Definite integration | Year 12

01 / LEARN

The idea, step by step

Use the fundamental theorem

If F is an antiderivative of a continuous f on [a,b], then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ . Parentheses around each endpoint value prevent sign errors. Reversing bounds negates the integral.

A definite integral is a signed accumulation. Regions below the horizontal axis subtract from those above it. A negative answer is possible even when the interval length is positive.

Check the conditions and magnitude

Inspect continuity before applying endpoint subtraction. A denominator zero inside the interval cannot be ignored. A plausible-looking endpoint value does not justify using the theorem across a singularity.

Symmetry can simplify exact calculation: odd functions integrate to zero over symmetric intervals, and even functions give twice the integral over the positive half. Use the graph or simple bounds to test the sign and approximate scale of a result.

A common trap

A definite integral is always a positive area.

Instead: It is signed accumulation. Geometric area requires splitting or an absolute value.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

Evaluate $\int_{0}^{2} (3 x^{2} - 2) d x$ .

An antiderivative is x cubed-2x.
Evaluate (8-4)-0.

4

Example 2

Worked example 2

Evaluate $\int_{- 2}^{2} x^{3} d x$ .

The integrand is odd on a symmetric interval.
The negative and positive signed regions cancel.

0

Check your understanding

How does reversing bounds change an integral?

Show answer

It negates its value.

The endpoint subtraction order is reversed.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Evaluate $\int_{1}^{e} 1 / x d x$ .
2 marks
Evaluate $\int_{0}^{π} sin x d x$ .
2 marks
Evaluate $\int_{1}^{2} (2 x + 1) d x$ .
2 marks
If $\int_{2}^{5} f (x) d x = 7$ , find $\int_{5}^{2} f (x) d x$ .
2 marks
Explain why ordinary endpoint subtraction is invalid for $\int_{- 1}^{1} x^{- 2} d x$ .
2 marks
Evaluate $\int_{- 1}^{1} (x^{2} + 2) d x$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Evaluate $\int_{1}^{e} 1 / x d x$ .
2 marks
Use ln x on the positive interval.
Subtract ln 1 from ln e.
$1$ .
Evaluate $\int_{0}^{π} sin x d x$ .
2 marks
Use antiderivative -cos x.
The endpoint difference is 1-(-1).
$2$ .
Evaluate $\int_{1}^{2} (2 x + 1) d x$ .
2 marks
Use x squared+x.
Subtract 2 from 6.
$4$ .
If $\int_{2}^{5} f (x) d x = 7$ , find $\int_{5}^{2} f (x) d x$ .
2 marks
Reversing limits changes the sign.
$- 7$ .
Explain why ordinary endpoint subtraction is invalid for $\int_{- 1}^{1} x^{- 2} d x$ .
2 marks
The continuity condition fails.
Treating the two endpoints alone would overlook the singularity.
The integrand is undefined at x=0 inside the interval.
Evaluate $\int_{- 1}^{1} (x^{2} + 2) d x$ .
2 marks
The integrand is even, so double the integral from 0 to 1.
$2 [1 / 3 + 2] = 14 / 3$ .
$14 / 3$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics integration techniques and applications, including inverse-trigonometric patterns, substitution, parts, simple partial fractions, areas and solids of revolution.

Official curriculum source ↗

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