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

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 $[a, b]$:

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:

Area between two curves

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

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:

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

Average value of a function

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

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.

Reversing each link:

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

For motion on $[t_1, t_2]$:

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.

Worked examples

Area under a curve

Find the area under $y = e^x$ from $x = 0$ to $x = 1$.

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

Area between curves

Find the area enclosed by $y = x$ and $y = x^3$ in the first quadrant.

Intersections at $x = x^3 \implies x (1 - x^2) = 0 \implies x = 0, 1$ (first quadrant).

On $[0, 1]$, $x \geq x^3$.

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

Average value

Find the average value of $f(x) = \sin x$ on $[0, \pi]$.

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

Kinematics: from acceleration to displacement

A particle starts at the origin with velocity $4$ m/s and is subject to acceleration $a(t) = -2$ m/s$^2$.

$v(t) = \int -2 \, dt = -2 t + C_1$. From $v(0) = 4$, $C_1 = 4$, so $v(t) = 4 - 2 t$.

$s(t) = \int (4 - 2 t) \, dt = 4 t - t^2 + C_2$. From $s(0) = 0$, $C_2 = 0$, so $s(t) = 4 t - t^2$.

The particle reaches maximum displacement when $v = 0$, at $t = 2$, where $s = 4$. After $t = 2$ the particle moves back towards the origin.

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.

In one sentence

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.