The Trapezium Rule

Approximating definite integrals using trapeziums when analytical integration is not possible

SPEC 9.4 • NUMERICAL METHODS

Why Numerical Integration?

Some integrals cannot be found analytically. There is no antiderivative we can write down using standard functions.

Examples of Non-Integrable Functions

\int e^{- x^{2}} d x

\int \frac{s i n x}{x} d x

\int 1 + x^{3} d x

For these functions, we need numerical methods to approximate the definite integral (the area under the curve).

Key Idea

The trapezium rule approximates the area under a curve by dividing it into trapeziums. The sum of these trapezium areas gives us an estimate of the integral.

The Geometric Basis

To approximate

\int_{a}^{b} f (x) d x

, we divide the interval [a, b] into n equal strips and form trapeziums:

Key Insight

The tops of the trapeziums connect points on the curve. Each strip width is

h = \frac{b - a}{n}

. The straight lines approximate the curve between ordinates.

The Trapezium Rule Formula

The Formula

\int_{a}^{b} y d x \approx \frac{h}{2} [(y_{0} + y_{n}) + 2 (y_{1} + y_{2} + \dots + y_{n - 1})]

SymbolMeaning

h

Strip width:

h = \frac{b - a}{n}

n

Number of strips

y_{0}, y_{1}, \dots, y_{n}

The y-values (ordinates) at equally spaced x-values

Critical

n strips require (n + 1) y-values! The first and last ordinates (

y_{0}

and

y_{n}

) appear once. All middle ordinates are doubled.

Calculating Strip Width

Strip Width Formula

h = \frac{b - a}{n}

Once you have h, you can find all the x-values:

The x-values

x_{0} = a

x_{1} = a + h

x_{2} = a + 2 h

⋮

x_{n} = a + nh = b

Example

For

\int_{1}^{3} f (x) d x

with 4 strips:

h = \frac{3 - 1}{4} = 0.5

x-values: 1, 1.5, 2, 2.5, 3 (that's 5 values for 4 strips)

01Worked Example

Use the trapezium rule with 4 strips to estimate $\int_{0}^{2} x^{2} d x$ , given the following table of values:

x00.511.52

y00.2512.254

1

Identify the values: a = 0, b = 2, n = 4 strips

2

Calculate strip width:

h = \frac{2 - 0}{4} = 0.5

3

Identify ordinates:

y_{0} = 0, y_{1} = 0.25, y_{2} = 1, y_{3} = 2.25, y_{4} = 4

4

Apply the formula:

\frac{h}{2} [(y_{0} + y_{4}) + 2 (y_{1} + y_{2} + y_{3})]

= \frac{0.5}{2} [(0 + 4) + 2 (0.25 + 1 + 2.25)]

= 0.25 [4 + 2 (3.5)]

= 0.25 [4 + 7]

= 0.25 \times 11 = 2.75

Answer

\int_{0}^{2} x^{2} d x \approx 2.75

Worked Example02

Use the trapezium rule with 3 strips to estimate $\int_{1}^{4} \frac{1}{x} d x$ . Give your answer to 3 decimal places.

1

Calculate strip width:

h = \frac{4 - 1}{3} = 1

2

Find x-values:

x_{0} = 1, x_{1} = 2, x_{2} = 3, x_{3} = 4

3

Calculate y-values using

y = \frac{1}{x}

:

y_{0} = 1, y_{1} = 0.5, y_{2} = 0.333..., y_{3} = 0.25

4

Apply the formula:

\frac{1}{2} [(1 + 0.25) + 2 (0.5 + 0.333...)]

= 0.5 [1.25 + 2 (0.833...)]

= 0.5 [1.25 + 1.666...]

= 0.5 \times 2.916... = 1.458...

Answer1.458 (3 d.p.)

Over or Underestimate?

Whether the trapezium rule gives an overestimate or underestimate depends on the concavity of the curve:

Convex (f''(x) > 0)

Chord lies above curve

Trapeziums include extra area

→ OVERESTIMATE

Concave (f''(x) < 0)

Chord lies below curve

Trapeziums miss area

→ UNDERESTIMATE

Worked Example03

Explain whether the trapezium rule gives an overestimate or underestimate for $\int_{1}^{3} ln x d x$ .

1

Let

f (x) = ln x

2

Find the first derivative:

f^{'} (x) = \frac{1}{x}

3

Find the second derivative:

f^{''} (x) = - \frac{1}{x ^{2}}

4

For all x in [1, 3], we have

x^{2} > 0

, so

f^{''} (x) = - \frac{1}{x ^{2}} < 0

5

Since

f^{''} (x) < 0

, the curve y = ln x is concave on this interval.

