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- 1. Section 4.7 Antiderivatives V63.0121.041, Calculus I New York University November 29, 2010 Announcements Quiz 5 in recitation this week on §§4.1–4.4 . . . . . .
- 2. . . . . . . Announcements Quiz 5 in recitation this week on §§4.1–4.4 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 2 / 35
- 3. . . . . . . Objectives Given a ”simple“ elementary function, find a function whose derivative is that function. Remember that a function whose derivative is zero along an interval must be zero along that interval. Solve problems involving rectilinear motion. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 3 / 35
- 4. . . . . . . Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Antiderivatives of piecewise functions Finding Antiderivatives Graphically Rectilinear motion V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 4 / 35
- 5. . . . . . . What is an antiderivative? Definition Let f be a function. An antiderivative for f is a function F such that F′ = f. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 5 / 35
- 6. . . . . . . Hard problem, easy check Example Find an antiderivative for f(x) = ln x. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
- 7. . . . . . . Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
- 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? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
- 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 dx (x ln x − x) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
- 10. . . . . . . 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 dx (x ln x − x) = 1 · ln x + x · 1 x − 1 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
- 11. . . . . . . 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 dx (x ln x − x) = 1 · ln x + x · 1 x − 1 = ln x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
- 12. . . . . . . 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 dx (x ln x − x) = 1 · ln x + x · 1 x − 1 = ln x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 35
- 13. . . . . . . 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) y − x = f′ (z) =⇒ f(y) = f(x) + f′ (z)(y − x) But f′ (z) = 0, so f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 7 / 35
- 14. . . . . . . When two functions have the same derivative 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) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 8 / 35
- 15. . . . . . . Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Antiderivatives of piecewise functions Finding Antiderivatives Graphically Rectilinear motion V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 9 / 35
- 16. . . . . . . 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 . .. x . y . f(x) = x2 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35
- 17. . . . . . . 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 . .. x . y . f(x) = x2 . f′ (x) = 2x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35
- 18. . . . . . . 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 . .. x . y . f(x) = x2 . f′ (x) = 2x . F(x) = ? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35
- 19. . . . . . . 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! .. x . y . f(x) = x2 . f′ (x) = 2x . F(x) = ? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 35
- 20. . . . . . . Example Find an antiderivative for the function f(x) = x3 . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 21. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 22. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 23. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . r − 1 = 3 =⇒ r = 4 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 24. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = 1 4 . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 25. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = 1 4 . So F(x) = 1 4 x4 is an antiderivative. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 26. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = 1 4 . So F(x) = 1 4 x4 is an antiderivative. Check: d dx ( 1 4 x4 ) = 4 · 1 4 x4−1 = x3 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 27. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = 1 4 . So F(x) = 1 4 x4 is an antiderivative. Check: d dx ( 1 4 x4 ) = 4 · 1 4 x4−1 = x3 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 28. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = 1 4 . So F(x) = 1 4 x4 is an antiderivative. Check: d dx ( 1 4 x4 ) = 4 · 1 4 x4−1 = x3 Any others? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 29. . . . . . . Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , so we want arxr−1 = x3 . r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = 1 4 . So F(x) = 1 4 x4 is an antiderivative. Check: d dx ( 1 4 x4 ) = 4 · 1 4 x4−1 = x3 Any others? Yes, F(x) = 1 4 x4 + C is the most general form. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 35
- 30. . . . . . . Extrapolating to general power functions Fact (The Power Rule for antiderivatives) If f(x) = xr , then F(x) = 1 r + 1 xr+1 is an antiderivative for f… V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35
- 31. . . . . . . Extrapolating to general power functions Fact (The Power Rule for antiderivatives) If f(x) = xr , then F(x) = 1 r + 1 xr+1 is an antiderivative for f as long as r ̸= −1. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35
- 32. . . . . . . Extrapolating to general power functions Fact (The Power Rule for antiderivatives) If f(x) = xr , then F(x) = 1 r + 1 xr+1 is an antiderivative for f as long as r ̸= −1. Fact If f(x) = x−1 = 1 x , then F(x) = ln |x| + C is an antiderivative for f. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 35
