Chapter6 Differential Equations and Mathematical Modeling One wa to measure how light in the ocean diminishes as water depth increases involves using a Secchi disk. This white disk is centimeters in diameter, and is lowered into the ocean until it disappears from view. The depth of this point (in meters), divided into.7, ields the coefficient k used in the equation l l e k. This equation estimates the intensit l of light at depth using l, the intensit of light at the surface. In an ocean eperiment, if the Secchi disk disappears at 55 meters, at what depth will onl % of surface radiation remain? Section 6.4 will help ou answer this question.
Section 6. Slope Fields and Euler s Method Chapter 6 Overview One of the earl accomplishments of calculus was predicting the future position of a planet from its present position and velocit. Toda this is just one of a number of occasions on which we deduce everthing we need to know about a function from one of its known values and its rate of change. From this kind of information, we can tell how long a sample of radioactive polonium will last; whether, given current trends, a population will grow or become etinct; and how large major league baseball salaries are likel to be in the ear. In this chapter, we eamine the analtic, graphical, and numerical techniques on which such predictions are based. 6. What ou ll learn about Differential Equations Slope Fields Euler s Method... and wh Differential equations have alwas been a prime motivation for the stud of calculus and remain so to this da. Slope Fields and Euler s Method Differential Equations We have alread seen how the discover of calculus enabled mathematicians to solve problems that had befuddled them for centuries because the problems involved moving objects. Leibniz and Newton were able to model these problems of motion b using equations involving derivatives what we call differential equations toda, after the notation of Leibniz. Much energ and creativit has been spent over the ears on techniques for solving such equations, which continue to arise in all areas of applied mathematics. DEFINITION Differential Equation An equation involving a derivative is called a differential equation. The order of a differential equation is the order of the highest derivative involved in the equation. EXAMPLE Solving a Differential Equation Find all functions that satisf d sec 5. We first encountered this sort of differential equation (called eact because it gives the derivative eactl) in Chapter 4. The solution can be an antiderivative of sec 5, which can be an function of the form tan 5 C. That famil of functions is the general solution to the differential equation. Now tr Eercise. Notice that we cannot find a unique solution to a differential equation unless we are given further information. If the general solution to a first-order differential equation is continuous, the onl additional information needed is the value of the function at a single point, called an initial condition. A differential equation with an initial condition is called an initial value problem. It has a unique solution, called the particular solution to the differential equation. EXAMPLE Solving an Initial Value Problem Find the particular solution to the equation d e 6 whose graph passes through the point (, ). The general solution is e C. Appling the initial condition, we have e C, from which we conclude that C e. Therefore, the particular solution is e e. Now tr Eercise.
Chapter 6 Differential Equations and Mathematical Modeling An initial condition determines a particular solution b requiring that a solution curve pass through a given point. If the curve is continuous, this pins down the solution on the entire domain. If the curve is discontinuous, the initial condition onl pins down the continuous piece of the curve that passes through the given point. In this case, the domain of the solution must be specified. EXAMPLE Handling Discontinuit in an Initial Value Problem Find the particular solution to the equation d sec whose graph passes through the point (, ). The general solution is tan C. Appling the initial condition, we have C, from which we conclude that C. Therefore, the particular solution is tan. Since the point (, ) onl pins down the continuous piece of the general solution over the interval (p, p), we add the domain stipulation p p/. Now tr Eercise 5. Sometimes we are unable to find an antiderivative to solve an initial value problem, but we can still find a solution using the Fundamental Theorem of Calculus. EXAMPLE 4 Using the Fundamental Theorem to Solve an Initial Value Problem Find the solution to the differential equation f () e for which f (7). This almost seems too simple, but f () has both of the necessar properties! 7 et Clearl, f (7) 7, and f () e 7 et b the Fundamental Theorem. The integral form of the solution in Eample 4 might seem less desirable than the eplicit form of the solutions in Eamples and, but (thanks to modern technolog) it does enable us to find f () for an. For eample, f () e t 7 fnint (e^(t ), t, 7,).7. Now tr Eercise. [ π, π] b [ 4, 4] Figure 6. A graph of the famil of functions Y sin() L, where L {,,,,,, }. This graph shows some of the functions that satisf the differential equation d cos. (Eample 5) EXAMPLE 5 Graphing a General Solution Graph the famil of functions that solve the differential equation d/ cos. An function of the form sin C solves the differential equation. We cannot graph them all, but we can graph enough of them to see what a famil of solutions would look like. The command {,,,,,,,} L stores seven values of C in the list L. Figure 6. shows the result of graphing the function Y sin() L. Now tr Eercises 5 8. Notice that the graph in Figure 6. consists of a famil of parallel curves. This should come as no surprise, since functions of the form sin() C are all vertical translations of the basic sine curve. It might be less obvious that we could have predicted the appearance of this famil of curves from the differential equation itself. Eploration gives ou a new wa to look at the solution graph.
