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Lesson 23: Antiderivatives

This document discusses antiderivatives in calculus, highlighting key theorems and properties, including the Mean Value Theorem and rules for finding antiderivatives of power, exponential, and trigonometric functions. It also provides examples and outlines the combinations of antiderivatives, emphasizing applications such as rectilinear motion. Important facts include the power rule for antiderivatives and the significance of constants of integration.

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Section 4.7 Antiderivatives V63.0121.027, Calculus I November 19, 2009 Announcements Next written assignment will be due Wednesday, Nov 25 next and last quiz will be the week after Thanksgiving (4.1–4.4, 4.7) Final Exam: Friday, December 18, 2:00–3:50pm . . Image credit: Ian Hampton . . . . . .
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) = 0. y−x So f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant. . . . . . .
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) . . . . . .
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 . . . . . .
Hard problem, easy check Example Find an antiderivative for f(x) = ln x. . . . . . .
Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? . . . . . .
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? . . . . . .
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 = ln x . . . . . .
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 = ln x Yes! . . . . . .
Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
Antiderivatives of power functions Recall that the derivative of a power function is a power function. Fact The Power Rule If f(x) = xr , then f′ (x) = rxr−1 . . . . . . .
Antiderivatives of power functions Recall that the derivative of a power function is a power function. Fact The Power Rule If f(x) = xr , then f′ (x) = rxr−1 . So in looking for antiderivatives of power functions, try power functions! . . . . . .
Example Find an antiderivative for the function f(x) = x3 . . . . . . .
Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr . . . . . .
Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . . . . . . .
Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 . . . . . .
Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 . . . . . .
Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, 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 · x 4 −1 = x 3 dx 4 4 . . . . . .
Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, 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 · x 4 −1 = x 3 dx 4 4 Any others? . . . . . .
Example Find an antiderivative for the function f(x) = x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, 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 · x 4 −1 = x 3 dx 4 4 1 4 Any others? Yes, F(x) = x + C is the most general form. 4 . . . . . .
Fact (The Power Rule for antiderivatives) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an antiderivative for f… . . . . . .
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 (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. . . . . . .
What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only defined on positive numbers. . . . . . .
What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only defined on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) . . . . . .
What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only defined on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) For negative numbers d d 1 1 ln |x| = ln(−x) = · (−1) = dx dx −x x . . . . . .
What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only defined on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) For negative numbers d d 1 1 ln |x| = ln(−x) = · (−1) = dx dx −x x We prefer the antiderivative with the larger domain. . . . . . .
Graph of ln |x| y . . f . (x ) = 1 /x x . . . . . . .
Graph of ln |x| y . . (x) = ln |x| F . f . (x ) = 1 /x x . . . . . . .
Graph of ln |x| y . . (x) = ln |x| F . f . (x ) = 1 /x x . . (x) = ln |x| F . . . . . .
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. . . . . . .
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 Again, if F′ = f, (cF)′ = cF′ = cf . . . . . .
Example Find an antiderivative for f(x) = 16x + 5 . . . . . .
Example Find an antiderivative for f(x) = 16x + 5 Solution The expression 8x2 is an antiderivative for 16x, and 5x is an antiderivative for 5. So F(x) = 8x2 + 5x + C is the antiderivative of f. . . . . . .
Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . . . . . . .
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 . . . . . .
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. . . . . . .
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. . . . . . .
Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. . . . . . .
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. . . . . . .
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 that ln a 1 F(x) = (x ln x − x) + C ln a is the antiderivative of f(x) = loga (x). . . . . . .
Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx . . . . . .
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. . . . . . .
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. . . . . . .
Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
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 . . . . . . . .
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? . . . . . .
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. . . . . . .
Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
Say what? “Rectlinear 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. . . . . . .
Example: Dead Reckoning . . . . . .
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). . . . . . .
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 F a constant acceleration. So a(t) = a = . m . . . . . .
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 F a 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. . . . . . .
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 F a 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 . . . . . .
Example Drop a ball off the roof of the Silver Center. What is its velocity when it hits the ground? . . . . . .
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. . . . . . .
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? . . . . . .
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) We know that the car is decelerated by a(t) = −20 We know that when s(t) = 160, v(t) = 0. . . . . . .
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) We know that the car is decelerated by a(t) = −20 We know that when s(t) = 160, v(t) = 0. We want to know v(0) = v0 . . . . . . .
Solution (Implementation) 1 2 In general, s(t) = s0 + v0 t + at , so we have 2 s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. . . . . . .
