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y = √(5x + 2) (a) Complete the table below, giving the values of y to 3 decimal places - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 2
Question 4
y = √(5x + 2) (a) Complete the table below, giving the values of y to 3 decimal places. | x | y | |---|---| | 0 | | | 0.5 | | | 1 | | | 1.5 | | | 2 | | ... show full transcript
Worked Solution & Example Answer:y = √(5x + 2) (a) Complete the table below, giving the values of y to 3 decimal places - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 2
Step 1
Complete the table: y values
Answer
To find the values of y for different x values, apply the function ( y = \sqrt{5x + 2} ) for each x:
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For ( x = 0 ): [ y = \sqrt{5(0) + 2} = \sqrt{2} \approx 1.732 ]
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For ( x = 0.5 ): [ y = \sqrt{5(0.5) + 2} = \sqrt{5(0.5) + 2} = \sqrt{2.5 + 2} = \sqrt{4.5} \approx 2.121 ]
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For ( x = 1 ): [ y = \sqrt{5(1) + 2} = \sqrt{5 + 2} = \sqrt{7} \approx 2.646 ]
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For ( x = 1.5 ): [ y = \sqrt{5(1.5) + 2} = \sqrt{7.5 + 2} = \sqrt{9.5} \approx 3.082 ]
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For ( x = 2 ): [ y = \sqrt{5(2) + 2} = \sqrt{10 + 2} = \sqrt{12} \approx 3.464 ]
The completed table is:
| x | y |
|---|---|
| 0 | 1.732 |
| 0.5 | 2.121 |
| 1 | 2.646 |
| 1.5 | 3.082 |
| 2 | 3.464 |
Step 2
Use the trapezium rule: Approximation
Answer
To find the approximate integral ( \int_0^2 \sqrt{5x + 2} ,dx ) using the trapezium rule, we use the values from the table:
- Trapezium rule formula for n intervals: [ \text{Area} = \frac{h}{2} [y_0 + 2y_1 + 2y_2 + 2y_3 + y_4] ]
where ( h = 0.5 ), ( y_0 = 1.732 ), ( y_1 = 2.121 ), ( y_2 = 2.646 ), ( y_3 = 3.082 ), and ( y_4 = 3.464 ).
- Plug the values into the formula: [ \text{Area} = \frac{0.5}{2} [1.732 + 2(2.121) + 2(2.646) + 2(3.082) + 3.464] ] [ = 0.25 [1.732 + 4.242 + 5.292 + 6.164 + 3.464] ] [ = 0.25 [20.894] ] [ = 5.2235 \approx 5.224 ]
Thus, the approximation for the value of ( \int_0^2 \sqrt{5x + 2} ,dx ) is approximately 5.224.