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Lesson 23: Antiderivatives (Section 041 handout)

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At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.

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Lesson 23: Antiderivatives (Section 041 handout)

  1. 1. Section 4.7 Antiderivatives V63.0121.041, Calculus I New York University November 29, 2010 Announcements Quiz 5 in recitation this week on §§4.1–4.4 Announcements Quiz 5 in recitation this week on §§4.1–4.4 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 2 / 33 Objectives Given a ”simple“ elementary function, find a function whose derivative is that function. Remember that a function whose derivative is zero along an interval must be zero along that interval. Solve problems involving rectilinear motion. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 3 / 33 Notes Notes Notes 1 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  2. 2. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 4 / 33 What is an antiderivative? Definition Let f be a function. An antiderivative for f is a function F such that F = f . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 5 / 33 Hard problem, easy check Example Find an antiderivative for f (x) = ln x. Solution ??? Example is F(x) = x ln x − x an antiderivative for f (x) = ln x? Solution d dx (x ln x − x) = 1 · ln x + x · 1 x − 1 = ln x " V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 6 / 33 Notes Notes Notes 2 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  3. 3. 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) y − x = f (z) =⇒ f (y) = f (x) + f (z)(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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 7 / 33 When two functions have the same derivative Theorem Suppose f and g are two differentiable functions on (a, b) with f = g . Then f and g differ by a constant. That is, there exists a constant C such that f (x) = g(x) + C. Proof. Let h(x) = f (x) − g(x) Then h (x) = f (x) − g (x) = 0 on (a, b) So h(x) = C, a constant This means f (x) − g(x) = C on (a, b) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 8 / 33 Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 9 / 33 Notes Notes Notes 3 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  4. 4. 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! x y f (x) = x2 f (x) = 2x F(x) = ? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 10 / 33 Example Find an antiderivative for the function f (x) = x3 . Solution Try a power function F(x) = axr Then F (x) = arxr−1 , so we want arxr−1 = x3 . r − 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a = 1 4 . So F(x) = 1 4 x4 is an antiderivative. Check: d dx 1 4 x4 = 4 · 1 4 x4−1 = x3 " Any others? Yes, F(x) = 1 4 x4 + C is the most general form. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 11 / 33 Extrapolating to general power functions Fact (The Power Rule for antiderivatives) If f (x) = xr , then F(x) = 1 r + 1 xr+1 is an antiderivative for f . . . as long as r = −1. Fact If f (x) = x−1 = 1 x , then F(x) = ln |x| + C is an antiderivative for f . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 12 / 33 Notes Notes Notes 4 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  5. 5. What’s with the absolute value? F(x) = ln |x| = ln(x) if x > 0; 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 13 / 33 Graph of ln |x| x y f (x) = 1/x F(x) = ln |x| V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 14 / 33 Combinations of antiderivatives Fact (Sum and Constant Multiple Rule for Antiderivatives) If F is an antiderivative of f and G is an antiderivative of g, then F + G is an antiderivative of f + g. If F is an antiderivative of f and c is a constant, then cF is an antiderivative of cf . 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 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 15 / 33 Notes Notes Notes 5 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  6. 6. Antiderivatives of Polynomials Example Find an antiderivative for f (x) = 16x + 5. Solution The expression 1 2 x2 is an antiderivative for x, and x is an antiderivative for 1. So F(x) = 16 · 1 2 x2 + 5 · x + C = 8x2 + 5x + C is the antiderivative of f . Question Why do we not need two C’s? Answer V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 16 / 33 Exponential Functions Fact If f (x) = ax , f (x) = (ln a)ax . Accordingly, Fact If f (x) = ax , then F(x) = 1 ln a ax + C is the antiderivative of f . Proof. Check it yourself. In particular, Fact If f (x) = ex , then F(x) = ex + C is the antiderivative of f . V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 17 / 33 Logarithmic functions? Remember we found F(x) = x ln x − x is an antiderivative of f (x) = ln x. This is not obvious. See Calc II for the full story. However, using the fact that loga x = ln x ln a , we get: Fact If f (x) = loga(x) F(x) = 1 ln a (x ln x − x) + C = x loga x − 1 ln a x + C is the antiderivative of f (x). V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 18 / 33 Notes Notes Notes 6 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  7. 7. Trigonometric functions Fact d dx sin x = cos x d dx cos x = − sin x 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 19 / 33 More Trig Example Find an antiderivative of f (x) = tan x. Solution ??? Answer F(x) = ln(sec x). Check d dx = 1 sec x · d dx sec x = 1 sec x · sec x tan x = tan x " More about this later. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 20 / 33 Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 21 / 33 Notes Notes Notes 7 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  8. 8. Finding Antiderivatives Graphically Problem Below is the graph of a function f . Draw the graph of an antiderivative for f . x y 1 2 3 4 5 6 y = f (x) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 22 / 33 Using f to make a sign chart for F Assuming F = f , we can make a sign chart for f and f to find the intervals of monotonicity and concavity for F: x y 1 2 3 4 5 6 f = F F1 2 3 4 5 6 + + − − + max min f = F F1 2 3 4 5 6 ++ −− −− ++ ++ IP IP F shape1 2 3 4 5 6 ? ? ? ? ? ? The only question left is: What are the function values? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 23 / 33 Could you repeat the question? Problem Below is the graph of a function f . Draw the graph of the antiderivative for f with F(1) = 0. Solution We start with F(1) = 0. Using the sign chart, we draw arcs with the specified monotonicity and concavity It’s harder to tell if/when F crosses the axis; more about that later. x y 1 2 3 4 5 6 f F shape1 2 3 4 5 6 IP max IP min V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 24 / 33 Notes Notes Notes 8 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  9. 9. Outline What is an antiderivative? Tabulating Antiderivatives Power functions Combinations Exponential functions Trigonometric functions Finding Antiderivatives Graphically Rectilinear motion V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 25 / 33 Say what? “Rectilinear motion” just means motion along a line. Often we are given information about the velocity or acceleration of a moving particle and we want to know the equations of motion. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 26 / 33 Application: Dead Reckoning V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 27 / 33 Notes Notes Notes 9 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  10. 10. 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 constant acceleration. So a(t) = a = F 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 s(t) = 1 2 at2 + v0t + C = 1 2 at2 + v0t + s0 V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 28 / 33 An earlier Hatsumon Example Drop a ball off the roof of the Silver Center. What is its velocity when it hits the ground? 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. V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 29 / 33 Finding initial velocity from stopping distance Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 ft before it came to a stop. Suppose that the car in question has a constant deceleration of 20 ft/s2 under the conditions of the skid. How fast was the car traveling when its brakes were first applied? Solution (Setup) V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 30 / 33 Notes Notes Notes 10 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010
  11. 11. Implementing the Solution V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 31 / 33 Solving V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 32 / 33 Summary Antiderivatives are a useful concept, especially in motion We can graph an antiderivative from the graph of a function We can compute antiderivatives, but not always x y 1 2 3 4 5 6 f F f (x) = e−x2 f (x) = ??? V63.0121.041, Calculus I (NYU) Section 4.7 Antiderivatives November 29, 2010 33 / 33 Notes Notes Notes 11 Section 4.7 : AntiderivativesV63.0121.041, Calculus I November 29, 2010

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