Solution (Implementation) 1 2 In general, s(t) = s0 + v0 t + at , so we have 2 s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. If t1 is the time it took for the car to stop, 160 = v0 t1 − 10t2 1 0 = v0 − 20t1 We need to solve these two equations. . . . . . .
We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 . . . . . .
We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the first: v0 ( v )2 0 v0 · − 10 = 160 20 20 or v2 0 10v2 0 − = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0 . . . . . .
We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the first: v0 ( v )2 0 v0 · − 10 = 160 20 20 or v2 0 10v2 0 − = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0 So v0 = 80 ft/s ≈ 55 mi/hr . . . . . .

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Lesson 23: Antiderivatives

  • 1.
    Section 4.7 Antiderivatives V63.0121.027, Calculus I November 19, 2009 Announcements Next written assignment will be due Wednesday, Nov 25 next and last quiz will be the week after Thanksgiving (4.1–4.4, 4.7) Final Exam: Friday, December 18, 2:00–3:50pm . . Image credit: Ian Hampton . . . . . .
  • 2.
    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) = 0. y−x So f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant. . . . . . .
  • 3.
    Theorem Suppose f andg 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) . . . . . .
  • 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 . . . . . .
  • 5.
    Hard problem, easy check Example Find an antiderivative for f(x) = ln x. . . . . . .
  • 6.
    Hard problem, easy check Example Find an antiderivative for f(x) = ln x. Solution ??? . . . . . .
  • 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? . . . . . .
  • 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 1 (x ln x − x) = 1 · ln x + x · − 1 dx x = ln x . . . . . .
  • 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 = ln x Yes! . . . . . .
  • 10.
    Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
  • 11.
    Antiderivatives of power functions Recall that the derivative of a power function is a power function. Fact The Power Rule If f(x) = xr , then f′ (x) = rxr−1 . . . . . . .
  • 12.
    Antiderivatives of power functions Recall that the derivative of a power function is a power function. Fact The Power Rule If f(x) = xr , then f′ (x) = rxr−1 . So in looking for antiderivatives of power functions, try power functions! . . . . . .
  • 13.
  • 14.
    Example Find an antiderivative for the function f(x) =x3 . Solution Try a power function F(x) = axr . . . . . .
  • 15.
    Example Find an antiderivative for the function f(x) =x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . . . . . . .
  • 16.
    Example Find an antiderivative for the function f(x) =x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 . . . . . .
  • 17.
    Example Find an antiderivative for the function f(x) =x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = . 4 1 4 So F(x) = x is an antiderivative. 4 . . . . . .
  • 18.
    Example Find an antiderivative for the function f(x) =x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, 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 · x 4 −1 = x 3 dx 4 4 . . . . . .
  • 19.
    Example Find an antiderivative for the function f(x) =x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, 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 · x 4 −1 = x 3 dx 4 4 Any others? . . . . . .
  • 20.
    Example Find an antiderivative for the function f(x) =x3 . Solution Try a power function F(x) = axr Then F′ (x) = arxr−1 , and we want this to be equal to x3 . 1 Apparently, 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 · x 4 −1 = x 3 dx 4 4 1 4 Any others? Yes, F(x) = x + C is the most general form. 4 . . . . . .
  • 21.
    Fact (The Power Rule for antiderivatives) If f(x) =xr , then 1 r+1 F(x) = x r+1 is an antiderivative for f… . . . . . .
  • 22.
    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. . . . . . .
  • 23.
    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. . . . . . .
  • 24.
    What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only defined on positive numbers. . . . . . .
  • 25.
    What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only defined on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) . . . . . .
  • 26.
    What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only defined on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) For negative numbers d d 1 1 ln |x| = ln(−x) = · (−1) = dx dx −x x . . . . . .
  • 27.
    What’s with the absolute value? F(x) = ln |x| has domain all nonzero numbers, while ln x is only defined on positive numbers. For positive numbers x, d d ln |x| = ln x dx dx (which we knew) For negative numbers d d 1 1 ln |x| = ln(−x) = · (−1) = dx dx −x x We prefer the antiderivative with the larger domain. . . . . . .
  • 28.
    Graph of ln |x| y . . f . (x ) = 1 /x x . . . . . . .
  • 29.
    Graph of ln |x| y . . (x) = ln |x| F . f . (x ) = 1 /x x . . . . . . .
  • 30.
    Graph of ln |x| y . . (x) = ln |x| F . f . (x ) = 1 /x x . . (x) = ln |x| F . . . . . .
  • 31.
    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. . . . . . .
  • 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 Again, if F′ = f, (cF)′ = cF′ = cf . . . . . .
  • 33.
  • 34.
