What is simpson's rule strip count?

Simpson's rule requires an even number of strips (an odd number of ordinates); an odd number of strips cannot be paired into parabolas.

Pearson Edexcel Edexcel 2021-style practice: Use Simpson's rule with four strips to estimate _0^2 sqrt(1 + x^3)\,dx, giving your answer to three decimal places.

Compute the five ordinates at strip width h = 0.5, then apply Simpson's rule. h = (2 - 0)/(4) = 0.5. Ordinates of y = sqrt(1 + x^3) at x = 0, 0.5, 1, 1.5, 2: y_0 = 1, y_1 = sqrt(1.125) ≈ 1.0607, y_2 = sqrt(2) ≈ 1.4142, y_3 = sqrt(4.375) ≈ 2.0917, y_4 = sqrt(9) = 3 (M1 A1 for ordinates). Simpson's rule: int ≈ (h)/(3)[y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)] (M1). = (0.5)/(3)[1 + 3 + 4(1.0607 + 2.0917) + 2(1.4142)] (A1) = (0.5)/(3)[4 + 12.6096 + 2.8284] = (0.5)/(3)(19.438) (A1) ≈ 3.240 (A1).

EnglandFurther MathsSyllabus dot point

How do you approximate roots and integrals numerically when no exact method works?

Solving equations numerically by interval bisection, linear interpolation and the Newton-Raphson method, and approximating definite integrals using Simpson's rule and the mid-ordinate rule.

A focused answer to the Edexcel A-Level Further Mathematics Further Pure numerical methods content, covering solving equations by interval bisection, linear interpolation and the Newton-Raphson method, and approximating definite integrals using Simpson's rule and the mid-ordinate rule.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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What this dot point is asking
Root finding by sign change
The Newton-Raphson method
Numerical integration
Examples in context
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What this dot point is asking

Edexcel Further Pure wants you to find roots of equations numerically using interval bisection, linear interpolation and the Newton-Raphson iteration, and to approximate definite integrals using Simpson's rule and the mid-ordinate rule, applying each method carefully and commenting on accuracy. Clear, tabulated working and the correct application of the rule formulae are what earn the marks.

Root finding by sign change

A continuous function with $f(a)$ and $f(b)$ of opposite signs must cross zero somewhere in $[a, b]$ , so there is a root in the interval. Bisection and linear interpolation refine that bracket. Always confirm the sign change before claiming a root, and watch out for an even number of roots in the interval, which would not produce a sign change.

The Newton-Raphson method

When you can differentiate $f$ , the tangent line at a point gives a fast iteration towards a nearby root. Each step follows the tangent down to the $x$ -axis and uses that intercept as the next estimate. Near a simple root convergence is rapid (roughly doubling the number of correct digits each step), but a poor starting point or a small $f'(x_n)$ can make the iteration diverge.

One Newton-Raphson step

Apply one step of Newton-Raphson to $f(x) = x^2 - 2$ from $x_0 = 1.5$ (so the root is $\sqrt{2}$ ).

Step 1: Compute the derivative

The Newton-Raphson iteration uses the tangent line at $x_0$ to find a better approximation to the root. To draw that tangent we need the slope, which is $f'(x_0)$ .

f'(x) = 2x, \qquad f'(1.5) = 3.

Step 2: Evaluate $f$ at the starting point

We also need the function value at $x_0$ to know how far we are from zero. A positive $f(x_0)$ tells us the curve sits above the axis, so the tangent will step to the left.

f(1.5) = 1.5^2 - 2 = 2.25 - 2 = 0.25.

Step 3: Apply the Newton-Raphson formula

The iteration follows the tangent line down to the $x$ -axis. Dividing $f(x_0)$ by $f'(x_0)$ gives the horizontal distance from $x_0$ to that intercept.

x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1.5 - \frac{0.25}{3} = 1.5 - 0.08\overline{3} = 1.41\overline{6}.

Final answer: $x_1 \approx 1.41667$ . This is already correct to three decimal places ( $\sqrt{2} \approx 1.41421$ ), demonstrating how quickly Newton-Raphson converges near a simple root.

Numerical integration

When an integral has no elementary antiderivative, approximate it by summing strip areas. The mid-ordinate rule uses the height of each strip at its midpoint; Simpson's rule fits a parabola over each pair of strips and is markedly more accurate for smooth integrands.

Simpson's rule with two strips

Estimate $\int_0^1 \sqrt{1 + x^2}\,dx$ using Simpson's rule with two strips.

