y = \sqrt{3 + x}
(a) Complete the table below, giving the values of y to 3 decimal places - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 3
Question 9
y = \sqrt{3 + x}
(a) Complete the table below, giving the values of y to 3 decimal places.
| x | 0 | 0.25 | 0.5 | 0.75 | 1 |
|-----|------|-------|----... show full transcript
Worked Solution & Example Answer:y = \sqrt{3 + x}
(a) Complete the table below, giving the values of y to 3 decimal places - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 3
Step 1
(a) Complete the table below, giving the values of y to 3 decimal places.
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Answer
To complete the table, we calculate the values of y using the function:
For x = 0:
y = \sqrt{3 + 0} = \sqrt{3} \approx 1.000
For x = 0.25:
y = \sqrt{3 + 0.25} = \sqrt{3.25} \approx 1.251
For x = 0.5:
y = \sqrt{3 + 0.5} = \sqrt{3.5} \approx 1.495
For x = 0.75:
y = \sqrt{3 + 0.75} = \sqrt{3.75} \approx 1.936
For x = 1:
y = \sqrt{3 + 1} = \sqrt{4} = 2.000
The final table will be:
x
0
0.25
0.5
0.75
1
y
1
1.251
1.495
1.936
2
Step 2
(b) Use the trapezium rule with all the values of y from your table to find an approximation for the value of \int_0^1 \sqrt{3 + x} \, dx
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Answer
To apply the trapezium rule, we use the formula:
I≈2h(y0+2y1+2y2+2y3+yn)
Where:
h is the width of each interval.
y_0, y_1, y_2, y_3, and y_n are the values of y from the table.
Here, the interval from 0 to 1 is divided into 4 parts:
h = (1 - 0) / 4 = 0.25
Plugging in values into the formula:
I≈20.25(1+2(1.251)+2(1.495)+2(1.936)+2)
=0.125(1+2.502+2.990+3.872+2)
=0.125(12.364)≈1.5455
Therefore, the approximation for the value of \int_0^1 \sqrt{3 + x} , dx \approx 1.5455.