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The table below shows corresponding values of x and y for y = log₂x The values of y are given to 2 decimal places as appropriate - Edexcel - A-Level Maths Pure - Question 6 - 2022 - Paper 2

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The table below shows corresponding values of x and y for y = log₂x The values of y are given to 2 decimal places as appropriate. x | 3 | 4.5 | 6 | 7.5 | 8 ---|---|... show full transcript

Worked Solution & Example Answer:The table below shows corresponding values of x and y for y = log₂x The values of y are given to 2 decimal places as appropriate - Edexcel - A-Level Maths Pure - Question 6 - 2022 - Paper 2

Step 1

Using the trapezium rule with all the values of y in the table, find an estimate for $$\int_{3}^{8} log_{2} x \, dx$$

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Answer

To apply the trapezium rule, we first identify the intervals and the widths between the points on the x-axis:

  • The intervals: [3, 4.5], [4.5, 6], [6, 7.5], and [7.5, 8]
  • The widths of these intervals (h) are: h = 1.5, 1.5, 1.5, and 0.5 respectively.

Using the trapezium rule: extArea=h2(y1+2y2+2y3+2y4+y5) ext{Area} = \frac{h}{2} \left( y_1 + 2y_2 + 2y_3 + 2y_4 + y_5 \right) Where:

  • y1=1.63y_1 = 1.63, y2=2.26y_2 = 2.26, y3=2.46y_3 = 2.46, y4=2.63y_4 = 2.63, and y5=2.63y_5 = 2.63.

Calculating: Area1.52(1.63+2(2.26)+2(2.46)+2.63)\text{Area} \approx \frac{1.5}{2} \left( 1.63 + 2(2.26) + 2(2.46) + 2.63 \right)

= 1.52(1.63+4.52+4.92+2.63)=1.52(13.8)10.35\frac{1.5}{2} \left( 1.63 + 4.52 + 4.92 + 2.63 \right) = \frac{1.5}{2} \left( 13.8 \right) \approx 10.35.

Thus, the estimate for 38log2xdx\int_{3}^{8} log_{2} x \, dx is approximately 10.35.

Step 2

Estimate $$\int_{3}^{8} log_{2} (2x)^{10} \ dx$$

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Answer

From part (a), we know: 38log22x dx=38log22+38log2x dx\int_{3}^{8} log_{2} 2x \ dx = \int_{3}^{8} log_{2} 2 + \int_{3}^{8} log_{2} x \ dx

Using properties of logarithms: log2(2x)10=10log2(2x)=10(log22+log2x)=10(1+log2x).log_{2} (2x)^{10} = 10 \, log_{2} (2x) = 10 (log_{2} 2 + log_{2} x) = 10 (1 + log_{2} x).

Thus, the integral becomes: 38log2(2x)10 dx=1038log2xdx+10×5=10(10.35)+50=153.5.\int_{3}^{8} log_{2} (2x)^{10} \ dx = 10 \int_{3}^{8} log_{2} x \, dx + 10 \times 5 = 10(10.35) + 50 = 153.5.

Step 3

Estimate $$\int_{3}^{8} log_{2} 18x \ dx$$

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Answer

Using linearity of integrals: 38log218x dx=38log218+38log2x dx\int_{3}^{8} log_{2} 18x \ dx = \int_{3}^{8} log_{2} 18 + \int_{3}^{8} log_{2} x \ dx

We know: 38log218=log218×(83)=log218×5\int_{3}^{8} log_{2} 18 = log_{2} 18 \times (8-3) = log_{2} 18 \times 5

The value of log218log_{2} 18 is approximately: log218log224.17ext(usingacalculator)log_{2} 18 \approx log_{2} 2^{4.17} ext{ (using a calculator)}

Thus, we can compute: =4.17×5+10.35=20.85+10.35=31.2.= \approx 4.17 \times 5 + 10.35 = 20.85 + 10.35 = 31.2.

Therefore, the estimate for 38log218x dx\int_{3}^{8} log_{2} 18x \ dx is approximately 31.2.

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