Antiderivatives and the Fundamental Theorem of Calculus (QCE Mathematical Methods Unit 3)

Syllabus dot point

“Find antiderivatives of standard functions including polynomial, exponential and trigonometric forms, evaluate definite integrals using the Fundamental Theorem of Calculus, and recognise the definite integral as the limit of a Riemann sum”

9 min readUpdated 7 Oct 2026

Quick answer

Antidifferentiation reverses the standard derivatives (with linear inside arguments handled by dividing by the coefficient), and the Fundamental Theorem of Calculus evaluates the definite integral $\int_a^b f(x) \, dx$ as $F(b) - F(a)$ for any antiderivative $F$ , with the definite integral itself defined as the limiting Riemann sum that geometrically gives signed area.

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What this dot point is asking
The answer
Exam-style questions

Key fact

In the 2025 syllabus, Unit 3 introduces integration through antidifferentiation, and the definite integral and the Fundamental Theorem of Calculus are developed in Unit 4 (Topic 1: Further integration). This page stays in Unit 3 so its link does not change.

What this dot point is asking

QCAA wants you to recognise integration as the reverse of differentiation, find antiderivatives of all standard Methods functions, evaluate definite integrals using the Fundamental Theorem of Calculus (FTC), and connect the definite integral to the limit of a Riemann sum. Integration underlies area, average value and kinematics, and appears on most Paper 1 and Paper 2 papers.

The answer

Standard antiderivatives

The constant $C$ is required on every indefinite integral.

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

\int \frac{1}{x} \, dx = \ln |x| + C

\int e^x \, dx = e^x + C

\int \sin x \, dx = -\cos x + C \qquad \int \cos x \, dx = \sin x + C

The minus sign on $\int \sin x \, dx$ is the Paper 1 trap that mirrors the minus sign on $\frac{d}{dx} \cos x$ . These two are paired; remember them together.

Linear inside argument

If the argument is linear ( $ax + b$ ), divide by the coefficient of $x$ .

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

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

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

\int (ax + b)^n \, dx = \frac{(ax + b)^{n + 1}}{a (n + 1)} + C \quad (n \neq -1)

These are not new rules. They are the chain rule run in reverse, with the linear inside argument simple enough that the $\frac{1}{a}$ factor is the only adjustment needed.

The Fundamental Theorem of Calculus

If $F$ is any antiderivative of $f$ (so $F'(x) = f(x)$ ), then

\int_a^b f(x) \, dx = F(b) - F(a).

A second statement: if $G(x) = \displaystyle \int_a^x f(t) \, dt$ , then $G'(x) = f(x)$ . In short, differentiation and integration are inverse operations.

The FTC turns the geometric problem (area under a curve) into the algebraic problem (evaluate an antiderivative at two points and subtract).

The Riemann sum definition

The definite integral $\int_a^b f(x) \, dx$ is defined as the limit of a Riemann sum:

\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i = 1}^{n} f(x_i^*) \, \Delta x

where the interval $[a, b]$ is split into $n$ subintervals of width $\Delta x = \frac{b - a}{n}$ and $x_i^*$ is a sample point in the $i$ -th subinterval. When $f(x) \geq 0$ , this limit equals the area under the curve from $x = a$ to $x = b$ . When $f(x)$ is negative, the integral counts that area as negative.

For Methods, QCAA expects you to recognise this definition and to use it to interpret what a definite integral represents (an accumulation), without needing to compute Riemann sums by hand at scale.

Properties of the definite integral

These properties speed up Paper 1 evaluation.

Linearity: $\int_a^b \bigl( c_1 f(x) + c_2 g(x) \bigr) dx = c_1 \int_a^b f + c_2 \int_a^b g$ .
Reversed limits: $\int_b^a f(x) \, dx = -\int_a^b f(x) \, dx$ .
Splitting: $\int_a^c f(x) \, dx = \int_a^b f + \int_b^c f$ .
Zero-width: $\int_a^a f(x) \, dx = 0$ .

Four integration techniques in practice

These four short examples cover the main antiderivative skills tested on QCAA Paper 1.

Step 1: Standard polynomial indefinite integral

To antidifferentiate $4x^3 - 6x + 2$ , apply the power rule in reverse to each term: raise the index by one and divide by the new index. The constant of integration $C$ is always required for an indefinite integral.

