Signed area versus geometric area | Year 12

01 / LEARN

The idea, step by step

The integral keeps the sign

A definite integral adds positive contributions above the $x$ -axis and negative contributions below it. Equal-sized regions can cancel, so a zero integral does not imply that no region exists.

Geometric area counts every piece positively. Find zeros in the interval, split there, and reverse the sign of a negative contribution. State square units when $x$ and $y$ both represent lengths.

Use a sign chart before integrating

For each interval between successive zeros, test a point or inspect the graph to decide the sign of $f (x)$ . Then calculate an ordinary integral on each piece. Keep exact fractions until the final sum.

You can write the area as $\int_{a}^{b} ∣ f (x) ∣ d x$ , but that notation does not remove the need to understand the sign changes when working by hand. A sketch is a useful check on both the zeros and the expected sizes.

A common trap

Taking the absolute value of the final signed integral even when the graph changes sign.

Instead: Take absolute values of the separate signed contributions, not only their combined total.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Find both $\int_{0}^{3} (x - 1) d x$ and the area between the graph and axis.

${[x^{2} / 2 - x]}_{0}^{3} = 3 / 2$ .
The zero is $x = 1$ . The integral from 0 to 1 is $- 1 / 2$ ; from 1 to 3 is 2.
Area $= 1 / 2 + 2 = 5 / 2$ .

Signed integral

3 / 2

; area

5 / 2

Example 2

Worked example

Find the area between $y = x^{2} - 4$ and the axis for $- 2 \leq x \leq 2$ .

The curve is nonpositive on this interval.
Area $= - \int_{- 2}^{2} (x^{2} - 4) d x = \int_{- 2}^{2} (4 - x^{2}) d x$ .
Evaluating gives $32 / 3$ .

32 / 3

Check your understanding

For $y = x$ on $[- 1, 1]$ , what are the signed integral and geometric area?

Show answer

$0$ and $1$

The two triangles cancel in the integral but add for area.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Find the area between $y = x - 2$ and the axis on $[0, 4]$ .
3 marks
Find $\int_{- 2}^{2} x d x$ and the corresponding geometric area.
3 marks
Find the area under $y = 4 - x^{2}$ on $[- 2, 2]$ .
3 marks
Find the area between $y = x (x - 2)$ and the axis on $[0, 3]$ .
3 marks
Find the area between $y = sin x$ and the axis on $[0, 2 π]$ .
3 marks
A student says an integral of 0 proves that an area is 0. Give a counterexample.
3 marks

Read the worked solutions

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

Find the area between $y = x - 2$ and the axis on $[0, 4]$ .
3 marks
Split at 2.
Two triangles each have area 2.
$4$
Find $\int_{- 2}^{2} x d x$ and the corresponding geometric area.
3 marks
The signed contributions are $- 2$ and 2.
Their absolute values sum to 4.
$0$ and $4$
Find the area under $y = 4 - x^{2}$ on $[- 2, 2]$ .
3 marks
The function is nonnegative, so integrate it directly.
$32 / 3$
Find the area between $y = x (x - 2)$ and the axis on $[0, 3]$ .
3 marks
An antiderivative is $x^{3} / 3 - x^{2}$ .
Contributions on $[0, 2]$ and $[2, 3]$ are $- 4 / 3$ and $4 / 3$ .
Add their magnitudes.
$8 / 3$
Find the area between $y = sin x$ and the axis on $[0, 2 π]$ .
3 marks
On $[0, π]$ the area is 2; on $[π, 2 π]$ the signed integral is $- 2$ .
Add 2 and 2.
$4$
A student says an integral of 0 proves that an area is 0. Give a counterexample.
3 marks
The integral is 0 but the two triangular regions have total area 1.
$f (x) = x$ on $[- 1, 1]$ .

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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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