What are area between two curves?

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

What is 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.

QCAA 2023 QCAA-style P2-style practice: Find the exact area enclosed between the curves y = x^2 and y = 4 - x^2.

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 2 x^2 = 4 x = sqrt(2). Top minus bottom: on [-sqrt(2), sqrt(2)], 4 - x^2 ≥ x^2, so the integrand is (4 - x^2) - x^2 = 4 - 2 x^2. Area = _-sqrt(2)^sqrt(2) (4 - 2 x^2) \, dx. By symmetry (even integrand), = 2 _0^sqrt(2) (4 - 2 x^2) \, dx = 2 [ 4 x - (2 x^3)/(3) ]_0^sqrt(2). = 2 ( 4 sqrt(2) - (2 (sqrt(2))^3)/(3) ) = 2 ( 4 sqrt(2) - (4 sqrt(2))/(3) ) = 2 · (12 sqrt(2) - 4 sqrt(2))/(3) = (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.

QCAA 2022 QCAA-style P2-style practice: A particle moves in a straight line with velocity v(t) = 3 t^2 - 12 t + 9 m/s for 0 ≤ t ≤ 4 seconds. (a) Find the displacement of the particle over the interval. (b) Find the total distance travelled.

A 4-mark kinematics answer must distinguish displacement (signed) from total distance (unsigned). (a) Displacement = _0^4 v(t) \, dt = _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 (3, 4]. Distance = _0^1 v \, dt - _1^3 v \, dt + _3^4 v \, dt = 4 - (-4) + 4 = 8 m. (Using _0^1 = 4, _1^3 = -4, _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 8 m as distinct from displacement of 4 m.

§-Syllabus dot point

QLDMath MethodsSyllabus dot point

Topic 2: Integrals

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

A focused answer to the QCE Mathematical Methods Unit 3 dot point on the applications of integration. Covers area under a curve, area between two curves (including curves that cross), the average value of a function, and the kinematics chain (integrate acceleration for velocity, integrate velocity for displacement), with worked Paper 2 and PSMT-style examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

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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 most of the Topic 2 marks in IA2 and the EA, and they are the most common Topic 2 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 practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 QCAA-style P25 marksFind 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.

2022 QCAA-style P24 marksA 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 $= \int_0^1 v \, dt - \int_1^3 v \, dt + \int_3^4 v \, dt = 4 - (-4) + 4 = 8$ 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 $8$ m as distinct from displacement of $4$ m.