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# Lesson 23: Antiderivatives

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An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.

An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.

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• 1. Section 4.7 Antiderivatives V63.0121.034, Calculus I November 18, 2009 Announcements Wednesday, November 25 is a regular class day next and last quiz will be the week after Thanksgiving (4.1–4.4, 4.7) Final Exam: Friday, December 18, 2:00–3:50pm . . Image credit: Ian Hampton . . . . . .
• 2. Why the MVT is the MITC Most Important Theorem In Calculus! Theorem Let f′ = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x < y. Then f is continuous on [x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y) such that f(y) − f(x) = f′ (z) = 0. y−x So f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant. . . . . . .
• 3. Theorem Suppose f and g are two differentiable functions on (a, b) with f′ = g′ . Then f and g differ by a constant. That is, there exists a constant C such that f(x) = g(x) + C. Proof. Let h(x) = f(x) − g(x) Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b) So h(x) = C, a constant This means f(x) − g(x) = C on (a, b) . . . . . .
• 4. Objectives Given an expression for function f, ﬁnd a differentiable function F such that F′ = f (F is called an antiderivative for f). Given the graph of a function f, ﬁnd a differentiable function F such that F′ = f Use antiderivatives to solve problems in rectilinear motion . . . . . .
• 5. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. . . . . . .
• 6. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? . . . . . .
• 7. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? Example is F(x) = x ln x − x an antiderivative for f(x) = ln x? . . . . . .
• 8. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? Example is F(x) = x ln x − x an antiderivative for f(x) = ln x? Solution d 1 (x ln x − x) = 1 · ln x + x · − 1 dx x = ln x . . . . . .
• 9. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? Example is F(x) = x ln x − x an antiderivative for f(x) = ln x? Solution d 1 (x ln x − x) = 1 · ln x + x · − 1 dx x = ln x Yes! . . . . . .
• 10. Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
• 11. Antiderivatives of power functions Recall that the derivative of a power function is a power function. Fact The Power Rule If f(x) = xr , then f′ (x) = rxr−1 . . . . . . .
• 12. Antiderivatives of power functions Recall that the derivative of a power function is a power function. Fact The Power Rule If f(x) = xr , then f′ (x) = rxr−1 . So in looking for antiderivatives of power functions, try power functions! . . . . . .
• 13. Example Find an antiderivative for the function f(x) = x3 . . . . . . .
• 14. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr . . . . . .
• 15. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . . . . . . .
• 16. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 . . . . . .
• 17. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 . . . . . .
• 18. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 Check: ( ) d 1 4 1 x = 4 · x 4 −1 = x 3 dx 4 4 . . . . . .
• 19. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 Check: ( ) d 1 4 1 x = 4 · x 4 −1 = x 3 dx 4 4 Any others? . . . . . .
• 20. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 Check: ( ) d 1 4 1 x = 4 · x 4 −1 = x 3 dx 4 4 1 4 Any others? Yes, F(x) = x + C is the most general form. 4 . . . . . .
• 21. Fact (The Power Rule for antiderivatives) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an antiderivative for f… . . . . . .
• 22. Fact (The Power Rule for antiderivatives) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an antiderivative for f as long as r ̸= −1. . . . . . .
• 23. Fact (The Power Rule for antiderivatives) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an antiderivative for f as long as r ̸= −1. Fact 1 If f(x) = x−1 = , then x F(x) = ln |x| + C is an antiderivative for f. . . . . . .
• 24. What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only deﬁned on positive numbers. . . . . . .
• 25. What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only deﬁned on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) . . . . . .
• 26. What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only deﬁned on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) For negative numbers d d 1 1 ln |x| = ln(−x) = · (−1) = dx dx −x x . . . . . .
• 27. What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only deﬁned on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) For negative numbers d d 1 1 ln |x| = ln(−x) = · (−1) = dx dx −x x We prefer the antiderivative with the larger domain. . . . . . .
• 28. Graph of ln |x| y . . f . (x ) = 1 /x x . . . . . . .
• 29. Graph of ln |x| y . . (x) = ln |x| F . f . (x ) = 1 /x x . . . . . . .
• 30. Graph of ln |x| y . . (x) = ln |x| F . f . (x ) = 1 /x x . . (x) = ln |x| F . . . . . .
