Figure 4 shows a sketch of part of the curve C with equation y = \frac{x^2 \ln x}{3} - 2x + 5, \quad x > 0 The finite region S, shown shaded in Figure 4, is bounded by the curve C, the line with equation x = 1, the x-axis and the line with equation x = 3. The table below shows corresponding values of x and y with the values of y given to 4 decimal places as appropriate. | x | 1 | 1.5 | 2.5 | 3 | |-----|-----|-----|-----|-----| | y | 2.3041 | 1.9242 | 1.9089 | 2.2958 | (a) Use the trapezium rule, with all the values of y in the table, to obtain an estimate for the area of S, giving your answer to 3 decimal places. (b) Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of S. (c) Show that the exact area of S can be written in the form $ \frac{a}{b} + \ln c $, where a, b and c are integers to be found.

A-Level Edexcel Maths Pure Differentiation: Figure 4 shows a sketch of part

Figure 4 shows a sketch of part of the curve C with equation y = \frac{x^2 \ln x}{3} - 2x + 5, \quad x > 0 The finite region S, shown shaded in Figure 4, is bounded by the curve C, the line with equation x = 1, the x-axis and the line with equation x = 3 - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Question 15

$Figure-4-shows-a-sketch-of-part-of-the-curve-C-with-equation--y-=-\frac{x^2-\ln-x}{3}---2x-+-5,-\quad-x->-0--The-finite-region-S,-shown-shaded-in-Figure-4,-is-bounded-by-the-curve-C,-the-line-with-equation-x-=-1,-the-x-axis-and-the-line-with-equation-x-=-3-Edexcel-A-Level Maths Pure-Question 15-2017-Paper 1.png$

Figure 4 shows a sketch of part of the curve C with equation y = \frac{x^2 \ln x}{3} - 2x + 5, \quad x > 0 The finite region S, shown shaded in Figure 4, is bounde... show full transcript

Worked Solution & Example Answer:Figure 4 shows a sketch of part of the curve C with equation y = \frac{x^2 \ln x}{3} - 2x + 5, \quad x > 0 The finite region S, shown shaded in Figure 4, is bounded by the curve C, the line with equation x = 1, the x-axis and the line with equation x = 3 - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Step 1

Use the trapezium rule, with all the values of y in the table, to obtain an estimate for the area of S, giving your answer to 3 decimal places.

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Answer

To apply the trapezium rule, we first determine the width of each strip (h):

$h = \frac{3 - 1}{4} = 0.5$

Using the trapezium rule formula:

$A = \frac{h}{2} \left( y_0 + 2 \sum_{i=1}^{n-1} y_i + y_n \right)$

Here, the values of y are:

$y_0 = 2.3041$ (when $x = 1$ )
$y_1 = 1.9242$ (when $x = 1.5$ )
$y_2 = 1.9089$ (when $x = 2.5$ )
$y_3 = 2.2958$ (when $x = 3$ )

Plugging in these values:

$A = \frac{0.5}{2} \left( 2.3041 + 2(1.9242 + 1.9089) + 2.2958 \right)$

Calculating this gives: $A = 0.25 \left( 2.3041 + 2(3.8331) + 2.2958 \right)$ $A = 0.25 \left( 2.3041 + 7.6662 + 2.2958 \right)$ $A = 0.25 \left( 12.2661 \right) \approx 3.0665$

Therefore, the estimated area of region S is approximately 3.067.

Step 2

Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of S.

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Answer

To achieve a more accurate estimate for the area of S using the trapezium rule, we can:

Increase the number of strips: By dividing the interval [1, 3] into more strips, we can calculate more y values, which will refine the estimate.
Decrease the width of the strips: A smaller strip width means that the trapezoids better approximate the curve's shape, leading to a more precise area calculation.
Use more trapezia: Employing multiple trapezia within the interval allows for better representation of the curve, thus improving the accuracy of the estimate.

Step 3

Show that the exact area of S can be written in the form $ \frac{a}{b} + \ln c $, where a, b and c are integers to be found.

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Answer

To find the exact area S, we need to evaluate the integral:

$A = \int_1^3 \left( \frac{x^2 \ln x}{3} - 2x + 5 \right) dx$

We can split this into three separate integrals:

$A = \int_1^3 \frac{x^2 \ln x}{3} dx - \int_1^3 2x dx + \int_1^3 5 dx$

Using integration by parts for the first integral: Letting ( u = \ln x ) and ( dv = x^2 dx ):
Then, ( du = \frac{1}{x}dx ) and ( v = \frac{x^3}{3} ).

Thus,

$\int \frac{x^2 \ln x}{3} dx = \frac{x^3 \ln x}{9} - \int \frac{x^3}{9} \cdot \frac{1}{x}dx \ $$$$= \frac{x^3 \ln x}{9} - \frac{x^4}{36}.$

Evaluating from 1 to 3:

Calculate ( \int_1^3 2x dx = \left[ x^2 \right]_1^3 = 9 - 1 = 8 )

Calculate ( \int_1^3 5 dx = \left[ 5x \right]_1^3 = 15 - 5 = 10 )

Thus:

The area becomes:

$A = \left( \frac{27 \ln 3}{9} - \frac{81}{36} + 10 - 8 \right)$

Simplifying gives: $= \frac{27 \ln 3}{9} - \frac{9}{4} + 2 \approx \frac{28}{27} + \ln 3$

This means that a = 28, b = 27, c = 3.

Use the trapezium rule, with all the values of y in the table, to obtain an estimate for the area of S, giving your answer to 3 decimal places.

Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of S.

Show that the exact area of S can be written in the form \( \frac{a}{b} + \ln c \), where a, b and c are integers to be found.

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