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Indefinite and Definite Integrals Using the Substitution Method

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Indefinite and Definite Integrals Using the Substitution Method

  1. 1. The Substitution Method Thursday, 5 December 13
  2. 2. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Thursday, 5 December 13
  3. 3. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example Integrate Thursday, 5 December 13 ∫ (1+ x ) 3 5 2 x dx
  4. 4. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example Integrate ∫ (1+ x ) 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . Thursday, 5 December 13
  5. 5. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x 3 Thursday, 5 December 13
  6. 6. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x du 2 = 3x dx 3 Thursday, 5 December 13
  7. 7. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x du 2 = 3x dx 2 du = 3x dx 3 Thursday, 5 December 13
  8. 8. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x du 2 = 3x dx 2 du = 3x dx du 2 = x dx 3 3 Thursday, 5 December 13
  9. 9. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x ∴∫ (1+ x 3 )5 x 2 dx du 2 = 3x dx 2 du = 3x dx du 2 = x dx 3 3 Thursday, 5 December 13
  10. 10. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x ∴∫ (1+ x 3 )5 x 2 dx du 2 = 3x dx u 2 du = 3x dx du 2 = x dx 3 3 Thursday, 5 December 13
  11. 11. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x ∴∫ (1+ x 3 )5 x 2 dx du 2 = 3x dx u 2 du du = 3x dx 3 du 2 = x dx 3 3 Thursday, 5 December 13
  12. 12. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . 5 du 3 5 2 u = 1+ x ∴∫ (1+ x ) x dx = ∫ u 3 du 2 = 3x dx u 2 du du = 3x dx 3 du 2 = x dx 3 3 Thursday, 5 December 13
  13. 13. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . 5 du 3 5 2 u = 1+ x ∴∫ (1+ x ) x dx = ∫ u 3 du 2 = 3x 1 5 = ∫ u du dx 3 u 2 du du = 3x dx 3 du 2 = x dx 3 3 Thursday, 5 December 13
  14. 14. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . 5 du 3 5 2 u = 1+ x ∴∫ (1+ x ) x dx = ∫ u 3 du 2 = 3x 1 5 = ∫ u du dx 3 u 6 2 du 1u du = 3x dx = +c 3 3 6 du 2 = x dx 3 3 Thursday, 5 December 13
  15. 15. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x ∴∫ (1+ x 3 )5 x 2 dx du 2 = 3x dx u 2 du du = 3x dx 3 du 2 = x dx 3 3 Thursday, 5 December 13 du = ∫u 3 1 5 = ∫ u du 3 6 1u = +c 3 6 6 u = +c 18 5
  16. 16. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x ∴∫ (1+ x 3 )5 x 2 dx du 2 = 3x dx u 2 du du = 3x dx 3 du 2 = x dx 3 3 Thursday, 5 December 13 du = ∫u 3 1 5 = ∫ u du 3 6 1u = +c 3 6 6 u = +c 18 5 but u = 1+ x 3
  17. 17. The Substitution Method This section describes the substitution method that changes the form of an integrand, preferably to one that we can integrate more easily. Example ∫ (1+ x ) Integrate 3 5 2 x dx Solution: 2 2 3 We notice the derivative of (1+ x ) is 3x , which differs from x in the 3 integrand only by a factor of 3. So let u = 1+ x . u = 1+ x ∴∫ (1+ x 3 )5 x 2 dx du 2 = 3x dx u 2 du du = 3x dx 3 du 2 = x dx 3 3 Thursday, 5 December 13 du 3 but u = 1+ x = ∫u 3 3 6 (1+ x ) 3 5 2 ∴ ∫ (1+ x ) x dx = +c 1 5 = ∫ u du 18 3 6 1u = +c 3 6 6 u = +c 18 5
  18. 18. Substitution in a Definite Integral Thursday, 5 December 13
  19. 19. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Thursday, 5 December 13
  20. 20. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 0 Thursday, 5 December 13 2x + 1 dx using u = 2x + 1
  21. 21. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 0 Solution: Thursday, 5 December 13 2x + 1 dx using u = 2x + 1
  22. 22. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 0 Solution: u = 2x + 1 Thursday, 5 December 13 2x + 1 dx using u = 2x + 1
