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AS Maths Masterclass

AS Maths Masterclass. Lesson 4: Areas by integration and the Trapezium rule. Learning objectives. The student should be able to: understand why integration is concerned with finding areas under curves; apply the integration rule to simple polynomials to find areas under curves;

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AS Maths Masterclass

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  1. AS Maths Masterclass Lesson 4: Areas by integration and the Trapezium rule

  2. Learning objectives The student should be able to: • understand why integration is concerned with finding areas under curves; • apply the integration rule to simple polynomials to find areas under curves; • evaluate approximate areas using the trapezium rule;

  3. Revision of C1 integration Can you still remember how to integrate? Recall that where c is a constant of integration. You should also remember that integration is the reverse process of differentiation. Click here for some revision questions: integration to find the constant.

  4. A first principles approach to area. Consider finding the area under the graph f (x) = 2x + 3, between the lines x = 0 and x = 10. We can divide the area into n strips, and so the width of each strip will be Now, and so on For the last strip we have f (x) = 2x + 3 23 3 10

  5. Finding the sum If we now calculate the sum of these n terms we get: And so: Did you notice the AP ? a = 3,

  6. Using the AP formula If we now use the AP formula: a=3, We have Hence, So

  7. Validation of the area We conclude that the area is 130 square units. Of course, this area could have been calculated from a single trapezium, but when the function f (x) is curved, a trapezium would only provide an approximation to the precise area. Extension activity: Now take and perform a first principles approach for the area under the curve between x = 1 and x = 5.

  8. Connecting up the results In the first case we saw that when f (x) = 2x+3 we obtained an area of 130 square units. But If we define Then F ( 10 ) = 130, F ( 0 ) = 0 and F ( 10 ) – F ( 0 ) = 130 In the second case we saw that when we obtained an area of square units. And again, defining then F ( 5 ) = and F ( 1 ) = So, as before F ( 5 ) – F ( 1 ) =

  9. Definite integration practice The three links below will help you practice evaluating a range of definite integrals: • Click here for the integration of quadratics • Click here for the integration of polynomials • Click here for ones with negative indices and surds

  10. Turning our attention to area … Click on this link to see a range of integration approaches in practice We shall talk about the trapezium rule later, but for now let’s investigate finding the area under a given curve y = f ( x ) between x = a and x = b.

  11. First principles in general form: If we take a small rectangular strip MNQP where P is the point (x, y), and N has x co-ordinate for small , then as gets smaller, the rectangle will approximate to the area under the curve and between x = M and x = N. The small section of the area can be called Now, if we divide up the whole required area that falls between x = a and x = b into many strips, then for each strip: y P Q x a b M N

  12. Integration connections Hence, total area = for small So taking even smaller, Area = But earlier we agreed that so Now, taking the limit once again, But by definition, and so Hence, because we know from previous work that integration and differentiation are reverse processes.

  13. The area function and finding areas between line and curve. Click here to see how an Area function can be obtained Click here to practice finding the area between oblique lines and curves Click here for more practice

  14. Towards the Trapezium rule As a crude first approximation to the area bounded by the lines x = a, x = b and the curve y = f (x), we could take an estimate: either the rectangle ABCD or the rectangle ABEF. With a little thought, however, both these approximations can be improved upon by taking the trapezium ABED. Hence, area Clearly, the smaller the width of the interval, the better the approximation to the precise area. We can therefore take advantage of this fact and divide the interval into two halves. If we let m denote the midpoint of the interval (a,b) ….

  15. Finding the formula Then area where h = m – a = b – m And by repeated subdivision into n trapezia (activity ?) we can easily obtain Area

  16. The trapezium rule This rule is easier to understand if we change the notation as shown in the diagram This is what we mean when we talk about the trapezium rule. Area where

  17. How it works in practice Click here to see the trapezium rule in practice Click here to see a comparison of numerical methods

  18. An example Let’s evaluate using the trapezium rule with n = 4 divisions. Integral 1.4463 This estimate is an over-estimate of the precise area 1.442695…in this particular case.(why ?)

  19. Tip of the day It is sensible to tabulate the working in order that any human errors in the calculation can be minimised. e.g. Using six equally spaced ordinates with the trapezium rule, find an estimate for Now, 6 ordinates means 5 trapezia so n = 5, a = 1, b = 6 and We let and form the table:

  20. Table method

  21. Final calculation Hence, Area 0.88213 + 5.70274 ) = 3.2924 Click here to practice questions on the trapezium rule Click here for more practice

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