What is wrong number of ordinates in the trapezium rule?

With n strips there are n + 1 ordinates; the first and last are not doubled, the interior ones are.

CCEA CCEA 2019-style practice: Use the trapezium rule with four strips to estimate _0^2 sqrt(1 + x^2)\,dx. Give your answer to three significant figures.

Four strips on [0, 2] give width h = 0.5 and x-values 0, 0.5, 1, 1.5, 2. The y-values sqrt(1 + x^2) are: 1, 1.118, 1.414, 1.803, 2.236. Trapezium rule: int ≈ (h)/(2)[y_0 + y_4 + 2(y_1 + y_2 + y_3)]. = (0.5)/(2)[1 + 2.236 + 2(1.118 + 1.414 + 1.803)] = 0.25[3.236 + 2(4.335)] = 0.25(11.906) = 2.98. Markers reward the strip width, the five ordinates, the trapezium-rule formula, and the estimate 2.98.

Northern IrelandMathsSyllabus dot point

How do we locate roots and estimate areas when exact methods fail?

Locating roots by a sign change, iterative methods including the Newton-Raphson method and fixed-point iteration, and the trapezium rule for estimating a definite integral.

A CCEA A-Level Mathematics answer on locating roots by a change of sign, iterative methods including fixed-point iteration and the Newton-Raphson method, and using the trapezium rule to estimate the value of a definite integral.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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Examples in context
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What this dot point is asking

CCEA wants you to locate a root of an equation by a change of sign, use iterative methods (fixed-point iteration and the Newton-Raphson method) to refine an estimate, and use the trapezium rule to estimate a definite integral. Numerical methods give approximate answers when an exact solution is impossible or impractical.

The answer

Locating roots by a sign change

Fixed-point iteration

The Newton-Raphson method

The trapezium rule

The Newton-Raphson method needs a starting value reasonably close to the root and a non-zero derivative there; a poor start can send successive estimates away from the root or oscillate. Quoting the iteration to a stated number of decimal places, and stopping when two successive values agree to that accuracy, is the expected exam practice.

Worked example: one iteration step

Improve a root estimate by iteration

The equation $x^3 + x - 3 = 0$ is rearranged as $x = (3 - x)^{1/3}$ . Starting from $x_0 = 1.2$ , find $x_1$ and $x_2$ .

step Apply the iteration once

x_1 = (3 - x_0)^{1/3} = (3 - 1.2)^{1/3} = (1.8)^{1/3} = 1.2164.

step Apply it again

x_2 = (3 - x_1)^{1/3} = (3 - 1.2164)^{1/3} = (1.7836)^{1/3} = 1.2127.

step Interpret

The values are settling near $1.213$ , so the iteration is converging to the root. Repeating until two successive values agree to the required accuracy gives the root, which a sign change can confirm lies in the interval.

Examples in context

Example 1. An equation with no algebraic solution. The equation $x = \cos x$ cannot be solved by algebra, but iterating $x_{n+1} = \cos x_n$ from $x_0 = 0.7$ converges to about $0.739$ . Iteration is the practical route to roots of transcendental equations.

Example 2. Estimating an awkward area. The integral $\int_0^1 e^{-x^2}\,dx$ has no elementary antiderivative, so the trapezium rule (or finer methods) estimates it numerically. Numerical integration is essential where exact integration is impossible, as in statistics.

Try this

Q1. $f(x) = x^2 - 7$ . Show a root lies between $2$ and $3$ . [2 marks]

Cue. $f(2) = -3$ , $f(3) = 2$ ; the sign change with continuity gives a root between.

Q2. State the Newton-Raphson iteration formula. [1 mark]

Cue. $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ .

Q3. Using two strips, how many ordinates does the trapezium rule need? [1 mark]

Cue. Three ordinates ( $n + 1$ for $n = 2$ ).

Exam-style practice questions

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

CCEA 20216 marksShow that the equation

x^3 - 5x - 1 = 0

has a root between

x = 2

and

x = 3

. Use the Newton-Raphson method once, starting from

x_0 = 2.3

, to find a better approximation.

Show worked answer →

Let $f(x) = x^3 - 5x - 1$ . Then $f(2) = 8 - 10 - 1 = -3$ and $f(3) = 27 - 15 - 1 = 11$ .

The sign changes from negative to positive, and $f$ is continuous, so there is a root between $2$ and $3$ .

Newton-Raphson: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$ with $f'(x) = 3x^2 - 5$ .

$f(2.3) = 12.167 - 11.5 - 1 = -0.333$ ; $f'(2.3) = 3(5.29) - 5 = 10.87$ .

$x_1 = 2.3 - \frac{-0.333}{10.87} = 2.3 + 0.0306 = 2.331.$

Markers reward the sign change with a continuity statement, the derivative, the Newton-Raphson formula, and the improved estimate.

CCEA 20196 marksUse the trapezium rule with four strips to estimate

\int_0^2 \sqrt{1 + x^2}\,dx

. Give your answer to three significant figures.

Show worked answer →

Four strips on $[0, 2]$ give width $h = 0.5$ and $x$ -values $0, 0.5, 1, 1.5, 2$ .

The $y$ -values $\sqrt{1 + x^2}$ are: $1$ , $1.118$ , $1.414$ , $1.803$ , $2.236$ .

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

$= \frac{0.5}{2}\left[1 + 2.236 + 2(1.118 + 1.414 + 1.803)\right] = 0.25\left[3.236 + 2(4.335)\right] = 0.25(11.906) = 2.98.$

Markers reward the strip width, the five ordinates, the trapezium-rule formula, and the estimate $2.98$ .

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)