Section 6. Slope Fields and Euler s Method EXPLORATION Seeing the Slopes Figure 6. shows the general solution to the eact differential equation d cos. [ π, π] b [ 4, 4] (a). Since cos at odd multiples of p, we should see that d at the odd multiples of p in Figure 6.. Is that true? How can ou tell?. Algebraicall, the -coordinate does not affect the value of d cos. Wh not?. Does the graph show that the -coordinate does not affect the value of d? How can ou tell? 4. According to the differential equation d cos, what should be the slope of the solution curves when? Can ou see this in the graph? 5. According to the differential equation d cos, what should be the slope of the solution curves when p? Can ou see this in the graph? 6. Since cos is an even function, the slope at an point should be the same as the slope at its reflection across the -ais. Is this true? How can ou tell? Eploration suggests the interesting possibilit that we could have produced the famil of curves in Figure 6. without even solving the differential equation, simpl b looking carefull at slopes. That is eactl the idea behind slope fields. Slope Fields [ π, π] b [ 4, 4] (b) Suppose we want to produce Figure 6. without actuall solving the differential equation d cos. Since the differential equation gives the slope at an point (, ), we can use that information to draw a small piece of the linearization at that point, which (thanks to local linearit) approimates the solution curve that passes through that point. Repeating that process at man points ields an approimation of Figure 6. called a slope field. Eample 6 shows how this is done. EXAMPLE 6 Constructing a Slope Field Construct a slope field for the differential equation d cos. [ π, π] b [ 4, 4] (c) We know that the slope at an point (, ) will be cos, so we can start b drawing tin segments with slope at several points along the -ais (Figure 6.a). Then, since the slope at an point (p, ) or (p, ) will be, we can draw tin segments with slope at several points along the vertical lines p and p (Figure 6.b). The slope at all odd multiples of p will be zero, so we draw tin horizontal segments along the lines p and p (Figure 6.c). Finall, we add tin segments of slope along the lines p (Figure 6.d). Now tr Eercise 9. [ π, π] b [ 4, 4] (d) Figure 6. The steps in constructing a slope field for the differential equation d cos. (Eample 6) To illustrate how a famil of solution curves conforms to a slope field, we superimpose the solutions in Figure 6. on the slope field in Figure 6.d. The result is shown in Figure 6. on the net page. We could get a smoother-looking slope field b drawing shorter line segments at more points, but that can get tedious. Happil, the algorithm is simple enough to be programmed into a graphing calculator. One such program, using a lattice of 5 sample points, produced in a matter of seconds the graph in Figure 6.4 on the net page.
4 Chapter 6 Differential Equations and Mathematical Modeling Differential Equation Mode If our calculator has a differential equation mode for graphing, it is intended for graphing slope fields. The usual Y turns into a d screen, and ou can enter a function of and/or. The grapher draws a slope field for the differential equation when ou press the GRAPH button. [ π, π] b [ 4, 4] Figure 6. The graph of the general solution in Figure 6. conforms nicel to the slope field of the differential equation. (Eample 6) [ π, π] b [ 4, 4] Figure 6.4 A slope field produced b a graphing calculator program. It is also possible to produce slope fields for differential equations that are not of the form d f (). We will stud analtic techniques for solving certain tpes of these noneact differential equations later in this chapter, but ou should keep in mind that ou can graph the general solution with a slope field even if ou cannot find it analticall. Can We Solve the Differential Equation in Eample 7? Although it looks harmless enough, the differential equation d is not eas to solve until ou have seen how it is done. It is an eample of a firstorder linear differential equation, and its general solution is Ce (which ou can easil check b verifing that d ). We will defer the analtic solution of such equations to a later course. EXAMPLE 7 Constructing a Slope Field for a Noneact Differential Equation Use a calculator to construct a slope field for the differential equation d and sketch a graph of the particular solution that passes through the point (, ). The calculator produces a graph like the one in Figure 6.5a. Notice the following properties of the graph, all of them easil predictable from the differential equation:. The slopes are zero along the line.. The slopes are along the line.. The slopes get steeper as increases. 4. The slopes get steeper as increases. The particular solution can be found b drawing a smooth curve through the point (, ) that follows the slopes in the slope field, as shown in Figure 6.5b. [ 4.7, 4.7] b [.,.] (a) [ 4.7, 4.7] b [.,.] (b) Figure 6.5 (a) A slope field for the differential equation d/, and (b) the same slope field with the graph of the particular solution through (, ) superimposed. (Eample 7) Now tr Eercise 5.