• 31. Combinations of antiderivatives Fact (Sum and Constant Multiple Rule for Antiderivatives) If F is an antiderivative of f and G is an antiderivative of g, then F + G is an antiderivative of f + g. If F is an antiderivative of f and c is a constant, then cF is an antiderivative of cf. . . . . . .
• 32. Combinations of antiderivatives Fact (Sum and Constant Multiple Rule for Antiderivatives) If F is an antiderivative of f and G is an antiderivative of g, then F + G is an antiderivative of f + g. If F is an antiderivative of f and c is a constant, then cF is an antiderivative of cf. Proof. These follow from the sum and constant multiple rule for derivatives: If F′ = f and G′ = g, then (F + G)′ = F′ + G′ = f + g Again, if F′ = f, (cF)′ = cF′ = cf . . . . . .
• 33. Example Find an antiderivative for f(x) = 16x + 5 . . . . . .
• 34. Example Find an antiderivative for f(x) = 16x + 5 Solution The expression 8x2 is an antiderivative for 16x, and 5x is an antiderivative for 5. So F(x) = 8x2 + 5x + C is the antiderivative of f. . . . . . .
• 35. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . . . . . . .
• 36. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . Accordingly, Fact 1 x If f(x) = ax , then F(x) = a + C is the antiderivative of f. ln a . . . . . .
• 37. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . Accordingly, Fact 1 x If f(x) = ax , then F(x) = a + C is the antiderivative of f. ln a Proof. Check it yourself. . . . . . .
• 38. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . Accordingly, Fact 1 x If f(x) = ax , then F(x) = a + C is the antiderivative of f. ln a Proof. Check it yourself. In particular, Fact If f(x) = ex , then F(x) = ex + C is the antiderivative of F. . . . . . .
• 39. Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. . . . . . .
• 40. Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. This is not obvious. See Calc II for the full story. . . . . . .
• 41. Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. This is not obvious. See Calc II for the full story. ln x However, using the fact that loga x = , we get that ln a 1 F(x) = (x ln x − x) + C ln a is the antiderivative of f(x) = loga (x). . . . . . .
• 42. Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx . . . . . .
• 43. Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx So to turn these around, Fact The function F(x) = − cos x + C is the antiderivative of f(x) = sin x. . . . . . .
• 44. Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx So to turn these around, Fact The function F(x) = − cos x + C is the antiderivative of f(x) = sin x. The function F(x) = sin x + C is the antiderivative of f(x) = cos x. . . . . . .
• 45. Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
• 46. Problem Below is the graph of a function f. Draw the graph of an antiderivative for F. y . . . . . = f(x) y . . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6 . . . . . . . .
• 47. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: ′ . . . . . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 48. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: ′ . .. . + . . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 49. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: ′ . .. .. . + + . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 50. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − ′ . .. .. .. . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 51. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − ′ . .. .. .. .. . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 52. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 53. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2 . . . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 54. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3 . . . . . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 55. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3↘4 . . . . . . . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 56. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3↘4↘5 . . . . . . . . . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 57. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . . . . . . . . . .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 58. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . . . . . . . max . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 59. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
• 60. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . . . . . . f′ .. = F ′′ . . . . . . . 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 61. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . .. + . + . . . f′ .. = F ′′ . . . . . . . 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 62. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . .. + .. − . + − . . f′ .. = F ′′ . . . . . . . 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 63. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . .. + .. − .. − . + − − . f′ .. = F ′′ . . . . . . . 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 64. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . .. + .. − .. − .. + . + − − + f′ .. = F ′′ . . . . . . . 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 65. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 66. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F . ⌣ 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 67. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 68. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 69. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
• 70. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . .F 6 . . . . . . .
• 71. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 . 3 . 4 . 5 . .F 6 IP . . . . . . .
• 72. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . .
• 73. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . F .. 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
• 74. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . F .. . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
• 75. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . F .. . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
• 76. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . F .. . . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
• 77. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . F .. . . . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
• 78. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . F .. . . . . . . hape 1 . 2 . 3 . 4 . 5 . .s 6 . . . . . .
• 79. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to ﬁnd the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . ? .. ? .. ? .. ? .. ? .. ?F .. . . . . . . . hape 1 . 2 . 3 . 4 . 5 . .s 6 The only question left is: What are the function values? . . . . . .
