Areas between curves | Year 12

01 / LEARN

Understand the idea.

Area is a difference of heights

The vertical height between two graphs is upper minus lower at the same input. Integrating this nonnegative gap adds thin strips across the region. Neither boundary needs to be the axis, and either graph may lie below it.

When bounds are not supplied, find intersections by solving $f (x) = g (x)$ . Their $x$ -coordinates give potential integration limits. Identify which intersections enclose the requested region before assuming a particular pair of bounds.

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

The line y=2x and parabola y=x squared are drawn from x=0 to 2.3. Their bounded region from (0,0) to (2,4) is shaded between the two curves. The line is above the parabola on this interval.

Check the region before calculating

Determine curve order on each interval between intersections. A test input works when there are no further crossings in that interval. If the graphs exchange order, split the area calculation and choose upper minus lower separately on each piece.

Subtract the entire lower expression using brackets before simplifying. A negative result is a reason to check curve order, bounds and arithmetic. Geometric area between distinct boundaries over a nonzero interval must be positive.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

Worked solution

1
Find the intersection inputs.
$\begin{aligned} x^{2} & = 2 x \\ x (x - 2) & = 0 \\ x & = 0, 2 \end{aligned}$
The closed region begins and ends where the two curves meet.
2
Determine the upper curve.
$2 (1) = 2 > 1^{2} = 1$
There are no further intersections between 0 and 2, so the line stays above the parabola inside this interval.
3
Integrate the vertical gap.
$A = \int_{0}^{2} (2 x - x^{2}) d x = {[x^{2} - \frac{x^{3}}{3}]}_{0}^{2}$
The integrand is upper height minus lower height.
4
Evaluate the area.
$A = 4 - \frac{8}{3} = \frac{4}{3}$
The positive value agrees with the order of the two boundaries.

Answer

$4 / 3$

EXAMPLE 02

Worked example

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

The curves y=4-x squared and y=x+2. A light region shows the bounded set between their intersections at x=-2 and x=1. No numerical area is supplied.

Worked solution

1
Find both intersections.
$\begin{aligned} 4 - x^{2} & = x + 2 \\ x^{2} + x - 2 & = 0 \\ (x + 2) (x - 1) & = 0 \\ x & = - 2, 1 \end{aligned}$
The intersection inputs bound the enclosed region.
2
Choose and simplify upper minus lower.
$(4 - x^{2}) - (x + 2) = 2 - x - x^{2}$
At $x = 0$ the parabola has height 4 and the line height 2; there are no internal crossings.
3
Integrate the gap.
$A = {[2 x - \frac{x^{2}}{2} - \frac{x^{3}}{3}]}_{- 2}^{1}$
The gap remains nonnegative between the two intersections.
4
Subtract the endpoint values.
$\begin{aligned} A & = (2 - \frac{1}{2} - \frac{1}{3}) - (- 4 - 2 + \frac{8}{3}) \\ = \frac{7}{6} - (- \frac{10}{3}) \\ = \frac{9}{2} \end{aligned}$
The complete lower endpoint value is subtracted, including its negative sign.

Answer

$9 / 2$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$6$

The vertical gap is $(x + 2) - x = 2$ throughout $[0, 3]$ . Therefore $A = \int_{0}^{3} 2 d x = 2 (3 - 0) = 6$ .

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6 worksheet questions · 18 marks

Open worksheet 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 $x - x^{2}$ .
${[\frac{x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{1} = \frac{1}{6}$ .
$\frac{1}{6}$
Find the bounded area between $y = 4$ and $y = x^{2}$ .
3 marks
Intersections at $\pm 2$ .
Integrate $4 - x^{2}$ on $[- 2, 2]$ .
$\frac{32}{3}$
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.
$\int_{0}^{3} (3 x - x^{2}) d x = \frac{27}{2} - 9 = \frac{9}{2}$ .
$\frac{9}{2}$
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 $- \frac{4}{3}$ ; the geometric area is $\frac{4}{3}$ .
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 ↗

In this lesson

Understand the idea.

Area is a difference of heights

Check the region before calculating

See the method in action.

Worked example

Find the intersection inputs.

Determine the upper curve.

Integrate the vertical gap.

Evaluate the area.

Worked example

Find both intersections.

Choose and simplify upper minus lower.

Integrate the gap.

Subtract the endpoint values.

Try it yourself.

Check your understanding

Keep practising