SACE Board SACE 2024-style practice: Calculator-assumed. Car A has velocity v(t) = 3 - 3e^-1.5t metres per second for 0 t 2. Given the displacement s(t) satisfies s(0) = 0, show that s(t) = 3t + 2e^-1.5t - 2.

Displacement is the antiderivative of velocity, so antidifferentiate v(t) = 3 - 3e^-1.5t term by term. The antiderivative of 3 is 3t. For -3e^-1.5t, note ddte^-1.5t = -1.5e^-1.5t, so the antiderivative of e^-1.5t is 1-1.5e^-1.5t = -23e^-1.5t. Therefore the antiderivative of -3e^-1.5t is -3 × -23e^-1.5t = 2e^-1.5t. So s(t) = 3t + 2e^-1.5t + C. Apply s(0) = 0: 0 = 0 + 2e^0 + C = 2 + C, so C = -2. Hence s(t) = 3t + 2e^-1.5t - 2, as required. Marks: one for the antiderivative including C, one for using s(0) = 0 to evaluate C.

§-Syllabus dot point

SAMath MethodsSyllabus dot point

How do we reverse differentiation to recover a function from its rate of change?

Antidifferentiation reverses differentiation to find the family of functions whose derivative is a given function, always including a constant of integration.

How to antidifferentiate power, exponential and reciprocal functions, why the +C constant is essential, and how to find a particular antiderivative from a boundary condition.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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

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What this dot point is asking
The standard antiderivatives
Linearity
Negative and fractional powers
Finding a particular antiderivative
Common errors
Antidifferentiating with the chain rule in reverse
Why it matters

What this dot point is asking

To antidifferentiate is to undo differentiation - to find a function $F(x)$ whose derivative is the given function $f(x)$ . Because the derivative of any constant is zero, there is a whole family of antiderivatives differing only by a constant, written $F(x)+C$ .

The standard antiderivatives

The power rule says "add one to the index, divide by the new index". It works for every power except $n=-1$ , where it would divide by zero - that case gives the logarithm instead.

A useful way to remember the logarithm case is to ask which function has derivative $\dfrac{1}{x}$ : it is $\ln x$ , so the antiderivative of $\dfrac{1}{x}$ must be $\ln|x|$ . The absolute value is needed because $\dfrac{1}{x}$ is defined for negative $x$ too, and $\ln|x|$ extends the antiderivative across both branches.

Linearity

Integration is linear, so you can integrate term by term and pull constants out:

\int \big(af(x)+bg(x)\big)\,dx=a\int f(x)\,dx+b\int g(x)\,dx.

Negative and fractional powers

Rewrite roots and reciprocals as powers first, then apply the rule:

\int \frac{1}{x^2}\,dx=\int x^{-2}\,dx=\frac{x^{-1}}{-1}+C=-\frac{1}{x}+C,

\int \sqrt{x}\,dx=\int x^{1/2}\,dx=\frac{x^{3/2}}{3/2}+C=\frac{2}{3}x^{3/2}+C.

Finding a particular antiderivative

A boundary condition pins down the constant $C$ , selecting one curve from the family.

Recovering a function from its gradient

A curve has gradient $\dfrac{dy}{dx} = 3x^2 - 2$ and passes through $(1, 4)$ . Find $y$ .

Step 1: Antidifferentiate to find the family of curves

The gradient tells us how $y$ changes with $x$ , so to recover $y$ itself we reverse the differentiation. Applying the power rule term by term:

y = \int (3x^2 - 2)\,dx = x^3 - 2x + C.

The constant $C$ is essential here: there are infinitely many curves with this gradient, shifted vertically by different amounts. The given point will select exactly one of them.

Step 2: Use the boundary condition to find $C$

The curve passes through $(1, 4)$ , meaning $y = 4$ when $x = 1$ . Substitute these values into the general family.

4 = (1)^3 - 2(1) + C = 1 - 2 + C = -1 + C \quad\Rightarrow\quad C = 5.

Step 3: Write the particular solution

Replacing $C$ with $5$ gives the unique curve through the given point.

y = x^3 - 2x + 5.

Final answer: $y = x^3 - 2x + 5$ .

Common errors

Antidifferentiating with the chain rule in reverse

Many exam integrands are built by a chain rule and must be undone by recognising the pattern. The two most common are $\int (ax+b)^n\,dx$ and $\int e^{ax+b}\,dx$ . For a linear inside function you simply divide by the derivative of the inside:

\int (ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}+C\ (n\ne -1), \qquad \int e^{ax+b}\,dx=\frac{1}{a}e^{ax+b}+C.

For instance $\int (2x-1)^4\,dx=\dfrac{(2x-1)^5}{2\times 5}+C=\dfrac{(2x-1)^5}{10}+C$ . The $\div a$ factor is exactly the reverse of the chain rule multiplying by $a$ when you differentiate. This shortcut works only when the inside is linear; non-linear inside functions need the formal substitution method studied in Specialist Mathematics.

