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Given the equation: y = 5x + ext{log}(x + 1), ext{ for } 0 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } 0 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } < x < 2 Complete the table below, by giving the value of y when x = 1 | x | 0 | 0.5 | 1 | 1.5 | 2 | |-----|-----|-----|------|------|------| | y | 1 | 2.821 | 12.502 | 26.585 | | (b) Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for \[ \int_{0}^{2} (5 + 5x + ext{log}(x + 1)) \, dx \] giving your answer to 2 decimal places - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 3
Question 5
![Given-the-equation:--y-=-5x-+--ext{log}(x-+-1),--ext{-for-}-0--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}---ext{-}--ext{-}--ext{-}-0--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}--ext{-}-<-x-<-2--Complete-the-table-below,-by-giving-the-value-of-y-when-x-=-1--|-x---|-0---|-0.5-|-1----|-1.5--|-2----|-|-----|-----|-----|------|------|------|-|-y---|-1---|-2.821-|-12.502-|-26.585-|--|--(b)-Use-the-trapezium-rule,-with-all-the-values-of-y-from-the-completed-table,-to-find-an-approximate-value-for--\[-\int_{0}^{2}-(5-+-5x-+--ext{log}(x-+-1))-\,-dx-\]-giving-your-answer-to-2-decimal-places-Edexcel-A-Level Maths Pure-Question 5-2017-Paper 3.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2017_3_1_QP_3_2017 .jpg)
Given the equation: y = 5x + ext{log}(x + 1), ext{ for } 0 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ }... show full transcript
Worked Solution & Example Answer:Given the equation: y = 5x + ext{log}(x + 1), ext{ for } 0 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } 0 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } < x < 2 Complete the table below, by giving the value of y when x = 1 | x | 0 | 0.5 | 1 | 1.5 | 2 | |-----|-----|-----|------|------|------| | y | 1 | 2.821 | 12.502 | 26.585 | | (b) Use the trapezium rule, with all the values of y from the completed table, to find an approximate value for \[ \int_{0}^{2} (5 + 5x + ext{log}(x + 1)) \, dx \] giving your answer to 2 decimal places - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 3
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Step 2
Use the trapezium rule to find \( \int_{0}^{2} (5 + 5x + \text{log}(x + 1))\, dx \)
Answer
Using the trapezium rule, we calculate the area under the curve:
The formula for the trapezium rule is: [ A \approx \frac{h}{2} (y_0 + 2y_1 + 2y_2 + ... + 2y_{n-1} + y_n) ] where ( h ) is the width of each interval.
In this case, we have:
- ( h = 0.5 ) (since ( 0 ) to ( 2 ) in 4 equal parts)
- Values from the table: ( y_0 = 1, y_1 = 2.821, y_2 = 12.502, y_3 = 26.585 )
Now plugging in these values: [ A \approx \frac{0.5}{2} (1 + 2.8212 + 12.5022 + 26.585) ] Calculating gives: [ A \approx 0.25(1 + 5.642 + 25.004 + 26.585) \approx 0.25(58.231) \approx 14.55775 \approx 17.57 ] Thus the approximate value is ( 17.57 ) for ( \int_{0}^{2} (5 + 5x + \text{log}(x + 1)), dx ).
Step 3