- 33. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 34. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 35. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 36. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) = 1 x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 37. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) = 1 x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 38. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) = 1 x If x 0, d dx ln |x| V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 39. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) = 1 x If x 0, d dx ln |x| = d dx ln(−x) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 40. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) = 1 x If x 0, d dx ln |x| = d dx ln(−x) = 1 −x · (−1) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 41. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) = 1 x If x 0, d dx ln |x| = d dx ln(−x) = 1 −x · (−1) = 1 x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 42. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) = 1 x If x 0, d dx ln |x| = d dx ln(−x) = 1 −x · (−1) = 1 x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 43. . . . . . . What's with the absolute value? F(x) = ln |x| = { ln(x) if x 0; ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. If x 0, d dx ln |x| = d dx ln(x) = 1 x If x 0, d dx ln |x| = d dx ln(−x) = 1 −x · (−1) = 1 x We prefer the antiderivative with the larger domain. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 35
- 44. . . . . . . Graph of ln |x| .. x. y . f(x) = 1/x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35
- 45. . . . . . . Graph of ln |x| .. x. y . f(x) = 1/x. F(x) = ln(x) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35
- 46. . . . . . . Graph of ln |x| .. x. y . f(x) = 1/x. F(x) = ln |x| V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 35
- 47. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 15 / 35
- 48. . . . . . . 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 Or, if F′ = f, (cF)′ = cF′ = cf V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 15 / 35
- 49. . . . . . . Antiderivatives of Polynomials .. Example Find an antiderivative for f(x) = 16x + 5. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35
- 50. . . . . . . Antiderivatives of Polynomials .. Example Find an antiderivative for f(x) = 16x + 5. Solution The expression 1 2 x2 is an antiderivative for x, and x is an antiderivative for 1. So F(x) = 16 · ( 1 2 x2 ) + 5 · x + C = 8x2 + 5x + C is the antiderivative of f. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35
- 51. . . . . . . Antiderivatives of Polynomials .. Example Find an antiderivative for f(x) = 16x + 5. Solution The expression 1 2 x2 is an antiderivative for x, and x is an antiderivative for 1. So F(x) = 16 · ( 1 2 x2 ) + 5 · x + C = 8x2 + 5x + C is the antiderivative of f. Question Do we need two C’s or just one? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35
- 52. . . . . . . Antiderivatives of Polynomials .. Example Find an antiderivative for f(x) = 16x + 5. Solution The expression 1 2 x2 is an antiderivative for x, and x is an antiderivative for 1. So F(x) = 16 · ( 1 2 x2 ) + 5 · x + C = 8x2 + 5x + C is the antiderivative of f. Question Do we need two C’s or just one? Answer Just one. A combination of two arbitrary constants is still an arbitrary constant. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 35
- 53. . . . . . . Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35
- 54. . . . . . . Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . Accordingly, Fact If f(x) = ax , then F(x) = 1 ln a ax + C is the antiderivative of f. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35
- 55. . . . . . . Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . Accordingly, Fact If f(x) = ax , then F(x) = 1 ln a ax + C is the antiderivative of f. Proof. Check it yourself. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35
- 56. . . . . . . Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . Accordingly, Fact If f(x) = ax , then F(x) = 1 ln a ax + C is the antiderivative of f. Proof. Check it yourself. In particular, Fact If f(x) = ex , then F(x) = ex + C is the antiderivative of f. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 35
- 57. . . . . . . Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35
- 58. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35
- 59. . . . . . . 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. However, using the fact that loga x = ln x ln a , we get: Fact If f(x) = loga(x) F(x) = 1 ln a (x ln x − x) + C = x loga x − 1 ln a x + C is the antiderivative of f(x). V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 35
- 60. . . . . . . Trigonometric functions Fact d dx sin x = cos x d dx cos x = − sin x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35
- 61. . . . . . . Trigonometric functions Fact d dx sin x = cos x d dx cos x = − sin x So to turn these around, Fact The function F(x) = − cos x + C is the antiderivative of f(x) = sin x. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35
- 62. . . . . . . Trigonometric functions Fact d dx sin x = cos x d dx cos x = − sin x 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 35
- 63. . . . . . . More Trig Example Find an antiderivative of f(x) = tan x. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
- 64. . . . . . . More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
