The curve shown in Figure 1 has equation $y = e^{rac{1}{2}}( ext{sin} \, x)$, $0 \leq x \leq \pi$. The finite region $R$ bounded by the curve and the x-axis is shown shaded in Figure 1. (a) Complete the table below with the values of $y$ corresponding to $x = 0$, $\frac{\pi}{4}$ and $\frac{3\pi}{2}$, giving your answers to 5 decimal places. | $x$ | 0 | $\frac{\pi}{4}$ | $\frac{3\pi}{2}$ | $\pi$ | |------------------|------------------|------------------|------------------|------------------| | $y$ | 0 | 1.84432 | 4.81047 | 8.87207 | 0 |

A-Level Edexcel Maths Pure Integration: The curve shown in Figure 1 has

The curve shown in Figure 1 has equation $y = e^{rac{1}{2}}( ext{sin} \, x)$, $0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 8

Question 3

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The curve shown in Figure 1 has equation $y = e^{rac{1}{2}}( ext{sin} \, x)$, $0 \leq x \leq \pi$. The finite region $R$ bounded by the curve and the x-axis is show... show full transcript

Worked Solution & Example Answer:The curve shown in Figure 1 has equation $y = e^{rac{1}{2}}( ext{sin} \, x)$, $0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 8

Step 1

Complete the table with the values of $y$ corresponding to $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{2}$

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Answer

To find the values of $y$ for $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{2}$ :

For $x = \frac{\pi}{4}$ :

y = e^{\frac{1}{2}}(\text{sin}(\frac{\pi}{4})) = e^{\frac{1}{2}} \times \frac{\sqrt{2}}{2} \approx 1.84432

For $x = \frac{3\pi}{2}$ :

y = e^{\frac{1}{2}}(\text{sin}(\frac{3\pi}{2})) = e^{\frac{1}{2}} \times (-1) \approx 4.81047

Thus, the completed table is:

$x$	0	$\frac{\pi}{4}$	$\frac{3\pi}{2}$	$\pi$
$y$	0	1.84432	4.81047	8.87207

Step 2

Use the trapezium rule, with all the values in the completed table, to obtain an estimate for the area of the region $R$.

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Answer

Using the trapezium rule, we can estimate the area as follows:

The formula for the trapezium rule is:

$A = \frac{1}{2} h (y_0 + 2y_1 + 2y_2 + y_n)$

where $h$ is the width between the x-values and $y_n$ corresponds to each function value from the table.

In this case:

$y_0 = 0$
$y_1 = 1.84432$
$y_2 = 4.81047$
$y_3 = 8.87207$
$h = \frac{\pi - 0}{3} = \frac{\pi}{3}$

Substituting these values into the formula:

$A = \frac{1}{2} \times \frac{\pi}{3} \left(0 + 2(1.84432) + 2(4.81047) + 8.87207\right)$

Calculating the expression yields:

$A \approx 12.1948$

Thus, the estimated area of the region $R$ is approximately 12.1948.

The curve shown in Figure 1 has equation $y = e^{ rac{1}{2}}( ext{sin} \, x)$, $0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 8

Worked Solution & Example Answer:The curve shown in Figure 1 has equation $y = e^{ rac{1}{2}}( ext{sin} \, x)$, $0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 8

Complete the table with the values of $y$ corresponding to $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{2}$

Use the trapezium rule, with all the values in the completed table, to obtain an estimate for the area of the region $R$.

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The curve shown in Figure 1 has equation $y = e^{rac{1}{2}}( ext{sin} \, x)$, $0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 8

Worked Solution & Example Answer:The curve shown in Figure 1 has equation $y = e^{rac{1}{2}}( ext{sin} \, x)$, $0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 8