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).

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.

Examples in context

Example 1. Net change in a savings balance. A fund's rate of change is modelled by $r(t) = 3t^2 - 12$ dollars per month. The net change over the first $3$ months is $\int_0^3 (3t^2 - 12)\,dt = [t^3 - 12t]_0^3 = (27 - 36) - 0 = -9$ dollars: a net fall of $\$9$, since the rate is negative for most of the interval (the FTC turns the rate function into the total change).

Example 2. Using symmetry to save work. Evaluating $\int_{-2}^{2}(x^3 + 5x)\,dx$: the integrand is odd ($f(-x) = -f(x)$) and the interval is symmetric about zero, so the integral is $0$ without computation. By contrast $\int_{-2}^{2} x^2\,dx = 2\int_0^2 x^2\,dx = 2 \times \frac{8}{3} = \frac{16}{3}$, using the even-function shortcut.

Try this

Q1. Evaluate $\displaystyle\int_1^3 (2x + 1)\,dx$. [2 marks]

• Cue. $[x^2 + x]_1^3 = 12 - 2 = 10$.

Q2. Evaluate $\displaystyle\int_0^{\pi} \sin x\,dx$. [2 marks]

• Cue. $[-\cos x]_0^{\pi} = -(-1) - (-1) = 2$.

Q3. Given $\displaystyle\int_0^4 f(x)\,dx = 10$ and $\displaystyle\int_0^1 f(x)\,dx = 3$, find $\displaystyle\int_1^4 f(x)\,dx$. [2 marks]

• Cue. Interval splitting: $10 = 3 + \int_1^4 f$, so $\int_1^4 f(x)\,dx = 7$.