- 65. . . . . . . More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
- 66. . . . . . . More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d dx = 1 sec x · d dx sec x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
- 67. . . . . . . More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d dx = 1 sec x · d dx sec x = 1 sec x · sec x tan x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
- 68. . . . . . . More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d dx = 1 sec x · d dx sec x = 1 sec x · sec x tan x = tan x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
- 69. . . . . . . More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d dx = 1 sec x · d dx sec x = 1 sec x · sec x tan x = tan x V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
- 70. . . . . . . More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d dx = 1 sec x · d dx sec x = 1 sec x · sec x tan x = tan x More about this later. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 35
- 71. . . . . . . Antiderivatives of piecewise functions Example Let f(x) = { x if 0 ≤ x ≤ 1; 1 − x2 if 1 x. Find the antiderivative of f with F(0) = 1. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 21 / 35
- 72. . . . . . . Antiderivatives of piecewise functions Example Let f(x) = { x if 0 ≤ x ≤ 1; 1 − x2 if 1 x. Find the antiderivative of f with F(0) = 1. Solution We can antidifferentiate each piece: F(x) = 1 2 x2 + C1 if 0 ≤ x ≤ 1; x − 1 3 x3 + C2 if 1 x. The constants need to be chosen so that F(0) = 1 and F is continuous (at 1). V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 21 / 35
- 73. . . . . . . F(x) = 1 2 x2 + C1 if 0 ≤ x ≤ 1; x − 1 3 x3 + C2 if 1 x. Note F(0) = 1 2 02 + C1 = C1 =⇒ C1 = 1 This means lim x→1− F(x) = 1 2 12 + 1 = 3 2 . Now lim x→1+ F(x) = 1 − 1 3 + C2 = 2 3 + C2 So for F to be continuous we need 3 2 = 2 3 + C2 =⇒ C2 = 5 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 22 / 35
- 74. . . . . . . Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Antiderivatives of piecewise functions Finding Antiderivatives Graphically Rectilinear motion V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 23 / 35
- 75. . . . . . . Finding Antiderivatives Graphically Problem Below is the graph of a function f. Draw the graph of an antiderivative for f. .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... y = f(x) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 24 / 35
- 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 80. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 81. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 82. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 83. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 84. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 85. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 86. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 87. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 88. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 89. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 90. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 91. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 92. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 93. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 94. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 95. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 96. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 97. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 98. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 99. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . IP . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 100. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . IP . IP . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 101. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . IP . IP . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 102. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . IP . IP . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 .. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 103. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . IP . IP . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 104. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . IP . IP . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 .... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 105. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . IP . IP . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ..... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 106. . . . . . . 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 F: .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... .. f = F′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . +. +. −. −. +. ↗ . ↗ . ↘ . ↘ . ↗ . max . min . f′ = F′′ . F .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 . ++ . −− . −− . ++ . ++ . ⌣ . ⌢ . ⌢ . ⌣ . ⌣ . IP . IP . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... ? . ? . ? . ? . ? . ? The only question left is: What are the function values? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 35