Section 6. Slope Fields and Euler s Method 5 EXAMPLE 8 Matching Slope Fields with Differential Equations Use slope analsis to match each of the following differential equations with one of the slope fields (a) through (d). (Do not use our graphing calculator.). d. d. d 4. d (a) (b) (c) (d) To match Equation, we look for a graph that has zero slope along the line. That is graph (b). To match Equation, we look for a graph that has zero slope along both aes. That is graph (d). To match Equation, we look for a graph that has horizontal segments when and vertical segments when. That is graph (a). To match Equation 4, we look for a graph that has vertical segments when and horizontal segments when. That is graph (c). Now tr Eercise 9. Euler s Method In Eample 7 we graphed the particular solution to an initial value problem b first producing a slope field and then finding a smooth curve through the slope field that passed through the given point. In fact, we could have graphed the particular solution directl, b starting at the given point and piecing together little line segments to build a continuous approimation of the curve. This clever application of local linearit to graph a solution without knowing its equation is called Euler s Method. Slope d/ (, ) Δ ( + Δ, + Δ) Δ = (d/) Δ Figure 6.6 How Euler s Method moves along the linearization at the point (, ) to define a new point ( Δ, Δ). The process is then repeated, starting with the new point. Euler s Method For Graphing a Solution to an Initial Value Problem. Begin at the point (, ) specified b the initial condition. This point will be on the graph, as required.. Use the differential equation to find the slope d at the point.. Increase b a small amount Δ. Increase b a small amount Δ, where Δ (d/)δ. This defines a new point ( Δ, Δ) that lies along the linearization (Figure 6.6). 4. Using this new point, return to step. Repeating the process constructs the graph to the right of the initial point. 5. To construct the graph moving to the left from the initial point, repeat the process using negative values for Δ. We illustrate the method in Eample 9.
6 Chapter 6 Differential Equations and Mathematical Modeling 4..4.6.8 Figure 6.7 Euler s Method is used to construct an approimate solution to an initial value problem between and. (Eample 9) EXAMPLE 9 Appling Euler s Method Let f be the function that satisfies the initial value problem in Eample 6 (that is, d and f () ). Use Euler s method and increments of Δ. to approimate f (). We use Euler s Method to construct an approimation of the curve from to, pasting together five small linearization segments (Figure 6.7). Each segment will etend from a point (, ) to a point ( Δ, Δ), where Δ. and Δ (d)δ.the following table shows how we construct each new point from the previous one. (, ) d Δ Δ (d) ( Δ, Δ) (, )..4 (.,.4) (.,.4).6..5 (.4,.9) (.4,.9)...664 (.6,.584) (.6,.584) 4.84..868 (.8,.48) (.8,.48) 5.8..446 (,.46496) Euler s Method leads us to an approimation f ().46496, which we would more reasonabl report as f ().465. Now tr Eercise 4. You can see from Figure 6.7 that Euler s Method leads to an underestimate when the curve is concave up, just as it will lead to an overestimate when the curve is concave down. You can also see that the error increases as the distance from the original point increases. In fact, the true value of f () is about 4.55, so the approimation error is about 6.6%. We could increase the accurac b taking smaller increments; a reasonable option if we have a calculator program to do the work. For eample, increments of. give an estimate of 4.44, cutting the error to about %. EXAMPLE Moving Backward with Euler s Method If d and if when, use Euler s Method with five equal steps to approimate when.5. Starting at, we need five equal steps of Δ.. (, ) (, ) d/ Δ Δ (d)δ ( Δ, Δ) (, ).. (.9,.9) (.9,.9).9..9 (.8,.8) (.8,.8).79..79 (.7,.7) (.7,.7).669..669 (.6,.664) (.6,.664).559..559 (.5,.65) [, 4] b [, 6] Figure 6.8 A grapher program using Euler s Method and increments of. produced this approimation to the solution curve for the initial value problem in Eample. The actual solution curve is shown in red. The value at.5 is approimatel.6. (The actual value is about.649, so the percentage error in this case is about.4%.) Now tr Eercise 45. If we program a grapher to do the work of finding the points, Euler s Method can be used to graph (approimatel) the solution to an initial value problem without actuall solving it. For eample, a graphing calculator program starting with the initial value problem in Eample 9 produced the graph in Figure 6.8 using increments of.. The graph of the actual solution is shown in red. Notice that Euler s Method does a better job of approimating the curve when the curve is nearl straight, as should be epected.
