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y = \sqrt{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 = \sqrt{5x + 2} (a) Complete the table below, giving the values of y to 3 decimal places. \begin{array}{|c|c|} \hline x & y \\ \hline 0 & \\ 0.5 & \\ 1 & \\ 1... show full transcript
Worked Solution & Example Answer:y = \sqrt{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 below, giving the values of y to 3 decimal places.
Answer
To complete the table, we need to calculate the values of ( y ) for each ( x ) value given in the table using the formula ( y = \sqrt{5x + 2} ).
-
For ( x = 0 ):
[ y = \sqrt{5(0) + 2} = \sqrt{2} \approx 1.732 ]
So, ( y \approx 1.732 ). -
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 ]
So, ( y \approx 2.121 ). -
For ( x = 1 ):
[ y = \sqrt{5(1) + 2} = \sqrt{5 + 2} = \sqrt{7} \approx 2.646 ]
So, ( y \approx 2.646 ). -
For ( x = 1.5 ):
[ y = \sqrt{5(1.5) + 2} = \sqrt{7.5 + 2} = \sqrt{9.5} \approx 3.082 ]
So, ( y \approx 3.082 ). -
For ( x = 2 ):
[ y = \sqrt{5(2) + 2} = \sqrt{10 + 2} = \sqrt{12} \approx 3.464 ]
So, ( y \approx 3.464 ).
The completed table will look like:
\begin{array}{|c|c|} \hline x & y \ \hline 0 & 1.732 \ 0.5 & 2.121 \ 1 & 2.646 \ 1.5 & 3.082 \ 2 & 3.464 \ \hline \end{array}
Step 2
Use the trapezium rule, with all the values of y from your table, to find an approximation for the value of \int_{0}^{2} \sqrt{5x + 2} \, dx.
Answer
To use the trapezium rule, we can apply the formula:
[ \text{Approximation} = \frac{(b-a)}{2n} (f(a) + 2\sum_{i=1}^{n-1} f(x_i) + f(b)) ]
where:
- ( a = 0 ) and ( b = 2 )
- ( n = 4 ) (the number of intervals, since there are 5 values of ( x ))
The values of ( y ) based on the table:
- ( y(0) = 1.732 )
- ( y(0.5) = 2.121 )
- ( y(1) = 2.646 )
- ( y(1.5) = 3.082 )
- ( y(2) = 3.464 )
Substituting into the formula we have,
[ \text{Approximation} = \frac{(2-0)}{2 \cdot 4} (1.732 + 2 \cdot (2.121 + 2.646 + 3.082) + 3.464) ] [ = \frac{2}{8} (1.732 + 2 \cdot 7.849 + 3.464) ] [ = \frac{2}{8} (1.732 + 15.698 + 3.464) ] [ = \frac{2}{8} (20.894) ] [ = \frac{20.894}{4} \approx 5.2235 ]
Thus, the approximate value of the integral is ( \approx 5.224 ).