Antidifferentiation (SACE Stage 2 Mathematical Methods)

Syllabus dot point

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

8 min readUpdated 2 Jun 2026

Quick answer

Antidifferentiation (indefinite integration) reverses differentiation: $\int f(x)\,dx=F(x)+C$ where $F'(x)=f(x)$ . The key power rule is $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ for $n\ne -1$ , and you must always include the constant of integration $C$ .

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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
Exam-style questions

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 $F(x)+C$ .

Indefinite integral

\int f(x)\,dx=F(x)+C \quad\text{means}\quad F'(x)=f(x),

where

C

is the arbitrary constant of integration.

The standard antiderivatives

Core antidifferentiation rules

\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\ (n\ne -1), \qquad \int \frac{1}{x}\,dx=\ln|x|+C,

\int e^{x}\,dx=e^{x}+C, \qquad \int e^{kx}\,dx=\frac{1}{k}e^{kx}+C.

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.

Antidifferentiating a polynomial

Worked example

Find $\displaystyle\int \left(6x^2-4x+5\right)dx$ .

Integrate each term with the power rule:

\int 6x^2\,dx=6\cdot\frac{x^3}{3}=2x^3, \qquad \int -4x\,dx=-4\cdot\frac{x^2}{2}=-2x^2, \qquad \int 5\,dx=5x.

Combine and add one constant:

\int \left(6x^2-4x+5\right)dx=2x^3-2x^2+5x+C.

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

Omitting +C or mishandling the index −1

The most penalised error in indefinite integration is leaving off the constant $C$ - it is mathematically required and worth a mark. The second is applying the power rule to $\int x^{-1}\,dx$ , which gives $\tfrac{x^0}{0}$ (undefined); that case is $\ln|x|+C$ . Also, when integrating $e^{kx}$ remember to divide by $k$ , not multiply.

Check by differentiating

After integrating, differentiate your answer mentally. If you get back the original integrand, the antiderivative is correct. This self-check catches index errors and missing coefficients in seconds.

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 questions

Questions in the style of SACE Board 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.

SACE 2024-style2 marks

Calculator-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 2022-style2 marks

Calculator-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 2021-style3 marks

Calculator-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 $C = 0$ .

Practise this

Sources & how we know this

Stage 2 Mathematical Methods subject outline (2025 onwards) — SACE Board
Stage 2 Mathematical Methods: external assessment — SACE Board

What this dot point is asking

The standard antiderivatives

Linearity

Worked example

Negative and fractional powers

Finding a particular antiderivative

Step 1: Antidifferentiate to find the family of curves

Step 2: Use the boundary condition to find CCC

Step 3: Write the particular solution

Common errors

Antidifferentiating with the chain rule in reverse

Step 1: Antidifferentiate the exponential rate

Step 2: Apply the initial condition to find CCC

Step 3: Write the population model

Why it matters

Exam-style questions

Practise this

Sources & how we know this

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

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