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# Lesson 23: Antiderivatives

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## Lesson 23: AntiderivativesPresentation Transcript

• Section 4.7 Antiderivatives V63.0121.002.2010Su, Calculus I New York University June 16, 2010Announcements Quiz 4 Thursday on 4.1–4.4 . . . . . .
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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
• 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 . .F 6 . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . .F 6 IP . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . .F 6 IP IP . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . .F 6 IP IP . . . . . . .. F . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . .F 6 IP IP . . . . . . .. F . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . .F 6 IP IP . . . . . . .. F . . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . .F 6 IP IP . . . . . . .. F . . . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . .F 6 IP IP . . . . . . ..F . . . . . . hape 1 . 2 . 3 . 4 . 5 . .s 6 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . .F 6 IP IP . ? .. ? .. ? .. ? .. ? .. ?F .. . . . . . . . hape 1 . 2 . 3 . 4 . 5 . .s 6 The only question left is: What are the function values? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 23 / 33
• 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 . . .. f . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .. F . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .. F . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 . . . . . 1 2 3 4 5 . . . . . 6s . . hape IP . max . IP . min . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 24 / 33
• 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 25 / 33
• 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 26 / 33
• Application: Dead Reckoning . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 27 / 33
• Application: Dead Reckoning . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 27 / 33
• 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). . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 28 / 33
• 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 F constant acceleration. So a(t) = a = . m . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 28 / 33
• 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 F 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 28 / 33
• 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 F 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 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 28 / 33
• An earlier Hatsumon Example Drop a ball off the roof of the Silver Center. What is its velocity when it hits the ground? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 29 / 33
• 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 29 / 33
• 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? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 30 / 33
• 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 breaking, the car has acceleration a(t) = −20 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 30 / 33
• 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 breaking, 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. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 30 / 33
• 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 breaking, 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 . . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 30 / 33
• Implementing the Solution In general, 1 s(t) = s0 + v0 t + at2 2 Since s0 = 0 and a = −20, we have s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 31 / 33
• Implementing the Solution In general, 1 s(t) = s0 + v0 t + at2 2 Since s0 = 0 and a = −20, we have s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. Plugging in t = t1 , 160 = v0 t1 − 10t2 1 0 = v0 − 20t1 We need to solve these two equations. . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 31 / 33
• Solving We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 32 / 33
• Solving We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the first: v0 ( v )2 v0 · − 10 0 = 160 20 20 or v2 10v2 0 − 0 = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0 . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 32 / 33
• Solving We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the first: v0 ( v )2 v0 · − 10 0 = 160 20 20 or v2 10v2 0 − 0 = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0 So v0 = 80 ft/s ≈ 55 mi/hr . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 32 / 33
• Summary Antiderivatives are a useful concept, especially in motion y . . We can graph an . . . . .. f antiderivative from the . . . . . . . graph of a function x . . . . . .. . F 1 2 3 4 5 6 We can compute antiderivatives, but not . always f(x) = e−x 2 f′ (x) = ??? . . . . . .V63.0121.002.2010Su, Calculus I (NYU) Section 4.7 Antiderivatives June 16, 2010 33 / 33