Antiderivatives and Indefinite Integration

Transcription

1 60_00.q //0 : PM Page 8 8 CHAPTER Integration Section. EXPLORATION Fining Antierivatives For each erivative, escribe the original function F. a. F b. F c. F. F e. F f. F cos What strateg i ou use to fin F? Antierivatives an Inefinite Integration Write the general solution of a ifferential equation. Use inefinite integral notation for antierivatives. Use basic integration rules to fin antierivatives. Fin a particular solution of a ifferential equation. Antierivatives Suppose ou were aske to fin a function F whose erivative is f. From our knowlege of erivatives, ou woul probabl sa that F because. The function F is an antierivative of f. Definition of an Antierivative A function F is an antierivative of f on an interval I if F f for all in I. Note that F is calle an antierivative of f, rather than the antierivative of f. To see wh, observe that F F an F, 5, 97 are all antierivatives of f. In fact, for an constant C, the function given b F C is an antierivative of f. THEOREM. Representation of Antierivatives If F is an antierivative of f on an interval I, then G is an antierivative of f on the interval I if an onl if G is of the form G F C, for all in I where C is a constant. Proof The proof of Theorem. in one irection is straightforwar. That is, if G F C, F f, an C is a constant, then G F C F 0 f. To prove this theorem in the other irection, assume that G is an antierivative of f. Define a function H such that H G( F. If H is not constant on the interval I, there must eist a an b a < b in the interval such that H a H b. Moreover, because H is ifferentiable on a, b, ou can appl the Mean Value Theorem to conclue that there eists some c in a, b such that H c H b H a. b a Because H b H a, it follows that H c 0. However, because G c F c, ou know that H c G c F c 0, which contraicts the fact that H c 0. Consequentl, ou can conclue that H is a constant, C. So, G F C an it follows that G F( C.

2 60_00.q //0 : PM Page 9 SECTION. Antierivatives an Inefinite Integration 9 Using Theorem., ou can represent the entire famil of antierivatives of a function b aing a constant to a known antierivative. For eample, knowing that D, ou can represent the famil of all antierivatives of f b G C Famil of all antierivatives of f ( where C is a constant. The constant C is calle the constant of integration. The famil of functions represente b G is the general antierivative of f, an G( C is the general solution of the ifferential equation G. Differential equation A ifferential equation in an is an equation that involves,, an erivatives of. For instance, an are eamples of ifferential equations. EXAMPLE Solving a Differential Equation C = 0 Fin the general solution of the ifferential equation. C = Functions of the form Figure. C = C Solution To begin, ou nee to fin a function whose erivative is. One such function is. is an antierivative of. Now, ou can use Theorem. to conclue that the general solution of the ifferential equation is C. General solution The graphs of several functions of the form C are shown in Figure.. Notation for Antierivatives When solving a ifferential equation of the form f it is convenient to write it in the equivalent ifferential form f. The operation of fining all solutions of this equation is calle antiifferentiation (or inefinite integration) an is enote b an integral sign. The general solution is enote b Variable of integration Constant of integration f F C. Integran NOTE In this tet, the notation f F C means that F is an antierivative of f on an interval. The epression f is rea as the antierivative of f with respect to. So, the ifferential serves to ientif as the variable of integration. The term inefinite integral is a snonm for antierivative.

3 60_00.q //0 : PM Page CHAPTER Integration Basic Integration Rules The inverse nature of integration an ifferentiation can be verifie b substituting F for f in the inefinite integration efinition to obtain F F C. Integration is the inverse of ifferentiation. Moreover, if f F C, then f f. Differentiation is the inverse of integration. These two equations allow ou to obtain integration formulas irectl from ifferentiation formulas, as shown in the following summar. Basic Integration Rules Differentiation Formula C 0 k k kf kf f ± g f ± g n n n sin cos cos sin tan sec sec sec tan cot csc csc csc cot Integration Formula 0 C k k C kf k f f ± g f ± g n n C, n cos sin C sin cos C sec tan C sec tan sec C csc cot C n csc cot csc C Power Rule NOTE Note that the Power Rule for Integration has the restriction that n. The evaluation of must wait until the introuction of the natural logarithm function in Chapter 5.