AnswerThe trapezium rule gives an underestimate because the curve is concave (the chord lies below the curve, so we miss some area).

Worked Example04

Use the trapezium rule with 4 strips to estimate $\int_{0}^{2} x^{2} d x$ . Compare your answer to the exact value and calculate the percentage error.

1

From Example 01, the trapezium estimate is 2.75

2

Find the exact value by integration:

\int_{0}^{2} x^{2} d x = [\frac{x ^{3}}{3}]_{0}^{2} = \frac{8}{3} - 0 = \frac{8}{3} = 2.6 \overline{6}

3

Calculate the error:

Error = Estimate − Exact = 2.75 − 2.6̄ = 0.083̄

4

Calculate the percentage error:

Percentage error = \frac{∣ Error ∣}{∣ Exact ∣} \times 100%

= \frac{0.083 3}{2.6 6} \times 100%

= \frac{1/12}{8/3} \times 100% = \frac{3}{96} \times 100% = 3.125%

5

This is an overestimate because

f^{''} (x) = 2 > 0

(curve is convex).

AnswerEstimate = 2.75, Exact = 8/3 ≈ 2.667. Percentage error = 3.125% (overestimate).

Improving Accuracy

Key Principle

More strips = Better approximation

Why It Works

With more strips, each chord segment is shorter and follows the curve more closely. The error gaps become much smaller. Doubling the number of strips roughly halves the error.

Practice

Question 1: Use the trapezium rule with 4 strips to estimate

\int_{0}^{4} (x^{2} + 1) d x

.

Step 1: h = (4 − 0)/4 = 1

Step 2: x-values: 0, 1, 2, 3, 4

Step 3: y-values: y = x² + 1

y₀ = 1, y₁ = 2, y₂ = 5, y₃ = 10, y₄ = 17

Step 4: Apply formula:

(1/2)[(1 + 17) + 2(2 + 5 + 10)]

= 0.5[18 + 2(17)]

= 0.5[18 + 34] = 0.5 × 52 = 26

Practice

Question 2:

x00.30.60.91.2

y22.83.43.73.9

Step 1: h = 0.3 (from the table)

Step 2: Apply formula:

(0.3/2)[(2 + 3.9) + 2(2.8 + 3.4 + 3.7)]

= 0.15[5.9 + 2(9.9)]

= 0.15[5.9 + 19.8]

= 0.15 × 25.7 = 3.855

Step 3: Since f''(x) < 0, the curve is concave, so the trapezium rule gives an underestimate.

Practice

Question 3: Use the trapezium rule with 6 strips to estimate

\int_{1}^{4} x d x

. Compare your answer to the exact value and calculate the percentage error.

Step 1: h = (4 − 1)/6 = 0.5

Step 2: x-values: 1, 1.5, 2, 2.5, 3, 3.5, 4

Step 3: y-values:

y₀ = 1, y₁ = 1.225, y₂ = 1.414, y₃ = 1.581, y₄ = 1.732, y₅ = 1.871, y₆ = 2

Step 4: Apply formula:

(0.5/2)[(1 + 2) + 2(1.225 + 1.414 + 1.581 + 1.732 + 1.871)]

= 0.25[3 + 2(7.823)]

= 0.25[3 + 15.646] = 0.25 × 18.646 = 4.662

Exact value: ∫₁⁴ √x dx = [⅔x^(3/2)]₁⁴ = ⅔(8) − ⅔(1) = 14/3 ≈ 4.667

Percentage error: |4.662 − 4.667|/4.667 × 100% = 0.11%

This is an underestimate because f''(x) = −¼x^(−3/2) < 0 (concave).

Summary

Key Concepts

ConceptKey Point

The Formula

\int_{a}^{b} y d x \approx \frac{h}{2} [(y_{0} + y_{n}) + 2 (y_{1} + ... + y_{n - 1})]

Strip Width

h = \frac{b - a}{n}

Number of Valuesn strips require (n + 1) y-values

OverestimateWhen curve is convex (f''(x) > 0)

UnderestimateWhen curve is concave (f''(x) < 0)

AccuracyMore strips → smaller error

Exam Tips

Set up a table for x and y values to stay organised

Keep extra decimal places during working; round at the end

Always check: did you use (n + 1) values for n strips?

For over/underestimate questions, find f''(x) and determine its sign