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Lesson 23: Antiderivatives

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Matthew Leingang
Matthew Leingang

This document is from a Calculus I class at New York University and covers antiderivatives. It begins with announcements about an upcoming quiz. The objectives are to find antiderivatives of simple functions, remember that a function whose derivative is zero must be constant, and solve rectilinear motion problems. It then outlines finding antiderivatives through tabulation, graphically, and with rectilinear motion examples. Examples are provided of finding the antiderivative of power functions like x^3 through identifying the power rule relationship between a function and its derivative.

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Section 4.7 Antiderivatives V63.0121.002.2010Su, Calculus I New York University June 16, 2010 Announcements 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 MITC Most Important Theorem In Calculus! Theorem Let f′ = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x y. Then f is continuous on [x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y) such that f(y) − f(x) = f′ (z) =⇒ 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
What's 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
What's 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
What's 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
What's 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
What's 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
What's 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
What's 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
What's 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
What's 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
What's 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
What's 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 / 33 V63.0121.002.2010Su, 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
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives
Lesson 23: Antiderivatives

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Lesson 23: Antiderivatives

  • 1. Section 4.7 Antiderivatives V63.0121.002.2010Su, Calculus I New York University June 16, 2010 Announcements 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 MITC Most Important Theorem In Calculus! Theorem Let f′ = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x y. Then f is continuous on [x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y) such that f(y) − f(x) = f′ (z) =⇒ 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. What's 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. What's 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. What's 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. What's 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. What's 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. What's 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. What's 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. What's 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. What's 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. What's 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. What's 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 / 33 V63.0121.002.2010Su, 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
  • 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: + + − − + 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