The trapezium rule · Integration

Some integrals are hard or impossible to work out exactly, and exam questions often say “use the trapezium rule” anyway. The rule estimates the area under a curve by splitting it into strips and treating each strip as a trapezium.

The rule

Split the interval from $a$ to $b$ into $n$ equal strips of width $h = \frac{b - a}{n}$ . The heights of the curve at the edges of the strips are the ordinates $y_0, y_1, \ldots, y_n$ :

Strips as trapeziumsEach strip is about a trapezium: ½h(yᵢ + yᵢ₊₁)

Adding up the trapeziums, every ordinate except the first and last is shared by two strips, so it is counted twice:

\begin{aligned} \int_a^b y\,dx &\approx \frac h2\,[\,y_0 + y_n \\ &\qquad + 2(y_1 + y_2 + \cdots + y_{n - 1})\,] \end{aligned}

Using the rule

Make a table of $x$ and $y$ , keep four decimal places, and round only at the end.

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q2

Using the trapezium rule with five ordinates, evaluate, correct to two decimal places, $\displaystyle\int_0^2 \frac{4}{1 + x^2}\,dx$ .

Approximate value

Strips and width
- Five ordinates means four strips.
- ${h = \frac{2 - 0}{4} = 0.5}$ .
Think first. Five ordinates from 0 to 2. How many strips? What is h?
The ordinates
- ${x = 0}$ : ${y_0 = 4}$ .
- ${x = 0.5}$ : ${y_1 = \frac{4}{1.25} = 3.2}$ .
- ${x = 1}$ : ${y_2 = 2}$ .
- ${x = 1.5}$ : ${y_3 = \frac{4}{3.25} = 1.2308}$ .
- ${x = 2}$ : ${y_4 = 0.8}$ .
Think first. Work out y = 4 ÷ (1 + x²) at x = 0, 0.5, 1, 1.5, 2.
Apply the rule
- First and last: ${4 + 0.8 = 4.8}$ .
- Twice the rest: ${2(3.2 + 2 + 1.2308) = 12.8615}$ .
- ${\frac{0.5}{2}(4.8 + 12.8615) = 0.25 \times 17.6615}$ .
- ${= 4.4154}$ , to four decimal places.
- So the integral is about ${4.42}$ .

Worked example · NECO 2023

NECO 2023 · Paper 2 · Q4

Using the trapezium rule with 7 ordinates $x = 2.0, 2.5, 3.0, 3.5, 4.0, 4.5$ and $5.0$ , estimate the value of $\displaystyle\int_2^5 (2x + 3)^2\,dx$ .

Strips and width
- Seven ordinates means six strips, and ${h = 0.5}$ .
Think first. Seven ordinates from 2 to 5. How many strips?
The ordinates
- ${(2x + 3)^2}$ at ${x = 2, 2.5, \ldots, 5}$ : ${49,\ 64,\ 81,\ 100,\ 121,\ 144,\ 169}$ .
Apply the rule
- First and last: ${49 + 169 = 218}$ .
- Twice the rest: ${2(64 + 81 + 100 + 121 + 144) = 1020}$ .
- ${\frac{0.5}{2}(218 + 1020) = 0.25 \times 1238 = 309.5}$ .

Too big or too small?

Try more strips, and compare the estimate with the exact value:

The trapezium rulePick an integral, change the strips

4.75estimate, 4 strips, h = 0.54.6667exact value

strips = 4

5 ordinates: 1, 1.25, 2, 3.25, 5. Estimate = ½ × 0.5 × [1 + 5 + 2 × (6.5)] = 4.75. The curve bends upwards, so each trapezium sits slightly above it: an overestimate.

Each trapezium has a straight top. If the curve bends upwards (like $x^2$ ), the straight tops lie above it and the rule overestimates; if it bends downwards, the rule underestimates. More strips always bring the estimate closer.

More: the trapezium rule

Your turn

WAEC 2022 · Paper 2 · Q3

Integration

(a)
Using the trapezium rule with five ordinates, evaluate, correct to two decimal places, $\displaystyle\int_1^3 \frac{2}{x + 4}\,dx$ .

Worked solution (try it first)

Five ordinates means four strips, so $h = \dfrac{3 - 1}{4} = 0.5$ .
The ordinates of $\dfrac{2}{x + 4}$ at $x = 1, 1.5, 2, 2.5, 3$ are $0.4,\ 0.3636,\ 0.3333,\ 0.3077,\ 0.2857$ .
First and last: $0.6857$ .
Twice the rest: $2(0.3636 + 0.3333 + 0.3077) = 2.0093$ .
Trapezium rule: $\dfrac{0.5}{2}(0.6857 + 2.0093) = 0.25 \times 2.6950$
$= 0.6738$ , about $0.67$ .

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