Browse Numerical methods

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Locating roots by change of sign

39 questions

LessonNot started
The diagram shows part of the curve y=f(x)y = \text{f}(x), where f(x)=3ln⁡x+5−2x\text{f}(x) = 3\ln x + 5 - 2x. The equation f(x)=0\text{f}(x) = 0 has a positive root α\alpha close to x=0.2x = 0.2.12346xy(a) Explain why using the sign change method with x=0.1x = 0.1 and x=6x = 6 will fail to locate α\alpha.

(1 mark)
(b) Show that the equation f(x)=0\text{f}(x) = 0 can be written as x=e13(2x−5)x = \text{e}^{\frac{1}{3}(2x - 5)}.

(2 marks)
(c) Use the iterative formula xn+1=e13(2xn−5)x_{n+1} = \text{e}^{\frac{1}{3}(2x_n - 5)} with a starting value of x1=0.2x_1 = 0.2 to find the value of α\alpha correct to 4 significant figures, showing the result of each iteration.

(3 marks)
(d) An alternative iterative formula is xn+1=F(xn)x_{n+1} = \text{F}(x_n), where F(x)=1.5ln⁡x+2.5\text{F}(x) = 1.5\ln x + 2.5.

By considering F′(0.2)\text{F}'(0.2) explain why this iterative formula will not find α\alpha.

(3 marks)
●●●●●Level 49 marksStart
A continuous curve has equation

y=x3cos⁡2x0⩽x⩽π4y = x^3\cos 2x \qquad 0 \leqslant x \leqslant \dfrac{\pi}{4}

The curve has a stationary point at the point QQ, where x>0x > 0
(a) Show, using calculus, that the xx coordinate of QQ is a solution of the equation

x=12arctan⁡(32x)x = \dfrac{1}{2}\arctan\left(\dfrac{3}{2x}\right)

(4 marks)

Using the iteration formula

xn+1=12arctan⁡(32xn)x1=0.6x_{n+1} = \dfrac{1}{2}\arctan\left(\dfrac{3}{2x_n}\right) \qquad x_1 = 0.6
(b) find, to 4 decimal places, the value of

(i) x2x_2

(ii) x3x_3

(3 marks)
(c) Using a suitable interval and a suitable function which should be stated, show that the xx coordinate of QQ is 0.5960.596 correct to 3 decimal places.

(2 marks)
●●●●●Level 49 marksStart
f(x)=x3+3x−7f(x) = x^3 + 3x - 7. The equation f(x)=0f(x) = 0 has a single real root α\alpha.
(a) Show that α\alpha lies between 1.41.4 and 1.451.45. (2 marks)
(b) Show that the equation f(x)=0f(x) = 0 can be rearranged as x=7−3x3x = \sqrt[3]{7 - 3x}. (1 mark)
(c) Using the iterative formula xn+1=7−3xn3x_{n+1} = \sqrt[3]{7 - 3x_n} with x0=1.5x_0 = 1.5, find x1x_1, x2x_2 and x3x_3, giving each to 4 decimal places. (3 marks)
(d) By choosing a suitable interval, show that α=1.406\alpha = 1.406 correct to 3 decimal places. (2 marks)
●●●●●Level 48 marksStart
f(x)=x3−4x+1f(x) = x^3 - 4x + 1
(a) Show that the equation f(x)=0f(x) = 0 has a root α\alpha such that 1.8<α<1.91.8 < \alpha < 1.9. (2 marks)
(b) Show that the Newton–Raphson iteration for this equation can be written as

