SACE Board SACE 2024-style practice: Calculator-assumed. Vehicles enter a car park at a rate E(t) vehicles per hour, with the car park empty at t = 0. A table gives _0^t E(t)\,dt at various times, including the value 471 at t = 24. (a) State how many vehicles entered during the 24-hour period. (b) Explain what _8^24 E(t)\,dt represents.

(a) The integral of a rate accumulates the total quantity. Since E(t) is the entry rate (vehicles per hour), _0^24 E(t)\,dt is the total number that entered over the day. From the table this is 471 vehicles. (1 mark) (b) _8^24 E(t)\,dt is the number of vehicles that entered between t = 8 and t = 24 hours, found by F(24) - F(8) from the table. (1 mark) This is the Fundamental Theorem idea that the definite integral of a rate gives net accumulated change over the interval.

SACE Board SACE 2021-style practice: Calculator-free. Evaluate _0^2 (3x^2 - 4x + 1)\,dx.

Antidifferentiate to F(x) = x^3 - 2x^2 + x, then apply the limits: $_0^2 (3x^2 - 4x + 1)\,dx = [x^3 - 2x^2 + x]_0^2 = (8 - 8 + 2) - 0 = 2.$ Marks: one for the antiderivative, one for the correct evaluation F(2) - F(0) = 2.

§-Syllabus dot point

SAMath MethodsSyllabus dot point

How does the definite integral connect antidifferentiation to accumulated change?

The Fundamental Theorem of Calculus evaluates a definite integral as the difference of an antiderivative at the two limits.

How to evaluate definite integrals using the Fundamental Theorem of Calculus, the key properties of definite integrals, and how they represent accumulated change.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The Fundamental Theorem of Calculus
Properties of definite integrals
Net accumulated change
Common errors
The two forms of the theorem
A worked accumulated-change problem
Definite integrals on the calculator
Why it matters

What this dot point is asking

A definite integral $\int_a^b f(x)\,dx$ has lower limit $a$ and upper limit $b$ . Unlike an indefinite integral (a family of functions), a definite integral evaluates to a single number.

The Fundamental Theorem of Calculus

The theorem links the two halves of calculus: to accumulate a rate of change $f$ over $[a,b]$ , find an antiderivative $F$ and subtract its values at the endpoints. The constant $C$ cancels in the subtraction, so it is omitted.

Evaluating a polynomial integral

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

Step 1: Find an antiderivative

The Fundamental Theorem requires an antiderivative $F(x)$ such that $F'(x) = 2x + 1$ . Applying the power rule term by term gives:

F(x) = x^2 + x.

No constant of integration is needed here because it cancels when we subtract $F(a)$ from $F(b)$ .

Step 2: Apply the limits using $F(b) - F(a)$

Evaluate $F$ at the upper limit $b = 3$ and the lower limit $a = 1$ , then subtract.

\int_1^3 (2x+1)\,dx = \Big[x^2 + x\Big]_1^3 = \big(3^2 + 3\big) - \big(1^2 + 1\big) = 12 - 2 = 10.

Final answer: $\displaystyle\int_1^3 (2x+1)\,dx = 10$ .

Properties of definite integrals

These properties simplify many problems:

Swapping the limits flips the sign; splitting the interval at an intermediate point $c$ lets you integrate piecewise. The additivity property is especially useful when a function is defined in pieces, or when a curve crosses the $x$ -axis and you want to handle the positive and negative parts separately.

An exponential definite integral

Evaluate $\displaystyle\int_0^2 e^{0.5x}\,dx$ .

Step 1: Find the antiderivative of $e^{0.5x}$

For an exponential of the form $e^{kx}$ , the antiderivative is $\dfrac{1}{k}e^{kx}$ , because differentiating $e^{kx}$ multiplies by $k$ , so the antiderivative must divide by $k$ to cancel that factor. Here $k = 0.5$ , so:

F(x) = \frac{1}{0.5}e^{0.5x} = 2e^{0.5x}.

Step 2: Evaluate at each limit and subtract

Apply the Fundamental Theorem by substituting the upper limit $x = 2$ and the lower limit $x = 0$ .

\int_0^2 e^{0.5x}\,dx = \Big[2e^{0.5x}\Big]_0^2 = 2e^{0.5(2)} - 2e^{0.5(0)} = 2e^{1} - 2e^{0} = 2e - 2.

