Figure 1 shows a sketch of part of the curve with equation $y = \sqrt{x^2 + 1}$, $x \geq 0$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the x-axis and the lines $x = 1$ and $x = 2$. The table below shows corresponding values for $x$ and $y$ for $y = \sqrt{x^2 + 1}$. | x | 1 | 1.25 | 1.5 | 1.75 | 2 | |-----|------|-------|-------|-------|------| | y | 1.414| 1.601 | 1.803 | 2.016 | 2.236| (a) Complete the table above, giving the missing value of $y$ to 3 decimal places. (b) Use the trapezium rule, with all the values of $y$ in the completed table, to find an approximate value for the area of $R$, giving your answer to 2 decimal places.

A-Level Edexcel Maths Pure Binomial Expansion: Figure 1 shows a sketch of part

Figure 1 shows a sketch of part of the curve with equation $y = \sqrt{x^2 + 1}$, $x \geq 0$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 1

Question 3

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Figure 1 shows a sketch of part of the curve with equation $y = \sqrt{x^2 + 1}$, $x \geq 0$. The finite region $R$, shown shaded in Figure 1, is bounded by the curv... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of part of the curve with equation $y = \sqrt{x^2 + 1}$, $x \geq 0$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 1

Step 1

Complete the table above, giving the missing value of y to 3 decimal places.

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Answer

To find the value of $y$ when $x = 1.25$ :

Calculate $y = \sqrt{x^2 + 1}$ for $x = 1.25$ :

$y = \sqrt{(1.25)^2 + 1} = \sqrt{1.5625 + 1} = \sqrt{2.5625} \approx 1.601$
Update the table:

x	1	1.25	1.5	1.75	2
y	1.414	1.601	1.803	2.016	2.236

Step 2

Use the trapezium rule, with all the values of y in the completed table, to find an approximate value for the area of R, giving your answer to 2 decimal places.

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Answer

To calculate the area $A$ using the trapezium rule:

Apply the formula:

$A \approx \frac{1}{2} \times h (y_0 + 2y_1 + 2y_2 + 2y_3 + y_4)$ where $h$ is the width of each segment (in this case, $h = 0.25$ ).

Here $y_0 = 1.414$ , $y_1 = 1.601$ , $y_2 = 1.803$ , $y_3 = 2.016$ , and $y_4 = 2.236$ .
Substitute the values into the formula:

$A \approx \frac{1}{2} \times 0.25 \times (1.414 + 2(1.601) + 2(1.803) + 2.236)$ Evaluating further, we get:

$A \approx \frac{1}{2} \times 0.25 \times (1.414 + 3.202 + 3.606 + 2.236) = \frac{1}{2} \times 0.25 \times 10.458$

$A \approx \frac{0.25 \times 10.458}{2} = 1.3067$
Round the result to 2 decimal places:

$A \approx 1.31$

Figure 1 shows a sketch of part of the curve with equation $y = \sqrt{x^2 + 1}$, $x \geq 0$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 1

Worked Solution & Example Answer:Figure 1 shows a sketch of part of the curve with equation $y = \sqrt{x^2 + 1}$, $x \geq 0$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 1

Complete the table above, giving the missing value of y to 3 decimal places.

Use the trapezium rule, with all the values of y in the completed table, to find an approximate value for the area of R, giving your answer to 2 decimal places.

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