The curve shown in Figure 1 has equation $y = e^{rac{1}{2}}(sin x), \, 0 \leq x \leq \pi$. 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 $x = 0$, $\frac{\pi}{4}$, $\frac{3\pi}{4}$ and $\pi$, giving your answers to 5 decimal places. | $x$ | $0$ | $\frac{\pi}{4}$ | $\frac{3\pi}{4}$ | $\pi$ | |-------------|-----------|-------------------|--------------------|---------| | $y$ | $0$ | $4.810477381...$ | $4.810477381...$ | $0$ | (b) Use the trapezium rule, with all the values in the completed table, to obtain an estimate for the area of the region $R$. Give your answer to 4 decimal places.

A-Level Edexcel Maths Pure Modelling with Functions: The curve shown in Figure 1 has

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

Question 2

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

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

Step 1

Complete the table below with the values of $y$ corresponding to $x = 0$, $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, giving your answers to 5 decimal places.

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Answer

To find the values of $y$ , we will substitute each $x$ into the equation:

For $x = 0$ :

$y = e^{0}(sin(0)) = 0$
So, $y = 0$ .
For $x = \frac{\pi}{4}$ :
$= e^{\frac{1}{2}} \cdot \frac{\sqrt{2}}{2} \approx 4.810477381$$ Therefore, $y \approx 4.81048$ (to 5 decimal places).$
For $x = \frac{3\pi}{4}$ :

$y = e^{\frac{1}{2}}(sin(\frac{3\pi}{4})) = e^{\frac{1}{2}} \cdot \frac{\sqrt{2}}{2} \approx 4.810477381$
Hence, $y \approx 4.81048$ (to 5 decimal places).
For $x = \pi$ :

$y = e^{\frac{1}{2}}(sin(\pi)) = 0$
Thus, $y = 0$ .

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

To estimate the area under the curve using the trapezium rule, we apply the formula:

$\text{Area} \approx \frac{h}{2} \cdot (y_0 + 2y_1 + 2y_2 + y_3)$

where:

$y_0 = 0$
$y_1 = 4.81048$
$y_2 = 4.81048$
$y_3 = 0$
The width of each interval, $h$ , is:

$h = \frac{\pi - 0}{3} = \frac{\pi}{3} \\ \approx 1.0472$

Now substituting:

= \frac{1.0472}{2} \cdot (2(4.81048 + 4.81048)) \\ = \frac{1.0472}{2} \cdot (2(9.62096)) \\ = \frac{1.0472 \cdot 19.24192}{2} = 12.1948$$ Thus, the estimated area of region $R$ is approximately $\mathbf{12.1948}$ (to 4 decimal places).

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

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

Complete the table below with the values of $y$ corresponding to $x = 0$, $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, giving your answers to 5 decimal places.

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 $y = e^{rac{1}{2}}(sin x), \, 0 \leq x \leq \pi$ - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 8

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