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

Step 2: Polynomial with a linear inside argument

For $\int (2x + 1)^4 \, dx$ , the argument is linear in $x$ , so the chain rule in reverse applies. Raise the power by one and divide by the new power, then also divide by the coefficient of $x$ inside the bracket (here $2$ ):

\int (2 x + 1)^4 \, dx = \frac{(2 x + 1)^5}{2 \cdot 5} + C = \frac{(2 x + 1)^5}{10} + C.

Step 3: Exponential with a linear argument

For $\int e^{-3x} \, dx$ , the antiderivative of $e^{kx}$ is $\dfrac{e^{kx}}{k}$ . Dividing by the coefficient $k = -3$ introduces a factor of $-\tfrac{1}{3}$ :

\int e^{-3 x} \, dx = \frac{e^{-3 x}}{-3} + C = -\frac{1}{3} e^{-3 x} + C.

Step 4: Evaluating a definite integral with the FTC

To evaluate $\int_1^3 (x^2 + 2x) \, dx$ , first find an antiderivative, then apply $F(b) - F(a)$ . No constant $C$ is needed for definite integrals because it cancels in the subtraction.

Antiderivative: $F(x) = \dfrac{x^3}{3} + x^2$ .

F(3) - F(1) = \bigl(9 + 9\bigr) - \bigl(\tfrac{1}{3} + 1\bigr) = 18 - \tfrac{4}{3} = \frac{50}{3}.

Final answers: $x^4 - 3x^2 + 2x + C$ ; $\dfrac{(2x+1)^5}{10} + C$ ; $-\dfrac{1}{3}e^{-3x} + C$ ; $\dfrac{50}{3}$ .

Common traps

Forgetting $+ C$ on an indefinite integral: Always include the constant of integration. It is a marked step.
Wrong division on linear inside argument: $\int e^{2x} \, dx = \frac{1}{2} e^{2x} + C$ , not $2 e^{2x}$ . Divide by the coefficient, do not multiply.
Sign error on $\int \sin x \, dx$: The antiderivative is $-\cos x$ , mirroring $\frac{d}{dx} \cos x = -\sin x$ .
Reading off wrong limits: $\int_a^b f \, dx = F(b) - F(a)$ , in that order. Reversing gives the negative.
Treating the integral as signed area for a negative function: $\int_0^{\pi} (-\sin x) \, dx$ counts the area below the axis as negative. If the geometric area is wanted, take absolute values or split the integral at the zeros.
Using the FTC over a discontinuity: $\int_{-1}^{1} \frac{1}{x^2} \, dx$ is improper because $\frac{1}{x^2}$ has a vertical asymptote at $x = 0$ . Methods does not formally cover improper integrals; if QCAA asks an evaluation question, the integrand will be continuous on the interval.

Exam-style questions

Questions in the style of QCAA 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.

Practice4 marks

Evaluate

\int_0^{\pi/2} \bigl( 3 \sin x + 2 \cos x \bigr) \, dx

exactly.

Show worked answer →

Find the antiderivative term by term.

$\int 3 \sin x \, dx = -3 \cos x$ and $\int 2 \cos x \, dx = 2 \sin x$ .

Antiderivative: $F(x) = -3 \cos x + 2 \sin x$ .

Apply the Fundamental Theorem of Calculus.

$F(\pi/2) - F(0) = (-3 \cdot 0 + 2 \cdot 1) - (-3 \cdot 1 + 2 \cdot 0) = 2 - (-3) = 5.$

Markers reward the explicit antiderivative (with correct minus sign on the cosine antiderivative), the use of exact trig values without a calculator, and the final answer of $5$ .

Practice3 marks

Find

\int (2 e^{4x} + \frac{1}{x}) \, dx

Show worked answer →

Antidifferentiate term by term.

$\int 2 e^{4x} \, dx = 2 \cdot \frac{e^{4x}}{4} = \frac{e^{4x}}{2}$ . The factor of $1/4$ comes from dividing by the coefficient of $x$ in the exponent.

$\int \frac{1}{x} \, dx = \ln |x|$ .

$\int \bigl( 2 e^{4x} + \frac{1}{x} \bigr) \, dx = \frac{e^{4x}}{2} + \ln |x| + C.$

Markers reward the divide-by-coefficient on the exponential, $\ln |x|$ rather than $\ln x$ (to handle the full domain), and the constant of integration $+ C$ . Forgetting the $+ C$ is the single most common Paper 1 indefinite-integral mistake.

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Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA