Antidifferentiation introduction: VCE Math Methods Unit 2 Year 11

Syllabus dot point

“Antidifferentiation as the reverse of differentiation, the antiderivative of polynomial functions via the power rule, the constant of integration, and the use of an initial condition to determine a specific antiderivative”

8 min readUpdated 15 Jun 2026

Quick answer

Antidifferentiation in Unit 2 is the reverse of differentiation, with the power rule $\int x^n \, dx = x^{n+1}/(n+1) + C$ for polynomial functions (excluding $n = -1$ ), linearity allowing term-by-term integration, and the constant of integration $C$ always required unless an initial condition determines it; the dot point is the foundation for Unit 4's definite integration and applications.

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What this dot point is asking
The reverse of differentiation
The reverse power rule
Linearity
The constant of integration
Connection to Unit 4
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to recognise antidifferentiation as the reverse of differentiation, apply the power rule to find antiderivatives of polynomial functions, include the constant of integration, and use an initial condition to determine the specific antiderivative. The dot point is the introduction to integration that Unit 4 will extend.

The reverse of differentiation

If $F'(x) = f(x)$ , then $F(x)$ is an antiderivative of $f(x)$ .

The general antiderivative is written:

\int f(x) \, dx = F(x) + C

where $C$ is the constant of integration. Because differentiation kills constants, two antiderivatives of the same function can differ by any constant, so $C$ must always be included unless an initial condition fixes it.

The reverse power rule

For polynomial functions, the reverse of $\frac{d}{dx}(x^n) = n x^{n-1}$ is:

\int x^n \, dx = \frac{x^{n+1}}{n + 1} + C, \quad n \neq -1

The reverse procedure: add 1 to the exponent, divide by the new exponent.

The $n = -1$ case ( $\int x^{-1} \, dx$ ) is excluded because dividing by zero is undefined; the antiderivative of $1/x$ is $\ln|x|$ , which is introduced in Unit 4.

Linearity

Antidifferentiation distributes over sums and pulls out constants:

\int [a f(x) + b g(x)] \, dx = a \int f(x) \, dx + b \int g(x) \, dx

So you can antidifferentiate term by term.

The constant of integration

Every antiderivative requires $+ C$ unless an initial condition fixes it.

Initial value problem. Given $f'(x) = \ldots$ and a specific value $f(a) = b$ :

Antidifferentiate to get $f(x) = F(x) + C$ .
Substitute $x = a$ : $F(a) + C = b$ .
Solve for $C$ .
Write the specific $f(x)$ .

The constant is then determined; the general antiderivative has become a specific function.

Worked example

Step 1: Antidifferentiate a simple polynomial term by term

Given $f'(x) = 3x^2 - 2x$ , we reverse the power rule on each term separately. Reversing the power rule means increasing the exponent by one and dividing by the new exponent, so each term is handled independently:

\int 3x^2 \, dx = \frac{3x^3}{3} = x^3

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

Because differentiation destroys any constant, we must add the constant of integration $C$ to represent the full family of antiderivatives:

f(x) = x^3 - x^2 + C

Step 2: Use an initial condition to determine the constant

Given $f'(x) = 4x^3 + 6x$ and the condition $f(0) = 5$ , we first antidifferentiate term by term to find the general antiderivative. Applying the reverse power rule to each term:

f(x) = x^4 + 3x^2 + C

Now we substitute the initial condition $x = 0$ , $f(0) = 5$ to pin down $C$ . Substituting gives $0 + 0 + C = 5$ , so $C = 5$ , and the specific antiderivative is:

f(x) = x^4 + 3x^2 + 5

Step 3: Apply antidifferentiation to a kinematics problem

A particle has velocity $v(t) = 3t^2 + 2$ m/s, and at $t = 0$ it is at the origin, so $x(0) = 0$ . Position is the antiderivative of velocity with respect to time, so we antidifferentiate $v(t)$ term by term:

x(t) = \int v(t) \, dt = t^3 + 2t + C

Applying the initial condition $x(0) = 0$ gives $0 + 0 + C = 0$ , so $C = 0$ and the specific position function is:

x(t) = t^3 + 2t \text{ m}

Step 4: Evaluate at a specific time

To find the position at $t = 3$ s, substitute directly into the position function. Each term is evaluated at $t = 3$ :

x(3) = 3^3 + 2(3) = 27 + 6 = 33 \text{ m}

Final answer: $x(t) = t^3 + 2t$ m; at $t = 3$ s the particle is $33$ m from the origin.

