A-Level Edexcel Maths Pure Rational Expressions: Figure 1 shows a sketch of part

Figure 1 shows a sketch of part of the curve with equation y = \frac{(x + 2)^{\frac{3}{2}}}{4} x > -2 The finite region R, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line with equation x = 10 The table below shows corresponding values of x and y for y = \frac{(x + 2)^{\frac{3}{2}}}{4} | x | -2 | 2 | 6 | 10 | |-------|------|-----|-----|------| | y | 0 | 2 | 4\sqrt{2} | 6\sqrt{3} | (a) Complete the table, giving values of y corresponding to x = 2 and x = 6 (b) Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for the area of R, giving your answer to 3 decimal places. - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 4

Question 3

$Figure-1-shows-a-sketch-of-part-of-the-curve-with-equation--y-=-\frac{(x-+-2)^{\frac{3}{2}}}{4}---x->--2--The-finite-region-R,-shown-shaded-in-Figure-1,-is-bounded-by-the-curve,-the-x-axis-and-the-line-with-equation-x-=-10--The-table-below-shows-corresponding-values-of-x-and-y-for-y-=-\frac{(x-+-2)^{\frac{3}{2}}}{4}--|---x---|---2--|--2--|--6--|--10--|-|-------|------|-----|-----|------|-|---y---|--0---|--2--|--4\sqrt{2}--|--6\sqrt{3}--|--(a)-Complete-the-table,-giving-values-of-y-corresponding-to-x-=-2-and-x-=-6--(b)-Use-the-trapezium-rule,-with-all-the-values-of-y-from-the-completed-table,-to-find-an-approximate-value-for-the-area-of-R,-giving-your-answer-to-3-decimal-places.-Edexcel-A-Level Maths Pure-Question 3-2018-Paper 4.png$

Figure 1 shows a sketch of part of the curve with equation y = \frac{(x + 2)^{\frac{3}{2}}}{4} x > -2 The finite region R, shown shaded in Figure 1, is bounded b... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of part of the curve with equation y = \frac{(x + 2)^{\frac{3}{2}}}{4} x > -2 The finite region R, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line with equation x = 10 The table below shows corresponding values of x and y for y = \frac{(x + 2)^{\frac{3}{2}}}{4} | x | -2 | 2 | 6 | 10 | |-------|------|-----|-----|------| | y | 0 | 2 | 4\sqrt{2} | 6\sqrt{3} | (a) Complete the table, giving values of y corresponding to x = 2 and x = 6 (b) Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for the area of R, giving your answer to 3 decimal places. - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 4

Step 1

Complete the table, giving values of y corresponding to x = 2 and x = 6

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Answer

To compute the values of y for the provided x values, we substitute into the equation:

For x = 2:

$y = \frac{(2 + 2)^{\frac{3}{2}}}{4} = \frac{(4)^{\frac{3}{2}}}{4} = \frac{8}{4} = 2$
For x = 6:

$y = \frac{(6 + 2)^{\frac{3}{2}}}{4} = \frac{(8)^{\frac{3}{2}}}{4} = \frac{4\sqrt{2} \cdot 8}{4} = 4\sqrt{2}$

Thus, the completed table is:

x	-2	2	6	10
y	0	2	4\sqrt{2}	6\sqrt{3}

Step 2

Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for the area of R

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Answer

To estimate the area of region R using the trapezium rule, we use the formula:

$A = \frac{1}{2} (b_1 + b_2)(h)$

Where:

$b_1$ is the first segment's y-value
$b_2$ is the last segment's y-value
$h$ is the width of each segment

The intervals are:

From x = -2 to x = 10 with corresponding y-values: 0, 2, 4\sqrt{2}, and 6\sqrt{3}.

Calculating:

Area using trapezium rule:

$A \approx \frac{1}{2} \times (0 + 6\sqrt{3}) \times 8 = \frac{1}{2} \times 0 + 2 + 2(4\sqrt{2}) + 6\sqrt{3} \times 4 \equiv 41.412$

The area of region R is approximately 41.412.

Complete the table, giving values of y corresponding to x = 2 and x = 6

Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for the area of R

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