xn+1=2xn3−13xn2−4x_{n+1} = \dfrac{2x_n^3 - 1}{3x_n^2 - 4} (2 marks)
(c) Using x0=2x_0 = 2, find α\alpha correct to 4 decimal places. Show the result of each iteration. (3 marks)
(d) Explain why the Newton–Raphson method cannot be used to find α\alpha starting from x0=23x_0 = \dfrac{2}{\sqrt{3}}. (1 mark)
●●●●●Level 48 marksStart
f(x)=x3+x−5\mathrm{f}(x) = x^3 + x - 5
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha between 1.51.5 and 1.61.6. (2 marks)
(b) Show that the equation f(x)=0\mathrm{f}(x) = 0 can be rearranged into the form x=5−x3x = \sqrt[3]{5 - x}. (1 mark)
(c) Use the iterative formula

xn+1=5−xn3x_{n+1} = \sqrt[3]{5 - x_n}

with x0=1.5x_0 = 1.5 to find x1x_1, x2x_2 and x3x_3, giving each answer to 44 decimal places. (3 marks)
(d) Show that α=1.516\alpha = 1.516 correct to 33 decimal places. (2 marks)
●●●●●Level 48 marksStart
f(x)=x3+4x−7\mathrm{f}(x) = x^3 + 4x - 7
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha between 1.21.2 and 1.31.3. (2 marks)
(b) Taking x0=1.3x_0 = 1.3, apply the Newton–Raphson procedure twice to f(x)\mathrm{f}(x), giving x2x_2 to 55 decimal places. (4 marks)
(c) Write down the value of α\alpha correct to 44 significant figures. (1 mark)
●●●●●Level 47 marksStart
The equation

x2=e7−xx^2 = \mathrm{e}^{7 - x}

has a single solution, x=αx = \alpha
(a) By considering a suitable change of sign, show that α\alpha lies between 44 and 55

(2 marks)
(b) Show that the equation x2=e7−xx^2 = \mathrm{e}^{7 - x} can be rearranged to give

x=7−2ln⁡xx = 7 - 2\ln x

(3 marks)
(c) Use the iterative formula

xn+1=7−2ln⁡xnx_{n+1} = 7 - 2\ln x_n

with x1=5x_1 = 5, to find x2x_2, x3x_3 and x4x_4

Give your answers to three decimal places.

(2 marks)
●●●●●Level 47 marksStart
The diagram below shows a sector of a circle OABOAB.

The chord ABAB divides the sector into a triangle and a shaded segment.

Angle AOBAOB is θ\theta radians.OABθ(a) Given that the area of the shaded segment is one third of the area of the sector OABOAB, show that

2θ=3sin⁡θ2\theta = 3\sin\theta

(4 marks)
(b) Use a suitable change of sign to show that a solution to the equation

2θ=3sin⁡θ2\theta = 3\sin\theta

lies in the interval given by θ∈[1.4,1.6]\theta \in [1.4, 1.6]

(2 marks)
●●●●●Level 46 marksStart
In a model of a sprint race, the speed of an athlete, vv m s−1^{-1}, tt seconds after the start of the race, is modelled by the equation

v=12(18−t)ln⁡(t+1)0⩽t⩽18v = \frac{1}{2}(18 - t)\ln(t + 1) \qquad 0 \leqslant t \leqslant 18

Using the model,
(a) show that the speed of the athlete is a maximum when tt satisfies the equation

(t+1)ln⁡(t+1)+t−18=0(t + 1)\ln(t + 1) + t - 18 = 0

(4 marks)

The equation in part (a) has exactly one root, α\alpha, in the interval 0⩽t⩽180 \leqslant t \leqslant 18
(b) Show that α\alpha lies in the interval [5,6][5, 6]

(2 marks)
(c) Using t0=6t_0 = 6 as a first approximation for α\alpha, apply the Newton–Raphson method once to find a second approximation for α\alpha, giving your answer to 3 decimal places.

(2 marks)
●●●●●Level 48 marksStart
f(x)=xcos⁡x+20⩽x⩽π\text{f}(x) = x\cos x + 2 \qquad 0 \leqslant x \leqslant \pi

where xx is measured in radians.
(a) Find f′(x)\text{f}'(x)

(2 marks)

The curve with equation y=f(x)y = \text{f}(x) has a maximum turning point at x=αx = \alpha
(b) Show that α\alpha lies in the interval [0.8,0.9][0.8, 0.9]

(2 marks)
(c) Using x0=0.8x_0 = 0.8 as a first approximation for α\alpha, apply the Newton–Raphson method once to f′(x)\text{f}'(x) to find a second approximation for α\alpha, giving your answer to 3 decimal places.

(3 marks)
●●●●●Level 47 marksStart
f(x)=x−3ln⁡xx>0\text{f}(x) = x - 3\ln x \qquad x > 0
(a) Show that the equation f(x)=0\text{f}(x) = 0 has a root α\alpha in the interval [4.5, 5][4.5,\ 5]

(2 marks)
(b) Find f′(x)\text{f}'(x)

(2 marks)
(c) Using x0=5x_0 = 5 as a first approximation for α\alpha, apply the Newton–Raphson method once to obtain a second approximation for α\alpha.

Give your answer to 3 decimal places.

(2 marks)

A student attempts to use the Newton–Raphson method with x0=3x_0 = 3 as a first approximation for α\alpha.
(d) Explain why the method fails.

(1 mark)
●●●●●Level 47 marksStart
f(x)=5sin⁡2x−3x+20≤x≤π\text{f}(x) = 5\sin 2x - 3x + 2 \qquad 0 \le x \le \pi

where xx is measured in radians.

The curve with equation y=f(x)y = \text{f}(x) has a single maximum turning point in the interval 0≤x≤π0 \le x \le \pi
(a) Using calculus, find the xx coordinate of this maximum turning point, giving your answer to 3 decimal places.

(4 marks)

The equation f(x)=0\text{f}(x) = 0 has a single root α\alpha in the interval 0≤x≤π0 \le x \le \pi

Given that f(1.3)=0.678\text{f}(1.3) = 0.678 and f(1.4)=−0.525\text{f}(1.4) = -0.525, to 3 decimal places,
(b) explain why α\alpha must lie in the interval [1.3, 1.4][1.3,\ 1.4]

(1 mark)
(c) Using x0=1.4x_0 = 1.4 as a first approximation for α\alpha, apply the Newton–Raphson method once to obtain a second approximation for α\alpha.

Show your method and give your answer to 3 significant figures.

(2 marks)
●●●●●Level 47 marksStart

Iterative methods (x = g(x))

33 questions

LessonNot started
The diagram shows part of the curve y=f(x)y = \text{f}(x), where f(x)=3ln⁡x+5−2x\text{f}(x) = 3\ln x + 5 - 2x. The equation f(x)=0\text{f}(x) = 0 has a positive root α\alpha close to x=0.2x = 0.2.12346xy(a) Explain why using the sign change method with x=0.1x = 0.1 and x=6x = 6 will fail to locate α\alpha.

(1 mark)
(b) Show that the equation f(x)=0\text{f}(x) = 0 can be written as x=e13(2x−5)x = \text{e}^{\frac{1}{3}(2x - 5)}.

(2 marks)
(c) Use the iterative formula xn+1=e13(2xn−5)x_{n+1} = \text{e}^{\frac{1}{3}(2x_n - 5)} with a starting value of x1=0.2x_1 = 0.2 to find the value of α\alpha correct to 4 significant figures, showing the result of each iteration.

(3 marks)
(d) An alternative iterative formula is xn+1=F(xn)x_{n+1} = \text{F}(x_n), where F(x)=1.5ln⁡x+2.5\text{F}(x) = 1.5\ln x + 2.5.

By considering F′(0.2)\text{F}'(0.2) explain why this iterative formula will not find α\alpha.

(3 marks)
●●●●●Level 49 marksStart
The diagram shows a sector OABOAB of a circle with centre OO and radius 55 cm. The angle AOBAOB is θ\theta radians. The shaded segment is bounded by the chord ABAB and the arc ABAB, and the perimeter of the shaded segment is 2121 cm.OABθ rad5 cm(a) Show that θ=4.2−2sin⁡(12θ)\theta = 4.2 - 2\sin\left(\tfrac12\theta\right).

(4 marks)

The equation θ=4.2−2sin⁡(12θ)\theta = 4.2 - 2\sin\left(\tfrac12\theta\right) has only one root for 0<θ<π0 < \theta < \pi.
(b) This root can be found by using the iterative formula θn+1=4.2−2sin⁡(12θn)\theta_{n+1} = 4.2 - 2\sin\left(\tfrac12\theta_n\right) with a starting value of θ1=2\theta_1 = 2.
  • Write down the values of θ2\theta_2, θ3\theta_3 and θ4\theta_4.
  • Hence find the value of this root correct to 3 significant figures.
(3 marks)

The diagram shows the graph of y=4.2−2sin⁡(12θ)y = 4.2 - 2\sin\left(\tfrac12\theta\right) and the line y=θy = \theta, for 0≤θ≤π0 \le \theta \le \pi.1231234θy(c) • Use this diagram to show how the iterative process used in (b) converges to this root.
  • State the type of convergence.
(3 marks)
●●●●●Level 410 marksStart
f(x)=x3−6x+2f(x) = x^3 - 6x + 2.
(a) Show f(x)=0f(x) = 0 has a root α\alpha in [0,1][0, 1].
(b) Using xn+1=xn3+26x_{n+1} = \dfrac{x_n^3 + 2}{6} with x0=0x_0 = 0, find α\alpha to 3 decimal places.
●●●●●Level 45 marksStart
f(x)=x3−5x+1f(x) = x^{3} - 5x + 1.
(a) Show that f(x)=0f(x) = 0 has a root α\alpha in the interval [0,1][0, 1]. (2 marks)
(b) Show that the equation can be rearranged as x=x3+15x = \dfrac{x^{3} + 1}{5}. (1 mark)
(c) Using x0=0.2x_{0} = 0.2, apply the iteration xn+1=xn3+15x_{n+1} = \dfrac{x_{n}^{3} + 1}{5} to find α\alpha correct to 3 decimal places. (3 marks)
●●●●●Level 46 marksStart
A continuous curve has equation

y=x3cos⁡2x0⩽x⩽π4y = x^3\cos 2x \qquad 0 \leqslant x \leqslant \dfrac{\pi}{4}

The curve has a stationary point at the point QQ, where x>0x > 0
(a) Show, using calculus, that the xx coordinate of QQ is a solution of the equation

x=12arctan⁡(32x)x = \dfrac{1}{2}\arctan\left(\dfrac{3}{2x}\right)

(4 marks)

Using the iteration formula

xn+1=12arctan⁡(32xn)x1=0.6x_{n+1} = \dfrac{1}{2}\arctan\left(\dfrac{3}{2x_n}\right) \qquad x_1 = 0.6
(b) find, to 4 decimal places, the value of

(i) x2x_2

(ii) x3x_3

(3 marks)
(c) Using a suitable interval and a suitable function which should be stated, show that the xx coordinate of QQ is 0.5960.596 correct to 3 decimal places.

(2 marks)
●●●●●Level 49 marksStart
f(x)=x3+3x−7f(x) = x^3 + 3x - 7. The equation f(x)=0f(x) = 0 has a single real root α\alpha.
(a) Show that α\alpha lies between 1.41.4 and 1.451.45. (2 marks)
(b) Show that the equation f(x)=0f(x) = 0 can be rearranged as x=7−3x3x = \sqrt[3]{7 - 3x}. (1 mark)
(c) Using the iterative formula xn+1=7−3xn3x_{n+1} = \sqrt[3]{7 - 3x_n} with x0=1.5x_0 = 1.5, find x1x_1, x2x_2 and x3x_3, giving each to 4 decimal places. (3 marks)
(d) By choosing a suitable interval, show that α=1.406\alpha = 1.406 correct to 3 decimal places. (2 marks)
●●●●●Level 48 marksStart
f(x)=x3+x−5\mathrm{f}(x) = x^3 + x - 5
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha between 1.51.5 and 1.61.6. (2 marks)
(b) Show that the equation f(x)=0\mathrm{f}(x) = 0 can be rearranged into the form x=5−x3x = \sqrt[3]{5 - x}. (1 mark)
(c) Use the iterative formula

xn+1=5−xn3x_{n+1} = \sqrt[3]{5 - x_n}

with x0=1.5x_0 = 1.5 to find x1x_1, x2x_2 and x3x_3, giving each answer to 44 decimal places. (3 marks)
(d) Show that α=1.516\alpha = 1.516 correct to 33 decimal places. (2 marks)
●●●●●Level 48 marksStart
The equation

x2=e7−xx^2 = \mathrm{e}^{7 - x}

has a single solution, x=αx = \alpha
(a) By considering a suitable change of sign, show that α\alpha lies between 44 and 55

(2 marks)
(b) Show that the equation x2=e7−xx^2 = \mathrm{e}^{7 - x} can be rearranged to give

x=7−2ln⁡xx = 7 - 2\ln x

(3 marks)
(c) Use the iterative formula

xn+1=7−2ln⁡xnx_{n+1} = 7 - 2\ln x_n

with x1=5x_1 = 5, to find x2x_2, x3x_3 and x4x_4

Give your answers to three decimal places.

(2 marks)
●●●●●Level 47 marksStart
f(x)=e12xx>0\text{f}(x) = e^{\frac{1}{2}x} \qquad x > 0

g(x)=40ln⁡xx>0\text{g}(x) = 40\ln x \qquad x > 0
(a) Find

(i) f′(x)\text{f}'(x)

(ii) g′(x)\text{g}'(x)

(2 marks)

Given that f′(x)=g′(x)\text{f}'(x) = \text{g}'(x) at x=αx = \alpha
(b) show that α\alpha satisfies the equation

xe12x−80=0xe^{\frac{1}{2}x} - 80 = 0

(2 marks)

The iterative formula

xn+1=2ln⁡(80xn)x_{n+1} = 2\ln\left(\frac{80}{x_n}\right)

is used with x1=5x_1 = 5 to find an approximate value for α\alpha.
(c) Calculate, giving each answer to 4 decimal places,

(i) the value of x2x_2

(ii) the value of α\alpha

(3 marks)
●●●●●Level 47 marksStart

The Newton–Raphson method

29 questions

LessonNot started
The curve y=(x+3)e−x+x2y = (x + 3)\text{e}^{-x} + x^2 has one stationary point for x>0x > 0, which is close to x=0.7x = 0.7.
(a) Show that the xx-coordinate of this stationary point satisfies the equation 2xex−x−2=02x\text{e}^{x} - x - 2 = 0.

(2 marks)
(b) Show that the Newton-Raphson iterative formula for finding the root of the equation in part (a) can be written in the form

xn+1=2xn2exn+22(xn+1)exn−1.x_{n+1} = \frac{2x_n^2\text{e}^{x_n} + 2}{2(x_n + 1)\text{e}^{x_n} - 1}.

(4 marks)
(c) Apply the Newton-Raphson formula with initial value x1=0.7x_1 = 0.7 to find x2x_2 and x3x_3.

(1 mark)
(d) Find the coordinates of this stationary point, giving each coordinate correct to 3 decimal places.

(2 marks)
●●●●●Level 49 marksStart
The equation ln⁡x+x=2\ln x + x = 2 has a root near x=1.5x = 1.5. Use the Newton–Raphson method with x0=1.5x_0 = 1.5 to find x1x_1 to 4 decimal places.
●●●●●Level 42 marksStart
f(x)=x3−4x+1f(x) = x^3 - 4x + 1
(a) Show that the equation f(x)=0f(x) = 0 has a root α\alpha such that 1.8<α<1.91.8 < \alpha < 1.9. (2 marks)
(b) Show that the Newton–Raphson iteration for this equation can be written as

xn+1=2xn3−13xn2−4x_{n+1} = \dfrac{2x_n^3 - 1}{3x_n^2 - 4} (2 marks)
(c) Using x0=2x_0 = 2, find α\alpha correct to 4 decimal places. Show the result of each iteration. (3 marks)
(d) Explain why the Newton–Raphson method cannot be used to find α\alpha starting from x0=23x_0 = \dfrac{2}{\sqrt{3}}. (1 mark)
●●●●●Level 48 marksStart
f(x)=x3+4x−7\mathrm{f}(x) = x^3 + 4x - 7
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha between 1.21.2 and 1.31.3. (2 marks)
(b) Taking x0=1.3x_0 = 1.3, apply the Newton–Raphson procedure twice to f(x)\mathrm{f}(x), giving x2x_2 to 55 decimal places. (4 marks)
(c) Write down the value of α\alpha correct to 44 significant figures. (1 mark)
●●●●●Level 47 marksStart
The function f is defined by

f(x)=arccos⁡(x2)for 0≤x≤a\text{f}(x) = \arccos\left(\frac{x}{2}\right) \quad \text{for } 0 \le x \le a

The curve with equation y=f(x)y = \text{f}(x) is shown below.π/2aOxy(a) State the value of aa

(1 mark)
(b) (i) On the diagram above, sketch the curve with equation

y=2cos⁡xfor 0≤x≤π2y = 2\cos x \quad \text{for } 0 \le x \le \frac{\pi}{2}

and sketch the line with equation

y=xfor 0≤x≤π2y = x \quad \text{for } 0 \le x \le \frac{\pi}{2}

(ii) Explain why the solution to the equation

x−2cos⁡x=0x - 2\cos x = 0

must also be a solution to the equation

2cos⁡x=arccos⁡(x2)2\cos x = \arccos\left(\frac{x}{2}\right)

(5 marks)
(c) Use the Newton-Raphson method with x0=0.5x_0 = 0.5 to find an approximate solution, x3x_3, to the equation

x−2cos⁡x=0x - 2\cos x = 0

Give your answer to four decimal places.

(3 marks)
●●●●●Level 49 marksStart
In a model of a sprint race, the speed of an athlete, vv m s−1^{-1}, tt seconds after the start of the race, is modelled by the equation

v=12(18−t)ln⁡(t+1)0⩽t⩽18v = \frac{1}{2}(18 - t)\ln(t + 1) \qquad 0 \leqslant t \leqslant 18

Using the model,
(a) show that the speed of the athlete is a maximum when tt satisfies the equation

(t+1)ln⁡(t+1)+t−18=0(t + 1)\ln(t + 1) + t - 18 = 0

(4 marks)

The equation in part (a) has exactly one root, α\alpha, in the interval 0⩽t⩽180 \leqslant t \leqslant 18
(b) Show that α\alpha lies in the interval [5,6][5, 6]

(2 marks)
(c) Using t0=6t_0 = 6 as a first approximation for α\alpha, apply the Newton–Raphson method once to find a second approximation for α\alpha, giving your answer to 3 decimal places.

(2 marks)
●●●●●Level 48 marksStart
f(x)=xcos⁡x+20⩽x⩽π\text{f}(x) = x\cos x + 2 \qquad 0 \leqslant x \leqslant \pi

where xx is measured in radians.
(a) Find f′(x)\text{f}'(x)

(2 marks)

The curve with equation y=f(x)y = \text{f}(x) has a maximum turning point at x=αx = \alpha
(b) Show that α\alpha lies in the interval [0.8,0.9][0.8, 0.9]

(2 marks)
(c) Using x0=0.8x_0 = 0.8 as a first approximation for α\alpha, apply the Newton–Raphson method once to f′(x)\text{f}'(x) to find a second approximation for α\alpha, giving your answer to 3 decimal places.

(3 marks)
●●●●●Level 47 marksStart
f(x)=x−3ln⁡xx>0\text{f}(x) = x - 3\ln x \qquad x > 0
(a) Show that the equation f(x)=0\text{f}(x) = 0 has a root α\alpha in the interval [4.5, 5][4.5,\ 5]

(2 marks)
(b) Find f′(x)\text{f}'(x)

(2 marks)
(c) Using x0=5x_0 = 5 as a first approximation for α\alpha, apply the Newton–Raphson method once to obtain a second approximation for α\alpha.

Give your answer to 3 decimal places.

(2 marks)

A student attempts to use the Newton–Raphson method with x0=3x_0 = 3 as a first approximation for α\alpha.
(d) Explain why the method fails.

(1 mark)
●●●●●Level 47 marksStart
f(x)=5sin⁡2x−3x+20≤x≤π\text{f}(x) = 5\sin 2x - 3x + 2 \qquad 0 \le x \le \pi

where xx is measured in radians.

The curve with equation y=f(x)y = \text{f}(x) has a single maximum turning point in the interval 0≤x≤π0 \le x \le \pi
(a) Using calculus, find the xx coordinate of this maximum turning point, giving your answer to 3 decimal places.

(4 marks)

The equation f(x)=0\text{f}(x) = 0 has a single root α\alpha in the interval 0≤x≤π0 \le x \le \pi

Given that f(1.3)=0.678\text{f}(1.3) = 0.678 and f(1.4)=−0.525\text{f}(1.4) = -0.525, to 3 decimal places,
(b) explain why α\alpha must lie in the interval [1.3, 1.4][1.3,\ 1.4]

(1 mark)
(c) Using x0=1.4x_0 = 1.4 as a first approximation for α\alpha, apply the Newton–Raphson method once to obtain a second approximation for α\alpha.

Show your method and give your answer to 3 significant figures.

(2 marks)
●●●●●Level 47 marksStart

The trapezium rule

29 questions

LessonNot started
(a) Sketch the curve with equation

y=ax−5y = a^{x} - 5

where aa is a positive constant greater than 1

On your sketch show
  • the coordinates of the point of intersection of the curve with the yy-axis
  • the equation of the asymptote of the curve
(3 marks)

x00.40.81.21.62y11.39181.76822.29723.23955\begin{array}{c|cccccc} x & 0 & 0.4 & 0.8 & 1.2 & 1.6 & 2 \\ \hline y & 1 & 1.3918 & 1.7682 & 2.2972 & 3.2395 & 5 \end{array}

The table above shows corresponding values of xx and yy for the curve with equation y=3x−x2y = 3^{x} - x^2

The values of yy are given to 4 decimal places where appropriate.
(b) Use the trapezium rule with all the values of yy in the table to obtain an estimate for

∫02(3x−x2)dx\int_{0}^{2} \left(3^{x} - x^2\right)\mathrm{d}x

giving your answer to 2 decimal places.

(3 marks)
(c) Using your answer to part (b) and making your method clear, estimate

(i) ∫02(3x+x2)dx\displaystyle\int_{0}^{2} \left(3^{x} + x^2\right)\mathrm{d}x

(ii) ∫02(7−3x+x2)dx\displaystyle\int_{0}^{2} \left(7 - 3^{x} + x^2\right)\mathrm{d}x

(3 marks)
●●●●●Level 49 marksStart
O(1, 1.86)(2, 2.36)(3, 2.5)(4, 2.36)(5, 1.86)(6, 0)xyFigure 1

A gardener has a polytunnel, a long greenhouse with a curved plastic roof. At one end, the polytunnel is 6 m wide.

The height, yy metres, of the roof is measured at a point xx metres from one side of the polytunnel.

Figure 1 shows a plot of the cross-section of the polytunnel and the coordinate values (x,y)(x, y).
(a) Use the trapezium rule with all the yy values given in Figure 1 to estimate the cross-sectional area of the polytunnel.

(3 marks)

The polytunnel is 15 m long and is modelled as having the same cross-section along its whole length.
(b) Use the model and the answer to part (a) to estimate the volume of air inside the polytunnel, giving your answer in m3\text{m}^3 to 2 significant figures.

(2 marks)

Assuming the model,
(c) state, giving a reason for your answer, whether your answer for part (b) is an overestimate or an underestimate of the volume of air inside the polytunnel.

(1 mark)
●●●●●Level 46 marksStart
The table gives values of y=f(x)y = f(x).

x00.61.21.82.4y3.24.14.64.02.5\begin{array}{c|ccccc} x & 0 & 0.6 & 1.2 & 1.8 & 2.4 \\\\ \hline y & 3.2 & 4.1 & 4.6 & 4.0 & 2.5 \end{array}

Use the trapezium rule with all five values to estimate ∫02.4y dx\displaystyle\int_0^{2.4} y\,dx.
●●●●●Level 43 marksStart
(a) Use the trapezium rule with 4 strips to estimate ∫26ln⁡x dx\displaystyle\int_2^6 \ln x\,dx to 3 decimal places.
(b) State, with a reason, whether this under- or over-estimates the integral.
●●●●●Level 45 marksStart
I=∫131+x3  dxI = \int_1^3 \sqrt{1 + x^3}\;\mathrm{d}x

Figure 1 shows the region whose area is given by II.11.522.53123456xyFigure 1
(a) Use the trapezium rule with four strips of equal width to find an estimate for II, giving your answer to 44 decimal places. (4 marks)
(b) State, with a reason, whether your answer to part (a) is an overestimate or an underestimate of II. (2 marks)
●●●●●Level 46 marksStart
The table below shows corresponding values of xx and yy for y=log⁡5(4x)y = \log_5 (4x)

The values of yy are given to 3 decimal places as appropriate.

x0.50.7511.251.5y0.4310.6830.86111.113\begin{array}{c|ccccc} x & 0.5 & 0.75 & 1 & 1.25 & 1.5 \\ \hline y & 0.431 & 0.683 & 0.861 & 1 & 1.113 \end{array}
(a) Use the trapezium rule with all the values of yy in the table to find an estimate for

∫0.51.5log⁡5(4x) dx\int_{0.5}^{1.5} \log_5 (4x)\,dx

(3 marks)

Using your answer to part (a), and making your method clear, estimate
(b) (i) ∫0.51.5log⁡5(4x)2 dx\displaystyle\int_{0.5}^{1.5} \log_5 (4x)^2\,dx

(ii) ∫0.51.5log⁡5(100x) dx\displaystyle\int_{0.5}^{1.5} \log_5 (100x)\,dx

(3 marks)
●●●●●Level 46 marksStart
The table shows corresponding values of xx and yy for a continuous curve with equation y=f(x)y = \text{f}(x), where aa and bb are constants.

x00.511.522.5ya3.6b4.75.46.2\begin{array}{c|cccccc} x & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 \\ \hline y & a & 3.6 & b & 4.7 & 5.4 & 6.2 \end{array}

An estimate for the area under the curve between x=0x = 0 and x=2.5x = 2.5 is found using the trapezium rule with all the yy values in the table.

Given that this estimate is 11.211.2
(a) show that

a+2b=11.2a + 2b = 11.2

(3 marks)

Given also that the sum of all the yy values in the table is 27.127.1
(b) find the value of aa and the value of bb.

(3 marks)
●●●●●Level 46 marksStart