Numerical area approximation and the integral idea | Year 12

01 / LEARN

Understand the idea.

Build area from narrow strips

Numerical integration adds the contributions of narrow strips. A rectangle uses one sampled height; a trapezium averages its two endpoint heights. Multiply by the actual strip width before adding: heights alone are not areas.

For equal spacing $h$ , an interior height belongs to two neighbouring trapeziums, while each endpoint belongs to one. This explains the weights in the trapezium formula. Integrating speed in metres per second over seconds estimates distance in metres.

T = \frac{h}{2} (y_{0} + 2 y_{1} + \dots + 2 y_{n - 1} + y_{n})

Interpret the approximation

An approximation depends on how the curve behaves between samples. Straight segments give the trapezium model. Increasing positive functions give left-rectangle underestimates and right-rectangle overestimates, but that information alone does not determine the trapezium error.

Curvature matters for trapeziums: chords lie above a convex curve, producing an overestimate. Finer spacing often improves estimates for smooth functions, but a few measured values do not establish exactness or justify inventing unsampled values.

Speed recorded at equally spaced times.
Time (s)	0	1	2	3
Speed (m/s)	0	2	3	6

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

For $f (x) = x^{2}$ on $[0, 2]$ , use two trapeziums of width 1.

Worked solution

1
List the strip endpoints and heights.
$\begin{aligned} h & = 1 \\ f (0) & = 0, f (1) = 1, f (2) = 4 \end{aligned}$
Two equal strips cover the interval of length 2.
2
Calculate the two trapezium areas.
$\begin{aligned} T_{1} & = \frac{1}{2} (0 + 1) = \frac{1}{2} \\ T_{2} & = \frac{1}{2} (1 + 4) = \frac{5}{2} \end{aligned}$
Each area is strip width multiplied by the mean of its endpoint heights.
3
Add the estimates.
$T = \frac{1}{2} + \frac{5}{2} = 3$
The strip areas cover the whole interval without overlap.
4
Compare with the exact integral.
$\int_{0}^{2} x^{2} d x = {[\frac{x^{3}}{3}]}_{0}^{2} = \frac{8}{3} < 3$
The trapeziums overestimate; the convex graph lies below each joining chord.

Answer

$T = 3$

EXAMPLE 02

Worked example

Use the supplied speed table to estimate distance by the trapezium rule.

Speed recorded at equally spaced times.
Time (s)	0	1	2	3
Speed (m/s)	0	2	3	6

Worked solution

1
Use the measured time spacing.
$h = 1 s$
Successive speed measurements are one second apart.
2
Estimate distance in each interval.
$\begin{aligned} d_{1} & = \frac{1}{2} (0 + 2) = 1 m \\ d_{2} & = \frac{1}{2} (2 + 3) = \frac{5}{2} m \\ d_{3} & = \frac{1}{2} (3 + 6) = \frac{9}{2} m \end{aligned}$
Mean speed times the one-second interval gives each trapezium distance.
3
Add the three distances.
$d \approx 1 + \frac{5}{2} + \frac{9}{2} = 8 m$
The integral of speed is total distance, and the table supplies an approximation to it.
4
State the model limitation.
The value is exact for straight speed segments between the samples; the measurements alone do not establish that model.

Answer

8 m

03 / YOUR TURN

Try it yourself.

Check your understanding

Use two left rectangles for $y = x^{2}$ on $[0, 2]$ .

Show answer & reasoning

$1$

Two left rectangles have width 1 and use heights $f (0) = 0$ and $f (1) = 1$ . Their total is $1 (0 + 1) = 1$ ; the height at the rightmost endpoint is not used.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

For $y = x^{2}$ on $[0, 2]$ , use two right rectangles.
3 marks
For $y = x^{2}$ on $[0, 2]$ , use midpoint rectangles of width 1.
3 marks
For $y = x^{2}$ on $[0, 2]$ , use four trapeziums.
3 marks
A rate is 4,6 and 10 L/min at 0,2 and 4 min. Estimate total volume by trapeziums.
3 marks
The function $f (x) = 2 x + 1$ is sampled at 0,1 and 2. Is the trapezium estimate exact? Explain.
3 marks
Why does the trapezium rule overestimate $\int_{0}^{2} x^{2} d x$ ?
3 marks

Read the worked solutions

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

For $y = x^{2}$ on $[0, 2]$ , use two right rectangles.
3 marks
Width 1; heights 1 and 4 give 5.
$5$
For $y = x^{2}$ on $[0, 2]$ , use midpoint rectangles of width 1.
3 marks
Heights $f (0.5) = 0.25$ and $f (1.5) = 2.25$ sum to 2.5.
$\frac{5}{2}$
For $y = x^{2}$ on $[0, 2]$ , use four trapeziums.
3 marks
$h = 0.5$ ; heights 0,0.25,1,2.25,4.
$T = 0.25 [0 + 2 (0.25 + 1 + 2.25) + 4] = 2.75$ .
$\frac{11}{4}$
A rate is 4,6 and 10 L/min at 0,2 and 4 min. Estimate total volume by trapeziums.
3 marks
$T = \frac{2 (4 + 6)}{2} + \frac{2 (6 + 10)}{2} = 10 + 16 = 26$ .
26 L
The function $f (x) = 2 x + 1$ is sampled at 0,1 and 2. Is the trapezium estimate exact? Explain.
3 marks
A straight line matches each trapezium top exactly.
$T = \frac{1 + 3}{2} + \frac{3 + 5}{2} = 6$ .
Yes; its value is 6.
Why does the trapezium rule overestimate $\int_{0}^{2} x^{2} d x$ ?
3 marks
Each chord between neighbouring endpoints lies above the curve, so the trapezium includes extra area.
The graph is convex.

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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.

Build area from narrow strips

Interpret the approximation

See the method in action.

Worked example

List the strip endpoints and heights.

Calculate the two trapezium areas.

Add the estimates.

Compare with the exact integral.

Worked example

Use the measured time spacing.

Estimate distance in each interval.

Add the three distances.

State the model limitation.

Try it yourself.

Check your understanding

Keep practising