Use the trapezium rule with two strips to estimate _0^2 x^2\,dx. [3 marks]

Pearson Edexcel Edexcel 2019-style practice: f(x) = x^3 + 2x - 5. Show that f(x) = 0 has a root α in the interval [1, 2], and use the iteration x_n+1 = [3]5 - 2x_n with x_0 = 1.3 to find x_1 and x_2 to four decimal places.

Evaluate at the ends (M1): f(1) = 1 + 2 - 5 = -2 and f(2) = 8 + 4 - 5 = 7. Since f(1) 0 and f is continuous, a root lies in [1, 2] (A1). Apply the iteration (M1): x_1 = [3]5 - 2(1.3) = [3]2.4 ≈ 1.3389 (A1). x_2 = [3]5 - 2(1.3389) = [3]2.3222 ≈ 1.3243 (A1). Markers reward the sign-change argument with continuity stated, and the two iterates to four decimal places.

Pearson Edexcel Edexcel 2022-style practice: Use the trapezium rule with four strips to estimate _0^2 sqrt(1 + x^2)\,dx, giving your answer to three decimal places.

With four strips on [0, 2], h = 2 - 04 = 0.5 and the ordinates are at x = 0, 0.5, 1, 1.5, 2 (M1). Compute y = sqrt(1 + x^2): y_0 = 1, y_1 = sqrt(1.25) ≈ 1.1180, y_2 = sqrt(2) ≈ 1.4142, y_3 = sqrt(3.25) ≈ 1.8028, y_4 = sqrt(5) ≈ 2.2361 (A1). Apply the rule (M1): int ≈ 0.52[y_0 + y_4 + 2(y_1 + y_2 + y_3)] = 0.25[3.2361 + 2(4.3350)] (A1). = 0.25[3.2361 + 8.6700] = 0.25 × 11.9061 ≈ 2.977 (A1). Markers reward the strip width, correct ordinates, the bracket structure with only the interior ordinates doubled, and the final value.

EnglandMathsSyllabus dot point

How do you find roots and areas approximately when an exact answer is not available?

Locating roots by change of sign, iterative methods including the Newton-Raphson method, the trapezium rule for numerical integration, and the conditions under which these methods fail.

A focused answer to the Edexcel A-Level Mathematics numerical methods content, covering locating roots by change of sign, iterative methods, the Newton-Raphson method, the trapezium rule for numerical integration, and when these methods fail.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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What this dot point is asking

Edexcel wants you to locate roots of an equation by showing a change of sign, use iterative formulae of the form $x_{n+1} = g(x_n)$ , apply the Newton-Raphson method, estimate definite integrals using the trapezium rule, and explain when each method fails.

The answer

Change of sign

Iteration

An equation rearranged into the form $x = g(x)$ gives an iteration $x_{n+1} = g(x_n)$ . Starting from a sensible estimate, repeated application converges to a root when the iteration is suitable, often shown as a staircase or cobweb diagram. The same equation can usually be rearranged into $x = g(x)$ in several ways, and only some of them converge: convergence happens when the gradient of $g$ near the root has magnitude less than $1$ . A staircase diagram appears when the iterates approach the root from one side, and a cobweb diagram when they alternate either side as they close in. To show a root is correct to a given number of decimal places, evaluate $f$ at the two ends of the rounding interval and check for a sign change.

The Newton-Raphson method

The trapezium rule

The rule approximates the area under the curve by a row of trapezia, so it is exact only when the curve is a straight line. For a curve that bends upward (concave up) the trapezia lie above the curve, so the rule overestimates the area; for a curve that bends downward it underestimates. Using more strips makes each trapezium hug the curve more closely and reduces the error.

