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Lesson 21: Antiderivatives (slides)

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Matthew Leingang

This document provides an overview of antiderivatives in calculus, focusing on definitions, examples, and important theorems related to antiderivatives of various functions. It outlines methods for finding antiderivatives, especially for power functions, and elucidates on essential properties applicable in rectilinear motion. Key topics include the relationship between derivatives and antiderivatives, and specific examples illustrate the concepts being taught.

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Section 4.7 Antiderivatives V63.0121.006/016, Calculus I New York University April 8, 2010 Announcements Quiz April 16 on §§4.1–4.4 Final Exam: Monday, May 10, 10:00am . . Image credit: Ian Hampton . . . . . .
Announcements Quiz April 16 on §§4.1–4.4 Final Exam: Monday, May 10, 10:00am . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 2 / 32
Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 3 / 32
Objectives Given an expression for function f, find a differentiable function F such that F′ = f (F is called an antiderivative for f). Given the graph of a function f, find a differentiable function F such that F′ = f Use antiderivatives to solve problems in rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 4 / 32
Hard problem, easy check Example Find an antiderivative for f(x) = ln x. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 6 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 7 / 32
Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 8 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
Example Find an antiderivative for the function f(x) = x3 . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 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… . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 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. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
Graph of ln |x| y . . f .(x) = 1/x x . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
Graph of ln |x| y . F . (x) = ln(x) . f .(x) = 1/x x . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
Graph of ln |x| y . . (x) = ln |x| F . f .(x) = 1/x x . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
Combinations of antiderivatives Fact (Sum and Constant Multiple Rule for Antiderivatives) If F is an antiderivative of f and G is an antiderivative of g, then F + G is an antiderivative of f + g. If F is an antiderivative of f and c is a constant, then cF is an antiderivative of cf. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 14 / 32
Combinations of antiderivatives Fact (Sum and Constant Multiple Rule for Antiderivatives) If F is an antiderivative of f and G is an antiderivative of g, then F + G is an antiderivative of f + g. If F is an antiderivative of f and c is a constant, then cF is an antiderivative of cf. Proof. These follow from the sum and constant multiple rule for derivatives: If F′ = f and G′ = g, then (F + G)′ = F′ + G′ = f + g Or, if F′ = f, (cF)′ = cF′ = cf . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 14 / 32
Antiderivatives of Polynomials Example Find an antiderivative for f(x) = 16x + 5. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
Antiderivatives of Polynomials Example Find an antiderivative for f(x) = 16x + 5. Solution ( ) 1 2 F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C 2 Question Why do we not need two C’s? Answer A combination of two arbitrary constants is still an arbitrary constant. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
More Trig Example Find an antiderivative of f(x) = tan x. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 20 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 21 / 32
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: ′ . . . . . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: ′ . .. . + . . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: ′ . .. .. . + + . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − ′ . .. .. .. . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − ′ . .. .. .. .. . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2 . . . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3 . . . . . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3↘4 . . . . . . . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3↘4↘5 . . . . . . . . . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . . . . . . . . . .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . . . . . . . max . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . ′ ′′ . . . . . . . . . . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . ′ ′′ . . . . . . . .. + . + . . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . ′ ′′ . . . . . . . .. + .. − . + − . . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . ′ ′′ . . . . . . . .. + .. − .. − . + − − . .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . ′ ′′ . . . . . . . .. + .. − .. − .. + . + − − + .. = F f 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F . ⌣ 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . 1 . 2 . 3 . 4 . 5 . .F 6 . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 . 3 . 4 . 5 . .F 6 IP . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . .. F 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . .. F . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . .. F . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . .. F . . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . .. F . . . . 1 . 2 . 3 . 4 . 5 . 6s . . hape . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . . . . . . ..F . . . . . . hape 1 . 2 . 3 . 4 . 5 . .s 6 . . . . . .
Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . + − − + + f′ ′′ . . . . . . . .. + .. − .. − .. + .. + . . = F ⌣ . . ⌢ . ⌢ . ⌣ . ⌣ . 1 2 3 4 5 6 . . . . . . x . .. 1 2 .. . 3 4 . 5 . .F 6 IP IP . ? .. ? .. ? .. ? .. ? .. ?F .. . . . . . . . hape 1 . 2 . 3 . 4 . 5 . .s 6 The only question left is: What are the function values? . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
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 . . . . . . .
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 . . . . . . .
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 . . . . . . .
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 . . . . . . .
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 . . . . . . .
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 . . . . . . .
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 . . . . . . .
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 . . . . . . .
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 . . . . . . .
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 . . . . . . .
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 24 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 25 / 32
Application: Dead Reckoning . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 26 / 32
Application: Dead Reckoning . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 26 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 28 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 28 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 30 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 30 / 32
Solving We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
What have we learned today? 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 32 / 32

