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# Lesson 23: Antiderivatives (slides)

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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 (slides)

1. 1. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Sec on 4.7 An deriva ves V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 19, 2011 . . Notes Announcements Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm I am teaching Calc II MW 2:00pm and Calc III TR 2:00pm both Fall ’11 and Spring ’12 . . Notes Objectives Given a ”simple“ elementary func on, ﬁnd a func on whose deriva ve is that func on. Remember that a func on whose deriva ve is zero along an interval must be zero along that interval. Solve problems involving rec linear mo on. . . . 1.
2. 2. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Outline What is an an deriva ve? Tabula ng An deriva ves Power func ons Combina ons Exponen al func ons Trigonometric func ons An deriva ves of piecewise func ons Finding An deriva ves Graphically Rec linear mo on . . Notes What is an antiderivative? Deﬁni on Let f be a func on. An an deriva ve for f is a func on F such that F′ = f. . . Notes Who cares? Ques on Why would we want the an deriva ve of a func on? Answers For the challenge of it For applica ons when the deriva ve of a func on is known but the original func on is not Biggest applica on will be a er the Fundamental Theorem of Calculus (Chapter 5) . . . 2.
3. 3. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Hard problem, easy check Example Find an an deriva ve for f(x) = ln x. Solu on ??? . . Notes Hard problem, easy check Example is F(x) = x ln x − x an an deriva ve for f(x) = ln x? Solu on d dx 1 (x ln x − x) = 1 · ln x + x · − 1 = ln x x . . Why the MVT is the MITC Notes 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. By MVT there exists a point z in (x, y) such that 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. . . . 3.
4. 4. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Functions with the same derivative Theorem Suppose f and g are two diﬀeren able func ons 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) . . Notes Outline What is an an deriva ve? Tabula ng An deriva ves Power func ons Combina ons Exponen al func ons Trigonometric func ons An deriva ves of piecewise func ons Finding An deriva ves Graphically Rec linear mo on . . Notes Antiderivatives of power functions ′ Recall that the deriva ve of a yf (x) = 2x power func on is a power f(x) = x2 func on. Fact (The Power Rule) F(x) = ? If f(x) = xr , then f′ (x) = rxr−1 . So in looking for an deriva ves of power . x func ons, try power func ons! . . . 4.
5. 5. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on Try a power func on 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 So F(x) = x4 is an an deriva ve. 4 . . Notes Antiderivatives of power functions Example Find an an deriva ve for the func on f(x) = x3 . Solu on 1 So F(x) = x4 is an an deriva ve. 4 ( ) Check: d 1 4 dx 4 1 x = 4 · x4−1 = x3 4 1 Any others? Yes, F(x) = x4 + C is the most general form. 4 . . Notes General power functions Fact (The Power Rule for an deriva ves) If f(x) = xr , then 1 r+1 F(x) = x r+1 is an an deriva ve for f… as long as r ̸= −1. Fact 1 If f(x) = x−1 = , then F(x) = ln |x| + C is an an deriva ve for f. x . . . 5.
6. 6. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes 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 posi ve 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 an deriva ve with the larger domain. . . Notes Graph of ln |x| y F(x) = ln |x| . x = 1/x f(x) . . Notes Combinations of antiderivatives Fact (Sum and Constant Mul ple Rule for An deriva ves) If F is an an deriva ve of f and G is an an deriva ve of g, then F + G is an an deriva ve of f + g. If F is an an deriva ve of f and c is a constant, then cF is an an deriva ve of cf. . . . 6.
7. 7. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Combinations of antiderivatives Proof. These follow from the sum and constant mul ple rule for deriva ves: If F′ = f and G′ = g, then (F + G)′ = F′ + G′ = f + g Or, if F′ = f, (cF)′ = cF′ = cf . . Notes Antiderivatives of Polynomials Example Find an an deriva ve for f(x) = 16x + 5. Solu on . . Notes Antiderivatives of Polynomials Ques on Do we need two C’s or just one? Answer . . . 7.
8. 8. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes 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 an deriva ve of f. ln a Proof. Check it yourself. . . Notes Exponential Functions In par cular, Fact If f(x) = ex , then F(x) = ex + C is the an deriva ve of f. . . Notes Logarithmic functions? Remember we found F(x) = x ln x − x is an an deriva ve 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 an deriva ve of f(x). . . . 8.