Step 1: Determine the strip width and the ordinates

Two strips over $[0, 1]$ means a strip width of $h = \frac{1 - 0}{2} = 0.5$ . We need three ordinates, one at each endpoint and one at the midpoint, evaluated from $y = \sqrt{1 + x^2}$ .

y_0 = \sqrt{1 + 0^2} = 1, \quad y_1 = \sqrt{1 + 0.5^2} = \sqrt{1.25} \approx 1.1180, \quad y_2 = \sqrt{1 + 1^2} = \sqrt{2} \approx 1.4142.

Step 2: Apply the Simpson's rule formula

Simpson's rule fits a parabola over each pair of strips. With two strips (one pair), the formula is $\frac{h}{3}[y_0 + y_2 + 4y_1]$ : the end ordinates receive weight $1$ , and the middle ordinate receives weight $4$ . This weighting comes from integrating the fitted parabola exactly.

\int_0^1 \sqrt{1 + x^2}\,dx \approx \frac{0.5}{3}\left[y_0 + y_2 + 4y_1\right] = \frac{0.5}{3}\left[1 + 1.4142 + 4(1.1180)\right] = \frac{0.5}{3}(6.886) \approx 1.1477.

Final answer: $\approx 1.1477$ . The exact value is about $1.1478$ , so even two strips give three-figure accuracy for this smooth integrand.

Examples in context

Numerical methods complement the exact techniques elsewhere in the course. The Newton-Raphson iteration is the practical way to find roots of the polynomials studied in further algebra when they do not factorise. Simpson's rule and the mid-ordinate rule estimate the improper and volume integrals of further calculus when no closed form exists. Taylor series provide an alternative route to approximating function values and underpin the error analysis of these methods (the truncation error of Simpson's rule depends on the fourth derivative). Numerical solutions of differential equations (Euler's method, in some specifications) extend the same step-by-step philosophy.

Try this

Q1. Write down the Newton-Raphson iteration formula. [1 mark]

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

Q2. Why must Simpson's rule use an even number of strips? [1 mark]

Cue. It fits a parabola over each pair of strips, so the strips must pair up.

Q3. Estimate the root of $f(x) = x^2 - 3$ near $x_0 = 1.7$ with one Newton-Raphson step. [3 marks]

Cue. $f(1.7) = -0.11$ , $f'(1.7) = 3.4$ , so $x_1 = 1.7 + \frac{0.11}{3.4} \approx 1.7324$ .

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 20186 marksThe equation

x^3 - 5x + 1 = 0

has a root

\alpha

in the interval

[2, 2.2]

. Apply the Newton-Raphson method once, starting from

x_0 = 2.1

, to find a better approximation to

\alpha

, giving your answer to four decimal places.

Show worked answer →

Differentiate, then apply one Newton-Raphson iteration.

$f(x) = x^3 - 5x + 1$ , so $f'(x) = 3x^2 - 5$ (M1).

At $x_0 = 2.1$ : $f(2.1) = 9.261 - 10.5 + 1 = -0.239$ (A1); $f'(2.1) = 3(4.41) - 5 = 13.23 - 5 = 8.23$ (A1).

Newton-Raphson: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2.1 - \frac{-0.239}{8.23} = 2.1 + 0.02904$ (M1 A1).

$x_1 \approx 2.1290$ to four decimal places (A1).

Edexcel 20216 marksUse Simpson's rule with four strips to estimate

\int_0^2 \sqrt{1 + x^3}\,dx

, giving your answer to three decimal places.

Show worked answer →

Compute the five ordinates at strip width $h = 0.5$ , then apply Simpson's rule.

$h = \frac{2 - 0}{4} = 0.5$ . Ordinates of $y = \sqrt{1 + x^3}$ at $x = 0, 0.5, 1, 1.5, 2$ : $y_0 = 1$ , $y_1 = \sqrt{1.125} \approx 1.0607$ , $y_2 = \sqrt{2} \approx 1.4142$ , $y_3 = \sqrt{4.375} \approx 2.0917$ , $y_4 = \sqrt{9} = 3$ (M1 A1 for ordinates).

Simpson's rule: $\int \approx \frac{h}{3}[y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$ (M1).

$= \frac{0.5}{3}[1 + 3 + 4(1.0607 + 2.0917) + 2(1.4142)]$ (A1) $= \frac{0.5}{3}[4 + 12.6096 + 2.8284] = \frac{0.5}{3}(19.438)$ (A1) $\approx 3.240$ (A1).

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Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)