Step 3: Compute the numerical value

2e - 2 \approx 2(2.7183) - 2 = 5.4366 - 2 = 3.437.

Final answer: $\displaystyle\int_0^2 e^{0.5x}\,dx = 2e - 2 \approx 3.437$ .

Net accumulated change

If $f(t)$ is a rate, the definite integral gives the net change in the quantity:

\int_a^b f'(t)\,dt=F(b)-F(a).

For example, if $v(t)$ is velocity, $\int_a^b v(t)\,dt$ is the displacement (net, signed) over the interval - distances travelled backward count as negative.

Common errors

The two forms of the theorem

The version above evaluates an integral, but the Fundamental Theorem also has a differentiation form: if $G(x)=\displaystyle\int_a^x f(t)\,dt$ , then $G'(x)=f(x)$ . In words, differentiating an integral with a variable upper limit just returns the integrand. This says that integration and differentiation are exact inverses. SACE questions occasionally use this to ask for the rate of change of an accumulated quantity, in which case the answer is simply the original rate function evaluated at the limit.

A worked accumulated-change problem

Definite integrals on the calculator

In calculator-assumed sections SACE permits evaluating $\int_a^b f(x)\,dx$ directly with a graphics or CAS calculator. The marks then sit in setting up the correct integral - the right integrand and the right limits - and in interpreting the number in context. For exact-value questions, though, you must show the antiderivative and the substitution by hand, because the calculator returns a decimal and an exact surd or logarithmic answer cannot be recovered from it. Read each question for the words "exact" or "to … decimal places" to decide which approach is required.

Why it matters

The Fundamental Theorem is the central result of Stage 2 integral calculus. It converts every area-under-a-curve and accumulated-change problem into a quick antidifferentiate-and-subtract, and it is examined directly in both the Skills and Applications Tasks and the external exam.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20233 marksCalculator-assumed. Given that

\int x e^x\,dx = x e^x - e^x + c

, find the exact value of the area bounded by

f(x) = x e^x

, the

x

-axis, and the vertical lines

x = 1

and

x = 5

Show worked answer →

Since $f(x) = x e^x$ is positive on $[1, 5]$ , the area equals the definite integral, evaluated with the Fundamental Theorem using $F(x) = x e^x - e^x$ .

\text{Area} = \int_1^5 x e^x\,dx = \Big[x e^x - e^x\Big]_1^5 = F(5) - F(1).

$F(5) = 5e^5 - e^5 = 4e^5$ and $F(1) = e - e = 0$ , so

\text{Area} = 4e^5 - 0 = 4e^5.

Marks: one for applying $F(b) - F(a)$ , one for substituting each limit, one for the exact answer $4e^5$ . Leaving it exact (not a decimal) is essential.

SACE 20242 marksCalculator-assumed. Vehicles enter a car park at a rate

E(t)

vehicles per hour, with the car park empty at

t = 0

. A table gives

\int_0^t E(t)\,dt

at various times, including the value

471

t = 24

. (a) State how many vehicles entered during the 24-hour period. (b) Explain what

\int_8^{24} E(t)\,dt

represents.

Show worked answer →

(a) The integral of a rate accumulates the total quantity. Since $E(t)$ is the entry rate (vehicles per hour), $\int_0^{24} E(t)\,dt$ is the total number that entered over the day. From the table this is $471$ vehicles. (1 mark)

(b) $\int_8^{24} E(t)\,dt$ is the number of vehicles that entered between $t = 8$ and $t = 24$ hours, found by $F(24) - F(8)$ from the table. (1 mark)

This is the Fundamental Theorem idea that the definite integral of a rate gives net accumulated change over the interval.

SACE 20212 marksCalculator-free. Evaluate

\int_0^2 (3x^2 - 4x + 1)\,dx

Show worked answer →

Antidifferentiate to $F(x) = x^3 - 2x^2 + x$ , then apply the limits:

\int_0^2 (3x^2 - 4x + 1)\,dx = \Big[x^3 - 2x^2 + x\Big]_0^2 = (8 - 8 + 2) - 0 = 2.

Marks: one for the antiderivative, one for the correct evaluation $F(2) - F(0) = 2$ .