Mathematical Methods · Units 3 & 4

Antidifferentiation

Learn antidifferentiation the easy way, with plain English intuition, worked examples and an auto marked practice test. Reverse differentiation, find antiderivatives and the plus c constant. VCE Maths Methods Units 3 and 4.

Rewind a tape and you undo whatever was recorded. Antidifferentiation is exactly that for calculus. Differentiation takes a function and tells you its gradient. Antidifferentiation runs the film backwards: you are handed the gradient and asked to rebuild the original function it came from. This matters because the real world usually hands you a rate of change, like a speed or a flow, and the thing you actually want is the total, like distance or volume. Antidifferentiation is the bridge from one to the other.

The whole topic in three chunks

1 · Reverse the power rule

Antidifferentiation undoes differentiation, rebuilding a function from its gradient
Reverse the power rule: add one to the power, then divide by that new power
$\displaystyle\int x^n \, dx = \dfrac{x^{n+1}}{n+1} + c$ for $n \neq -1$

2 · Always add a + c

Constants vanish when you differentiate, so the original constant is unknown
The $+\, c$ stands for the whole family of parallel curves
Leaving it off is the single most common mistake

3 · Standard results and pinning down c

Exponential $\int e^{kx}\,dx = \tfrac{1}{k}e^{kx}+c$ ; reciprocal $\int \tfrac{1}{x}\,dx = \log_e|x|+c$ ; and the $\sin$ / $\cos$ pair
A point on the curve or a starting value lets you solve for the exact $c$

What antidifferentiation undoes

Differentiation has a clear job. Give it $f(x) = x^3$ and it returns $f'(x) = 3x^2$ . Antidifferentiation asks the reverse question: “which function has $3x^2$ as its gradient?” The answer is $x^3$ , and we call $x^3$ an antiderivative of $3x^2$ .

To reverse the power rule, you simply flip its two steps. The power rule multiplies by the power then drops it by one. So going backwards you add one to the power first, then divide by that new power.

\int x^n \, dx = \frac{x^{n+1}}{n+1} + c, \qquad n \neq -1

The stretched S symbol $\int$ and the $dx$ at the end are the formal way of writing “antidifferentiate this with respect to $x$ ”. Read $\int 3x^2 \, dx$ as “the antiderivative of $3x^2$ ”.

Why there is always a plus c

Here is the twist that makes antidifferentiation different from differentiation. The functions $x^3$ , $x^3 + 7$ and $x^3 - 100$ all have the same derivative, $3x^2$ , because a constant has a gradient of zero and simply disappears when you differentiate.

So when you run the process backwards you cannot know which constant was there originally. To cover every possibility you write a $+\, c$ , the constant of integration. It stands for the whole family of parallel curves that share that gradient.

\int 3x^2 \, dx = x^3 + c

The curves y = x³, y = x³ + 4 and y = x³ − 4 are vertical shifts of one another, so they share the identical gradient 3x² at every x. Antidifferentiating 3x² cannot tell them apart, which is why the answer carries a + c.

The standard antiderivatives

Beyond powers, four results cover almost everything in this course. Each one is just a differentiation rule read backwards.

The exponential. Since $\frac{d}{dx}e^{kx} = ke^{kx}$ , going backwards you divide by the $k$ .

\int e^{kx} \, dx = \frac{1}{k}e^{kx} + c

The reciprocal $\frac{1}{x}$ . The power rule breaks here because dividing by $n + 1$ would mean dividing by zero. Instead this is the one antiderivative that produces a logarithm.

\int \frac{1}{x} \, dx = \log_e|x| + c

If there is a number on top, such as $\frac{3}{x}$ , that number stays in front as $3\log_e|x|$ . Dropping it is a classic error.

The circular functions. These come as a matched pair, and the minus sign is easy to misplace.

\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + c, \qquad \int \sin(kx) \, dx = -\frac{1}{k}\cos(kx) + c

When a bracket is raised to a power, like $(2x - 1)^5$ , raise the power and divide by the new power, then divide again by the coefficient of $x$ inside the bracket. For $(2x - 1)^5$ that means dividing by both $6$ and $2$ .

Pinning down the constant

A $+\, c$ is honest but vague. Often a question gives you one extra fact, such as a point the curve passes through or a starting value. That single condition is enough to find the exact value of $c$ and lock onto one curve from the family.

The recipe is short:

Antidifferentiate to get the general answer, complete with $+\, c$ .
Substitute the given condition, putting the known $x$ and $y$ values into the equation.
Solve the resulting equation for $c$ .
Write the final function with that exact constant in place.