4 60_00.q //0 : PM Page 5 SECTION. Antierivatives an Inefinite Integration 5 EXAMPLE Appling the Basic Integration Rules Describe the antierivatives of. Solution Constant Multiple Rule C C Rewrite as. Power Rule n Simplif. So, the antierivatives of are of the form C, where C is an constant. When inefinite integrals are evaluate, a strict application of the basic integration rules tens to prouce complicate constants of integration. For instance, in Eample, ou coul have written C C. However, because C represents an constant, it is both cumbersome an unnecessar to write C as the constant of integration. So, is written in the simpler form, C C. In Eample, note that the general pattern of integration is similar to that of ifferentiation. Original integral Rewrite Integrate Simplif EXAMPLE Rewriting Before Integrating TECHNOLOGY Some software programs, such as Derive, Maple, Mathca, Mathematica, an the TI-89, are capable of performing integration smbolicall. If ou have access to such a smbolic integration utilit, tr using it to evaluate the inefinite integrals in Eample. a. b. c. Original Integral sin Rewrite sin Integrate C C cos C Simplif C C cos C Remember that ou can check our answer to an antiifferentiation problem b ifferentiating. For instance, in Eample (b), ou can check that C is the correct antierivative b ifferentiating the answer to obtain D C. Use ifferentiation to check antierivative. inicates that in the HM mathspace CD-ROM an the online Euspace sstem for this tet, ou will fin an Open Eploration, which further eplores this eample using the computer algebra sstems Maple, Mathca, Mathematica, an Derive.

5 60_00.q //0 : PM Page 5 5 CHAPTER Integration The basic integration rules liste earlier in this section allow ou to integrate an polnomial function, as shown in Eample. EXAMPLE Integrating Polnomial Functions a. Integran is unerstoo to be. C Integrate. b. C C Integrate. C The secon line in the solution is usuall omitte c. Integrate. C C C C C Simplif. EXAMPLE 5 Rewriting Before Integrating C C Rewrite as two fractions. Rewrite with fractional eponents. Integrate. Simplif. C NOTE When integrating quotients, o not integrate the numerator an enominator separatel. This is no more vali in integration than it is in ifferentiation. For instance, in Eample 5, be sure ou unerstan that is not the same as C C. C EXAMPLE 6 sin Rewriting Before Integrating cos cos sin cos sec tan sec C Rewrite as a prouct. Rewrite using trigonometric ientities. Integrate.

6 60_00.q //0 : PM Page 5 SECTION. Antierivatives an Inefinite Integration 5 The particular solution that satisfies the initial conition F is F. Figure. C = C = C = C = C = 0 C = C = C = F() = + C C = (, ) Initial Conitions an Particular Solutions You have alrea seen that the equation f has man solutions (each iffering from the others b a constant). This means that the graphs of an two antierivatives of f are vertical translations of each other. For eample, Figure. shows the graphs of several antierivatives of the form C General solution for various integer values of C. Each of these antierivatives is a solution of the ifferential equation. In man applications of integration, ou are given enough information to etermine a particular solution. To o this, ou nee onl know the value of F for one value of. This information is calle an initial conition. For eample, in Figure., onl one curve passes through the point (,. To fin this curve, ou can use the following information. F C F General solution Initial conition B using the initial conition in the general solution, ou can etermine that F 8 C, which implies that C. So, ou obtain F. EXAMPLE 7 Fin the general solution of Fining a Particular Solution Particular solution F, > 0 (, 0) C = C = C = C = C = 0 an fin the particular solution that satisfies the initial conition F 0. Solution To fin the general solution, integrate to obtain F C F F Rewrite as a power. Integrate. C = C, > 0. General solution C = F() = + C C = Using the initial conition F 0, ou can solve for C as follows. F C 0 C The particular solution that satisfies the initial conition F 0 is F, > 0. Figure. So, the particular solution, as shown in Figure., is F, > 0. Particular solution