Connection to Unit 4

Unit 4 will extend antidifferentiation to:

The antiderivatives of $e^{kx}$ , $\frac{1}{x}$ , $\sin(kx)$ , $\cos(kx)$ .
The definite integral and the Fundamental Theorem of Calculus.
Area between curves.
Integration by substitution.

The Unit 2 foundation is the power rule and the discipline of including the constant of integration.

Common errors

Forgetting $+ C$: An indefinite integral without the constant of integration loses marks.
Power rule applied to $1/x$: The power rule for $\int x^n \, dx$ requires $n \neq -1$ . The antiderivative of $1/x$ is $\ln|x|$ (Unit 4).
Reversing the rule incorrectly: Differentiate: multiply by $n$ , decrease exponent. Antidifferentiate: increase exponent, divide by new exponent. The factor of $n$ goes opposite directions.
Substituting before antidifferentiating: When applying an initial condition, antidifferentiate the function first, then substitute.
Antiderivative as plural: A function has infinitely many antiderivatives differing by a constant. The general antiderivative captures all of them via $+ C$ .

Examples in context

Example 1. From acceleration to velocity. A train accelerates with $a(t) = 6 - 2t$ m/s $^2$ , starting from rest ( $v(0) = 0$ ). Antidifferentiating, $v(t) = 6t - t^2 + C$ . The initial condition $v(0) = 0$ gives $C = 0$ , so $v(t) = 6t - t^2$ . The train reaches maximum speed when $a(t) = 0$ , i.e. $t = 3$ s, giving $v(3) = 18 - 9 = 9$ m/s.

Example 2. Recovering total cost from marginal cost. A workshop's marginal cost (dollars per item) is $C'(x) = 3x^2 - 10x + 40$ , and producing nothing costs the fixed setup $C(0) = 500$ . Antidifferentiating, $C(x) = x^3 - 5x^2 + 40x + k$ . With $C(0) = 500$ , $k = 500$ , so $C(x) = x^3 - 5x^2 + 40x + 500$ . The cost of $10$ items is $C(10) = 1000 - 500 + 400 + 500 = \$1400$ .

Try this

Q1. Find the general antiderivative of $f'(x) = 8x^3 - 6x + 5$ . [2 marks]

Cue. $f(x) = 2x^4 - 3x^2 + 5x + C$ .

Q2. Given $f'(x) = 4x - 3$ and $f(2) = 1$ , find $f(x)$ . [3 marks]

Cue. $f(x) = 2x^2 - 3x + C$ ; $f(2) = 8 - 6 + C = 1 \Rightarrow C = -1$ ; so $f(x) = 2x^2 - 3x - 1$ .

Q3. A particle has velocity $v(t) = 3t^2 - 4t$ m/s and is at position $x = 5$ m when $t = 0$ . Find $x(t)$ and the position at $t = 2$ . [3 marks]

Cue. $x(t) = t^3 - 2t^2 + 5$ ; $x(2) = 8 - 8 + 5 = 5$ m.

Exam-style questions

Questions in the style of VCAA 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.

Year 11 SAC-style4 marks

f'(x) = 6 x^2 - 4 x + 1

and

f(1) = 2

, find

f(x)

Show worked answer →

Antidifferentiate term by term.

$\int 6 x^2 \, dx = 2 x^3$ .

$\int -4 x \, dx = -2 x^2$ .

$\int 1 \, dx = x$ .

Adding: $f(x) = 2 x^3 - 2 x^2 + x + C$ .

Apply $f(1) = 2$ : $2 - 2 + 1 + C = 2$ , so $C = 1$ .

Therefore $f(x) = 2 x^3 - 2 x^2 + x + 1$ .

Markers reward the term-by-term antiderivative, the constant of integration $C$ , and the use of the initial condition to determine $C$ .