Areas | Year 12 | Ascent Library

01 / LEARN

Understand the idea.

Set the region up first

For a region between two curves, a vertical strip has height upper minus lower. Solve for intersections, inspect the order between them, and integrate that height over the required interval. The strip height must remain nonnegative.

Split the integral wherever the curves switch order. For area between one curve and the axis, split at its zeros and choose the positive height on each subinterval. Sketch the boundaries so that the integral covers exactly the requested region.

Shaded region between y equals x and y equals x squared from x zero to one. The line is above the parabola and both meet at each endpoint.

Respect the chosen variable

A horizontal strip instead has width right boundary minus left boundary and thickness $d y$ . Express both boundaries as functions of $y$ and find the corresponding vertical limits. This can avoid several pieces when vertical slicing would be awkward.

Area has square units because strip height and thickness both measure length. If an unsplit calculation gives zero for a visible region, check for cancellation: signed integral and total geometric area answer different questions.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find the area between y=x and y=x squared from x=0 to 1.

Worked solution

1
Find the intersections and order.
$x = x^{2} \Rightarrow x (x - 1) = 0, x = 0, 1$
Between zero and one, x is at least x squared, so the line is the upper curve.
2
Integrate the positive strip height.
$A = \int_{0}^{1} (x - x^{2}) d x = {[\frac{x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{1}$
A vertical strip extends from the parabola to the line.
3
Evaluate the area.
$A = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} square units$
The difference is positive and uses square units, as a geometric area requires.

Answer

$1 / 6$ square units

EXAMPLE 02

Worked example 2

Find total area between y=x squared-1 and the x axis from -2 to 2.

Worked solution

1
Find zeros and use symmetry.
$x^{2} - 1 = 0 \Rightarrow x = \pm 1$
The function is even; it lies below the axis between the roots and above it outside them.
2
Write nonnegative heights on the positive half.
$A = 2 [\int_{0}^{1} (1 - x^{2}) d x + \int_{1}^{2} (x^{2} - 1) d x]$
Reflection doubles the right-hand area, with the sign reversed only where the curve lies below the axis.
3
Evaluate both pieces.
$\begin{aligned} \int_{0}^{1} (1 - x^{2}) d x & = {[x - \frac{x^{3}}{3}]}_{0}^{1} = \frac{2}{3} \\ \int_{1}^{2} (x^{2} - 1) d x & = {[\frac{x^{3}}{3} - x]}_{1}^{2} = \frac{4}{3} \end{aligned}$
Each expression integrates the positive distance from the curve to the axis.
4
Add the geometric areas.
$A = 2 (\frac{2}{3} + \frac{4}{3}) = 4 square units$
All four symmetric portions contribute positively to total area.

Answer

$4$ square units

03 / YOUR TURN

Try it yourself.

Check your understanding

What are the intersection x values for x squared=2x?

Show answer & reasoning

0 and 2.

Rearrange $x^{2} = 2 x$ as $x (x - 2) = 0$ . A product is zero when at least one factor is zero, giving intersection inputs $x = 0$ and $x = 2$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the area between $y = x^{2}$ and $y = 2 x$ .
2 marks
Find the area under $y = 4 - x^{2}$ above the $x$ axis.
2 marks
Find the area between $y = ∣ x ∣$ and the axis on $[- 1, 1]$ .
2 marks
Find the area between $y = x$ and $y = sin x$ on $[0, π]$ .
2 marks
Find area between $x = y^{2}$ and $x = 4$ for $0 \leq y \leq 2$ .
2 marks
A student reports area zero between $y = x^{3}$ and the $x$ axis on $[- 1, 1]$ . Correct the result.
2 marks

Read the worked solutions

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

Find the area between $y = x^{2}$ and $y = 2 x$ .
2 marks
The intersections are 0 and 2.
Integrate $2 x - x^{2}$ over that interval.
$\frac{4}{3}$ square units
Find the area under $y = 4 - x^{2}$ above the $x$ axis.
2 marks
Zeros are -2 and 2.
Integrate $4 - x^{2}$ between them.
$\frac{32}{3}$ square units
Find the area between $y = ∣ x ∣$ and the axis on $[- 1, 1]$ .
2 marks
Use two congruent right triangles.
Each has area $\frac{1}{2}$ .
$1$ square unit
Find the area between $y = x$ and $y = sin x$ on $[0, π]$ .
2 marks
$x$ is at least $sin x$ throughout this interval.
Integrate $x - sin x$ and evaluate endpoints.
$\frac{π^{2}}{2} - 2$ square units
Find area between $x = y^{2}$ and $x = 4$ for $0 \leq y \leq 2$ .
2 marks
Horizontal slices have width $4 - y^{2}$ .
Integrate from $y = 0$ to 2.
$\frac{16}{3}$ square units
A student reports area zero between $y = x^{3}$ and the $x$ axis on $[- 1, 1]$ . Correct the result.
2 marks
The signed integral is zero, but areas add.
Use twice the integral of $x^{3}$ from 0 to 1.
$\frac{1}{2}$ square unit

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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.

Set the region up first

Respect the chosen variable

See the method in action.

Worked example 1

Find the intersections and order.

Integrate the positive strip height.

Evaluate the area.

Worked example 2

Find zeros and use symmetry.

Write nonnegative heights on the positive half.

Evaluate both pieces.

Add the geometric areas.

Try it yourself.

Check your understanding

Keep practising