1. The FTC Part 2, Total Change/Area & U-Sub

2. Question from Test 1 Liquid drains into a tank at the rate 21e -3t units per minute. If the tank starts empty and can hold 6 units, at what time will it overflow? A. log(7)/3 B. (1/3)log(13/7) C. 3 log (13/7) D. 3log(7) E. Never

3. Question from Test 1 Liquid drains into a tank at the rate 21e -3t units per minute. If the tank starts empty and can hold 6 units, at what time will it overflow? A. log(7)/3

4. The Total Change Theorem The integral of a rate of change is the total change from a to b. (displacement) (still from last weeks notes)

5. The Total Change Theorem Ex: Given Find the displacement and total distance traveled from time 1 to time 6. Displacement: (negative area takes away from positive) Total Distance: (all area counted positive)

6. Total Area Find the area of the region bounded by the x-axis, y-axis and y = 2 – 2x. First find the bounds by setting 2 – 2x = 0 and by substitution 0 in for x

7. Total Area Ex. Find the area of the region bounded by the y-axis and the curve

8. Fundamental Theorem of Calculus (Part 1) (Chain Rule) If f is continuous on [a, b], then the function defined by is continuous on [a, b] and differentiable on (a, b) and

9. Fundamental Theorem of Calculus (Part 1)

10. Fundamental Theorem of Calculus (Part 2) If f is continuous on [a, b], then : Where F is any antiderivative of f. ( ) Helps us to more easily evaluate Definite Integrals in the same way we calculate the Indefinite!

11. Example

12. We have to find an antiderivative; evaluate at 3; evaluate at 2; subtract the results.

13. Example

14. This notation means: evaluate the function at 3 and 2, and subtract the results.

15. Example

18. Don’t need to include “+ C” in our antiderivative, because any antiderivative will work.

19. Example the “C’s” will cancel each other out.

20. Example

22. Alternate notation

23. Example

25. = –1

26. Example = –1= 1

27. Example

30. Given: Write a similar expression for the continuous function:

31. Fundamental Theorem of Calculus (Part 2) If f is continuous on [a, b], then : Where F is any antiderivative of f. ( )

32. Evaluate: Multiply out: Use FTC 2 to Evaluate:

33. What if instead? It would be tedious to use the same multiplication strategy! There is a better way! We’ll use the chain rule (backwards)

34. Chain Rule for Derivatives: Chain Rule backwards for Integration:

35. Look for:

36. Back to Our Example Let

37. Our Example as an Indefinite Integral With AND Without worrying about the bounds for now: Back to x (Indefinite):

38. The same substitution holds for the higher power! With Back to x (Indefinite):

39. Our Original Example of a Definite Integral: To make the substitution complete for a Definite Integral: We make a change of bounds using: When x = -1, u = 2(-1)+1 = -1 When x = 2, u = 2(2) + 1 = 5 The x-interval [-1,2] is transformed to the u-interval [-1, 5]

40. Substitution Rule for Indefinite Integrals If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then Substitution Rule for Definite Integrals If g’(x) is continuous on [a,b] and f is continuous on the range of u = g(x), then

41. Using the Chain Rule, we know that: Evaluate:

42. Looks almost like cos(x 2 ) 2x, which is the derivative of sin(x 2 ). Using the Chain Rule, we know that:

43. Evaluate: We will rearrange the integral to get an exact match: Using the Chain Rule, we know that:

44. Evaluate: We will rearrange the integral to get an exact match: Using the Chain Rule, we know that:

45. Evaluate: We will rearrange the integral to get an exact match: Using the Chain Rule, we know that:

46. Evaluate: We will rearrange the integral to get an exact match: We put in a 2 so the pattern will match. Using the Chain Rule, we know that:

47. Evaluate: We will rearrange the integral to get an exact match: We put in a 2 so the pattern will match. So we must also put in a 1/2 to keep the problem the same. Using the Chain Rule, we know that:

48. Evaluate: We will rearrange the integral to get an exact match: Using the Chain Rule, we know that:

49. Evaluate: We will rearrange the integral to get an exact match: Using the Chain Rule, we know that:

50. Check Answer:

51. Check: Check Answer:

52. Check: From the chain rule Check Answer:

53. 1)Choose u. Indefinite Integrals by Substitution

54. 1)Choose u. 2)Calculate du. Indefinite Integrals by Substitution

55. 1)Choose u. 2)Calculate du. 3)Substitute u. Arrange to have du in your integral also. (All xs and dxs must be replaced!) Indefinite Integrals by Substitution

56. 1)Choose u. 2)Calculate du. 3)Substitute u. Arrange to have du in your integral also. (All xs and dxs must be replaced!) 4)Solve the new integral. Indefinite Integrals by Substitution

57. 1)Choose u. 2)Calculate du. 3)Substitute u. Arrange to have du in your integral also. (All xs and dxs must be replaced!) 4)Solve the new integral. 5)Substitute back in to get x again.

58. Example A linear substitution: Let u = 3x + 2. Then du = 3dx.

59. Choosing u Try to choose u to be an inside function. (Think chain rule.) Try to choose u so that du is in the problem, except for a constant multiple.

60. Choosing u For u = 3x + 2 was a good choice because (1)3x + 2 is inside the exponential. (2)The derivative is 3, which is only a constant.

61. Practice Let u = du =

62. Practice Let u = x 2 + 1 du =

63. Practice Let u = x 2 + 1 du = 2x dx

64. Practice Let u = x 2 + 1 du = 2x dx

65. Practice Let u = x 2 + 1 du = 2x dx Make this a 2x dx and we’re set!

66. Practice Let u = x 2 + 1 du = 2x dx

67. Practice Let u = x 2 + 1 du = 2x dx

68. Practice Let u = x 2 + 1 du = 2x dx

69. Practice Let u = x 2 + 1 du = 2x dx

70. Practice Let u = du =

71. Practice Let u = sin(x) du =

72. Practice Let u = sin(x) du = cos(x) dx

73. Practice Let u = sin(x) du = cos(x) dx

74. Practice Let u = sin(x) du = cos(x) dx

75. Practice Let u = sin(x) du = cos(x) dx

76. Practice Let u = du = An alternate possibility:

77. Practice Let u = cos(x) du = –sin(x) dx An alternate possibility:

78. Practice Let u = cos(x) du = –sin(x) dx An alternate possibility:

79. Practice Let u = cos(x) du = –sin(x) dx An alternate possibility:

80. Practice Let u = cos(x) du = –sin(x) dx An alternate possibility:

81. Practice Note:

82. Practice Note:

83. Practice Note: What’s the difference?

84. Practice Note: What’s the difference?

85. Practice Note: What’s the difference?

86. Practice Note: What’s the difference? This is 1!

87. Practice Note: What’s the difference?

88. Practice Note: What’s the difference? That is, the difference is a constant.

89. In-Class Assignment Integrate using two different methods: 1st by multiplying out and integrating 2nd by u-substitution Do you get the same result? (Don’t just assume or claim you do; multiply out your results to show it!) If you don’t get exactly the same answer, is it a problem? Why or why not?