Upcoming SlideShare
×

# Lesson 21: Antiderivatives (notes)

2,469
-1

Published on

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.

Published in: Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total Views
2,469
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
35
0
Likes
0
Embeds 0
No embeds

No notes for slide

### 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