    Example Find an antiderivative for f(x) =16x + 5 Solution The expression 8x2 is an antiderivative for 16x, and 5x is an antiderivative for 5. So F(x) = 8x2 + 5x + C is the antiderivative of f. . . . . . .
  • 35.
    Exponential Functions Fact If f(x) = ax , f′ (x) = (ln a)ax . . . . . . .
  • 36.
    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 . . . . . .
  • 37.
    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. . . . . . .
  • 38.
    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. . . . . . .
  • 39.
    Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f(x) = ln x. . . . . . .
  • 40.
    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. . . . . . .
  • 41.
    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 that ln a 1 F(x) = (x ln x − x) + C ln a is the antiderivative of f(x) = loga (x). . . . . . .
  • 42.
    Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx . . . . . .
  • 43.
    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. . . . . . .
  • 44.
    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. . . . . . .
  • 45.
    Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
  • 46.
    Problem Below is the graph of a function f. Draw the graph of an antiderivative forF. y . . . . . = f(x) y . . . . . . . x . 1 . 2 . 3 . 4 . 5 . 6 . . . . . . . .
  • 47.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 48.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 49.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 50.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 51.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 52.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 53.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 54.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 55.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 56.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 57.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 58.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 59.
    Using f to make a sign chart forF 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 . . . . . . . . . . . . .
  • 60.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 61.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 62.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 63.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 64.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 65.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 66.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 67.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 68.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 69.
    Using f to make a sign chart forF 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 .. . . . . . . .
  • 70.
    Using f to make a sign chart forF 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 . . . . . . .
  • 71.
    Using f to make a sign chart forF 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 . . . . . . .
  • 72.
    Using f to make a sign chart forF 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 . . . . . . .
  • 73.
    Using f to make a sign chart forF 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 . . . . . .
  • 74.
    Using f to make a sign chart forF 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 . . . . . .
  • 75.
    Using f to make a sign chart forF 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 . . . . . .
  • 76.
    Using f to make a sign chart forF 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 . . . . . .
  • 77.
    Using f to make a sign chart forF 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 . . . . . .
  • 78.
    Using f to make a sign chart forF 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 . . . . . .
  • 79.
    Using f to make a sign chart forF 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? . . . . . .
  • 80.
    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. . . . . . .
  • 81.
    Outline Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion . . . . . .
  • 82.
    Say what? “Rectlinear 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. . . . . . .
  • 83.
  • 84.
    Problem Suppose a particle of mass m is acted upon by a constant forceF. Find the position function s(t), the velocity function v(t), and the acceleration function a(t). . . . . . .
  • 85.
    Problem Suppose a particle of mass m is acted upon by a constant forceF. 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 F a constant acceleration. So a(t) = a = . m . . . . . .
  • 86.
    Problem Suppose a particle of mass m is acted upon by a constant forceF. 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 F a 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. . . . . . .
  • 87.
    Problem Suppose a particle of mass m is acted upon by a constant forceF. 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 F a 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 . . . . . .
  • 88.
  • 89.
    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. . . . . . .
  • 90.
  • 91.
    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 of20 ft/s2 under the conditions of the skid. How fast was the car traveling when its brakes were first applied? Solution (Setup) We know that the car is decelerated by a(t) = −20 We know that when s(t) = 160, v(t) = 0. . . . . . .
  • 92.
    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 of20 ft/s2 under the conditions of the skid. How fast was the car traveling when its brakes were first applied? Solution (Setup) We know that the car is decelerated by a(t) = −20 We know that when s(t) = 160, v(t) = 0. We want to know v(0) = v0 . . . . . . .
  • 93.
    Solution (Implementation) 1 2 In general, s(t) = s0 + v0 t + at , so we have 2 s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. . . . . . .
  • 94.
    Solution (Implementation) 1 2 In general, s(t) = s0 + v0 t + at , so we have 2 s(t) = v0 t − 10t2 v(t) = v0 − 20t for all t. If t1 is the time it took for the car to stop, 160 = v0 t1 − 10t2 1 0 = v0 − 20t1 We need to solve these two equations. . . . . . .
  • 95.
    We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 . . . . . .
  • 96.
    We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the first: v0 ( v )2 0 v0 · − 10 = 160 20 20 or v2 0 10v2 0 − = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0 . . . . . .
  • 97.
    We have v0 t1 − 10t2 = 160 1 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the first: v0 ( v )2 0 v0 · − 10 = 160 20 20 or v2 0 10v2 0 − = 160 20 400 2v2 − v2 = 160 · 40 = 6400 0 0 So v0 = 80 ft/s ≈ 55 mi/hr . . . . . .