Examples in context

Two steps of Newton-Raphson on

f(x) = x^3 - 5

from

x_0 = 1.7

to estimate

\sqrt[3]{5}

step 1 Set up the formula

$f'(x) = 3x^2$ , so $x_{n+1} = x_n - \dfrac{x_n^3 - 5}{3x_n^2}$ .

step 2 First iterate

$x_1 = 1.7 - \dfrac{1.7^3 - 5}{3(1.7^2)} = 1.7 - \dfrac{4.913 - 5}{8.67} = 1.7 - \dfrac{-0.087}{8.67} \approx 1.7100$ .

step 3 Second iterate

$x_2 = 1.7100 - \dfrac{1.7100^3 - 5}{3(1.7100^2)} \approx 1.7100 - \dfrac{0.0003}{8.7723} \approx 1.7100$ , agreeing with $\sqrt[3]{5} \approx 1.7100$ .

Show that

x_{n+1} = \sqrt{\dfrac{6}{x_n + 1}}

has a fixed point that is a root of

x^3 + x^2 - 6 = 0

, and find

x_2

from

x_0 = 1.5

step 1 Check the fixed point

If the iteration settles at $x = g(x)$ , then $x = \sqrt{\dfrac{6}{x + 1}}$ , so $x^2(x + 1) = 6$ , that is $x^3 + x^2 - 6 = 0$ , as required.

step 2 First iterate

$x_1 = \sqrt{\dfrac{6}{1.5 + 1}} = \sqrt{\dfrac{6}{2.5}} = \sqrt{2.4} \approx 1.5492$ .

step 3 Second iterate

$x_2 = \sqrt{\dfrac{6}{1.5492 + 1}} = \sqrt{\dfrac{6}{2.5492}} = \sqrt{2.3537} \approx 1.5342$ .

Try this

Q1. Show that $f(x) = x^3 - x - 1$ has a root between $1$ and $2$ . [2 marks]

Cue. $f(1) = -1$ and $f(2) = 5$ have opposite signs, so a root lies between them.

Q2. Use the trapezium rule with two strips to estimate $\int_0^2 x^2\,dx$ . [3 marks]

Cue. Ordinates $0, 1, 4$ with $h = 1$ : $\tfrac{1}{2}(0 + 4 + 2(1)) = 3$ .

Exam-style practice questions

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

Edexcel 20195 marks

f(x) = x^3 + 2x - 5

. Show that

f(x) = 0

has a root

\alpha

in the interval

[1, 2]

, and use the iteration

x_{n+1} = \sqrt[3]{5 - 2x_n}

with

x_0 = 1.3

to find

x_1

and

x_2

to four decimal places.

Show worked answer →

Evaluate at the ends (M1): $f(1) = 1 + 2 - 5 = -2$ and $f(2) = 8 + 4 - 5 = 7$ . Since $f(1) < 0$ and $f(2) > 0$ and $f$ is continuous, a root lies in $[1, 2]$ (A1).

Apply the iteration (M1): $x_1 = \sqrt[3]{5 - 2(1.3)} = \sqrt[3]{2.4} \approx 1.3389$ (A1).

$x_2 = \sqrt[3]{5 - 2(1.3389)} = \sqrt[3]{2.3222} \approx 1.3243$ (A1).

Markers reward the sign-change argument with continuity stated, and the two iterates to four decimal places.

Edexcel 20225 marksUse the trapezium rule with four strips to estimate

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

, giving your answer to three decimal places.

Show worked answer →

With four strips on $[0, 2]$ , $h = \dfrac{2 - 0}{4} = 0.5$ and the ordinates are at $x = 0, 0.5, 1, 1.5, 2$ (M1).

Compute $y = \sqrt{1 + x^2}$ : $y_0 = 1$ , $y_1 = \sqrt{1.25} \approx 1.1180$ , $y_2 = \sqrt{2} \approx 1.4142$ , $y_3 = \sqrt{3.25} \approx 1.8028$ , $y_4 = \sqrt{5} \approx 2.2361$ (A1).

Apply the rule (M1): $\int \approx \dfrac{0.5}{2}\big[y_0 + y_4 + 2(y_1 + y_2 + y_3)\big] = 0.25\big[3.2361 + 2(4.3350)\big]$ (A1).

$= 0.25[3.2361 + 8.6700] = 0.25 \times 11.9061 \approx 2.977$ (A1).

Markers reward the strip width, correct ordinates, the bracket structure with only the interior ordinates doubled, and the final value.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Mathematics (9MA0) specification — Pearson Edexcel (2017)