Photo AI
Complete the table below, giving the values of y to 2 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 3
Question 8
Complete the table below, giving the values of y to 2 decimal places. | x | 2 | 2.25 | 2.5 | 2.75 | 3 | |----|------|------|------|------|-----| | y | 0.5 ... 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 8 - 2011 - Paper 3
Step 1
Complete the table below, giving the values of y to 2 decimal places.
Answer
To complete the table, we need to calculate the values of y for each corresponding x using the formula:
[ y = \frac{5}{3x^2 - 2} ]
-
For ( x = 2 ): [ y = \frac{5}{3(2^2) - 2} = \frac{5}{12 - 2} = \frac{5}{10} = 0.5 ]
-
For ( x = 2.25 ): [ y = \frac{5}{3(2.25^2) - 2} = \frac{5}{3(5.0625) - 2} = \frac{5}{15.1875 - 2} = \frac{5}{13.1875} \approx 0.38 ]
-
For ( x = 2.5 ): [ y = \frac{5}{3(2.5^2) - 2} = \frac{5}{3(6.25) - 2} = \frac{5}{18.75 - 2} = \frac{5}{16.75} \approx 0.30 ]
-
For ( x = 2.75 ): [ y = \frac{5}{3(2.75^2) - 2} = \frac{5}{3(7.5625) - 2} = \frac{5}{22.6875 - 2} = \frac{5}{20.6875} \approx 0.24 ]
-
For ( x = 3 ): [ y = \frac{5}{3(3^2) - 2} = \frac{5}{27 - 2} = \frac{5}{25} = 0.2 ]
The completed table is:
| x | 2 | 2.25 | 2.5 | 2.75 | 3 |
|---|---|---|---|---|---|
| y | 0.5 | 0.38 | 0.30 | 0.24 | 0.2 |
Step 2
Use the trapezium rule, with all the values of y from your table, to find an approximate value for \( \int_2^3 \frac{5}{3x^2-2} \, dx \).
Answer
The trapezium rule formula is given by:
[ \int_a^b f(x) , dx \approx \frac{h}{2} (f(a) + f(b) + 2 \sum_{i=1}^{n-1} f(x_i)) ]
Where ( a = 2 ), ( b = 3 ), and ( h = \frac{b-a}{n} = \frac{3-2}{4-1} = 0.25 ). Thus:
Values from the table:
- ( y(2) = 0.5 )
- ( y(2.25) = 0.38 )
- ( y(2.5) = 0.30 )
- ( y(2.75) = 0.24 )
- ( y(3) = 0.2 )
Now, applying the trapezium rule:
[ \int_2^3 \frac{5}{3x^2-2} , dx \approx \frac{0.25}{2} \left(0.5 + 0.2 + 2(0.38 + 0.30 + 0.24) \right) ]
[ = 0.125 (0.7 + 2(0.92)) ]
[ = 0.125 (0.7 + 1.84) ]
[ = 0.125 (2.54) \approx 0.3175 ]
Step 3
Use your answer to part (b) to find an approximate value for the area of S.
Answer
The area of region S can be approximated as:
[ \text{Area}(S) = \text{Area under curve from } x=2 \text{ to } x=3 - \text{Area of triangle} \]
The area under the curve, found using the trapezium rule, is approximately ( 0.3175 ) as calculated above.
To find the area of triangle S:
The base of triangle is from ( x=2 ) to ( x=3 ), so:
- Base = 1
- Height = value of the curve at ( x = 2 ), which is ( 0.5 )
Area of triangle:
[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 0.5 = 0.25 ]
Hence,
[ ext{Approximate Area of } S = 0.3175 - 0.25 = 0.0675 ]