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Module 3 · L14 of 15 ~35 min +95 XP available

The Trapezoidal Rule

Not all integrals can be found exactly. The trapezoidal rule approximates the area under a curve by dividing it into trapezoids, like estimating the area of an irregular field by splitting it into straight-edged strips.

Today's hook, A GPS records your elevation every 500 m along a hiking trail. You want the total vertical gain but you only have sampled values, not a formula. The trapezoidal rule turns these measurements into a reliable area estimate. Can you work out how?

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Orient to the trapezoidal rule

Work through the visible teaching and complete each embedded check.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall, your gut answer first

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Before we start, what do you already know about approximating areas under curves?

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The two moves

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There are only two moves in this lesson. Lock them in and the rest is arithmetic.

Move 1: Calculate $h$ and all $y$-values. Find the subinterval width $h = \frac{b-a}{n}$ and evaluate the function at each equally spaced point.

Move 2: Apply the formula. The first and last $y$-values get coefficient 1; all interior values get coefficient 2. Organise into a table first.

$$\int_a^b f(x) \, dx \approx \frac{h}{2}\left[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n\right]$$

$h$ formula

$h = \dfrac{b-a}{n}$ where $n$ is the number of strips (not the number of function values).

Coefficients

End values ($y_0$ and $y_n$): coefficient 1. Interior values ($y_1$ through $y_{n-1}$): coefficient 2.

Concavity check

Concave up → overestimate. Concave down → underestimate. Check $f''(x)$ to confirm.

What you will master

Know

Key facts

The trapezoidal rule formula and what each symbol means
That $h = \frac{b-a}{n}$ and $n$ strips require $n+1$ function values
When the rule overestimates vs underestimates

Understand

Concepts

Why trapezoids approximate the area under a curve
How more strips improve accuracy
The link between concavity and the direction of error

Can do

Skills

Apply the trapezoidal rule to approximate definite integrals
Calculate function values at equally spaced points
Assess whether the trapezoidal rule gives an overestimate or underestimate

Build the approximation language

Work through the visible teaching and complete each embedded check.

Key terms

Trapezoidal ruleAn approximation for definite integrals using trapezoids: $\frac{h}{2}[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n]$ where $h = \frac{b-a}{n}$.

SubintervalOne of the equal-width strips into which the interval $[a, b]$ is divided; width is $h$.

OverestimateThe trapezoidal rule overestimates when the curve is concave up (trapezoids sit above the curve).

UnderestimateThe trapezoidal rule underestimates when the curve is concave down (trapezoids sit below the curve).

Function values $y_i$The values $y_i = f(x_i)$ evaluated at each equally spaced point $x_i = a + ih$.

ConcavityWhether the curve bends upward ($f'' > 0$, concave up) or downward ($f'' < 0$, concave down).

Construct the trapezoidal rule

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Concept, the trapezoidal rule

core concept

The trapezoidal rule approximates the area under a curve by replacing the curve with straight line segments between points, forming trapezoids. The area of each trapezoid is the average of the two parallel sides times the width: $\frac{h}{2}(y_i + y_{i+1})$. Summing all trapezoid areas gives the approximation.

$$\int_a^b f(x) \, dx \approx \frac{h}{2}\left[y_0 + 2(y_1 + y_2 + \cdots + y_{n-1}) + y_n\right]$$

where $h = \dfrac{b-a}{n}$ and $y_i = f(x_i)$

More strips mean greater accuracy because the straight-line segments better approximate the curve. When the curve is concave up, the straight-line tops of the trapezoids lie above the curve, giving an overestimate. When concave down, the opposite occurs.

The coefficients: 1, 2, 2..., 2, 1. Notice the pattern, the two endpoint values $y_0$ and $y_n$ each appear in exactly one trapezoid (as one edge). Every interior value $y_i$ appears in two adjacent trapezoids (as a shared edge), hence the coefficient of 2. This is the key to the formula.

Trapezoidal rule: $\int_a^b f(x) \, dx \approx \frac{h}{2}[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n]$; $h = \frac{b-a}{n}$; $n$ strips require $n+1$ function values

Pause, copy the trapezoidal rule $\int_a^b f(x)\,dx \approx \dfrac{h}{2}[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n]$ where $h = \dfrac{b-a}{n}$ into your book.

Quick check: True or false, if you use 4 strips with the trapezoidal rule, you need exactly 4 function values.

Apply the rule

Work through the visible teaching and complete each embedded check.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC APPLICATION

Use the trapezoidal rule with 4 strips to approximate $\int_0^2 x^2 \, dx$.

$a = 0$, $b = 2$, $n = 4$, so $h = \dfrac{2-0}{4} = 0.5$

Calculate the width of each subinterval.

PROBLEM 2 · GIVEN NUMBER OF VALUES

Use the trapezoidal rule with 5 function values to approximate $\int_1^3 \frac{1}{x} \, dx$.

5 function values means 4 strips, so $n = 4$. $h = \dfrac{3-1}{4} = 0.5$

Number of strips = number of function values $- 1$.

PROBLEM 3 · FROM A TABLE

A curve has the following values. Approximate $\int_0^{1.2} y \, dx$ using the trapezoidal rule.

$x$	0	0.3	0.6	0.9	1.2
$y$	1.00	1.09	1.36	1.81	2.44

$h = 0.3$, $n = 4$ strips (5 values)

The subinterval width is constant at 0.3 from the table.

