Applications of integration: area, average value and kinematics (QCE Mathematical Methods Unit 3)

Syllabus dot point

“Apply the definite integral to find the area under a curve, the area between two curves, the average value of a function, and to solve kinematics problems involving displacement, velocity and acceleration”

9 min readUpdated 7 Oct 2026

Quick answer

The definite integral computes area under a curve, area between two curves (top minus bottom, split at intersections), the average value $\frac{1}{b - a} \int_a^b f \, dx$ , and the kinematic chain (integrate acceleration for velocity, integrate velocity for displacement, integrate the absolute value for total distance), making it the workhorse application tool of Unit 3.

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What this dot point is asking
The answer
Exam-style questions

Key fact

In the 2025 syllabus, Unit 3 introduces integration through antidifferentiation, and definite integrals and their applications are developed in Unit 4 (Topic 1: Further integration). This page stays in Unit 3 so its link does not change.

What this dot point is asking

QCAA wants you to apply the definite integral to find areas, average values and kinematic quantities. These three application types account for many of the integration marks in the EA and are common contexts for PSMTs.

The answer

Area under a single curve

For $f(x) \geq 0$ on $[a, b]$ :

\text{Area} = \int_a^b f(x) \, dx.

If $f(x)$ is negative on part of the interval, the definite integral subtracts that portion (counts it as negative). To find the geometric area in that case, split the integral at the zeros and take absolute values:

\text{Geometric area} = \int_a^c |f(x)| \, dx \quad \text{or equivalently} \quad \sum \left| \int_{\text{sign-piece}} f(x) \, dx \right|.

Area between two curves

For two curves $y = f(x)$ and $y = g(x)$ on $[a, b]$ where $f(x) \geq g(x)$ :

\text{Area} = \int_a^b \bigl( f(x) - g(x) \bigr) \, dx.

This is the "top minus bottom" rule. If the curves cross inside the interval, split the integral at each intersection and switch which curve is on top.

Method:

Find the intersection points by solving $f(x) = g(x)$ .
On each subinterval, identify which function is on top.
Integrate top minus bottom on each subinterval.
Add the pieces (all positive).

Average value of a function

The average value of $f$ on $[a, b]$ is

\bar{f} = \frac{1}{b - a} \int_a^b f(x) \, dx.

Interpretation: the constant height of a rectangle on $[a, b]$ that has the same area as the region under the curve. This is asked frequently in modelling contexts (average temperature, average concentration, average rate of demand over a day).

Kinematics

In rectilinear (straight-line) motion, displacement $s(t)$ , velocity $v(t)$ and acceleration $a(t)$ are linked by differentiation and integration.

v(t) = \frac{ds}{dt}, \qquad a(t) = \frac{dv}{dt} = \frac{d^2 s}{dt^2}

Reversing each link:

v(t) = \int a(t) \, dt + C_1, \qquad s(t) = \int v(t) \, dt + C_2.

The constants of integration are fixed by initial conditions (typically $v(0)$ and $s(0)$ ).

For motion on $[t_1, t_2]$ :

\text{Displacement} = \int_{t_1}^{t_2} v(t) \, dt, \qquad \text{Total distance} = \int_{t_1}^{t_2} |v(t)| \, dt.

The distinction matters. Displacement is the signed change in position. Total distance is the path length. They are equal only when $v(t)$ does not change sign.

Four applications of integration

These four short examples cover area, average value and kinematics, the main applications tested on Paper 2.

Step 1: Area under a single curve

Find the area under $y = e^x$ from $x = 0$ to $x = 1$ . Because $e^x > 0$ on the interval, the definite integral gives the geometric area directly:

\int_0^1 e^x \, dx = \bigl[e^x\bigr]_0^1 = e^1 - e^0 = e - 1 \approx 1.718.

Step 2: Area between two curves

Find the area enclosed by $y = x$ and $y = x^3$ in the first quadrant. First, find the intersection points so we know the limits of integration. Setting $x = x^3$ gives $x(1 - x^2) = 0$ , so $x = 0$ and $x = 1$ in the first quadrant. On $[0, 1]$ , the line $y = x$ lies above the cubic $y = x^3$ , so the integrand is top minus bottom:

\text{Area} = \int_0^1 (x - x^3) \, dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.

Step 3: Average value of a function

Find the average value of $f(x) = \sin x$ on $[0, \pi]$ . The average-value formula divides the integral by the length of the interval, giving the constant height of a rectangle with the same area as the region under the curve:

\bar{f} = \frac{1}{\pi - 0} \int_0^{\pi} \sin x \, dx = \frac{1}{\pi} \bigl[-\cos x\bigr]_0^{\pi} = \frac{1}{\pi}\bigl(-\cos\pi + \cos 0\bigr) = \frac{1}{\pi}(1 + 1) = \frac{2}{\pi} \approx 0.637.

