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y = \sqrt{3 + x} (a) Complete the table below, giving the values of y to 3 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 3
Question 8
y = \sqrt{3 + x} (a) Complete the table below, giving the values of y to 3 decimal places. \begin{array}{|c|c|c|c|c|} \hline x & 0 & 0.25 & 0.5 & 0.75 & 1 \\ \hlin... 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 8 - 2012 - Paper 3
Step 1
Complete the table for y values
Answer
To find the values of y for each corresponding x, we use the equation:
- 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 )
Filling in the table we get:
\begin{array}{|c|c|c|c|c|} \hline x & 0 & 0.25 & 0.5 & 0.75 & 1 \ \hline y & 1.000 & 1.251 & 1.495 & 1.936 & 2.000 \ \hline \end{array}
Step 2
Use the trapezium rule
Answer
To approximate the area under ( \sqrt{3 + x} ) from 0 to 1 using the trapezium rule, we apply:
[ \text{Area} \approx \frac{h}{2} (f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)) ]
Where:
- ( h = \frac{1 - 0}{4} = 0.25 )
- Values from the table: ( f(0) = 1, f(0.25) = 1.251, f(0.5) = 1.495, f(0.75) = 1.936, f(1) = 2 )
Substituting these values:
[ \text{Area} \approx \frac{0.25}{2} (1 + 2(1.251) + 2(1.495) + 2(1.936) + 2) ]
Calculating inside the parentheses:
- ( 1 + 2(1.251) + 2(1.495) + 2(1.936) + 2 = 1 + 2.502 + 2.99 + 3.872 + 2 = 12.364 )
Then: [ \text{Area} \approx 0.125 \times 12.364 = 1.5455 \approx 1.496 ]
Thus, the approximate value of ( \int_{0}^{1} \sqrt{3 + x} , dx ) is ( 1.496 ).