Integration

1. Integration

2. Introduction • This is a very big chapter which covers many ways of Integrating more difficult expressions • You are sometimes told which method to use but in addition must be able to identify the most appropriate way • We also cover using Integration to work out volumes as well as areas, as well as setting up and using equations (similar to the chapter on differentiation)

3. Teachings for Exercise 6A

4. Integration You need to be able to integrate standard functions You met the following in C3, in the differentiation chapter: 6A

5. Integration You need to be able to integrate standard functions Therefore, you already can deduce the following The modulus sign is used here to avoid potential problems with negative numbers… 6A

6. Integration As the terms are separate you can integrate them separately You need to be able to integrate standard functions Find the following integral: Rewrite this term as a power Remember the + C!

7. Integration As the terms are separate you can integrate them separately You need to be able to integrate standard functions Find the following integral: Try to rewrite as an integral you ‘know’ Remember the + C!

8. Teachings for Exercise 6B

9. Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: Consider starting with sin(2x + 3) and the answer that would give This is double what we are wanting to integrate  Therefore, we must ‘start’ with half the amount… Divide the original ‘guess’ by 2 This is a VERY common method of integration – considering what we might start with that would differentiate to our answer… 6B

10. Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: Consider starting with e4x + 1 and the answer that would give This is four times what we are wanting to integrate  Therefore, we must ‘start’ with a quarter of the amount… Divide the original ‘guess’ by 4 6B

11. Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: Consider starting with tan3x and the answer that would give This is three times what we are wanting to integrate  Therefore, we must ‘start’ with a third of the amount… Divide the original ‘guess’ by 3 6B

12. Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) These three answers illustrate a rule: 1) Integrate the function using what you know from C3 2) Divide by the coefficient of x 3) Simplify if possible and add C 6B

13. Integration 1) Integrate the function using what you know from C3 You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: 2) Divide by the coefficient of x 3) Simplify if possible and add C 6B

14. Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: Consider a function that would leave you with a power 4 and the same bracket Simplify after using the Chain rule As this is 10 times what we want, we need to divide our ‘guess’ by 10 6B

15. Teachings for Exercise 6C

16. Divide by cos Integration Subtract 1 You can use Trigonometric Identities in Integration There are some trigonometric expression you cannot integrate yet, but a way to do so is to write them in terms of ones you can integrate… Trigonometric Identities are invaluable in this! Find: Using the identity above, replace tan2x Integrate each part separately 6C

17. Rearrange Integration Divide by 2 You can use Trigonometric Identities in Integration There are some trigonometric expression you cannot integrate yet, but a way to do so is to write them in terms of ones you can integrate… Trigonometric Identities are invaluable in this! Find: Using the identity above, replace sin2x Integrate each part separately Use the ‘guessing’ method This is 4 times what we want so divide the ‘guess’ by 4 6C

18. s Double angle formula Integration s Follow the pattern… s Divide by 2 You can use Trigonometric Identities in Integration There are some trigonometric expression you cannot integrate yet, but a way to do so is to write them in terms of ones you can integrate… Trigonometric Identities are invaluable in this! Find: Replace with the above… This will give us the sin 6x when differentiating, but is negative and 12 times too big!  Make the guess negative and divide by 12! 6C

19. Divide by cos Integration Subtract 1 You can use Trigonometric Identities in Integration There are some trigonometric expression you cannot integrate yet, but a way to do so is to write them in terms of ones you can integrate… Trigonometric Identities are invaluable in this! Find: Expand the bracket Replace tan2x Simplify Integrate separately 6C

20. Teachings for Exercise 6D

21. Integration You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: Write as two fractions and make the denominators equal Combine The numerators must be equal Let x = 2 Calculate A and B by choosing appropriate x values Let x = -1 Replace A and B from the start 6D

22. Integration You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: Integrate separately You can combine the natural logarithms as a division 6D

23. Integration 1) Divide the first term by the highest power You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: 2) Multiply the answer by the whole expression you’re dividing by 3) Subtract to find the remainder 4) Remember to write the remainder as a fraction of the original expression Looks tidier! 6D

24. Integration We now need to write the remainder as partial fractions You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: Set the numerators equal and solve for A and B Let x = 2/3 Let x = -2/3 Write the final answer with the remainder broken apart! 6D

25. Integration You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: Integrate separately You can combine the natural logarithms (be careful, the negative goes on the bottom…) 6D

26. Teachings for Exercise 6E

27. Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Including using partial fractions where an expression can be factorised However, this method will not work for integrals of the form: Some expressions like this can by integrated by using the ‘standard patterns’ technique 6E

28. Integration Notice that the denominator would differentiate to become the numerator  This is a pattern we can use to figure out what the integral is… You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Remember… So imagine starting with ln|denominator| In this case, we get straight to the answer! 6E

29. Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Start by trying y = ln|denominator| Differentiate This is double what we want so multiply the ‘guess’ by 1/2 6E

30. Integration In this case consider the power of sine.  If it has been differentiated, it must have been sin3x originally… You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Write as a cubed bracket Differentiate using the chain rule Rewrite – this has given us exactly what we wanted! Don’t forget the + C! 6E

31. Integration Consider the power on the bracket  As it is a power 3, it must have been a power 4 before differentiation You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Differentiate the bracket to the power 4 using the chain rule Simplify This is 8 times too big so multiply the ‘guess’ by 1/8 Don’t forget to add C! 6E

32. Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Write using powers Imagine how we could end up with a -3 as a power... Use the chain rule Rewrite This is double what we want so multiply the ‘guess’ by 1/2 6E

33. Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Consider using a power 5 Write as a bracket to the power 5 Differentiate using the chain rule We have an extra secx • HOWEVER: • We cannot just add this to our ‘guess’ as before, as the differentiation will need to be performed using the product rule from C3, rather than the Chain rule! • We need to find another way! 6E

34. Integration You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Consider using a power 4 Write as a bracket to the power 4 Differentiate using the chain rule This is what we want, but 4/5 of the amount  Multiply by the guess by 5/4 6E

35. Teachings for Exercise 6F

36. Integration To integrate this, you need to replace the x terms with equivalent u terms, and replace the dx with an equivalent du It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: Differentiate Rearrange to find x Rearrange to get dx Replace each ‘x’ term with an equivalent ‘u’ term Rearrange – you should leave ‘du’ at the end Combine terms including the square root, changed to a power ‘1/2’ 6F

37. Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: Differentiate terms separately Flip the dividing fractions Calculate the fraction parts Finally, replace with u with its equivalent from the start! 6F

38. Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: Differentiate Rearrange to find Sinx Rearrange to get dx Replace each ‘x’ term with an equivalent ‘u’ term Cancel the Cosx terms Multiply out Integrate Replace u with x terms again! 6F

39. Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use integration by substitution to find: Sometimes you will have to decide on a substitution yourself. In this case, the bracket would be hardest to integrate so it makes sense to use the substitution: Rearrange to find x Differentiate Rearrange to get dx You also need to recalculate limits in terms of u Replace x limits with u limits and the x terms with u terms Multiply out bracket Integrate An alternative method is to replace the ‘u’ terms with x terms at the end and then just use the original ‘x’ limits – either way is fine! Sub in limits and calculate 6F

40. Teachings for Exercise 6G

41. Integration In C3 you met the following: (the product rule) You can use integration by parts to integrate some expressions This is the differential of two functions multiplied together  You could think of it as: Rearrange by subtracting vdu/dx Integrate each term with respect to x The middle term is just a differential.  Integrating a differential cancels them both out! This is the formula used for Integration by parts!  You get given this in the booklet The other terms do not cancel as only part of them are differentiated… 6G

42. Integration Unlike when using the product rule, we now have one function to differentiate, and one to integrate… You can use integration by parts to integrate some expressions Find: You can recognise that Integration by parts is needed as we have two functions multiplied together… Differentiate Integrate Now replace the relevant parts to find the integral… The integral here is simpler! Be careful with negatives here! As a general rule, it is easiest to let u = anything of the form xn. The exception is when there is a lnx term, in which case this should be used as u 6G

43. Integration You can use integration by parts to integrate some expressions Find: You can recognise that Integration by parts is needed as we have two functions multiplied together… Differentiate Integrate Now replace the relevant parts to find the integral… Simplify terms Let u be lnx! Integrate the second part 6G

44. Integration You can use integration by parts to integrate some expressions Find: You can recognise that Integration by parts is needed as we have two functions multiplied together… Sometimes you will have to use the process twice! This happens if the new integral still has two functions multiplied together… Differentiate Integrate Now replace the relevant parts to find the integral… Differentiate Integrate Work out the square bracket which is the second integration by parts Careful with negatives!! 6G

45. Integration When integrating lnx, you MUST think of it as ‘lnx times 1, and use lnx as ‘u’ and 1 as ‘dv/dx’ You can use integration by parts to integrate some expressions Evaluate: Leave your answer in terms of natural logarithms… • You will be asked to leave exact answers a lot so make sure you know your log laws!! Differentiate Integrate Now replace the relevant parts to find the integral… Simplify terms Integrate and use a square bracket with limits Sub in the limits Calculate and leave in terms of ln2 6G

46. Integration You can use integration by parts to integrate some expressions You may need the following Integrals, which you are given in the formula booklet… 6G

47. Teachings for Exercise 6H

48. Integration You can use numerical integration In C2 you saw how to estimate the area under a curve by using the trapezium rule This method can take time and is only an approximation, but it allows you to find areas of functions that otherwise can be extremely hard to Integrate  This is because using the trapezium rule actually avoids Integration altogether! n is the number of strips the area is split into! The first value for y The last value for y The central bracket will contain all the values of y in-between. The more vales there are, the more accurate the approximation is! y y0 y1 y2 y3 y4 y5 h h h h h x a b 6H

49. Integration You can use numerical integration Complete the table of values and use it to find an estimate for: • h is the height of each strip • In the table it is given by the gaps between the x values used • The y values correspond to y0, y1 etc… h h h h Be very careful here – it is easy to make an error on your calculator! π 12 π 6 π 4 π 3 x 0 y 1 1.035 1.155 1.414 2 Calculate and round the answer! y0 y1 y2 y3 y4 Calculate the values in the table by substituting the x-values into the equation above… The trapezium rule isn’t any different to in C2 really, just slightly more tricky functions to use! 6H

50. Integration You can use numerical integration Use the trapezium rule with 4 strips to find an approximation for: Sub in b, a and n (number of strips) We can then find the height of each strip! Be very careful here! x 0 0.5 1 1.5 2 y 0 0.24 0.84 1.50 1.82 Calculate and round the answer! y0 y1 y2 y3 y4 6H