Figure 1 shows part of the curve with equation $x=4e^{t/3} + 3$. The finite region $R$ shown shaded in Figure 1 is bounded by the curve, the x-axis, the t-axis and the line $t = 8$. (a) Complete the table with the value of $x$ corresponding to $t = 6$, giving your answer to 3 decimal places. | $t$ | 0 | 2 | 4 | 6 | 8 | |-----|---|---|---|---|---| | $x$ | 3 | 7.107 | 7.218 | | 5.223 | (b) Use the trapezium rule with all the values of $x$ in the completed table to obtain an estimate for the area of the region $R$, giving your answer to 2 decimal places. (c) Use calculus to find the exact value for the area of $R$. (d) Find the difference between the values obtained in part (b) and part (c), giving your answer to 2 decimal places.

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

Figure 1 shows part of the curve with equation $x=4e^{t/3} + 3$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 1

Question 7

$Figure-1-shows-part-of-the-curve-with-equation-$x=4e^{t/3}-+-3$-Edexcel-A-Level Maths Pure-Question 7-2013-Paper 1.png$

Figure 1 shows part of the curve with equation $x=4e^{t/3} + 3$. The finite region $R$ shown shaded in Figure 1 is bounded by the curve, the x-axis, the t-axis and t... show full transcript

Worked Solution & Example Answer:Figure 1 shows part of the curve with equation $x=4e^{t/3} + 3$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 1

Step 1

Complete the table with the value of $x$ corresponding to $t = 6$, giving your answer to 3 decimal places.

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Answer

To find the value of $x$ for $t = 6$ , we substitute $t$ into the equation:

$x = 4e^{6/3} + 3 = 4e^2 + 3$

Calculating this gives:

$x ext{ approximately equals } 4(7.389) + 3 ext{ which is } 29.556.$

Thus, $x$ corresponding to $t = 6$ is approximately 29.556.

Step 2

Use the trapezium rule with all the values of $x$ in the completed table to obtain an estimate for the area of the region $R$, giving your answer to 2 decimal places.

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Answer

Using the trapezium rule, the area can be estimated using the formula:

$ext{Area} = rac{h}{2} (y_0 + 2y_1 + 2y_2 + ... + 2y_{n-1} + y_n)$

where $h$ is the width of each interval. Here:

$h = 2$ (since $(8 - 0) / 4 = 2$ )
$y_0 = 3$ , $y_1 = 7.107$ , $y_2 = 7.218$ , $y_3 = 29.556$ , $y_4 = 5.223$ .

Calculating the area:

$ext{Area} = rac{2}{2} (3 + 2(7.107) + 2(7.218) + 2(29.556) + 5.223) = 49.37$

Thus, the estimated area is approximately 49.37.

Step 3

Use calculus to find the exact value for the area of $R$.

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Answer

To find the exact area under the curve, we evaluate the integral:

$A = ext{Area} = int_0^8 (4e^{t/3} + 3) \, dt$

Calculating gives:

$A = [12e^{t/3} + 3t]_{0}^{8} = 12e^{8/3} + 24 - (12 + 0) = 12(e^{8/3} - 1) + 24$

Evaluating this precisely yields approximately 49.369.

Step 4

Find the difference between the values obtained in part (b) and part (c), giving your answer to 2 decimal places.

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Answer

The estimated area from part (b) is 49.37 and the exact area from part (c) is approximately 49.369.

The difference is:

$49.37 - 49.369 = 0.001$

Thus, the difference is approximately 0.00 when rounded to 2 decimal places.

Figure 1 shows part of the curve with equation $x=4e^{t/3} + 3$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 1

Worked Solution & Example Answer:Figure 1 shows part of the curve with equation $x=4e^{t/3} + 3$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 1

Complete the table with the value of $x$ corresponding to $t = 6$, giving your answer to 3 decimal places.

Use the trapezium rule with all the values of $x$ in the completed table to obtain an estimate for the area of the region $R$, giving your answer to 2 decimal places.

Use calculus to find the exact value for the area of $R$.

Find the difference between the values obtained in part (b) and part (c), giving your answer to 2 decimal places.

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