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Lesson 21: Antiderivatives (slides)

  • 1. Section 4.7 Antiderivatives V63.0121.006/016, Calculus I New York University April 8, 2010 Announcements Quiz April 16 on §§4.1–4.4 Final Exam: Monday, May 10, 10:00am . . Image credit: Ian Hampton . . . . . .
  • 2. Announcements Quiz April 16 on §§4.1–4.4 Final Exam: Monday, May 10, 10:00am . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 2 / 32
  • 3. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 3 / 32
  • 4. Objectives Given an expression for function f, find a differentiable function F such that F′ = f (F is called an antiderivative for f). Given the graph of a function f, find a differentiable function F such that F′ = f Use antiderivatives to solve problems in rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 4 / 32
  • 5. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
  • 6. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
  • 7. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? Example is F(x) = x ln x − x an antiderivative for f(x) = ln x? . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
  • 8. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? Example is F(x) = x ln x − x an antiderivative for f(x) = ln x? Solution d (x ln x − x) dx . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
  • 9. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? Example is F(x) = x ln x − x an antiderivative for f(x) = ln x? Solution d 1 (x ln x − x) = 1 · ln x + x · − 1 dx x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
  • 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 = ln x dx x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
  • 11. Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? Example is F(x) = x ln x − x an antiderivative for f(x) = ln x? Solution d dx 1 (x ln x − x) = 1 · ln x + x · − 1 = ln x x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
  • 12. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 6 / 32
  • 13. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 7 / 32
  • 14. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 8 / 32
  • 15. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
  • 16. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
  • 17. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
  • 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 . So in looking for antiderivatives . of power functions, try power x . functions! . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
  • 19. Example Find an antiderivative for the function f(x) = x3 . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 20. Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 21. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 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 . r − 1 = 3 =⇒ r = 4 . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 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 . 1 r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 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 1 4 So F(x) = x is an antiderivative. 4 . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 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 Check: ( ) d 1 4 1 x = 4 · x4−1 = x3 dx 4 4 . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 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 dx 4 1 x = 4 · x4−1 = x3 4 . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 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 Any others? . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 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 1 4 Any others? Yes, F(x) = x + C is the most general form. 4 . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
  • 29. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 32
  • 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 as long as r ̸= −1. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 32
  • 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. Fact 1 If f(x) = x−1 = , then x F(x) = ln |x| + C is an antiderivative for f. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 32
  • 32. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 d ln |x| = ln(x) dx dx . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 1 ln |x| = ln(x) = dx dx x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 dx ln |x| = d dx ln(x) = 1 x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 If x 0, d ln |x| dx . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 d ln |x| = ln(−x) dx dx . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 1 ln |x| = ln(−x) = · (−1) dx dx −x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 1 ln |x| = ln(−x) = · (−1) = dx dx −x x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 dx ln |x| = d dx ln(−x) = 1 −x · (−1) = 1 x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 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 We prefer the antiderivative with the larger domain. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
  • 43. Graph of ln |x| y . . f .(x) = 1/x x . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
  • 44. Graph of ln |x| y . F . (x) = ln(x) . f .(x) = 1/x x . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
  • 45. Graph of ln |x| y . . (x) = ln |x| F . f .(x) = 1/x x . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
  • 46. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 14 / 32
  • 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. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 14 / 32
  • 48. Antiderivatives of Polynomials Example Find an antiderivative for f(x) = 16x + 5. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
  • 49. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
  • 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. Question Why do we not need two C’s? . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
  • 51. Antiderivatives of Polynomials Example Find an antiderivative for f(x) = 16x + 5. Solution ( ) 1 2 F(x) = 16 · x + 5 · x + C = 8x2 + 5x + C 2 Question Why do we not need two C’s? Answer A combination of two arbitrary constants is still an arbitrary constant. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
  • 52. Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
  • 53. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
  • 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 Proof. Check it yourself. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
  • 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. In particular, Fact If f(x) = ex , then F(x) = ex + C is the antiderivative of f. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
  • 56. Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
  • 57. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
  • 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. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
  • 59. Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
  • 60. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
  • 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. The function F(x) = sin x + C is the antiderivative of f(x) = cos x. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
  • 62. More Trig Example Find an antiderivative of f(x) = tan x. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
  • 63. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
  • 64. More Trig Example Find an antiderivative of f(x) = tan x. Solution ??? Answer F(x) = ln(sec x). . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
  • 65. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
  • 66. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