Quick check: Using 3 strips to approximate $\int_0^3 f(x) \, dx$, the value of $h$ is:

Avoid approximation traps

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Common errors · the 3 traps that cost marks

Trap 01

Confusing number of strips with number of function values

$n$ strips require $n + 1$ function values. If the question gives 5 function values, there are 4 strips. The formula uses $h = \frac{b-a}{n}$ where $n$ is the number of strips, not values.

Trap 02

Forgetting the factor of 2 for interior points

In the trapezoidal rule, the first and last $y$-values have coefficient 1, while all interior points have coefficient 2. Missing the factor of 2 is the most common arithmetic error.

Trap 03

Using $h = b - a$ instead of $\frac{b-a}{n}$

$h$ is the width of each strip, not the total interval. Always divide the total width by the number of strips to get $h$.

Odd one out: Three of the following are true about the trapezoidal rule. Which one is FALSE?

Interactive · Trapezoidal rule explorer

Try this: pick $\sqrt{x}$, look at the shading between the chords and the curve, and commit to overestimate or underestimate before the exact value unmasks. Then drag $n$ up and watch the error shrink without the direction ever flipping. Finish with $x^3 - 3x$, where the concavity changes sign inside the interval.

Use the explorer. Choose $\sqrt{x}$ on $[0, 4]$ with $n = 4$ and predict, then read the verdict. True or false: $T_4$ is an overestimate of $\int_0^4 \sqrt{x}\,dx$.

Use the explorer. Choose $\frac{1}{x}$ on $[1, 3]$ and set $n = 8$ strips. The value table shows function values, and the verdict says $T_8$ is an (one word) because $f''(x) = \frac{2}{x^3}$ is positive on the whole interval.

Practise and reflect

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Quick-fire practice · 5 problems

Use the trapezoidal rule with 4 strips to approximate $\int_0^4 x \, dx$.

Use 3 strips to approximate $\int_1^4 x^2 \, dx$.

A curve has values: $(0, 1)$, $(1, 3)$, $(2, 5)$, $(3, 7)$. Approximate $\int_0^3 y \, dx$.

Use 4 strips to approximate $\int_0^2 e^x \, dx$ (use $e^0 = 1$, $e^{0.5} \approx 1.65$, $e^1 \approx 2.72$, $e^{1.5} \approx 4.48$, $e^2 \approx 7.39$).

Explain whether the trapezoidal rule overestimates or underestimates $\int_0^1 x^3 \, dx$.

Fill the blanks: drag each token into the matching blank.

overestimate n+1 two h

The subinterval width is ___ $= \frac{b-a}{n}$. Using $n$ strips requires ___ function values. Interior values have coefficient ___. A concave-up curve gives an ___.

Teach it back: Explain in your own words why interior $y$-values in the trapezoidal rule formula have a coefficient of 2 while the endpoint values have a coefficient of 1.

Revisit your thinking

Earlier you were asked about approximating areas. The trapezoidal rule is a practical method when exact integration is difficult or impossible. It replaces curves with straight lines, trading exactness for computability. More strips give a better approximation because the straight-line segments more closely follow the actual curve.

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Complete the mixed practice

Complete the question-bank round and short-answer practice, then compare your responses with the comprehensive answers.

Multiple choice

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Short answer

ApplyBand 4

Q1. Use the trapezoidal rule with 4 strips to approximate $\int_0^2 x^2 \, dx$. Give your answer to 2 decimal places. Show all working. 3 MARKS

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ApplyBand 4

Q2. The following table gives values of $y = f(x)$:

$x$	0	0.5	1.0	1.5	2.0
$y$	1.00	1.41	2.00	2.83	4.00

Use the trapezoidal rule to approximate $\int_0^2 f(x) \, dx$. 3 MARKS

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AnalyseBand 5

Q3. Use the trapezoidal rule with 2 strips to approximate $\int_0^2 (x^3 - 3x) \, dx$. Determine whether your answer is an overestimate or underestimate, giving a reason. 4 MARKS

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📖 Comprehensive answers (click to reveal)

Q1 (3 marks): $h = 0.5$ [0.5]. $y$-values: 0, 0.25, 1, 2.25, 4 [0.5]. $\approx \frac{0.5}{2}[0 + 2(0.25 + 1 + 2.25) + 4] = 0.25 \times 11 = 2.75$ [2].

Q2 (3 marks): $h = 0.5$ (4 strips from 5 values) [0.5]. $\approx \frac{0.5}{2}[1.00 + 2(1.41 + 2.00 + 2.83) + 4.00]$ [0.5]. $= 0.25[1 + 12.48 + 4] = 0.25 \times 17.48 = 4.37$ [2].

Q3 (4 marks): $h = 1$ [0.5]. Values at $x = 0, 1, 2$: $0, -2, 2$ [0.5]. $\approx \frac{1}{2}[0 + 2(-2) + 2] = \frac{1}{2}(-2) = -1$ [1]. $f''(x) = 6x > 0$ for $x > 0$ on $(0,2)$, so the curve is predominantly concave up [1]. When concave up, trapezoids lie above the curve, so the trapezoidal rule overestimates [1].

Review and complete

Retrieve the central ideas, then mark the lesson complete or continue to the module quiz.

Retrieve the lesson before you leave

Write the trapezoidal rule and define every symbol.
Explain the relationship between strips and function values.
Describe how increasing the number of strips changes an approximation.

Answer from memory first, then return to the matching Learn checkpoint to check and correct your response.

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← Lesson 8 · Areas Checkpoint 3 · Further Calculus →

Module overview · Maths Advanced