7 60_00.q //0 : PM Page 5 5 CHAPTER Integration So far in this section ou have been using as the variable of integration. In applications, it is often convenient to use a ifferent variable. For instance, in the following eample involving time, the variable of integration is t. EXAMPLE 8 Solving a Vertical Motion Problem A ball is thrown upwar with an initial velocit of 6 feet per secon from an initial height of 80 feet. a. Fin the position function giving the height s as a function of the time t. b. When oes the ball hit the groun? Height (in feet) s t = 0 s(t) = 6t + 6t + 80 t = t = Height of a ball at time t Figure. t = t = 5 t = 5 Time (in secons) NOTE In Eample 8, note that the position function has the form s t gt v 0 t s 0 where g, v 0 is the initial velocit, an s 0 is the initial height, as presente in Section.. t Solution a. Let t 0 represent the initial time. The two given initial conitions can be written as follows. s 0 80 s 0 6 Initial height is 80 feet. Initial velocit is 6 feet per secon. Using feet per secon per secon as the acceleration ue to gravit, ou can write s t s t s t t t t C. Using the initial velocit, ou obtain s C, which implies that C 6. Net, b integrating s t, ou obtain s t s t t t 6 t 6t 6t C. Using the initial height, ou obtain s C which implies that C 80. So, the position function is s t 6t 6t 80. See Figure.. b. Using the position function foun in part (a), ou can fin the time that the ball hits the groun b solving the equation s t 0. s t 6t 6t t t 5 0 t, 5 Because t must be positive, ou can conclue that the ball hits the groun 5 secons after it was thrown. Eample 8 shows how to use calculus to analze vertical motion problems in which the acceleration is etermine b a gravitational force. You can use a similar strateg to analze other linear motion problems (vertical or horizontal) in which the acceleration (or eceleration) is the result of some other force, as ou will see in Eercises

8 60_00.q //0 : PM Page 55 SECTION. Antierivatives an Inefinite Integration 55 Before ou begin the eercise set, be sure ou realize that one of the most important steps in integration is rewriting the integran in a form that fits the basic integration rules. To illustrate this point further, here are some aitional eamples. Original Integral Rewrite Integrate Simplif C C t t t t t t 5 5 t t C C 5 t 5 t t C C 7 7 C 7 7 Eercises for Section. In Eercises, verif the statement b showing that the erivative of the right sie equals the integran of the left sie. See for worke-out solutions to o-numbere eercises. In Eercises 5, fin the inefinite integral an check the result b ifferentiation.. 9 C C In Eercises 5 8, fin the general solution of the ifferential equation an check the result b ifferentiation. r 5. t 6. t In Eercises 9, complete the table C C Original Integral Rewrite Integrate Simplif t t t.. t In Eercises 5, fin the inefinite integral an check the result b ifferentiation. 5. sin cos 6. t sin t t 7. csc t cot t t 8. t t sec 9. sec sin 0. sec tan sec.. tan cos cos

9 60_00.q //0 : PM Page CHAPTER Integration In Eercises 6, the graph of the erivative of a function is given. Sketch the graphs of two functions that have the given erivative. (There is more than one correct answer.) To print an enlarge cop of the graph, go to the website In Eercises 7 an 8, fin the equation for, given the erivative an the inicate point on the curve (, ) Slope Fiels In Eercises 9 5, a ifferential equation, a point, an a slope fiel are given. A slope fiel (or irection fiel) consists of line segments with slopes given b the ifferential equation. These line segments give a visual perspective of the slopes of the solutions of the ifferential equation. (a) Sketch two approimate solutions of the ifferential equation on the slope fiel, one of which passes through the inicate point. (To print an enlarge cop of the graph, go to the website (b) Use integration to fin the particular solution of the ifferential equation an use a graphing utilit to graph the solution. Compare the result with the sketches in part (a). (, ) 9.,, cos, 0, , > 0,, Slope Fiels In Eercises 5 an 5, (a) use a graphing utilit to graph a slope fiel for the ifferential equation, (b) use integration an the given point to fin the particular solution of the ifferential equation, an (c) graph the solution an the slope fiel in the same viewing winow. 5.,, 5.,, In Eercises 55 6, solve the ifferential equation. 55. f, f g 6, g h t 8t 5, h 58. f s 6s 8s, f f, f 5, f 0 f, f 0 6, f 0 f, f, f 0 0 f sin, f 0, f Tree Growth An evergreen nurser usuall sells a certain shrub after 6 ears of growth an shaping. The growth rate uring those 6 ears is approimate b h t.5t 5, where t is the time in ears an h is the height in centimeters. The seelings are centimeters tall when plante t 0. (a) Fin the height after t ears. (b) How tall are the shrubs when the are sol? 6. Population Growth The rate of growth P t of a population of bacteria is proportional to the square root of t, where P is the population size an t is the time in as 0 t 0. That is, P t k t. The initial size of the population is 500. After a the population has grown to 600. Estimate the population after 7 as.,, 7

