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- 1. The Substitution Method Thursday, 5 December 13
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Substitution in a Definite Integral Thursday, 5 December 13
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Summary This section introduced the most commonly used integration technique, ‘substitution’, which replaces Thursday, 5 December 13
- 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. 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. 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. 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. 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. 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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