  • 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 = tan x dx sec x dx sec x . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
  • 68. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
  • 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 More about this later. . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
  • 70. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 20 / 32
  • 71. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 21 / 32
  • 72. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: ′ . . . . . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 73. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: ′ . .. . + . . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 74. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: ′ . .. .. . + + . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 75. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − ′ . .. .. .. . . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 76. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − ′ . .. .. .. .. . .. = F f y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 77. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 . 2 . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 78. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2 . . . 3 . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 79. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3 . . . . . 4 . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 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 for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3↘4 . . . . . . . 5 . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 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 for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1↗2↗3↘4↘5 . . . . . . . . . 6F .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 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 for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . . . . . . . . . .. . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 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 for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . . . . . . . max . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 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 for F: + + − − + f ′ . . . . . . . . . . . .. = F y . 1 ↗ 2 ↗ 3 ↘ 4 ↘ 5 ↗ 6F . . .. . . . .. . . . . . max min . . . . . . . . . . x . 1 2 3 4 5 6 . . . . . . . . . . . . .
  • 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 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 .. . . . . . . .
  • 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 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 .. . . . . . . .
  • 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 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 .. . . . . . . .
  • 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 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 .. . . . . . . .
  • 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 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 .. . . . . . . .
  • 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 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 .. . . . . . . .
  • 91. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 .. . . . . . . .
  • 92. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 .. . . . . . . .
  • 93. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 .. . . . . . . .
  • 94. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 .. . . . . . . .
  • 95. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . . .
  • 96. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . . .
  • 97. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . . .
  • 98. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . .
  • 99. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . .
  • 100. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . .
  • 101. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . .
  • 102. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . .
  • 103. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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 . . . . . .
  • 104. Using f to make a sign chart for F Assuming F′ = f, we can make a sign chart for f and f′ to find the intervals of monotonicity and concavity for 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
  • 105. 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 . . . . . . .
  • 106. 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 . . . . . . .
  • 107. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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 . . . . . . .
  • 108. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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 . . . . . . .
  • 109. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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 . . . . . . .
  • 110. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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 . . . . . . .
  • 111. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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 . . . . . . .
  • 112. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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 . . . . . . .
  • 113. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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 . . . . . . .
  • 114. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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 . . . . . . .
  • 115. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
  • 116. Could you repeat the question? Problem Below is the graph of a function f. Draw the graph of the antiderivative for F with F(1) = 0. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
  • 117. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 24 / 32
  • 118. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 25 / 32
  • 119. Application: Dead Reckoning . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 26 / 32
  • 120. Application: Dead Reckoning . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 26 / 32
  • 121. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
  • 122. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
  • 123. Problem Suppose a particle of mass m is acted upon by a constant force F. Find the position function s(t), the velocity function v(t), and the acceleration function a(t). 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
  • 124. Problem Suppose a particle of mass m is acted upon by a constant force F. Find the position function s(t), the velocity function v(t), and the acceleration function a(t). Solution By Newton’s Second Law (F = ma) a constant force induces a 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
  • 125. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 28 / 32
  • 126. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 28 / 32
  • 127. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
  • 128. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
  • 129. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
  • 130. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
  • 131. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 30 / 32
  • 132. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 30 / 32
  • 133. Solving We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 . . . . . . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
  • 134. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
  • 135. 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
  • 136. What have we learned today? 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, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 32 / 32