Step 4: Kinematics from acceleration to displacement

A particle starts at the origin with initial velocity $4$ m/s and constant acceleration $a(t) = -2$ m/s $^2$ . Integrate acceleration to recover velocity, using the initial condition to fix the constant of integration:

v(t) = \int -2 \, dt = -2t + C_1. \quad v(0) = 4 \;\Rightarrow\; C_1 = 4, \quad \text{so } v(t) = 4 - 2t.

Integrate velocity to recover displacement:

s(t) = \int (4 - 2t) \, dt = 4t - t^2 + C_2. \quad s(0) = 0 \;\Rightarrow\; C_2 = 0, \quad \text{so } s(t) = 4t - t^2.

The particle reaches maximum displacement when $v = 0$ , i.e. at $t = 2$ , where $s(2) = 8 - 4 = 4$ m. After $t = 2$ the particle moves back towards the origin.

Final answers: area under $e^x$ : $e - 1$ ; enclosed area: $\tfrac{1}{4}$ ; average of $\sin x$ : $\tfrac{2}{\pi}$ ; maximum displacement $4$ m at $t = 2$ s.

Common traps

Top minus bottom inverted: If you integrate bottom minus top you get a negative result. Take absolute values or swap the order.
Forgetting to split at intersections: Curves that cross on the interval require multiple integrals with the order flipped on each piece.
Mixing displacement and distance: Displacement is the signed integral of velocity. Total distance is the integral of $|v(t)|$ . PSMT and EA markers test this distinction frequently.
Skipping initial conditions in kinematics: Both integration constants ( $C_1$ and $C_2$ ) must be evaluated from initial conditions, not left in the answer.
Wrong factor on the average value: The denominator is $b - a$ , not $b$ alone. A common slip on a $[0, b]$ interval gives the right answer by coincidence and then the wrong answer everywhere else.
Using the definite integral for area without checking sign: $\int_a^b f \, dx$ is signed area. For geometric area you may need to split at the zeros of $f$ and absolute-value each piece.

Exam-style questions

Questions in the style of QCAA 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.

Practice5 marks

Find the exact area enclosed between the curves

y = x^2

and

y = 4 - x^2

Show worked answer →

A 5-mark answer needs the intersection points, the top-minus-bottom integrand, the evaluation, and the simplified exact value.

Intersections: $x^2 = 4 - x^2 \implies 2 x^2 = 4 \implies x = \pm \sqrt{2}$ .

Top minus bottom: on $[-\sqrt{2}, \sqrt{2}]$ , $4 - x^2 \geq x^2$ , so the integrand is $(4 - x^2) - x^2 = 4 - 2 x^2$ .

Area $= \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2 x^2) \, dx$ .

By symmetry (even integrand), $= 2 \int_0^{\sqrt{2}} (4 - 2 x^2) \, dx = 2 \left[ 4 x - \frac{2 x^3}{3} \right]_0^{\sqrt{2}}$ .

$= 2 \left( 4 \sqrt{2} - \frac{2 (\sqrt{2})^3}{3} \right) = 2 \left( 4 \sqrt{2} - \frac{4 \sqrt{2}}{3} \right) = 2 \cdot \frac{12 \sqrt{2} - 4 \sqrt{2}}{3} = \frac{16 \sqrt{2}}{3}.$

Markers reward correct intersection points, the top-minus-bottom orientation (often inverted), use of symmetry to halve the work, and the simplified exact form.

Practice4 marks

A particle moves in a straight line with velocity

v(t) = 3 t^2 - 12 t + 9

m/s for

0 \leq t \leq 4

seconds. (a) Find the displacement of the particle over the interval. (b) Find the total distance travelled.

Show worked answer →

A 4-mark kinematics answer must distinguish displacement (signed) from total distance (unsigned).

(a) Displacement $= \int_0^4 v(t) \, dt = \int_0^4 (3 t^2 - 12 t + 9) \, dt = [t^3 - 6 t^2 + 9 t]_0^4 = (64 - 96 + 36) - 0 = 4$ m.

(b) Total distance requires identifying when $v$ changes sign. $v(t) = 3 (t^2 - 4 t + 3) = 3 (t - 1)(t - 3)$ . Zeros at $t = 1$ and $t = 3$ .

Sign: $v > 0$ on $[0, 1)$ , $v < 0$ on $(1, 3)$ , $v > 0$ on $(3, 4]$ .

Distance $= \int_0^1 v \, dt - \int_1^3 v \, dt + \int_3^4 v \, dt = 4 - (-4) + 4 = 12$ m. (Using $\int_0^1 = 4$ , $\int_1^3 = -4$ , $\int_3^4 = 4$ , by direct evaluation.)

Markers reward the displacement integral, identification of velocity zeros, splitting the integral at those zeros, and the absolute-value treatment that gives total distance of $12$ m as distinct from displacement of $4$ m.

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA