SEAB Original-style practice: Use the trapezium rule with four strips to estimate _0^1 (1)/(1 + x^2)\,dx, giving your answer to four decimal places.

Four strips on [0, 1] give strip width h = 1 - 04 = 0.25 and ordinates at x = 0, 0.25, 0.5, 0.75, 1. Compute y = 11 + x^2: y_0 = 1, y_1 = 11.0625 = 0.94118, y_2 = 11.25 = 0.8, y_3 = 11.5625 = 0.64, y_4 = 12 = 0.5. Trapezium rule: ≈ (h)/(2)[y_0 + y_4 + 2(y_1 + y_2 + y_3)]: $= (0.25)/(2)[1 + 0.5 + 2(0.94118 + 0.8 + 0.64)] = 0.125[1.5 + 2(2.38118)] = 0.125(6.26236) = 0.7828.$ (The exact value is 1 = π4 ≈ 0.7854.) Markers reward the strip width, the five ordinates, the trapezium formula with the doubled interior terms, and the estimate 0.7828.

SEAB Original-style practice: Use Simpson's rule with four strips to estimate _0^1 (1)/(1 + x^2)\,dx to four decimal places, and compare its accuracy with the trapezium rule.

With the same ordinates (h = 0.25): y_0 = 1, y_1 = 0.94118, y_2 = 0.8, y_3 = 0.64, y_4 = 0.5. Simpson's rule (four strips, an even number): ≈ (h)/(3)[y_0 + y_4 + 4(y_1 + y_3) + 2 y_2]: $= (0.25)/(3)[1 + 0.5 + 4(0.94118 + 0.64) + 2(0.8)] = (0.25)/(3)[1.5 + 4(1.58118) + 1.6].$ $= (0.25)/(3)[1.5 + 6.32472 + 1.6] = (0.25)/(3)(9.42472) = 0.78539.$ To four decimal places this is 0.7854, matching π4 = 0.7854. Simpson's rule is far more accurate than the trapezium estimate 0.7828, because it fits parabolas rather than straight lines. Markers reward the Simpson pattern of weights 1, 4, 2, 4, 1, the estimate 0.7854, and the comparison showing Simpson is more accurate.

SingaporeFurther MathsSyllabus dot point

How do the trapezium rule and Simpson's rule approximate a definite integral, and how accurate are they?

Approximate a definite integral using the trapezium rule and Simpson's rule and comment on the accuracy of the estimate

A focused answer to the H2 Further Mathematics outcome on numerical integration. The trapezium rule and Simpson's rule, the strip width and ordinates, applying each rule, the over- or under-estimate behaviour, and which rule is more accurate.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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What this dot point is asking

SEAB wants you to approximate a definite integral $\displaystyle\int_a^b \mathrm{f}(x)\,\mathrm{d}x$ when an exact antiderivative is unavailable or awkward, using the trapezium rule and Simpson's rule. You must set up the ordinates correctly, apply the right weights, and comment on the accuracy, including whether an estimate is an over- or under-estimate and which rule is better.

The answer

Strips and ordinates

Divide $[a, b]$ into $n$ equal strips of width

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

giving $n + 1$ ordinates $y_0, y_1, \dots, y_n$ at $x = a, a + h, \dots, b$ . Both rules combine these ordinates with fixed weights.

The trapezium rule

The trapezium rule joins consecutive ordinates with straight lines:

\int_a^b \mathrm{f}(x)\,\mathrm{d}x \approx \frac{h}{2}\Big[y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1})\Big].

The end ordinates have weight $1$ and the interior ones weight $2$ .

Over- or under-estimate

The trapezium rule's accuracy depends on the curvature. For a curve that is concave up (bending upward, like $\tfrac{1}{1 + x^2}$ is on parts of its range), the chords lie above the curve, so the trapezium rule over-estimates; for a concave-down curve it under-estimates. Sketching or considering $\mathrm{f}''$ tells you which.

Simpson's rule

Simpson's rule fits parabolas through successive triples of points and needs an even number of strips:

\int_a^b \mathrm{f}(x)\,\mathrm{d}x \approx \frac{h}{3}\Big[y_0 + y_n + 4(y_1 + y_3 + \cdots) + 2(y_2 + y_4 + \cdots)\Big].

The pattern of weights is $1, 4, 2, 4, 2, \dots, 4, 1$ : ends weight $1$ , odd-indexed ordinates weight $4$ , even-indexed (interior) weight $2$ .

Which is more accurate

Simpson's rule is much more accurate than the trapezium rule for the same number of strips, because parabolas match a smooth curve far better than straight lines. It is exact for any cubic. Increasing the number of strips improves both rules.

Worked example

Estimate $\displaystyle\int_1^{3} \ln x\,\mathrm{d}x$ using the trapezium rule with four strips.

Step 1: Find the strip width and ordinate positions

$h = \dfrac{3 - 1}{4} = 0.5$ , with ordinates at $x = 1, 1.5, 2, 2.5, 3$ .

Step 2: Evaluate the ordinates

$y_0 = \ln 1 = 0$ , $y_1 = \ln 1.5 = 0.405465$ , $y_2 = \ln 2 = 0.693147$ , $y_3 = \ln 2.5 = 0.916291$ , $y_4 = \ln 3 = 1.098612$ .

Step 3: Apply the trapezium formula

\int_1^3 \ln x\,\mathrm{d}x \approx \frac{0.5}{2}\Big[y_0 + y_4 + 2(y_1 + y_2 + y_3)\Big].

Step 4: Substitute

= 0.25\Big[0 + 1.098612 + 2(0.405465 + 0.693147 + 0.916291)\Big] = 0.25\big[1.098612 + 2(2.014903)\big].

= 0.25(1.098612 + 4.029806) = 0.25(5.128418) = 1.2821.

Step 5: Comment on accuracy

The exact value is $[x\ln x - x]_1^3 = (3\ln 3 - 3) - (0 - 1) = 1.2958$ . Since $\ln x$ is concave down, the trapezium chords lie below the curve, so the rule under-estimates, consistent with $1.2821 < 1.2958$ .

Examples in context

Example 1. Area under experimental data. When only measured values are available, with no formula to integrate, the trapezium and Simpson rules estimate quantities like distance from a velocity-time table, the everyday use of numerical integration in the laboratory.

Example 2. Integrals with no elementary antiderivative. The integral $\int_0^1 \mathrm{e}^{-x^2}\,\mathrm{d}x$ , central to the normal distribution, has no elementary closed form, so it is evaluated numerically by Simpson's rule, linking the technique directly to statistics.

Try this

Q1. For $n$ strips, how many ordinates are there? [1 mark]

Cue. $n + 1$ .

Q2. State the weight pattern for Simpson's rule. [1 mark]

Cue. $1, 4, 2, 4, 2, \dots, 4, 1$ (ends $1$ , odd ordinates $4$ , even interior $2$ ).

Q3. Why must Simpson's rule use an even number of strips? [2 marks]

Cue. It fits a parabola to each pair of strips, so the strips must group into pairs, requiring an even total.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original6 marksUse the trapezium rule with four strips to estimate

\displaystyle\int_0^{1} \frac{1}{1 + x^2}\,dx

, giving your answer to four decimal places.

Show worked answer →

Four strips on $[0, 1]$ give strip width $h = \dfrac{1 - 0}{4} = 0.25$ and ordinates at $x = 0, 0.25, 0.5, 0.75, 1$ .

Compute $y = \dfrac{1}{1 + x^2}$ : $y_0 = 1$ , $y_1 = \dfrac{1}{1.0625} = 0.94118$ , $y_2 = \dfrac{1}{1.25} = 0.8$ , $y_3 = \dfrac{1}{1.5625} = 0.64$ , $y_4 = \dfrac{1}{2} = 0.5$ .

Trapezium rule: $\displaystyle\int \approx \frac{h}{2}\left[y_0 + y_4 + 2(y_1 + y_2 + y_3)\right]$ :

= \frac{0.25}{2}\left[1 + 0.5 + 2(0.94118 + 0.8 + 0.64)\right] = 0.125\left[1.5 + 2(2.38118)\right] = 0.125(6.26236) = 0.7828.

(The exact value is $\arctan 1 = \tfrac{\pi}{4} \approx 0.7854$ .)

Markers reward the strip width, the five ordinates, the trapezium formula with the doubled interior terms, and the estimate $0.7828$ .

Original6 marksUse Simpson's rule with four strips to estimate

\displaystyle\int_0^{1} \frac{1}{1 + x^2}\,dx

to four decimal places, and compare its accuracy with the trapezium rule.

Show worked answer →

With the same ordinates ( $h = 0.25$ ): $y_0 = 1$ , $y_1 = 0.94118$ , $y_2 = 0.8$ , $y_3 = 0.64$ , $y_4 = 0.5$ .

Simpson's rule (four strips, an even number): $\displaystyle\int \approx \frac{h}{3}\left[y_0 + y_4 + 4(y_1 + y_3) + 2 y_2\right]$ :

= \frac{0.25}{3}\left[1 + 0.5 + 4(0.94118 + 0.64) + 2(0.8)\right] = \frac{0.25}{3}\left[1.5 + 4(1.58118) + 1.6\right].

= \frac{0.25}{3}\left[1.5 + 6.32472 + 1.6\right] = \frac{0.25}{3}(9.42472) = 0.78539.

To four decimal places this is $0.7854$ , matching $\tfrac{\pi}{4} = 0.7854$ . Simpson's rule is far more accurate than the trapezium estimate $0.7828$ , because it fits parabolas rather than straight lines.

Markers reward the Simpson pattern of weights $1, 4, 2, 4, 1$ , the estimate $0.7854$ , and the comparison showing Simpson is more accurate.