Definite integration | Year 12

01 / LEARN

Understand the idea.

Use the fundamental theorem

The fundamental theorem turns signed accumulation into endpoint subtraction: if $f$ is continuous on $[a, b]$ and $F^{'} = f$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ . Write each endpoint substitution in parentheses to protect its signs.

A definite integral is a number, so the arbitrary constant cancels. Contributions above the axis are positive and those below are negative when the bounds increase. Cancellation can produce zero even when substantial regions lie between the graph and the axis.

Check the conditions and magnitude

Check the integrand throughout the interval before using the theorem. A denominator zero can invalidate ordinary endpoint subtraction even if the endpoint values are finite. Splitting at a singularity leads to improper integrals, whose convergence must be assessed separately.

Use symmetry when the function and interval support it: an odd function integrates to zero on $[- a, a]$ , and an even function gives twice its integral on $[0, a]$ . A sign estimate or simple upper and lower bounds provides an independent plausibility check.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

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

Worked solution

1
Find an antiderivative.
$F (x) = x^{3} - 2 x, F^{'} (x) = 3 x^{2} - 2$
The integrand is a continuous polynomial on the whole interval.
2
Subtract the endpoint values.
$\int_{0}^{2} (3 x^{2} - 2) d x = {[x^{3} - 2 x]}_{0}^{2} = (8 - 4) - (0 - 0) = 4$
The upper endpoint value is reduced by the lower endpoint value.
3
Interpret the result.
$\int_{0}^{2} (3 x^{2} - 2) d x = 4$
This is net signed accumulation; it already includes the negative contribution near zero.

Answer

$4$ .

EXAMPLE 02

Worked example 2

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

Worked solution

1
Check odd symmetry.
$f (- x) = {(- x)}^{3} = - x^{3} = - f (x)$
The function is odd and continuous on the symmetric interval.
2
Use cancellation.
$\int_{- 2}^{2} x^{3} d x = 0$
The left and right signed contributions have equal magnitude and opposite signs.
3
Confirm with endpoint subtraction.
${[\frac{x^{4}}{4}]}_{- 2}^{2} = \frac{16}{4} - \frac{16}{4} = 0$
The even antiderivative takes the same value at both endpoints.

Answer

$0$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

How does reversing bounds change an integral?

Show answer & reasoning

It negates its value.

Reversing bounds reverses endpoint subtraction: $F (a) - F (b) = - (F (b) - F (a))$ . Therefore $\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Evaluate $\int_{1}^{e} \frac{1}{x} d x$ .
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 $\int_{1}^{e} \frac{1}{x} d x$ .
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 $x^{2} + x$ .
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.
$2 [\frac{1}{3} + 2] = \frac{14}{3}$ .
$\frac{14}{3}$ .

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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 ↗

In this lesson

Understand the idea.

Use the fundamental theorem

Check the conditions and magnitude

See the method in action.

Worked example 1

Find an antiderivative.

Subtract the endpoint values.

Interpret the result.

Worked example 2

Check odd symmetry.

Use cancellation.

Confirm with endpoint subtraction.

Try it yourself.

Check your understanding

Keep practising