Section 6. Slope Fields and Euler s Method 7 Euler s Method is one eample of a numerical method for solving differential equations. The table of values is the numerical solution. The analsis of error in a numerical solution and the investigation of methods to reduce it are important, but appropriate for a more advanced course (which would also describe more accurate numerical methods than the one shown here). Quick Review 6. In Eercises 8, determine whether or not the function satisfies the differential equation.. d e Yes. d 4 e 4 Yes. d e No 4. d e Yes 5. d e 5 No 6. d Yes 7. d tan sec Yes 8. d No In Eercises 9, find the constant C. 9. 4 C and when 5. sin cos C and 4 when 7. e sec C and 5 when. tan ln( ) C and p when p4 Section 6. Eercises In Eercises, find the general solution to the eact differential equation.. d 5 4 sec 5 tan C. d sec tan e sec e C. d sin e 8 cos e 4 C 4. d ( ) ln C 5. d 5 ln 5 5 tan C 6. d sin C 7. d t cos(t ) sin (t ) C 8. d (cos t) e sin t e sint C 9. d u (sec 5 )(5 4 ) u tan( 5 ) C d. 4(sin u) (cos u) d u (sinu) 4 C In Eercises, solve the initial value problem eplicitl.. d sin and when cos 5. d e cos and when e sin. d u 7 6 5 and u when u 7 5 4 4. d A 9 5 4 4 and A 6 when 5. d A 5 4 4 and when 6. d 5 sec ( ) and 7 when 7. d 5 tan 7 ( < p) t t ln and when t tan t t 8. d t t 6 and when t 9. d ln t t 6t 7 (t ) v 4 sec t tan t e t 6t and v 5 when t. d v 4 sec t e t t (p t p) (Note that C.) s t(t ) and s when t s t t (Note that C.) In Eercises 4, solve the initial value problem using the Fundamental Theorem. (Your answer will contain a definite integral.). d sin ( ) and 5 when sin (t ) 5. d u s co and u when u s cot e cos t 9. F() e cos and F() 9 F() 4. G(s) tan s and G() 4 G(s) s tan t 4
8 Chapter 6 Differential Equations and Mathematical Modeling In Eercises 5 8, match the differential equation with the graph of a famil of functions (a) (d) that solve it. Use slope analsis, not our graphing calculator. 5. d (sin ) Graph (b) 6. d (sin ) 7. d (cos ) Graph (a) 8. d (cos ) In Eercises 9 4, construct a slope field for the differential equation. In each case, cop the graph at the right and draw tin segments through the twelve lattice points shown in the graph. Use slope analsis, not our graphing calculator. 9. d (a) (c). d. d. d Graph (c) Graph (d). d 4. d In Eercises 5 4, match the differential equation with the appropriate slope field. Then use the slope field to sketch the graph of the particular solution through the highlighted point (, ). (All slope fields are shown in the window [6, 6] b [4, 4].) (a) (b) (d) (b) 5. d 7. d 9. d 6. d 8. d 4. d In Eercises 4 44, use Euler s Method with increments of Δ. to approimate the value of when.. 4. d and when. 4. d and when.66 4. d and when. 44. d and when.6 In Eercises 45 48, use Euler s Method with increments of Δ. to approimate the value of when.7. 45. d and when.97 46. d and when.7 47. d and when. 48. d and when. In Eercises 49 and 5, (a) determine which graph shows the solution of the initial value problem without actuall solving the problem. (b) Writing to Learn Eplain how ou eliminated two of the possibilities. (a) Graph (b) 49. d, () p (b) The slope is alwas positive, so graphs (a) and (c) can be ruled out. (a) π (e) (b) π (f) (c) (c) (d) π
