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The table below shows corresponding values of $x$ and $y$ for $y = ext{log}_2 x$ The values of $y$ are given to 2 decimal places as appropriate - Edexcel - A-Level Maths: Pure - Question 5 - 2022 - Paper 2

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The-table-below-shows-corresponding-values-of-$x$-and-$y$-for-$y-=--ext{log}_2-x$-The-values-of-$y$-are-given-to-2-decimal-places-as-appropriate-Edexcel-A-Level Maths: Pure-Question 5-2022-Paper 2.png

The table below shows corresponding values of $x$ and $y$ for $y = ext{log}_2 x$ The values of $y$ are given to 2 decimal places as appropriate. | $x$ | 3 | 4.5... show full transcript

Worked Solution & Example Answer:The table below shows corresponding values of $x$ and $y$ for $y = ext{log}_2 x$ The values of $y$ are given to 2 decimal places as appropriate - Edexcel - A-Level Maths: Pure - Question 5 - 2022 - Paper 2

Step 1

Using the trapezium rule with all the values of $y$ in the table, find an estimate for $$\int_3^8 \text{log}_2 2x \, dx$$

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Answer

To estimate the integral using the trapezium rule, we first identify the width of each segment, denoted as hh. For the given data, we have:

  • h=8−3=5h = 8 - 3 = 5 and there are 5 segments in total, thus: h=54=1.5h = \frac{5}{4} = 1.5.

Using the trapezium rule: ∫38log22x dx≈h2[y0+2(y1+y2+y3)+y4]\int_3^8 \text{log}_2 2x \, dx \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4] Substituting the values: =1.52[1.63+2(2.26+2.46+2.63)]= \frac{1.5}{2} [1.63 + 2(2.26 + 2.46 + 2.63)] Calculating the terms: =0.75[1.63+2(7.35)]= 0.75 [1.63 + 2(7.35)] =0.75[1.63+14.70]= 0.75 [1.63 + 14.70] =0.75×16.33=12.2475= 0.75 \times 16.33 = 12.2475. Hence, the estimated value is approximately 12.2512.25.

Step 2

Estimate $$\int_3^8 \text{log}_2 (2x)^{10} \, dx$$

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Answer

Using the result from part (a), we have: ∫38log2(2x)10 dx=10∫38log22x dx\int_3^8 \text{log}_2 (2x)^{10} \, dx = 10 \int_3^8 \text{log}_2 2x \, dx Thus: ≈10×12.25=122.5\approx 10 \times 12.25 = 122.5.

Step 3

Estimate $$\int_3^8 \text{log}_2 18x \, dx$$

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Answer

We can break this integral into parts: ∫38log218x dx=∫38log218 dx+∫38log2x dx.\int_3^8 \text{log}_2 18x \, dx = \int_3^8 \text{log}_2 18 \, dx + \int_3^8 \text{log}_2 x \, dx.

The first part simplifies as follows: ∫38log218 dx=log218∗(8−3)=log218∗5=5×log218.\int_3^8 \text{log}_2 18 \, dx = \text{log}_2 18 * (8 - 3) = \text{log}_2 18 * 5 = 5 \times \text{log}_2 18.

For the second part, we use the result from part (a): =5×log2x∣38+5(12.25).= 5 \times \text{log}_2 x |^8_3 + 5(12.25).

So: ∫38log218x dx≈5×4.169≈20.845+61.25=82.095.\int_3^8 \text{log}_2 18x \, dx \approx 5 \times 4.169 \approx 20.845 + 61.25 = 82.095.

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