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Figure 3 shows a sketch of part of the curve with equation $y = x^3 imes ext{ln} ext{2} x$ - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 7

Question 8

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

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation $y = x^3 imes ext{ln} ext{2} x$ - Edexcel - A-Level Maths Pure - Question 8 - 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 of region $R$ , we first need to determine the width of each strip. The interval from $x = 1$ to $x = 4$ has a total width of $4 - 1 = 3$ . Dividing this into 3 equal strips yields a width of:

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

Next, we evaluate the function at the endpoints and midpoints of each strip:

For $x = 1$ : $y = 0.6931$
For $x = 2$ : $y = \frac{\sqrt{2}}{\text{ln} 2}$ , which approximates to $1.9605$
For $x = 3$ : $y = \frac{\sqrt{3}}{6 \text{ln} 8}$ , which approximates to $3.1034$
For $x = 4$ : $y = 4.1589$

Using the trapezium rule, we find the area as:

$\text{Area} = \frac{1}{2} h (y_0 + 2y_1 + 2y_2 + y_3)$

Substituting the function values:

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

Calculating this gives:

$\text{Area} \approx \frac{1}{2} \cdot 1 \cdot 14.97989 \approx 7.49$

Step 2

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

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Answer

To find the integral, we can apply integration by parts:

Let:

$u = \text{ln} 2 x$
$dv = x^3 \, dx$

Then:

$du = \frac{1}{x} \, dx$ and $v = \frac{1}{4} x^4$ .

Using the integration by parts formula:

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

Substituting gives:

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

Simplifying further allows us to integrate:

$= \frac{1}{4} x^4 \text{ln} 2 x - \frac{1}{4} \int x^3 \, dx = \frac{1}{4} x^4 \text{ln} 2 x - \frac{1}{16} x^4 + C$

Evaluating from 1 to 4 will provide the definite integral value.

Step 3

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

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Answer

The expression for the integral yields:

$\int_1^4 x^3 \text{ln} 2 x \, dx = \left[ \frac{1}{4} (4^4) \text{ln} 2 (4) - \frac{1}{16} (4^4) \right] - \left[ \frac{1}{4} (1^4) \text{ln} 2 (1) - \frac{1}{16} (1^4) \right]$

Calculating further gives:

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

This simplifies to:

$16 \text{ln} 2 - 16 - \left( \frac{1}{4} \text{ln} 2 - \frac{1}{16} \right) = 16 \text{ln} 2 - \frac{1}{4} \text{ln} 2 + \frac{1}{16}$

Combining terms results in:

$\text{Area} = (16 - \frac{1}{4}) \text{ln} 2 + \left(\frac{1}{16} \right)$

Thus, the area of region $R$ can be expressed as: $a \text{ln} 2 + b$ , where $a = \frac{63}{4}$ and $b = \frac{1}{16}$ .

Figure 3 shows a sketch of part of the curve with equation $y = x^3 imes ext{ln} ext{2} x$ - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 7

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation $y = x^3 imes ext{ln} ext{2} x$ - Edexcel - A-Level Maths Pure - Question 8 - 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} 2 x \, dx.

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

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