  23. 23. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 0 Solution: u = 2x + 1 du =2 dx Thursday, 5 December 13 2x + 1 dx using u = 2x + 1
  24. 24. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 0 Solution: u = 2x + 1 du =2 dx du = 2dx Thursday, 5 December 13 2x + 1 dx using u = 2x + 1
  25. 25. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 2x + 1 dx using u = 2x + 1
  26. 26. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0
  27. 27. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 u =1
  28. 28. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 u =1
  29. 29. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 u =1 u=9
  30. 30. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 4 ∴∫ 2x + 1dx 0 u =1 u=9
  31. 31. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 4 ∴∫ 2x + 1dx 0 u u =1 u=9
  32. 32. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 4 ∴∫ 2x + 1dx 0 u du 2 u =1 u=9
  33. 33. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 u =1 u=9 4 9 0 1 ∴∫ 2x + 1dx = ∫ u du 2 du u 2
  34. 34. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 u =1 u=9 4 9 0 1 du u 2 ∴∫ 2x + 1dx = ∫ 9 u du 2 1 2 1 = ∫ u du 21
  35. 35. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 u =1 u=9 4 9 0 1 du u 2 ∴∫ 2x + 1dx = ∫ 9 u du 2 1 2 1 = ∫ u du 21 9 1 ⎡2 ⎤ = ⎢ u ⎥ 2 ⎣3 ⎦1 3 2
  36. 36. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 u =1 u=9 4 9 0 1 du u 2 ∴∫ 2x + 1dx = ∫ 9 u du 2 ⎞ 1⎛ = ⎜ 9 −1 ⎟ 3⎝ ⎠ 3 2 1 2 1 = ∫ u du 21 9 1 ⎡2 ⎤ = ⎢ u ⎥ 2 ⎣3 ⎦1 3 2 3 2
  37. 37. Substitution in a Definite Integral In evaluating a definite integral by use of change of variable, there must be a corresponding change in the limits of integration. Example Evaluate 4 ∫ 2x + 1 dx using u = 2x + 1 0 Solution: u = 2x + 1 du =2 dx du = 2dx du = dx 2 Thursday, 5 December 13 when x = 0 x=4 u =1 u=9 4 9 0 1 du u 2 ∴∫ 2x + 1dx = ∫ 9 u du 2 ⎞ 1⎛ = ⎜ 9 −1 ⎟ 3⎝ ⎠ 3 2 1 2 1 = ∫ u du 21 26 = 3 9 1 ⎡2 ⎤ = ⎢ u ⎥ 2 ⎣3 ⎦1 3 2 3 2
  38. 38. Summary This section introduced the most commonly used integration technique, ‘substitution’, which replaces Thursday, 5 December 13
  39. 39. Summary This section introduced the most commonly used integration technique, ‘substitution’, which replaces ∫ f (x)dx Thursday, 5 December 13 with ∫ g(u)du
  40. 40. Summary This section introduced the most commonly used integration technique, ‘substitution’, which replaces ∫ f (x)dx with u(b) b and ∫ a Thursday, 5 December 13 ∫ g(u)du f (x)dx with ∫ u(a) g(u)du
  41. 41. Summary This section introduced the most commonly used integration technique, ‘substitution’, which replaces ∫ f (x)dx with u(b) b and ∫ ∫ g(u)du f (x)dx with ∫ g(u)du u(a) a It is hoped that the problem of finding ∫ g(u)du Thursday, 5 December 13
  42. 42. Summary This section introduced the most commonly used integration technique, ‘substitution’, which replaces ∫ f (x)dx with u(b) b and ∫ ∫ g(u)du f (x)dx with ∫ g(u)du u(a) a It is hoped that the problem of finding ∫ g(u)du Thursday, 5 December 13 is easier to find than
  43. 43. Summary This section introduced the most commonly used integration technique, ‘substitution’, which replaces ∫ f (x)dx with u(b) b and ∫ ∫ g(u)du f (x)dx with ∫ g(u)du u(a) a It is hoped that the problem of finding ∫ g(u)du Thursday, 5 December 13 is easier to find than ∫ f (x)dx
  44. 44. Summary This section introduced the most commonly used integration technique, ‘substitution’, which replaces ∫ f (x)dx with u(b) b and ∫ ∫ g(u)du f (x)dx with ∫ g(u)du u(a) a It is hoped that the problem of finding ∫ g(u)du is easier to find than if it isn’t, try another substitution. Thursday, 5 December 13 ∫ f (x)dx

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