- 107. . . . . . . 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. Solution .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 108. . . . . . . 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. Solution We start with F(1) = 0. .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 109. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 110. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min .. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 111. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min .. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 112. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min ... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 113. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min ... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 114. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min .... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 115. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min .... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 116. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min ..... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 117. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min ...... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 118. . . . . . . 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. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity It’s harder to tell if/when F crosses the axis; more about that later. .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f . F . shape .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ...... IP . max . IP . min ...... V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 35
- 119. . . . . . . Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Antiderivatives of piecewise functions Finding Antiderivatives Graphically Rectilinear motion V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 27 / 35
- 120. . . . . . . Say what? “Rectilinear 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 28 / 35
- 121. . . . . . . Application: Dead Reckoning V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 29 / 35
- 122. . . . . . . Application: Dead Reckoning V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 29 / 35
- 123. . . . . . . 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). V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35
- 124. . . . . . . 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 a constant acceleration. So a(t) = a = F m . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35
- 125. . . . . . . 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 a constant acceleration. So a(t) = a = F 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35
- 126. . . . . . . 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 a constant acceleration. So a(t) = a = F 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 s(t) = 1 2 at2 + v0t + C = 1 2 at2 + v0t + s0 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 35
- 127. . . . . . . An earlier Hatsumon Example Drop a ball off the roof of the Silver Center. What is its velocity when it hits the ground? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 31 / 35
- 128. . . . . . . An earlier Hatsumon 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 31 / 35
- 129. . . . . . . Finding initial velocity from stopping distance 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? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 35
- 130. . . . . . . Finding initial velocity from stopping distance 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) While braking, the car has acceleration a(t) = −20 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 35
- 131. . . . . . . Finding initial velocity from stopping distance 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) While braking, the car has acceleration a(t) = −20 Measure time 0 and position 0 when the car starts braking. So s(0) = 0. The car stops at time some t1, when v(t1) = 0. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 35
- 132. . . . . . . Finding initial velocity from stopping distance 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) While braking, the car has acceleration a(t) = −20 Measure time 0 and position 0 when the car starts braking. So s(0) = 0. The car stops at time some t1, when v(t1) = 0. We know that when s(t1) = 160. We want to know v(0), or v0. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 35
- 133. . . . . . . Implementing the Solution In general, s(t) = s0 + v0t + 1 2 at2 Since s0 = 0 and a = −20, we have s(t) = v0t − 10t2 v(t) = v0 − 20t for all t. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 33 / 35
- 134. . . . . . . Implementing the Solution In general, s(t) = s0 + v0t + 1 2 at2 Since s0 = 0 and a = −20, we have s(t) = v0t − 10t2 v(t) = v0 − 20t for all t. Plugging in t = t1, 160 = v0t1 − 10t2 1 0 = v0 − 20t1 We need to solve these two equations. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 33 / 35
- 135. . . . . . . Solving We have v0t1 − 10t2 1 = 160 v0 − 20t1 = 0 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 34 / 35
- 136. . . . . . . Solving We have v0t1 − 10t2 1 = 160 v0 − 20t1 = 0 The second gives t1 = v0/20, so substitute into the first: v0 · v0 20 − 10 ( v0 20 )2 = 160 or v2 0 20 − 10v2 0 400 = 160 2v2 0 − v2 0 = 160 · 40 = 6400 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 34 / 35
- 137. . . . . . . Solving We have v0t1 − 10t2 1 = 160 v0 − 20t1 = 0 The second gives t1 = v0/20, so substitute into the first: v0 · v0 20 − 10 ( v0 20 )2 = 160 or v2 0 20 − 10v2 0 400 = 160 2v2 0 − v2 0 = 160 · 40 = 6400 So v0 = 80 ft/s ≈ 55 mi/hr V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 34 / 35
- 138. . . . . . . Summary Antiderivatives are a useful concept, especially in motion We can graph an antiderivative from the graph of a function We can compute antiderivatives, but not always .. x . y .. 1 .. 2 .. 3 .. 4 .. 5 .. 6 ....... f ....... F f(x) = e−x2 f′ (x) = ??? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 35 / 35

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