9. 9. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Trigonometric functions Fact d d sin x = cos x cos x = − sin x dx dx So to turn these around, Fact The func on F(x) = − cos x + C is the an deriva ve of f(x) = sin x. The func on F(x) = sin x + C is the an deriva ve of f(x) = cos x. . . Notes More Trig Example Find an an deriva ve of f(x) = tan x. Answer F(x) = ln | sec x|. Check d = 1 · d dx sec x dx sec x = 1 sec x · sec x tan x = tan x . More about this later. . Notes Antiderivatives of piecewise functions Example Let { x if 0 ≤ x ≤ 1; f(x) = 1 − x2 if 1 x. Find the an deriva ve of f with F(0) = 1. . . . 9.
10. 10. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Antiderivatives of piecewise functions Solu on . . Notes . . Notes Outline What is an an deriva ve? Tabula ng An deriva ves Power func ons Combina ons Exponen al func ons Trigonometric func ons An deriva ves of piecewise func ons Finding An deriva ves Graphically Rec linear mo on . . . 10.
11. 11. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Finding Antiderivatives Graphically y Problem Pictured is the graph of a y = f(x) func on f. Draw the graph of . an an deriva ve for f. x 1 2 3 4 5 6 . . Notes 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: ′ y . + + − − + f=F 1↗2↗3↘4↘5↗6 F max min ′ ′′ ++ −−−− ++ ++ f = F . ⌣ ⌢ ⌢ ⌣ ⌣ F x 1 2 3 4 5 6 1 2 3 4 5 6 IP IP ? ? ? ? ? ?F 1 2 3 4 5 6 shape The only ques on le is: What are the func on values? . . Notes Could you repeat the question? Problem Below is the graph of a func on f. Draw the graph of the an deriva ve for f with F(1) = 0. Solu on y We start with F(1) = 0. f . Using the sign chart, we draw arcs x 1 2 3 4 5 6 with the speciﬁed monotonicity and concavity F It’s harder to tell if/when F crosses 1 2 3 4 5 6 shape IP max IP min the axis; more about that later. . . . 11.
12. 12. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Outline What is an an deriva ve? Tabula ng An deriva ves Power func ons Combina ons Exponen al func ons Trigonometric func ons An deriva ves of piecewise func ons Finding An deriva ves Graphically Rec linear mo on . . Notes Say what? “Rec linear mo on” just means mo on along a line. O en we are given informa on about the velocity or accelera on of a moving par cle and we want to know the equa ons of mo on. . . Notes Application: Dead Reckoning . . . 12.
13. 13. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Problem Notes Suppose a par cle of mass m is acted upon by a constant force F. Find the posi on func on s(t), the velocity func on v(t), and the accelera on func on a(t). Solu on By Newton’s Second Law (F = ma) a constant force induces a F constant accelera on. So a(t) = a = . m Since v′ (t) = a(t), v(t) must be an an deriva ve of the constant func on a. So v(t) = at + C = at + v0 . where v0 is the ini al velocity. Since s′ (t) = v(t), s(t) must be an an deriva ve of v(t), . meaning 1 1 s(t) = at2 + v0 t + C = at2 + v0 t + s0 2 2 Notes An earlier Hatsumon Example Drop a ball oﬀ the roof of the Silver Center. What is its velocity when it hits the ground? Solu on 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. . Finding initial velocity from Notes stopping distance Example The skid marks made by an automobile indicate that its brakes were fully applied for a distance of 160 before it came to a stop. Suppose that the car in ques on has a constant decelera on of 20 ft/s2 under the condi ons of the skid. How fast was the car traveling when its brakes were ﬁrst applied? Solu on (Setup) . . . 13.
14. 14. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Implementing the Solution . . Notes Solving . . Notes Summary of Antiderivatives so far f(x) F(x) 1 r+1 x , r ̸= 1 r x +C r+1 1 −1 =x ln |x| + C x x e ex + C x 1 x a a +C ln a ln x x ln x − x + C x ln x − x loga x +C ln a sin x − cos x + C . cos x sin x + C tan x ln | tan x| + C . . 14.
15. 15. . V63.0121.001: Calculus I . Sec on 4.7: An deriva ves . April 19, 2011 Notes Final Thoughts An deriva ves are a useful concept, especially in mo on y We can graph an f an deriva ve from the . xF graph of a func on 123456 We can compute an deriva ves, but not f(x) = e−x 2 always . f′ (x) = ??? . Notes . . Notes . . . 15.