Definite integrals | Year 12

01 / LEARN

Understand the idea.

Upper value minus lower value

A definite integral gives a signed accumulated amount over fixed bounds. If $f$ is continuous on $[a, b]$ and $F^{'} = f$ , the fundamental theorem evaluates it as $F (b) - F (a)$ : upper endpoint value minus lower endpoint value.

Use brackets when subtracting the lower value, especially if it is negative. Any arbitrary constant cancels, so the final result is a number rather than a family. Retain exact fractions and logarithms unless an approximation is requested.

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

Use structure before calculating

Before calculating, notice useful structure. Reversing the limits reverses the sign; splitting at an interior point adds the two signed contributions. A constant factor can be taken outside the integral.

The sign has meaning. A negative result may describe loss or a graph lying below the axis. Check the integrand and interval before interpreting the number; a signed integral and a geometric area answer different questions.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

Worked solution

1
Find an antiderivative.
$F (x) = x^{3} + x$
Its derivative is the polynomial $3 x^{2} + 1$ .
2
Evaluate both endpoints.
$\begin{aligned} F (2) & = 2^{3} + 2 = 10 \\ F (0) & = 0^{3} + 0 = 0 \end{aligned}$
The definite integral requires both endpoint values.
3
Subtract lower from upper.
$\int_{0}^{2} (3 x^{2} + 1) d x = 10 - 0 = 10$
The fundamental theorem gives the signed accumulation over the stated interval.

Answer

$10$

EXAMPLE 02

Worked example

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

Worked solution

1
Check the interval and select the antiderivative.
$F (x) = 2 ln x (1 \leq x \leq 3)$
The reciprocal is continuous on this positive interval, so a real logarithm without absolute values is valid.
2
Substitute the endpoints.
$\int_{1}^{3} \frac{2}{x} d x = 2 ln 3 - 2 ln 1$
Upper minus lower evaluates the definite integral.
3
Simplify exactly.
$2 ln 3 - 2 \cdot 0 = 2 ln 3$
The logarithm of 1 is zero; the remaining logarithm is an exact value.

Answer

$2 ln 3$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$3$

An antiderivative of $2 x$ is $x^{2}$ . Thus $\int_{1}^{2} 2 x d x = 2^{2} - 1^{2} = 3$ , with no arbitrary constant in the final value.

Keep practising

6 worksheet questions · 18 marks

Open worksheet 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 $\int_{1}^{e} \frac{1}{x} d x$ .
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
${[\frac{x^{2}}{2} + x]}_{2}^{0} = 0 - 4 = - 4$ .
$- 4$
Evaluate $\int_{0}^{π} sin x d x$ .
3 marks
${[- cos x]}_{0}^{π} = 1 - (- 1) = 2$ .
$2$
Evaluate $\int_{1}^{e} \frac{1}{x} d x$ .
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 ↗

In this lesson

Understand the idea.

Upper value minus lower value

Use structure before calculating

See the method in action.

Worked example

Find an antiderivative.

Evaluate both endpoints.

Subtract lower from upper.

Worked example

Check the interval and select the antiderivative.

Substitute the endpoints.

Simplify exactly.

Try it yourself.

Check your understanding

Keep practising