Complete the table below, giving the values of y to 2 decimal places. $x$ | 1 | 1.4 | 1.8 | 2.2 | 2.6 | 3 ---|---|---|---|---|---|--- $y$ | 3 | 3.47 | | | | 4.39 Use the trapezium rule, with all the values of y from your table, to find an approximation for the value of $\\int_{1}^{3} \, \sqrt{10x - x^{2}} \, dx$.

A-Level Edexcel Maths Pure Equation of a Straight Line: Complete the table below, giving

Complete the table below, giving the values of y to 2 decimal places - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 2

Question 5

Complete the table below, giving the values of y to 2 decimal places. $x$ | 1 | 1.4 | 1.8 | 2.2 | 2.6 | 3 ---|---|---|---|---|---|--- $y$ | 3 | 3.47 | | | | 4... show full transcript

Worked Solution & Example Answer:Complete the table below, giving the values of y to 2 decimal places - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 2

Step 1

Complete the table below, giving the values of y to 2 decimal places.

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Answer

To find the values of y for given x, we will use the equation:
$y = \sqrt{10x - x^{2}}.$

Calculating each value:

For $x = 1$ :
$y = \sqrt{10 \cdot 1 - 1^{2}} = \sqrt{9} = 3.00$
For $x = 1.4$ : $y = \sqrt{10 \cdot 1.4 - (1.4)^{2}} = \sqrt{14 - 1.96} = \sqrt{12.04} \approx 3.47$
For $x = 1.8$ : $y = \sqrt{10 \cdot 1.8 - (1.8)^{2}} = \sqrt{18 - 3.24} = \sqrt{14.76} \approx 3.84$
For $x = 2.2$ : $y = \sqrt{10 \cdot 2.2 - (2.2)^{2}} = \sqrt{22 - 4.84} = \sqrt{17.16} \approx 4.14$
For $x = 2.6$ : $y = \sqrt{10 \cdot 2.6 - (2.6)^{2}} = \sqrt{26 - 6.76} = \sqrt{19.24} \approx 4.39$
For $x = 3$ :
$y = \sqrt{10 \cdot 3 - 3^{2}} = \sqrt{30 - 9} = \sqrt{21} \approx 4.58$

Thus, the completed table is:

$x$	1	1.4	1.8	2.2	2.6	3
$y$	3.00	3.47	3.84	4.14	4.39	4.58

Step 2

Use the trapezium rule, with all the values of y from your table, to find an approximation for the value of $ \int_{1}^{3} \sqrt{10x - x^{2}} \, dx $.

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Answer

The trapezium rule formula for approximating the integral is given by:

$\int_{a}^{b} f(x) \, dx \approx \frac{h}{2} \left( f(x_{0}) + 2f(x_{1}) + 2f(x_{2}) + \ldots + 2f(x_{n-1}) + f(x_{n}) \right)$

where:

$h = \frac{b-a}{n}$
$f(x_{i})$ are values of the function at the corresponding $x_{i}$ values.

In our case, $a = 1$ , $b = 3$ , and $(n = 5)$ so:

$h = \frac{3 - 1}{5} = 0.4$

The values of y from the completed table are:

$f(1) = 3.00$
$f(1.4) = 3.47$
$f(1.8) = 3.84$
$f(2.2) = 4.14$
$f(2.6) = 4.39$
$f(3) = 4.58$

Thus, substituting in the trapezium rule:

$\int_{1}^{3} \sqrt{10x - x^{2}} \, dx \approx \frac{0.4}{2} \left( 3.00 + 2 \cdot 3.47 + 2 \cdot 3.84 + 2 \cdot 4.14 + 2 \cdot 4.39 + 4.58 \right)$

Calculating this gives:

$\approx 0.2 \left( 3.00 + 6.94 + 7.68 + 8.28 + 8.78 + 4.58 \right)$ $= 0.2 \cdot 39.26 = 7.852$

Therefore, the approximation for the integral is approximately 7.85.

Complete the table below, giving the values of y to 2 decimal places - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 2

Worked Solution & Example Answer:Complete the table below, giving the values of y to 2 decimal places - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 2

Complete the table below, giving the values of y to 2 decimal places.

Use the trapezium rule, with all the values of y from your table, to find an approximation for the value of \( \int_{1}^{3} \sqrt{10x - x^{2}} \, dx \).

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