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y = 3^x + 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 = 3^x + 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 = 3^x + 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:
- Calculate y for each x using the formula ( y = 3^x + 2x ).
- For ( x = 0.2 ): ( y = 3^{0.2} + 2(0.2) \approx 1.25 )
- For ( x = 0.4 ): ( y = 3^{0.4} + 2(0.4) \approx 1.65 )
- For ( x = 0.6 ): ( y = 3^{0.6} + 2(0.6) \approx 2.08 )
- For ( x = 0.8 ): ( y = 3^{0.8} + 2(0.8) \approx 2.66 )
The completed table is:
| x | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1 |
|---|---|---|---|---|---|---|
| y | 1 | 1.25 | 1.65 | 2.08 | 2.66 | 5 |
Step 2
Use the trapezium rule, with all the values of y from your table, to find an approximate value for \( \int_0^1 (3^x + 2x) dx \).
Answer
To apply the trapezium rule:
- The formula for the trapezium rule is: [ T_n = \frac{h}{2} (y_0 + 2y_1 + 2y_2 + \ldots + 2y_{n-1} + y_n) ] where ( h = \frac{b-a}{n} ) and ( n ) is the number of trapezoids.
- Here, ( h = 0.2 ) since we are evaluating from 0 to 1 with 5 intervals.
- Substitute the y values: [ T_5 = \frac{0.2}{2} (1 + 2(1.25) + 2(1.65) + 2(2.08) + 2(2.66) + 5) ]
- Calculate:
- First calculate the sum: ( 1 + 2(1.25) + 2(1.65) + 2(2.08) + 2(2.66) + 5 = 1 + 2.5 + 3.3 + 4.16 + 5.32 + 5 = 21.28 )
- Then plug it back: ( T_5 = 0.1 \times 21.28 \approx 2.128 )
- Therefore, the approximate value of the integral is ( 2.13 ) (to two decimal places).