(t) = 50 + 12t − t2. Its derivative changes from positive to negative at t = 6, so V has an interior maximum there. Comparing V(0) = 50, V(6) = 86 and V(8) = 82 confirms that the maximum is 86 L. This assumes the tank can hold at least that amount and no overflow changes the model.
The increase before t = 6 is 36 L and the decrease after that time is 4 L. Thus the accumulated magnitude of net volume change is 40 L, while net change is 32 L. Neither figure determines total water entering through all pipes if simultaneous inflow and outflow are hidden within the net-rate model.
3. Displacement and distance use different totals
For straight-line motion, velocity is the rate of change of position. Its signed integral is displacement: final position minus initial position. Distance travelled adds the non-negative lengths of movement in either direction. If velocity changes sign, split at the turning times and add the magnitudes of the signed displacements on each segment.
A stationary time satisfies v(t) = 0, but direction reverses only when velocity changes sign. Touching zero and returning to the same sign gives a temporary stop without reversal. This is the motion version of a graph touching an axis without changing which side it lies on. Check signs in the intervals, rather than assuming every solution of v = 0 requires a change of direction.
Average velocity equals displacement divided by elapsed time. Average speed equals total distance divided by elapsed time. They coincide when motion stays in one positive direction; when the direction changes they generally differ. Average speed cannot be negative, whereas average velocity can. An unknown initial position affects final position but not displacement or distance over a specified interval.
Worked example 2: One journey, two different averages
A particle has v(t) = 3t2 − 12t + 9 m s−1 for 0 ≤ t ≤ 4. Factor v(t) = 3(t − 1)(t − 3). Its sign is positive on (0, 1), negative on (1, 3), then positive on (3, 4). An antiderivative is F(t) = t3 − 6t2 + 9t.
The values F(0), F(1), F(3), F(4) are 0, 4, 0, 4. Hence the segment displacements are +4 m, −4 m and +4 m. Total displacement is 4 m, while distance is 4 + 4 + 4 = 12 m. Average velocity is 4/4 = 1 m s−1; average speed is 12/4 = 3 m s−1.
If the initial position were 7 m, the final position would be 11 m. That initial value is added once to the net displacement; it is not added to each distance segment. The roots 1 and 3 are turning times because the velocity changes sign at both.
4. Average value is accumulated value divided by width
The average value of a continuous function f over [a, b], with a < b, is favg = [1/(b − a)]∫ab f(x) dx. Multiplying the average height by the interval width reproduces the signed integral. For a non-negative function, this gives a rectangle with the same area as the region under the curve.
This formula averages over the independent variable. An average over time need not equal an average over distance or volume, because equal increments of one variable may not correspond to equal increments of another. Always identify the variable with respect to which the average is requested. If a table samples unequal time intervals, simply averaging its listed values is not a time-weighted average.
Even on equal intervals, the average of just the two endpoint values is generally insufficient. That calculation ignores the function’s behaviour between the endpoints. It is exact for a linear function, but a curved graph may rise well above both endpoint values. The integral includes all the intermediate values with the appropriate interval weighting.
Worked example 3: Average temperature over a full daytime interval
A model gives T(t) = 18 + 6sin(πt/12) °C for 0 ≤ t ≤ 12, where t is hours after the start of observation. The mean over those twelve hours is (1/12)∫012[18 + 6sin(πt/12)] dt.
An antiderivative is 18t − (72/π)cos(πt/12). The integral is 216 + 144/π degree-hours. Dividing by 12 gives an average temperature of 18 + 12/π °C, approximately 21.82 °C. This lies between the minimum 18 °C and maximum 24 °C, a useful reasonableness check.
The endpoint temperatures are both 18 °C, so their arithmetic mean misses the warmer middle of the interval. The average value is a temperature, while the integral alone has degree-hour units. Using radians is essential: the model’s πt/12 argument makes the middle time t = 6 correspond to π/2.
5. Keep three common “averages” separate
For a quantity Q(t), average rate of change is [Q(b) − Q(a)]/(b − a). For its derivative r = Q′, average value of the rate is [1/(b − a)]∫r, which equals that same average rate of change by the fundamental theorem. However, the average value of Q itself is [1/(b − a)]∫Q, a different calculation with different units.
In a practical answer, state the quantity being averaged rather than writing only “average”. If asked for average population, integrate the population function; if asked for average population-growth rate, integrate its derivative or use change divided by time. The distinction prevents a correct integral from answering the wrong question.
Finish with the model assumptions. A formula may be valid only during a specified observation period, with no capacity limit reached and a fixed sign convention. Report a result within that period, attach units, and round only after evaluating the full expression. If the question asks when an accumulated amount reaches a target, solve the accumulation equation and retain only times in the given domain.
Practise, then check
Show the mathematical reason for each answer. Decide whether the result is a function, a signed quantity or a non-negative area before calculating.
- A reservoir starts with 100 L and has net inflow 3e−t/2 L min−1 for 0 ≤ t ≤ 4. Find its amount after four minutes.
- Using the same rate, find the average net inflow over those four minutes.
- Let A(t) = ∫2t(u − 4) du for 2 ≤ t ≤ 6. Find A(2), A′(t) and the time of its minimum.
- A particle has velocity v(t) = 2t − 4 m s−1 on [0, 5]. Find displacement and distance.
- Find the average value of f(x) = 3x2 + 2 on [0, 2]. Why is it different from averaging the endpoint values?
- A tank loses water at a positive loss rate of 2 L min−1 for three minutes, then 5 L min−1 for two minutes. Its initial amount is 40 L. Find the final amount and average loss rate.
- A student integrates a temperature function over time and reports the result directly in °C as the average temperature. Identify the missing operation and explain the units.
Reasoned answers
- The change is [−6e−t/2]04 = 6(1 − e−2) L. Add the initial amount: 106 − 6e−2 L, approximately 105.19 L. The model’s rate is positive throughout.
- Divide the accumulated inflow by four minutes: (3/2)(1 − e−2) L min−1, approximately 1.297 L min−1. The final volume divided by four would incorrectly include the initial water.
- A(2) = 0 and A′(t) = t − 4. The derivative is negative before 4 and positive after 4, so the minimum occurs at t = 4. In fact A(4) = −2, showing an accumulation function can be negative.
- Displacement = [t2 − 4t]05 = 5 m. Velocity changes sign at t = 2; the segment displacements are −4 m and +9 m. Distance is 13 m.
- The integral is [x3 + 2x]02 = 12, so the average is 6. Endpoint averaging gives (2 + 14)/2 = 8, which ignores the quadratic curve between them.
- The loss is 2 × 3 + 5 × 2 = 16 L, so 24 L remains. Average loss rate is 16/5 = 3.2 L min−1. The simple mean 3.5 gives equal weight to periods that have different lengths.
- Divide by the duration of the interval. Integrating temperature with respect to time gives temperature × time; division by time restores temperature units. Also ensure the bounds match the period whose average was requested.
Curriculum and sources
Prepared for the current VCE Mathematical Methods Units 3 and 4 course. Scope follows the VCAA Mathematics Study Design 2023–2027, Mathematical Methods Units 3 and 4, Area of Study 3: Calculus, with the current VCAA clarification. Curriculum checked 15 October 2026. Independent teaching material, not a VCAA publication. Teaching reference: Cambridge Senior Mathematics VCE Mathematical Methods 3&4, second edition (2023), Chapter 11; algorithm conventions also draw on the VCAA pseudocode guidance. All teaching prose, questions, numerical datasets and diagrams are original.