• 80. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. y . . Solution . . We start with F(1) = 0. . . .. f . . . . . . . Using the sign chart, we x . draw arcs with the . . . . .. . 1 2 3 4 5 6 speciﬁed monotonicity and concavity . . . . . . F .. It’s harder to tell if/when . . . . . F crosses the axis; more 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . about that later. . . . . . .
• 81. Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
• 82. Say what? “Rectlinear motion” just means motion along a line. Often we are given information about the velocity or acceleration of a moving particle and we want to know the equations of motion. . . . . . .
• 83. Example: Dead Reckoning . . . . . .
• 84. Problem Suppose a particle of mass m is acted upon by a constant force F. Find the position function s(t), the velocity function v(t), and the acceleration function a(t). . . . . . .
• 85. Problem Suppose a particle of mass m is acted upon by a constant force F. Find the position function s(t), the velocity function v(t), and the acceleration function a(t). Solution By Newton’s Second Law (F = ma) a constant force induces F a constant acceleration. So a(t) = a = . m . . . . . .
• 86. Problem Suppose a particle of mass m is acted upon by a constant force F. Find the position function s(t), the velocity function v(t), and the acceleration function a(t). Solution By Newton’s Second Law (F = ma) a constant force induces F a constant acceleration. So a(t) = a = . m Since v′ (t) = a(t), v(t) must be an antiderivative of the constant function a. So v(t) = at + C = at + v0 where v0 is the initial velocity. . . . . . .
• 87. Problem Suppose a particle of mass m is acted upon by a constant force F. Find the position function s(t), the velocity function v(t), and the acceleration function a(t). Solution By Newton’s Second Law (F = ma) a constant force induces F a constant acceleration. So a(t) = a = . m Since v′ (t) = a(t), v(t) must be an antiderivative of the constant function a. So v(t) = at + C = at + v0 where v0 is the initial velocity. Since s′ (t) = v(t), s(t) must be an antiderivative of v(t), meaning 1 2 1 s(t) = at + v0 t + C = at2 + v0 t + s0 2 2 . . . . . .
• 88. Example Drop a ball off the roof of the Silver Center. What is its velocity when it hits the ground? . . . . . .
• 89. Example Drop a ball off the roof of the Silver Center. What is its velocity when it hits the ground? Solution Assume s0 = 100 m, and v0 = 0. Approximate a = g ≈ −10. Then s(t) = 100 − 5t2 √ √ So s(t) = 0 when t = 20 = 2 5. Then v(t) = −10t, √ √ so the velocity at impact is v(2 5) = −20 5 m/s. . . . . . .
• 90. Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 ft before it came to a stop. Suppose that the car in question has a constant deceleration of 20 ft/s2 under the conditions of the skid. How fast was the car traveling when its brakes were ﬁrst applied? . . . . . .
• 91. Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 ft before it came to a stop. Suppose that the car in question has a constant deceleration of 20 ft/s2 under the conditions of the skid. How fast was the car traveling when its brakes were ﬁrst applied? Solution (Setup) We know that the car is decelerated by a(t) = −20 We know that when s(t) = 160, v(t) = 0. . . . . . .
• 92. Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 ft before it came to a stop. Suppose that the car in question has a constant deceleration of 20 ft/s2 under the conditions of the skid. How fast was the car traveling when its brakes were ﬁrst applied? Solution (Setup) We know that the car is decelerated by a(t) = −20 We know that when s(t) = 160, v(t) = 0. We want to know v(0) = v0 . . . . . . .
• 93. Solution (Implementation) 1 2 In general, s(t) = s0 + v0 t + at , so we have 2 s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. . . . . . .
• 94. Solution (Implementation) 1 2 In general, s(t) = s0 + v0 t + at , so we have 2 s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. If t1 is the time it took for the car to stop, 160 = v0 t1 − 10t2 1 0 = v0 − 20t1 We need to solve these two equations. . . . . . .
• 95. We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 . . . . . .
• 96. We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the ﬁrst: v0 ( v )2 0 v0 · − 10 = 160 20 20 or v2 0 10v2 0 − = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0 . . . . . .
• 97. We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the ﬁrst: v0 ( v )2 0 v0 · − 10 = 160 20 20 or v2 0 10v2 0 − = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0 So v0 = 80 ft/s ≈ 55 mi/hr . . . . . .