The definite integral, the fundamental theorem of calculus linking definite integration to antidifferentiation, and the properties of the definite integral over intervals

What this dot point is asking

VCAA wants by-hand evaluation of definite integrals using the fundamental theorem of calculus, plus the ability to manipulate definite integrals using the interval and linearity properties. Definite integration is high-yield: it appears in both Paper 1 (exact value) and Paper 2 (technology-assisted, calculator-active) every year.

The definite integral

The definite integral of $f$ from $a$ to $b$ is written:

It is a number, not a function. Conceptually it is the signed area between the graph of $f$, the $x$-axis, and the vertical lines $x = a$ and $x = b$. Areas above the $x$-axis count positively; areas below count negatively.

The Unit 3 / 4 syllabus does not require the Riemann-sum definition explicitly, but the intuition is needed when an integrand crosses zero on the interval.

The fundamental theorem of calculus

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

The notation $[F(x)]_{a}^{b}$ or $F(x) \Big\vert_{a}^{b}$ is shorthand for $F(b) - F(a)$.

This is the working tool. To evaluate a definite integral by hand:

1. Find any antiderivative $F$ of the integrand.
2. Compute $F(b)$ and $F(a)$.
3. Subtract.

The constant of integration cancels in step 3, so it does not need to be included for definite integrals.

Properties of the definite integral

Six properties VCAA expects you to use.

Linearity.

Same endpoint.

Reverse endpoints.

Splitting the interval.

This holds for any $b$, not just $b$ between $a$ and $c$.

Even and odd integrands over a symmetric interval.

If $f$ is even ($f(-x) = f(x)$): $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.

If $f$ is odd ($f(-x) = -f(x)$): $\int_{-a}^{a} f(x) \, dx = 0$.

VCAA Paper 1 occasionally asks for $\int_{-\pi}^{\pi} \sin(x) \, dx$ or similar, expecting you to recognise the odd symmetry and write $0$ without computation.

Sign of the integrand.

If $f(x) \geq 0$ on $[a, b]$, the definite integral is non-negative. If $f$ changes sign, the integral is the net signed area; absolute value bars or interval-splitting may be needed for "area" (covered in the area-under-curves dot point).

Worked examples

Example 1. Polynomial

$\int_{0}^{2} (x^3 - 3 x) \, dx$.

Antiderivative. $F(x) = \frac{x^4}{4} - \frac{3 x^2}{2}$.

Apply. $F(2) - F(0) = \left( \frac{16}{4} - \frac{12}{2} \right) - 0 = (4 - 6) - 0 = -2$.

Note the negative result: the integrand is negative on most of $[0, 2]$ (specifically on $(0, \sqrt{3})$), and the net signed area is negative.

Example 2. Exponential

$\int_{0}^{1} e^{2x} \, dx$.

Antiderivative. $F(x) = \frac{1}{2} e^{2x}$.

Apply. $F(1) - F(0) = \frac{1}{2} e^{2} - \frac{1}{2} \cdot 1 = \frac{e^2 - 1}{2}$.

Example 3. Logarithm

$\int_{1}^{2} \frac{1}{x} \, dx = [\ln\lvert x \rvert]_{1}^{2} = \ln 2 - \ln 1 = \ln 2$.

Example 4. Symmetric circular

$\int_{-\pi}^{\pi} \sin(x) \, dx$.

Sine is odd; the interval is symmetric about zero. The integral is $0$ by symmetry, no computation needed.

Verification. $[-\cos(x)]_{-\pi}^{\pi} = (-\cos \pi) - (-\cos(-\pi)) = -(-1) - (-(-1)) = 1 - 1 = 0$. Confirmed.

Example 5. Interval splitting

Given $\int_{0}^{5} f(x) \, dx = 12$ and $\int_{3}^{5} f(x) \, dx = 4$, find $\int_{0}^{3} f(x) \, dx$.

Use $\int_{0}^{5} = \int_{0}^{3} + \int_{3}^{5}$. So $12 = \int_{0}^{3} + 4$, giving $\int_{0}^{3} f(x) \, dx = 8$.

Definite integrals on the calculator

In Paper 2, VCAA expects calculator-active evaluation of definite integrals for integrands too cumbersome to integrate by hand. The TI-Nspire and Casio ClassPad both support the syntax int(f(x), x, a, b) directly. For exact-value Paper 1 questions, only by-hand evaluation is allowed.

Common errors

Forgetting to subtract. Writing $F(b) - F(a)$ as $F(b)$ alone or as $F(a) - F(b)$ (wrong sign). Always evaluate at upper minus lower.

Wrong antiderivative. Especially the $\frac{1}{k}$ factor for $e^{kx}$, $\sin(kx)$, $\cos(kx)$. Forgetting it makes the answer off by a factor.

Bracket discipline. $F(b) - F(a)$ where $F$ has multiple terms requires brackets around the substituted expressions. Without brackets, sign errors cascade.

Including the constant of integration. For definite integrals, $c$ cancels and is not written. Including it earns no marks and signals a procedural slip.

Treating signed area as area. $\int_{a}^{b} f(x) \, dx$ is signed area. For total area (always non-negative), see the area-under-curves dot point. Confusing the two is a common error.

Substituting $\pi$ as 3.14 in Paper 1. Paper 1 expects exact values. Use $\sin(\pi/2) = 1$, $\cos(\pi) = -1$ exactly.

In one sentence

The definite integral $\int_{a}^{b} f(x) \, dx$ is the signed area under $f$ on $[a, b]$ and is evaluated by the fundamental theorem of calculus as $F(b) - F(a)$ where $F$ is any antiderivative; linearity, interval splitting, and the odd / even symmetry properties are the four manipulation rules VCAA expects fluent use of.