The trap to avoid is rushing step 3. Substitute the values, then solve carefully for $c$ , and only then write the final answer with the real constant. Forgetting to update $c$ after substituting, or leaving the answer as the general family, throws away the mark.

Learn it, then fade the help

1 · I do — fully worked

Find c from a condition. A curve has gradient $\dfrac{dy}{dx} = 3x^2 - 2$ and passes through $(1, 4)$ . Antidifferentiate to get the family: $y = x^3 - 2x + c$ . Substitute the point: $4 = (1)^3 - 2(1) + c = -1 + c$ , so $c = 5$ . The curve is $y = x^3 - 2x + 5$ .

2 · We do — your turn, guided

Same steps, your turn. A curve has $\dfrac{dy}{dx} = 6x^2 - 4x + 5$ and passes through $(0, 3)$ . Antidifferentiate first: $y = 2x^3 - 2x^2 + 5x + c$ . Now substitute $x = 0$ , $y = 3$ : $3 = 2(0)^3 - 2(0)^2 + 5(0) + c = \;?$ .

Check your step

$3 = c$ , so $c = 3$ and the curve is $y = 2x^3 - 2x^2 + 5x + 3$ .

3 · You do — solo

No scaffolding now: a curve has $\dfrac{dy}{dx} = 4x - 3$ and passes through $(2, 1)$ . Find its equation.

Reveal solution

Antidifferentiate: $y = 2x^2 - 3x + c$ . Substitute $(2, 1)$ : $1 = 2(2)^2 - 3(2) + c = 8 - 6 + c = 2 + c$ , so $c = -1$ . The curve is $y = 2x^2 - 3x - 1$ .

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

State the reverse power rule for

\displaystyle\int x^n \, dx

Why does every indefinite antiderivative carry a $+\, c$ ?

What is

\displaystyle\int e^{kx} \, dx

What is

\displaystyle\int \dfrac{1}{x} \, dx

, and why isn’t it a power?

Antidifferentiate the circular pair

\cos(kx)

and

\sin(kx)

What are the four steps to pin down $c$ from a condition?

Recall · Derivatives of Exponential and Logarithmic Functions

What is

\dfrac{d}{dx}\,e^{kx}

, and how does it explain

\int e^{kx}\,dx

Recall · The Definite Integral and Area

What extra thing does a definite integral give that an indefinite one does not?

See these antiderivatives in action in the Worked Examples tab, then test yourself in Try It.

Worked Example 1An antiderivative through two points, from a real exam

Let $F$ be an antiderivative of $f(x) = \tfrac{x}{2} + 7$ that passes through $(0, c)$ , where $c \in R$ . Show that it is not possible for the graph of $y = F(x)$ to pass through both $(-12, 1)$ and $(2, 8)$ .

1
Antidifferentiate $f$ using the reverse power rule, carrying the constant $c$ .
$F(x) = \frac{x^2}{4} + 7x + c$
2
Substitute each point in turn and solve for $c$ .
$F(-12) = 1 \Rightarrow c = 49; \quad F(2) = 8 \Rightarrow c = -7$
3
A single antiderivative has one fixed value of $c$ , so it cannot equal both. Sign errors in the substitution are the main pitfall.
$49 \ne -7 \ \Rightarrow \ \text{impossible}$

Answer

F(x) = \dfrac{x^2}{4} + 7x + c; \text{ the two points force } c = 49 \text{ and } c = -7, \text{ a contradiction}

VCAA 2025 Mathematical Methods Exam 2, Section B Q2fi

Worked Example 2A reciprocal antiderivative with a condition, from a real exam

Let $g(x)$ be a function defined for $x > -\dfrac{3}{2}$ so that $g'(x) = \dfrac{1}{2x + 3}$ and $g(1) = 0$ . Find $g(x)$ .

1
Antidifferentiate, remembering the factor of $\tfrac{1}{2}$ that comes from the derivative of $2x + 3$ .
$g(x) = \frac{1}{2}\log_e(2x + 3) + c$
2
Apply $g(1) = 0$ to find $c$ .
$\frac{1}{2}\log_e(5) + c = 0 \implies c = -\frac{1}{2}\log_e(5)$
3
Combine the logarithms. The examiner report flags omitting the factor of $\tfrac{1}{2}$ as the common error.
$g(x) = \frac{1}{2}\log_e\!\left(\frac{2x + 3}{5}\right)$

Answer

g(x) = \dfrac{1}{2}\log_e(2x + 3) - \dfrac{1}{2}\log_e(5) = \dfrac{1}{2}\log_e\!\left(\dfrac{2x + 3}{5}\right)

VCAA 2025 Mathematical Methods Exam 1, Q2

Worked Example 3A power and the plus c

Find an antiderivative of $f(x) = 6x^2 - 4x + 5$ .

1
Reverse the power rule on each term: add one to the power, then divide by the new power.
$\int 6x^2 \, dx = \frac{6x^3}{3} = 2x^3$
2
Do the same to the other terms. The $5$ is $5x^0$ , so it becomes $5x$ .
$\int (-4x + 5) \, dx = -2x^2 + 5x$
3
Add the constant of integration, because any constant differentiates to zero.
$F(x) = 2x^3 - 2x^2 + 5x + c$

Answer

F(x) = 2x^3 - 2x^2 + 5x + c

Worked Example 4Exponential, logarithm and a circular function

Find $\displaystyle\int \left( e^{2x} + \frac{3}{x} + \sin(x) \right) dx$ .

1
For $e^{kx}$ divide by $k$ . Here $k = 2$ .
$\int e^{2x} \, dx = \frac{1}{2}e^{2x}$
2
The antiderivative of $\frac{a}{x}$ is $a\log_e|x|$ , so keep the factor of $3$ out the front.
$\int \frac{3}{x} \, dx = 3\log_e|x|$
3
The antiderivative of $\sin(x)$ is $-\cos(x)$ . Watch that minus sign.
$\int \sin(x) \, dx = -\cos(x)$
4
Combine the pieces and add one constant for the whole expression.
$\frac{1}{2}e^{2x} + 3\log_e|x| - \cos(x) + c$

Answer

F(x) = \frac{1}{2}e^{2x} + 3\log_e|x| - \cos(x) + c

Worked Example 5Finding c from a condition

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

1
Antidifferentiate the gradient to get a family of curves, all with the same shape.
$y = x^3 - 2x + c$
2
Use the known point. Substitute $x = 1$ and $y = 4$ to pin down which curve.
$4 = (1)^3 - 2(1) + c = -1 + c$
3
Solve for $c$ .
$c = 5$

Answer

y = x^3 - 2x + 5

Practice test

Try it yourself

7 questions, 9 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.A function $g : R \to R$ has the derivative $g'(x) = x^3 - x$ . Given that $g(0) = 5$ , the value of $g(2)$ is:

1mark

VCAA 2024 Mathematical Methods Exam 2, Section A Q2

Q2.An antiderivative of $f(x) = 8x^3$ is:

1mark

Need a hint?

Reverse the power rule: add one to the power, then divide by the new power, not the old one.

Q3. $\displaystyle\int \frac{4}{x} \, dx$ is equal to:

1mark

Need a hint?

This is the one antiderivative that gives a logarithm; the number on top stays in front.

Q4. $\displaystyle\int \cos(2x) \, dx$ is equal to:

1mark

Need a hint?

The antiderivative of

\cos(kx)

divides by

k

, and only

\sin

picks up a minus sign, not

\cos

Q5.A function $f$ has $f'(x) = e^{-3x}$ and $f(0) = 1$ . The value of $f(0)$ allows $c$ to be found, and $f(x)$ equals:

1mark

Need a hint?

Antidifferentiate first to get the

+c

, then substitute

x = 0

and solve for

c

before writing the final answer.

Q6.An antiderivative of $f(x) = (2x - 1)^5$ is:

1mark

Need a hint?

Raise the power and divide by the new power, then divide again by the coefficient of

x

inside the bracket.

Q7.The gradient of a curve is given by $\dfrac{dy}{dx} = 4x - \dfrac{2}{x}$ , for $x > 0$ . The curve passes through the point $(1, 3)$ . Find the equation of the curve.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

Why do I always have to add a plus c?

Because any constant disappears when you differentiate, so the gradient you started with could have come from infinitely many curves that differ only by a constant. The plus c stands for all of them, and you can only pin down its value if the question gives you an extra fact, like a point on the curve.

What is the difference between an antiderivative and an integral?

For this course they are essentially the same idea. An antiderivative is any function whose derivative gives you back what you started with, and the integral sign is just the formal notation for finding it. A definite integral, with numbers on the sign, goes one step further and gives an actual value rather than a family of functions.

How do I antidifferentiate something like one over x?

This is the one case where the reverse power rule breaks, because it would ask you to divide by zero. Instead, the antiderivative of one over x is the natural logarithm of the absolute value of x. If there is a number on top, keep that number out the front.

Antidifferentiation

Learn

What antidifferentiation undoes

Why there is always a plus c

The standard antiderivatives

Pinning down the constant

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions