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The region R enclosed by the lines $x = 1$, $x = 6$, $y = 0$ and the curve $y = ext{ln} (8 - x)$ is shown shaded in Figure 3 below - AQA - A-Level Maths Pure - Question 11 - 2020 - Paper 1

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Question 11

The-region-R-enclosed-by-the-lines-$x-=-1$,-$x-=-6$,-$y-=-0$-and-the-curve---$y-=--ext{ln}-(8---x)$---is-shown-shaded-in-Figure-3-below-AQA-A-Level Maths Pure-Question 11-2020-Paper 1.png

The region R enclosed by the lines $x = 1$, $x = 6$, $y = 0$ and the curve $y = ext{ln} (8 - x)$ is shown shaded in Figure 3 below. All distances are measured... show full transcript

Worked Solution & Example Answer:The region R enclosed by the lines $x = 1$, $x = 6$, $y = 0$ and the curve $y = ext{ln} (8 - x)$ is shown shaded in Figure 3 below - AQA - A-Level Maths Pure - Question 11 - 2020 - Paper 1

Step 1

Use a single trapezium to find an approximate value of the area of the shaded region

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Answer

To find the area of the shaded region R using the trapezium rule, we first need to evaluate the function at the endpoints and determine the height of the trapezium.

  1. Calculate the values:

    • For x=1x = 1: f(1)=extln(81)=extln(7)1.94591f(1) = ext{ln}(8 - 1) = ext{ln}(7) \approx 1.94591
    • For x=6x = 6: f(6)=extln(86)=extln(2)0.69315f(6) = ext{ln}(8 - 6) = ext{ln}(2) \approx 0.69315
  2. The trapezium rule formula for area A is:
    A=(ba)2(f(a)+f(b))A = \frac{(b-a)}{2}(f(a) + f(b))
    where a=1a = 1, b=6b = 6.
    Substituting the values:
    A=(61)2(f(1)+f(6))=52(1.94591+0.69315)=5×1.3195305=6.5976525A = \frac{(6-1)}{2}(f(1) + f(6)) = \frac{5}{2}(1.94591 + 0.69315) = 5 \times 1.3195305 = 6.5976525
    Rounding to two decimal places gives 6.606.60 cm².

Step 2

Use the trapezium rule with six ordinates to calculate an approximate value of the mass of Shape B

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Answer

To approximate the mass of Shape B, follow these steps:

  1. Identify the six ordinates using the function f(x)=extln(8x)f(x) = ext{ln}(8 - x) for x=1x = 1 to x=6x = 6. The values are:

    • f(1)1.94591f(1) \approx 1.94591
    • f(2)1.79176f(2) \approx 1.79176
    • f(3)1.09861f(3) \approx 1.09861
    • f(4)1.38629f(4) \approx 1.38629
    • f(5)0.69315f(5) \approx 0.69315
    • f(6)0.69315f(6) \approx 0.69315
      (repeating the last ordinate due to the nature of the trapezium rule).
  2. Calculate the area A under the curve using the trapezium rule:
    A=(ba)12(f0+2(f1+f2+f3+f4)+f5)A = \frac{(b-a)}{12}(f_0 + 2(f_1+f_2+f_3+f_4) + f_5)
    As A=7.205633A = 7.205633 cm². This area is the combined area for 4 regions, hence repeat for the shape B defined above.

  3. The volume of Shape B, given thickness of 2 mm (0.2 cm), is:
    V=A×thickness=7.205633×0.2=1.4411266 cm3V = A \times \text{thickness} = 7.205633 \times 0.2 = 1.4411266 \text{ cm}^3

  4. To find mass mm, use the formula m=density×volumem = \text{density} \times \text{volume}:
    m=10.5 g/cm3×1.4411266 cm3=15.105831 gm = 10.5 \text{ g/cm}^3 \times 1.4411266 \text{ cm}^3 = 15.105831 \text{ g}
    Rounding gives approximately 6161 g.

Step 3

Without further calculation, give one reason why the mass found in part (b) may be: an underestimate

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Answer

The trapezia are all below the curve, indicating that under the assumptions of the trapezium rule, the area calculated will be an underestimate.

Step 4

Without further calculation, give one reason why the mass found in part (b) may be: an overestimate

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Answer

The calculations round numbers, likely causing the final answer to be higher than the actual mass.

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