Areas between curves | Year 12

01 / LEARN

The idea, step by step

Area is a difference of heights

At a particular $x$ , the vertical height between two graphs is upper minus lower. Integrating that height across the interval adds thin strips to produce the area. The order must be determined on the actual interval, not from the names of the functions.

If the region has no stated bounds, solve the equation $f (x) = g (x)$ to locate intersections. Include both intersection coordinates when the question asks for them, although the $x$ -coordinates supply the integral limits.

A = \int_{a}^{b} (upper (x) - lower (x)) d x

Check the region before calculating

A rough sketch and a test input help identify the upper curve. If the graphs exchange order inside the chosen interval, split the integral there. An unexpectedly negative result usually signals reversed curve order or limits.

Keep curve labels on any diagram. Distinguish area between two curves from area between a curve and the axis: the lower boundary need not be zero, and a region below the axis can still have positive geometric area.

A common trap

Integrating only the upper function and ignoring a nonzero lower boundary.

Instead: Subtract the lower function first, including its brackets; then integrate the difference.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Find the bounded area between $y = 2 x$ and $y = x^{2}$ .

Solve $x^{2} = 2 x$ : intersections at $x = 0, 2$ .
For $x = 1$ , $2 x > x^{2}$ , so use $2 x - x^{2}$ .
${[x^{2} - x^{3} / 3]}_{0}^{2} = 4 - 8 / 3 = 4 / 3$ .

4 / 3

Example 2

Worked example

Find the bounded area between $y = 4 - x^{2}$ and $y = x + 2$ .

Solve $4 - x^{2} = x + 2$ : $(x + 2) (x - 1) = 0$ .
The parabola is above the line between $- 2$ and 1.
Integrate $2 - x - x^{2}$ from $- 2$ to 1 to obtain $9 / 2$ .

9 / 2

Check your understanding

Find the area between $y = x + 2$ and $y = x$ from 0 to 3.

Show answer

$6$

The vertical gap is 2, so area $= \int_{0}^{3} 2 d x = 6$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Find the bounded area between $y = x$ and $y = x^{2}$ .
3 marks
Find the bounded area between $y = 4$ and $y = x^{2}$ .
3 marks
Find the area between $y = 3 x$ and $y = x$ on $[1, 3]$ .
3 marks
Find the bounded area between $y = 3 x$ and $y = x^{2}$ .
3 marks
Find the area between $y = x$ and $y = - x$ on $[- 1, 1]$ .
3 marks
Explain why $\int_{0}^{2} (x^{2} - 2 x) d x$ is not the geometric area between $y = x^{2}$ and $y = 2 x$ .
3 marks

Read the worked solutions

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

Find the bounded area between $y = x$ and $y = x^{2}$ .
3 marks
Bounds 0 and 1; upper minus lower $x - x^{2}$ .
${[x^{2} / 2 - x^{3} / 3]}_{0}^{1} = 1 / 6$ .
$1 / 6$
Find the bounded area between $y = 4$ and $y = x^{2}$ .
3 marks
Intersections at $\pm 2$ .
Integrate $4 - x^{2}$ on $[- 2, 2]$ .
$32 / 3$
Find the area between $y = 3 x$ and $y = x$ on $[1, 3]$ .
3 marks
Integrate $2 x$ from 1 to 3: ${[x^{2}]}_{1}^{3} = 8$ .
$8$
Find the bounded area between $y = 3 x$ and $y = x^{2}$ .
3 marks
Bounds 0 and 3.
$\int_{0}^{3} (3 x - x^{2}) d x = 27 / 2 - 9 = 9 / 2$ .
$9 / 2$
Find the area between $y = x$ and $y = - x$ on $[- 1, 1]$ .
3 marks
The upper curve changes at 0.
$\int_{- 1}^{0} (- 2 x) d x + \int_{0}^{1} 2 x d x = 1 + 1 = 2$ .
$2$
Explain why $\int_{0}^{2} (x^{2} - 2 x) d x$ is not the geometric area between $y = x^{2}$ and $y = 2 x$ .
3 marks
It gives $- 4 / 3$ ; the geometric area is $4 / 3$ .
The integrand is lower minus upper.

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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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