A-Level Edexcel Maths Pure Differentiation: Figure 3 shows a sketch of part

Figure 3 shows a sketch of part of the curve with equation $y = x^3 \, ext{ln} \, 2x$ - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 7

Question 1

$Figure-3-shows-a-sketch-of-part-of-the-curve-with-equation-$y-=-x^3-\,--ext{ln}-\,-2x$-Edexcel-A-Level Maths Pure-Question 1-2012-Paper 7.png$

Figure 3 shows a sketch of part of the curve with equation $y = x^3 \, ext{ln} \, 2x$. The finite region $R$, shown shaded in Figure 3, is bounded by the curve, th... show full transcript

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation $y = x^3 \, ext{ln} \, 2x$ - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 7

Step 1

Use the trapezium rule, with 3 strips of equal width, to find an estimate for the area of R, giving your answer to 2 decimal places.

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Answer

To estimate the area under the curve using the trapezium rule, we divide the interval from $x=1$ to $x=4$ into 3 equal strips. The width of each strip is:

$h = \frac{4 - 1}{3} = 1$

Next, we find the function values at the endpoints and midpoints:

At $x=1$ : $y(1) = 0.6931$
At $x=2$ : $y(2) = 1.9605$
At $x=3$ : $y(3) = 3.1034$
At $x=4$ : $y(4) = 4.1589$

Applying the trapezium rule:

$\text{Area} = \frac{1}{2}h \left( f(1) + 2f(2) + 2f(3) + f(4) \right)$

Plugging in the values:

$\text{Area} = \frac{1}{2} \cdot 1 \left( 0.6931 + 2(1.9605) + 2(3.1034) + 4.1589 \right)$

Calculating gives:

$\text{Area} \approx 7.49$

Step 2

Find \int_1^4 x^3 \text{ln} 2x \, dx.

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Answer

To solve this integral, we use integration by parts. Let:

$u = \text{ln} \, 2x \Rightarrow du = \frac{1}{x}dx$
$dv = x^3 dx \Rightarrow v = \frac{1}{4}x^4$

Applying integration by parts:

$\int u \, dv = uv - \int v \, du$

Thus,

$\int x^3 \text{ln} 2x \, dx = \frac{1}{4} x^4 \text{ln} 2x \biggr|_1^4 - \int \frac{1}{4} x^4 \cdot \frac{1}{x} dx$

This simplifies to:

$= \frac{1}{4} x^4 \text{ln} 2x \biggr|_1^4 - \frac{1}{4} \int x^3 dx$

Calculating gives:

$= \frac{1}{4} \left( 64 \text{ln} (8) - 1 \cdot \text{ln} (2) \right) - \frac{1}{16} [x^4]_1^4 = ...$

Step 3

Hence find the exact area of R, giving your answer in the form a ln 2 + b, where a and b are exact constants.

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Answer

Using the results from part (b), we can derive the exact area of region $R$ . We have:

$= \left[ \frac{2}{3} \left(4 \ln8 - 1 \ln 2 \right) - \frac{1}{16} (256 - 1) \right]$

Simplifying further gives:

$= \frac{16}{3} \ln 2 + b$

where $a = \frac{16}{3}$ and we collect the constant term $b$ from our calculations.

Figure 3 shows a sketch of part of the curve with equation $y = x^3 \, ext{ln} \, 2x$ - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 7

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation $y = x^3 \, ext{ln} \, 2x$ - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 7

Use the trapezium rule, with 3 strips of equal width, to find an estimate for the area of R, giving your answer to 2 decimal places.

Find \int_1^4 x^3 \text{ln} 2x \, dx.

Hence find the exact area of R, giving your answer in the form a ln 2 + b, where a and b are exact constants.

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