Lesson 21: Antiderivatives (notes)

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An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.

This is the handout version to take notes on.

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Lesson 21: Antiderivatives (notes)

  1. 1. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 Section 4.7 Notes 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 Notes 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 Notes 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 1
  2. 2. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 Objectives Notes 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 Notes 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! Notes 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 2
  3. 3. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 When two functions have the same derivative Notes 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 Notes 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 Notes y f (x) = 2x f (x) = x 2 Recall that the derivative of a power function is a power function. F (x) = ? Fact (The Power Rule) If f (x) = x r , then f (x) = rx r −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 3
  4. 4. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 Example Notes Find an antiderivative for the function f (x) = x 3 . Solution Try a power function F (x) = ax r Then F (x) = arx r −1 , so we want arx r −1 = x 3 . 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 x 1 = 4 · x 4−1 = x 3 4 " 1 Any others? Yes, F (x) = x 4 + C is the most general form. 4 V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32 Notes Fact (The Power Rule for antiderivatives) If f (x) = x r , 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? Notes 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 4
  5. 5. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 Graph of ln |x| Notes y F (x) = ln(x) ln |x| f (x) = 1/x x V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32 Combinations of antiderivatives Notes 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 Notes Example Find an antiderivative for f (x) = 16x + 5. Solution 1 The expression x 2 is an antiderivative for x, and x is an antiderivative for 2 1. So 1 2 F (x) = 16 · x + 5 · x + C = 8x 2 + 5x + C 2 is the antiderivative of f . Question Why do we not need two C ’s? Answer A combination of two arbitrary constants is still an arbitrary constant. V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32 5
  6. 6. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 Exponential Functions Notes 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) = e x , then F (x) = e x + C is the antiderivative of f . V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32 Logarithmic functions? Notes 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 Notes 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 6
  7. 7. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 More Trig Notes 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 Notes 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 Notes F. y y = f (x) x 1 2 3 4 5 6 V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 21 / 32 7
  8. 8. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 Using f to make a sign chart for F Notes 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 IP IP ? ? ? ? ? ?F 1 2 3 4 5 6 shape 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? Notes Problem Below is the graph of a function f . Draw the graph of the antiderivative for F with F (1) = 0. y Solution f We start with F (1) = 0. Using the sign chart, we x 1 2 3 4 5 6 draw arcs with the specified monotonicity and concavity It’s harder to tell if/when F F crosses the axis; more about 1 2 3 4 5 6 shape IP max IP min that later. V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32 Outline Notes 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 8
  9. 9. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 Say what? Notes “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 Notes 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 Notes 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 1 s(t) = at 2 + v0 t + C = at 2 + v0 t + s0 2 2 V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32 9
  10. 10. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 An earlier Hatsumon Notes 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 − 5t 2 √ √ 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 Notes 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 Notes In general, 1 s(t) = s0 + v0 t + at 2 2 Since s0 = 0 and a = −20, we have s(t) = v0 t − 10t 2 v (t) = v0 − 20t for all t. Plugging in t = t1 , 2 160 = v0 t1 − 10t1 0 = v0 − 20t1 We need to solve these two equations. V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 30 / 32 10
  11. 11. V63.0121, Calculus I Section 4.7 : Antiderivatives April 8, 2010 Solving Notes We have 2 v0 t1 − 10t1 = 160 v0 − 20t1 = 0 The second gives t1 = v0 /20, so substitute into the first: v0 v0 2 v0 · − 10 = 160 20 20 or 2 v0 10v02 − = 160 20 400 2 2 2v0 − v0 = 160 · 40 = 6400 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? Notes Antiderivatives are a useful concept, especially in motion y We can graph an antiderivative from the f graph of a function xF We can compute 1 2 3 4 5 6 antiderivatives, but not always 2 f (x) = e −x f (x) = ??? V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 32 / 32 Notes 11

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