Definite integrals | Year 12

01 / LEARN

The idea, step by step

Upper value minus lower value

A definite integral is a number. For a continuous integrand, find an antiderivative $F$ , then subtract its value at the lower bound from its value at the upper bound. Brackets matter when $F$ contains several terms or negative values.

There is no arbitrary constant in the final answer: any constant cancels between the two endpoint values. Keep exact values such as fractions or logarithms until a decimal is explicitly requested.

\int_{a}^{b} f (x) d x = F (b) - F (a)

Use structure before calculating

Reversing the limits changes the sign. Splitting an interval at an intermediate point adds the contributions. Multiplying the integrand by a constant multiplies the integral by that constant.

A negative definite integral is not automatically an error: it may represent a negative net change or a region below the axis. The interpretation of geometric area is developed separately; first carry out the signed calculation correctly.

\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x

A common trap

Using only the upper endpoint or reversing the subtraction.

Instead: Write $F (b) - F (a)$ explicitly before simplifying. Do not append $+ C$ to the numerical result.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Evaluate $\int_{0}^{2} (3 x^{2} + 1) d x$ .

An antiderivative is $x^{3} + x$ .
${[x^{3} + x]}_{0}^{2} = (8 + 2) - 0 = 10$ .

10

Example 2

Worked example

Evaluate $\int_{1}^{3} \frac{2}{x} d x$ .

Use $2 ln x$ since the interval is positive.
$2 ln 3 - 2 ln 1 = 2 ln 3$ .

2 ln 3

Check your understanding

Evaluate $\int_{1}^{2} 2 x d x$ .

Show answer

$3$

${[x^{2}]}_{1}^{2} = 4 - 1 = 3$ .

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Evaluate $\int_{0}^{3} 2 x d x$ .
3 marks
Evaluate $\int_{1}^{2} 3 x^{2} d x$ .
3 marks
Evaluate $\int_{2}^{0} (x + 1) d x$ .
3 marks
Evaluate $\int_{0}^{π} sin x d x$ .
3 marks
Evaluate $\int_{1}^{e} 1 / x d x$ .
3 marks
If $\int_{0}^{2} f (x) d x = 5$ and $\int_{2}^{5} f (x) d x = - 1$ , find $\int_{0}^{5} 3 f (x) d x$ .
3 marks

Read the worked solutions

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

Evaluate $\int_{0}^{3} 2 x d x$ .
3 marks
${[x^{2}]}_{0}^{3} = 9$ .
$9$
Evaluate $\int_{1}^{2} 3 x^{2} d x$ .
3 marks
${[x^{3}]}_{1}^{2} = 8 - 1 = 7$ .
$7$
Evaluate $\int_{2}^{0} (x + 1) d x$ .
3 marks
${[x^{2} / 2 + x]}_{2}^{0} = 0 - 4 = - 4$ .
$- 4$
Evaluate $\int_{0}^{π} sin x d x$ .
3 marks
${[- cos x]}_{0}^{π} = 1 - (- 1) = 2$ .
$2$
Evaluate $\int_{1}^{e} 1 / x d x$ .
3 marks
${[ln x]}_{1}^{e} = 1 - 0 = 1$ .
$1$
If $\int_{0}^{2} f (x) d x = 5$ and $\int_{2}^{5} f (x) d x = - 1$ , find $\int_{0}^{5} 3 f (x) d x$ .
3 marks
Add the interval contributions: $5 + (- 1) = 4$ .
Multiply by 3 to obtain 12.
$12$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods integration and logarithmic antiderivatives; original learning and assessment material, with by-hand methods and explicitly described numerical models.

Official curriculum source ↗

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