10 60_00.q //0 : PM Page 57 SECTION. Antierivatives an Inefinite Integration 57 Writing About Concepts 65. Use the graph of shown in the figure to answer the following, given that f 0. (a) Approimate the slope of f at. Eplain. (b) Is it possible that f? Eplain. (c) Is f 5 f > 0? Eplain. () Approimate the value of where f is maimum. Eplain. (e) Approimate an intervals in which the graph of f is concave upwar an an intervals in which it is concave ownwar. Approimate the -coorinates of an points of inflection. (f) Approimate the -coorinate of the minimum of f. (g) Sketch an approimate graph of f. To print an enlarge cop of the graph, go to the website 5 Figure for 65 Figure for The graphs of f an each pass through the origin. Use the graph of shown in the figure to sketch the graphs of f an f. To print an enlarge cop of the graph, go to the website f f f Vertical Motion In Eercises 67 70, use a t feet per secon per secon as the acceleration ue to gravit. (Neglect air resistance.) 67. A ball is thrown verticall upwar from a height of 6 feet with an initial velocit of 60 feet per secon. How high will the ball go? 68. Show that the height above the groun of an object thrown upwar from a point s 0 feet above the groun with an initial velocit of v 0 feet per secon is given b the function f t 6t v 0 t s With what initial velocit must an object be thrown upwar (from groun level) to reach the top of the Washington Monument (approimatel 550 feet)? 70. A balloon, rising verticall with a velocit of 6 feet per secon, releases a sanbag at the instant it is 6 feet above the groun. (a) How man secons after its release will the bag strike the groun? (b) At what velocit will it hit the groun? Vertical Motion In Eercises 7 7, use a t 9.8 meters per secon per secon as the acceleration ue to gravit. (Neglect air resistance.) 7. Show that the height above the groun of an object thrown upwar from a point s 0 meters above the groun with an initial velocit of v 0 meters per secon is given b the function f t.9t v 0 t s The Gran Canon is 800 meters eep at its eepest point. A rock is roppe from the rim above this point. Write the height of the rock as a function of the time t in secons. How long will it take the rock to hit the canon floor? 7. A baseball is thrown upwar from a height of meters with an initial velocit of 0 meters per secon. Determine its maimum height. 7. With what initial velocit must an object be thrown upwar (from a height of meters) to reach a maimum height of 00 meters? 75. Lunar Gravit On the moon, the acceleration ue to gravit is.6 meters per secon per secon. A stone is roppe from a cliff on the moon an hits the surface of the moon 0 secons later. How far i it fall? What was its velocit at impact? 76. Escape Velocit The minimum velocit require for an object to escape Earth s gravitational pull is obtaine from the solution of the equation v v GM where v is the velocit of the object projecte from Earth, is the istance from the center of Earth, G is the gravitational constant, an M is the mass of Earth. Show that v an are relate b the equation v v 0 GM R where v 0 is the initial velocit of the object an R is the raius of Earth. Rectilinear Motion In Eercises 77 80, consier a particle moving along the -ais where t is the position of the particle at time t, t is its velocit, an t is its acceleration. 77. t t 6t 9t, 0 t 5 (a) Fin the velocit an acceleration of the particle. (b) Fin the open t-intervals on which the particle is moving to the right. (c) Fin the velocit of the particle when the acceleration is Repeat Eercise 77 for the position function t t t, 0 t A particle moves along the -ais at a velocit of v t t, t > 0. At time t, its position is. Fin the acceleration an position functions for the particle.

