Figure 1 shows part of the curve with equation $y = e^{0.5x}$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the x-axis, the y-axis and the line $x = 2$. (a) Complete the table with the values of $y$ corresponding to $x = 0.8$ and $x = 1.6$. | $x$ | 0 | 0.4 | 0.8 | 1.2 | 1.6 | 2 | |-------|-------|--------|--------|--------|--------|-------| | $y$ | $e^0$ | $e^{0.4}$ | $e^{0.8}$ | $e^{1.2}$ | $e^{1.6}$ | $e^2$ | (b) Use the trapezium rule with all the values in the table to find an approximate value for the area of $R$, giving your answer to 4 significant figures.

A-Level Edexcel Maths Pure Differentiation: Figure 1 shows part of the curve

Figure 1 shows part of the curve with equation $y = e^{0.5x}$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 7

Question 4

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Figure 1 shows part of the curve with equation $y = e^{0.5x}$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the x-axis, the y-axis and t... show full transcript

Worked Solution & Example Answer:Figure 1 shows part of the curve with equation $y = e^{0.5x}$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 7

Step 1

Complete the table with the values of $y$ corresponding to $x = 0.8$ and $x = 1.6$

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Answer

To find the values for $y$ when $x = 0.8$ and $x = 1.6$ , we can use the equation $y = e^{0.5x}$ :

For $x = 0.8$ : $y = e^{0.5 imes 0.8} = e^{0.4} \\ \approx 1.49182$ (rounded to 5 significant figures)

For $x = 1.6$ : $y = e^{0.5 imes 1.6} = e^{0.8} \\ \approx 2.22554$ (rounded to 5 significant figures)

Thus, the completed table with approximate values is:

$x$	0	0.4	0.8	1.2	1.6	2
$y$	$1.0$	$1.49182$	$1.64872$	$2.22554$	$3.68078$	$7.38906$

Step 2

Use the trapezium rule with all the values in the table to find an approximate value for the area of $R$

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Answer

To apply the trapezium rule, we need to calculate the area using the formula: $A = rac{1}{2}ht(a+b) + h \\sum_{i=1}^{n-1}y_i$ where $A$ is the area, $h$ is the width of the trapezium, $a$ and $b$ are the heights of the first and last sections respectively, and $y_i$ are the intermediate heights.

Width of each sub-interval (h):
- The interval is $[0, 2]$ , thus $h = 0.4$ .
- The values of $y$ from the table are approximately: $1.0, 1.49182, 1.64872, 2.22554, 3.68078, 7.38906$
Using the trapezium rule:
- Area: $A = rac{1}{2} (0.4) (1.0 + 7.38906) + 0.4 (1.49182 + 1.64872 + 2.22554 + 3.68078)$
- Calculating the sums: $= 0.2(8.38906) + 0.4(9.04686) = 1.677812 + 3.618744 = 5.296556$

Thus, the approximate value for the area of $R$ is: $A \\approx 5.2966 \\text{ (to 4 significant figures: } 5.297)$

Figure 1 shows part of the curve with equation $y = e^{0.5x}$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 7

Worked Solution & Example Answer:Figure 1 shows part of the curve with equation $y = e^{0.5x}$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 7

Complete the table with the values of $y$ corresponding to $x = 0.8$ and $x = 1.6$

Use the trapezium rule with all the values in the table to find an approximate value for the area of $R$

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