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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 | | ... 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.
Answer
To find the values of y, we will substitute each value of x into the equation ( y = \sqrt{10x - x^2} ).
-
For ( x = 1 ):
[ y = \sqrt{10(1) - (1)^2} = \sqrt{9} = 3.00 ] -
For ( x = 1.4 ):
[ y = \sqrt{10(1.4) - (1.4)^2} = \sqrt{14 - 1.96} = \sqrt{12.04} \approx 3.47 ]
(rounded to 2 decimal places) -
For ( x = 1.8 ):
[ y = \sqrt{10(1.8) - (1.8)^2} = \sqrt{18 - 3.24} = \sqrt{14.76} \approx 3.84 ]
(rounded to 2 decimal places) -
For ( x = 2.2 ):
[ y = \sqrt{10(2.2) - (2.2)^2} = \sqrt{22 - 4.84} = \sqrt{17.16} \approx 4.14 ]
(rounded to 2 decimal places) -
For ( x = 2.6 ):
[ y = \sqrt{10(2.6) - (2.6)^2} = \sqrt{26 - 6.76} = \sqrt{19.24} \approx 4.39 ]
(rounded to 2 decimal places) -
For ( x = 3 ):
[ y = \sqrt{10(3) - (3)^2} = \sqrt{30 - 9} = \sqrt{21} \approx 4.58 ]
(rounded to 2 decimal places)
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.
Answer
To apply the trapezium rule, we use the formula:
Where:
- ( h ) is the width of each interval, and
- ( y_i ) are the calculated values of y at each x.
-
Determine the width of each interval: [ h = \frac{b - a}{n} = \frac{3 - 1}{5} = 0.4 ]
(since n = 5 intervals from 1 to 3) -
Substitute the y values: [ y_0 = 3.00, \quad y_1 = 3.47, \quad y_2 = 3.84, \quad y_3 = 4.14, \quad y_4 = 4.39, \quad y_5 = 4.58 ]
-
Calculate: [ ext{Area} \approx \frac{0.4}{2} \times (3.00 + 2 \times 3.47 + 2 \times 3.84 + 2 \times 4.14 + 2 \times 4.39 + 4.58) ]
= [ 0.2 \times (3.00 + 6.94 + 7.68 + 8.28 + 8.78 + 4.58) ] = [ 0.2 \times 39.26 = 7.852 ]
Thus, the approximation for ( \int_1^3 \sqrt{(10x - x^2)} , dx ) is approximately ( 7.85 ).