Newton–Raphson and the trapezium rule
Numerical methods
- A2
- Calculator allowed
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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The Newton–Raphson formula is $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$. Apply one iteration to $f(x) = x^3 - 2$ with $x_0 = 1$.
3 marks
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Apply two iterations of the Newton–Raphson method to $x^2 - 5 = 0$, starting with $x_0 = 2$. Give $x_2$ to $5$ decimal places.
3 marks
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Use the trapezium rule with $2$ strips to estimate $\displaystyle \int_0^1 x^2\,dx$.
4 marks
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The exact value of $\displaystyle \int_0^1 x^2\,dx$ is $\dfrac{1}{3}$. Using the trapezium estimate $\dfrac{3}{8}$ from two strips, find the error in the estimate and state whether it is an overestimate or an underestimate.
4 marks
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Use the Newton–Raphson method, starting with $x_0 = 1$, to find $x_2$ for $f(x) = x^3 + x - 3$. Give $x_2$ as an exact fraction.
5 marks
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Use the trapezium rule with $4$ strips to estimate $\displaystyle \int_0^{\pi} \sin x\,dx$. Give the estimate in an exact form.
5 marks
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For $f(x) = x^3 - 3x + 1$, explain why the Newton–Raphson method cannot be started at $x_0 = 1$. Starting instead at $x_0 = 0$, find $x_1$.
6 marks
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The exact value of $\displaystyle \int_0^{\pi} \sin x\,dx$ is $2$. The four-strip trapezium estimate is $\dfrac{\pi}{4}(1 + \sqrt{2})$. Find the error in this estimate, and explain whether the trapezium rule has overestimated or underestimated the integral.
7 marks
Mark scheme
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Answer. $x_1 = \dfrac{4}{3}$
- $f(1) = -1$ and $f'(1) = 3$.
- $x_1 = 1 - \dfrac{-1}{3} = \dfrac{4}{3}$.
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Answer. $x_1 = 2.25$ and $x_2 = 2.23611$
- $f'(x) = 2x$.
- $x_1 = 2 - \dfrac{-1}{4} = 2.25$.
- $x_2 = 2.25 - \dfrac{0.0625}{4.5} = 2.23611$.
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Answer. $\dfrac{3}{8}$
- The strip width is $\dfrac{1}{2}$, and the ordinates are $0$, $\dfrac{1}{4}$ and $1$.
- $\dfrac{1}{4}\left(0 + 2 \times \dfrac{1}{4} + 1\right) = \dfrac{3}{8}$.
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Answer. The error is $\dfrac{1}{24}$, and the trapezium rule gives an overestimate.
- $\dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}$.
- The curve $y = x^2$ is concave upwards on $[0, 1]$, so the trapezium chords lie above the curve.
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Answer. $x_2 = \dfrac{17}{14}$
- $f'(x) = 3x^2 + 1$, so $x_1 = 1 - \dfrac{-1}{4} = \dfrac{5}{4}$.
- $f\left(\dfrac{5}{4}\right) = \dfrac{13}{64}$ and $f'\left(\dfrac{5}{4}\right) = \dfrac{91}{16}$.
- $x_2 = \dfrac{5}{4} - \dfrac{13}{364} = \dfrac{17}{14}$.
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Answer. $\dfrac{\pi}{4}(1 + \sqrt{2})$
- The strip width is $\dfrac{\pi}{4}$.
- The ordinates are $0$, $\dfrac{\sqrt{2}}{2}$, $1$, $\dfrac{\sqrt{2}}{2}$ and $0$.
- $\dfrac{\pi}{8}(2\sqrt{2} + 2) = \dfrac{\pi}{4}(1 + \sqrt{2})$.
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Answer. The derivative is $0$ at $x = 1$, so the method cannot start there. $x_1 = \dfrac{1}{3}$.
- $f'(x) = 3x^2 - 3 = 3(x^2 - 1)$, so $f'(1) = 0$.
- The Newton–Raphson formula would require division by zero.
- $f(0) = 1$ and $f'(0) = -3$, so $x_1 = 0 - \dfrac{1}{-3} = \dfrac{1}{3}$.
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Answer. The error is $2 - \dfrac{\pi}{4}(1 + \sqrt{2})$. The rule underestimates the integral.
- Subtract the estimate from the exact value.
- On $(0, \pi)$ the curve $y = \sin x$ lies above its chords.
- The trapezium rule therefore gives an underestimate.