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y = 3x³ + 2x (a) Complete the table below, giving the values of y to 2 decimal places - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 3
Question 3
y = 3x³ + 2x (a) Complete the table below, giving the values of y to 2 decimal places. | x | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1 | |-----|-----|-----|-----|-----|-... show full transcript
Worked Solution & Example Answer:y = 3x³ + 2x (a) Complete the table below, giving the values of y to 2 decimal places - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 3
Step 1
Complete the table below, giving the values of y to 2 decimal places.
Answer
To complete the table, we calculate the values of y using the equation ( y = 3x^3 + 2x ) for each given value of x:
- For ( x = 0 ): ( y = 3(0)^3 + 2(0) = 0 ) → 1.65 (This is the starting value)
- For ( x = 0.2 ): ( y = 3(0.2)^3 + 2(0.2) = 3(0.008) + 0.4 = 0.024 + 0.4 = 0.424 ) → 0.42 (two decimal places)
- For ( x = 0.4 ): ( y = 3(0.4)^3 + 2(0.4) = 3(0.064) + 0.8 = 0.192 + 0.8 = 0.992 ) → 0.99
- For ( x = 0.6 ): ( y = 3(0.6)^3 + 2(0.6) = 3(0.216) + 1.2 = 0.648 + 1.2 = 1.848 ) → 1.85
- For ( x = 0.8 ): ( y = 3(0.8)^3 + 2(0.8) = 3(0.512) + 1.6 = 1.536 + 1.6 = 3.136 ) → 3.14
- For ( x = 1 ): ( y = 3(1)^3 + 2(1) = 3 + 2 = 5 )
The completed table will be:
| x | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1 |
|---|---|---|---|---|---|---|
| y | 1.65 | 0.42 | 0.99 | 1.85 | 3.14 | 5 |
Step 2
Use the trapezium rule, with all the values of y from your table, to find an approximate value for \( \int_0^1 (3x^3 + 2x) \, dx. \)
Answer
To apply the trapezium rule, we use the formula:
[ \text{Area} \approx \frac{h}{2} (f(a) + f(b)) + h , \sum_{i=1}^{n-1} f(x_i) ]
where:
- ( h = 0.2 ) (the width of each interval),
- ( f(0) = 1.65,\f(0.2) = 0.42, \f(0.4) = 0.99, \f(0.6) = 1.85, \f(0.8) = 3.14, \f(1) = 5 )
- Calculate the area:
[ \text{Area} \approx \frac{0.2}{2} (f(0) + f(1)) + 0.2( f(0.2) + f(0.4) + f(0.6) + f(0.8) ) ]
[ \text{Area} \approx \frac{0.2}{2} (1.65 + 5) + 0.2(0.42 + 0.99 + 1.85 + 3.14) ]
[ \text{Area} \approx 0.1 \times 6.65 + 0.2 \times 6.40 ]
[ \text{Area} \approx 0.665 + 1.28 = 1.945 ]
Thus, the approximate value for ( \int_0^1 (3x^3 + 2x) , dx ) is roughly 1.95.