11 60_00.q //0 : PM Page CHAPTER Integration 80. A particle, initiall at rest, moves along the -ais such that its acceleration at time t > 0 is given b a t cos t. At the time t 0, its position is. (a) Fin the velocit an position functions for the particle. (b) Fin the values of t for which the particle is at rest. 8. Acceleration The maker of an automobile avertises that it takes secons to accelerate from 5 kilometers per hour to 80 kilometers per hour. Assuming constant acceleration, compute the following. (a) The acceleration in meters per secon per secon (b) The istance the car travels uring the secons 8. Deceleration A car traveling at 5 miles per hour is brought to a stop, at constant eceleration, feet from where the brakes are applie. (a) How far has the car move when its spee has been reuce to 0 miles per hour? (b) How far has the car move when its spee has been reuce to 5 miles per hour? (c) Draw the real number line from 0 to, an plot the points foun in parts (a) an (b). What can ou conclue? 8. Acceleration At the instant the traffic light turns green, a car that has been waiting at an intersection starts with a constant acceleration of 6 feet per secon per secon. At the same instant, a truck traveling with a constant velocit of 0 feet per secon passes the car. (a) How far beon its starting point will the car pass the truck? (b) How fast will the car be traveling when it passes the truck? 8. Moeling Data The table shows the velocities (in miles per hour) of two cars on an entrance ramp to an interstate highwa. The time t is in secons. (a) Assuming the eceleration of each airplane is constant, fin the position functions s an s for airplane A an airplane B. Let t 0 represent the times when the airplanes are 0 an 7 miles from the airport. (b) Use a graphing utilit to graph the position functions. (c) Fin a formula for the magnitue of the istance between the two airplanes as a function of t. Use a graphing utilit to graph. Is < for some time prior to the laning of airplane A? If so, fin that time. True or False? In Eercises 87 9, etermine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 87. Each antierivative of an nth-egree polnomial function is an n th-egree polnomial function. 88. If p is a polnomial function, then p has eactl one antierivative whose graph contains the origin. 89. If F an G are antierivatives of f, then F G C. 90. If f g, then g f C. 9. f g f g 9. The antierivative of f is unique. 9. Fin a function f such that the graph of f has a horizontal tangent at, 0 an f. 9. The graph of is shown. Sketch the graph of f given that f is continuous an f 0. f t v v (a) Rewrite the table converting miles per hour to feet per secon. (b) Use the regression capabilities of a graphing utilit to fin quaratic moels for the ata in part (a). (c) Approimate the istance travele b each car uring the 0 secons. Eplain the ifference in the istances. 85. Acceleration Assume that a full loae plane starting from rest has a constant acceleration while moving own a runwa. The plane requires 0.7 mile of runwa an a spee of 60 miles per hour in orer to lift off. What is the plane s acceleration? 86. Airplane Separation Two airplanes are in a straight-line laning pattern an, accoring to FAA regulations, must keep at least a three-mile separation. Airplane A is 0 miles from touchown an is grauall ecreasing its spee from 50 miles per hour to a laning spee of 00 miles per hour. Airplane B is 7 miles from touchown an is grauall ecreasing its spee from 50 miles per hour to a laning spee of 5 miles per hour., 0 < 95. If f f is continuous, an f,, 5, fin f. Is f ifferentiable at? 96. Let s an c be two functions satisfing s c an c s for all. If s 0 0 an c 0, prove that s c. Putnam Eam Challenge 97. Suppose f an g are nonconstant, ifferentiable, real-value functions on R. Furthermore, suppose that for each pair of real numbers an, f f f g g an g f g g f. If f 0 0, prove that f g for all. This problem was compose b the Committee on the Putnam Prize Competition. The Mathematical Association of America. All rights reserve.