Antidifferentiating an exponential model

A population grows so that its rate of change is $P'(t) = 200e^{0.1t}$ people per year. Given $P(0) = 5000$ , find $P(t)$ .

Step 1: Antidifferentiate the exponential rate

To recover the population $P(t)$ from its rate of change, antidifferentiate with respect to $t$ . For a function of the form $e^{kt}$ , the antiderivative is $\dfrac{1}{k}e^{kt}$ , because dividing by the coefficient $k$ exactly undoes the factor of $k$ that differentiation introduced.

P(t) = \int 200e^{0.1t}\,dt = \frac{200}{0.1}e^{0.1t} + C = 2000e^{0.1t} + C.

Step 2: Apply the initial condition to find $C$

The initial population $P(0) = 5000$ pins down which member of the family applies. Substitute $t = 0$ .

5000 = 2000e^{0} + C = 2000(1) + C = 2000 + C \quad\Rightarrow\quad C = 3000.

Step 3: Write the population model

Replacing $C$ with $3000$ gives the complete population function.

P(t) = 2000e^{0.1t} + 3000.

Final answer: $P(t) = 2000e^{0.1t} + 3000$ .

Why it matters

Antidifferentiation is the engine of all of Topic 3. The definite integral, the Fundamental Theorem of Calculus, and every area calculation rely on first finding an antiderivative. Recovering a function from its rate of change is also the basis of every kinematics and growth model you will integrate, and the boundary-condition technique reappears whenever you solve a differential equation.

Exam-style practice questions

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

SACE 20242 marksCalculator-assumed. Car A has velocity

v(t) = 3 - 3e^{-1.5t}

metres per second for

0 \le t \le 2

. Given the displacement

s(t)

satisfies

s(0) = 0

, show that

s(t) = 3t + 2e^{-1.5t} - 2

Show worked answer →

Displacement is the antiderivative of velocity, so antidifferentiate $v(t) = 3 - 3e^{-1.5t}$ term by term.

The antiderivative of $3$ is $3t$ . For $-3e^{-1.5t}$ , note $\dfrac{d}{dt}e^{-1.5t} = -1.5e^{-1.5t}$ , so the antiderivative of $e^{-1.5t}$ is $\dfrac{1}{-1.5}e^{-1.5t} = -\dfrac{2}{3}e^{-1.5t}$ . Therefore the antiderivative of $-3e^{-1.5t}$ is $-3 \times -\dfrac{2}{3}e^{-1.5t} = 2e^{-1.5t}$ .

So $s(t) = 3t + 2e^{-1.5t} + C$ .

Apply $s(0) = 0$ : $0 = 0 + 2e^0 + C = 2 + C$ , so $C = -2$ .

Hence $s(t) = 3t + 2e^{-1.5t} - 2$ , as required. Marks: one for the antiderivative including $C$ , one for using $s(0) = 0$ to evaluate $C$ .

SACE 20222 marksCalculator-assumed. The velocity of a braking vehicle is

v(t) = -6t + k

metres per second, where

k

is the initial velocity, and the vehicle stops when

v = 0

. Write an integral expression for the stopping distance

d

, and integrate it to find

d

in terms of

k

Show worked answer →

The vehicle stops when $v(t) = 0$ : $-6t + k = 0$ , giving $t = \dfrac{k}{6}$ . The distance is the integral of velocity from $t = 0$ to $t = \dfrac{k}{6}$ .

d = \int_0^{k/6} (-6t + k)\,dt.

Antidifferentiate to $-3t^2 + kt$ and evaluate:

d = \Big[-3t^2 + kt\Big]_0^{k/6} = -3\left(\frac{k}{6}\right)^2 + k\left(\frac{k}{6}\right) = -\frac{k^2}{12} + \frac{2k^2}{12} = \frac{k^2}{12}.

So $d = \dfrac{k^2}{12}$ . Marks: one for the integral expression, one for the integration and result.

SACE 20213 marksCalculator-assumed. A curve has gradient

\dfrac{dy}{dx} = \dfrac{4}{x} + 6x

for

x > 0

and passes through

(1, 3)

. Find the equation of the curve.

Show worked answer →

Antidifferentiate, using $\int \dfrac{1}{x}\,dx = \ln|x|$ :

y = \int \left(\frac{4}{x} + 6x\right) dx = 4\ln x + 3x^2 + C \quad (x > 0).

Apply the point $(1, 3)$ : $3 = 4\ln 1 + 3(1)^2 + C = 0 + 3 + C$ , so $C = 0$ .

Hence $y = 4\ln x + 3x^2$ . Marks: one for $4\ln x$ , one for $3x^2$ plus $C$ , and one for using the point to find $C = 0$ .