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- 1. Section 4.7 Antiderivatives V63.0121.002.2010Su, Calculus I New York University June 16, 2010Announcements Quiz 4 Thursday on 4.1–4.4 . . . . . .
- 2. Announcements Quiz 4 Thursday on 4.1–4.4 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 2 / 33
- 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 3 / 33
- 4. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 4 / 33
- 5. What is an antiderivative? Definition Let f be a function. An antiderivative for f is a function F such that F′ = f. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 5 / 33
- 6. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
- 7. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
- 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? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
- 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 (x ln x − x) dx . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
- 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 1 (x ln x − x) = 1 · ln x + x · − 1 dx x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
- 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 1 (x ln x − x) = 1 · ln x + x · − 1 = ln x dx x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
- 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 1 (x ln x − x) = 1 · ln x + x · − 1 = ln x x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 6 / 33
- 13. Why the MVT is the MITCMost 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) =⇒ f(y) = f(x) + f′ (z)(y − x) 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 7 / 33
- 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) . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 8 / 33
- 15. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 9 / 33
- 16. Antiderivatives of power functions y . .(x) = x2 f 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 . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 10 / 33
- 17. Antiderivatives of power functions ′ y f . . (x) = 2x .(x) = x2 f 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 . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 10 / 33
- 18. Antiderivatives of power functions ′ y f . . (x) = 2x .(x) = x2 f Recall that the derivative of a power function is a power function. F . (x) = ? Fact (The Power Rule) If f(x) = xr , then f′ (x) = rxr−1 . . x . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 10 / 33
- 19. Antiderivatives of power functions ′ y f . . (x) = 2x .(x) = x2 f Recall that the derivative of a power function is a power function. F . (x) = ? Fact (The Power Rule) If f(x) = xr , then f′ (x) = rxr−1 . So in looking for antiderivatives . of power functions, try power x . functions! . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 10 / 33
- 20. Example Find an antiderivative for the function f(x) = x3 . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 21. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 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 . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 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 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 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 . 1 r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 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 . 1 r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 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 . 1 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 · x4−1 = x3 dx 4 4 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 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 . 1 r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 Check: ( ) d 1 4 dx 4 1 x = 4 · x4−1 = x3 4 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 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 . 1 r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 Check: ( ) d 1 4 dx 4 1 x = 4 · x4−1 = x3 4 Any others? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 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 . 1 r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 Check: ( ) d 1 4 dx 4 1 x = 4 · x4−1 = x3 4 1 4 Any others? Yes, F(x) = x + C is the most general form. 4 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 11 / 33
- 30. Fact (The Power Rule for antiderivatives) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an antiderivative for f… . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 12 / 33
- 31. 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 12 / 33
- 32. 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 12 / 33
- 33. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = ln(−x) if x 0. The domain of F is all nonzero numbers, while ln x is only defined on positive numbers. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 34. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 ln |x| dx . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 35. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 d ln |x| = ln(x) dx dx . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 36. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 d 1 ln |x| = ln(x) = dx dx x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 37. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 38. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 ln |x| dx . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 39. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 d ln |x| = ln(−x) dx dx . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 40. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 d 1 ln |x| = ln(−x) = · (−1) dx dx −x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 41. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 d 1 1 ln |x| = ln(−x) = · (−1) = dx dx −x x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 42. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 43. Whats with the absolute value? { ln(x) if x 0; F(x) = ln |x| = 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 13 / 33
- 44. Graph of ln |x| y . . f .(x) = 1/x x . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 14 / 33
- 45. Graph of ln |x| y . F . (x) = ln(x) . f .(x) = 1/x x . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 14 / 33
- 46. Graph of ln |x| y . . (x) = ln |x| F . f .(x) = 1/x x . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 14 / 33
- 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 15 / 33
- 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 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 15 / 33
- 49. Antiderivatives of Polynomials Example Find an antiderivative for f(x) = 16x + 5. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 16 / 33
- 50. Antiderivatives of Polynomials Example Find an antiderivative for f(x) = 16x + 5. Solution 1 2 The expression x is an antiderivative for x, and x is an antiderivative 2 for 1. So ( ) 1 2 F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C 2 is the antiderivative of f. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 16 / 33
- 51. Antiderivatives of Polynomials Example Find an antiderivative for f(x) = 16x + 5. Solution 1 2 The expression x is an antiderivative for x, and x is an antiderivative 2 for 1. So ( ) 1 2 F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C 2 is the antiderivative of f. Question Why do we not need two C’s? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 16 / 33
- 52. Antiderivatives of Polynomials Example Find an antiderivative for f(x) = 16x + 5. Solution 1 2 The expression x is an antiderivative for x, and x is an antiderivative 2 for 1. So ( ) 1 2 F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C 2 is the antiderivative of f. Question Why do we not need two C’s? Answer . . . . . . A combination of two arbitrary constants is still an arbitrary constant.16 / 33V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010
- 53. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 17 / 33
- 54. 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 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 17 / 33
- 55. 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 17 / 33
- 56. 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 17 / 33
- 57. Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 18 / 33
- 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 18 / 33
- 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. ln x However, using the fact that loga x = , we get: ln a Fact If f(x) = loga (x) 1 1 F(x) = (x ln x − x) + C = x loga x − x+C ln a ln a is the antiderivative of f(x). . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 18 / 33
- 60. Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 19 / 33
- 61. 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 19 / 33
- 62. 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 19 / 33
- 63. More Trig Example Find an antiderivative of f(x) = tan x. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
- 64. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
- 65. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
- 66. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d 1 d = · sec x dx sec x dx . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
- 67. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d 1 d 1 = · sec x = · sec x tan x dx sec x dx sec x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
- 68. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d 1 d 1 = · sec x = · sec x tan x = tan x dx sec x dx sec x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
- 69. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d dx = 1 · d sec x dx sec x = 1 sec x · sec x tan x = tan x . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
- 70. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d dx = 1 · d sec x dx sec x = 1 sec x · sec x tan x = tan x More about this later. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 20 / 33
- 71. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 21 / 33
- 72. Finding Antiderivatives Graphically 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 . . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 22 / 33
- 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 F: ′ . . . . . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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 F: ′ . .. . + . . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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 F: ′ . .. .. . + + . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − ′ . .. .. .. . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − ′ . .. .. .. .. . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2 . . . 3 . 4 . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3 . . . . . 4 . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3↘4 . . . . . . . 5 . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3↘4↘5 . . . . . . . . . 6F .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . . . . . . . . . .. . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . . . . . . . max . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + 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 .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + 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 .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + 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 .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + 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 .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
- 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: + + − − + 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 .. . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33

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