d 5., See page. d (, ) (a) (, ) (b) (, ) 5. Writing to Learn Eplain wh could not be a solution to the differential equation with slope field shown below. (c) Section 6. Slope Fields and Euler s Method 9 Since the slopes must be negative reciprocals, g() cos. 56. Perpendicular Slope Fields If the slope fields for the differential equations d sec and d g() are perpendicular (as in Eercise 55), find g(). 57. Plowing Through a Slope Field The slope field for the differential equation d/ csc is shown below. Find a function that will be perpendicular to ever line it crosses in the slope field. (Hint: First find a differential equation that will produce a perpendicular slope field.) The perpendicular slope field would be produced b d sin, so cos C for an constant C. [ 4.7, 4.7] b [.,.] 5. Writing to Learn Eplain wh sin could not be a solution to the differential equation with slope field shown below. [ 4.7, 4.7] b [.,.] For one thing, there are positive slopes in the second quadrant of the slope field. The graph of has negative slopes in the second quadrant. For one thing, the slope of sin would be at the origin, while the slope field shows a slope of zero at ever point on the -ais. 5. Percentage Error Let f () be the solution to the initial value problem d such that f (). Find the percentage error if Euler s Method with Δ. is used to approimate f (.4). See page. 54. Percentage Error Let f () be the solution to the initial value problem d such that f (). Find the percentage error if Euler s Method with Δ. is used to approimate f (.6). See page. 55. Perpendicular Slope Fields The figure below shows the slope fields for the differential equations d e ()/ and d/ e ()/ superimposed on the same grid. It appears that the slope lines are perpendicular wherever the intersect. Prove algebraicall that this must be so. See page. [ 4.7, 4.7] b [.,.] 58. Plowing Through a Slope Field The slope field for the differential equation d/ / is shown below. Find a function that will be perpendicular to ever line it crosses in the slope field. (Hint: First find a differential equation that will produce a perpendicular slope field.) [ 4.7, 4.7] b [.,.] Standardized Test Questions You should solve the following problems without using a graphing calculator. 59. True or False An two solutions to the differential equation d 5 are parallel lines. Justif our answer. True. The are all lines of the form 5 C. 6. True or False If f () is a solution to d, then f () is a solution to d. Justif our answer. See page. 6. Multiple Choice A slope field for the differential equation d 4 will show C (A) a line with slope and -intercept 4. (B) a vertical asmptote at 4. (C) a horizontal asmptote at 4. (D) a famil of parabolas opening downward. (E) a famil of parabolas opening to the left. The perpendicular slope field would be produced b d, so.5 C for an constant C. 6. Multiple Choice For which of the following differential equations will a slope field show nothing but negative slopes in the fourth quadrant? E (A) d (D) d (B) d 5 (E) d (C) d
Chapter 6 Differential Equations and Mathematical Modeling 6. Multiple Choice If d and when, then B (A) (B) e (C) (D) (E) 64. Multiple Choice Which of the following differential equations would produce the slope field shown below? A (A) d (B) d (C) d (D) d (E) d Eplorations [, ] b [.98,.98] d 65. Solving Differential Equations Let d. (a) Find a solution to the differential equation in the interval, that satisfies., (b) Find a solution to the differential equation in the interval, that satisfies., (c) Show that the following piecewise function is a solution to the differential equation for an values of C and C. C, { C, /, /, (d) Choose values for C and C so that the solution in part (c) agrees with the solutions in parts (a) and (b). C, C (e) Choose values for C and C so that the solution in part (c) satisfies and. C, C 7 66. Solving Differential Equations Let d. (a) Show that ln C is a solution to the differential d equation in the interval,. (ln C) for d (b) Show that ln C is a solution to the differential equation in the interval,. d (ln () C) for Answers: d 5. (a) Graph (b) (b) The solution should have positive slope when is negative, zero slope when is zero and negative slope when is positive since slope d. Graphs (a) and (c) don t show this slope pattern. 5. Euler s Method gives an estimate f (.4) 4.. The solution to the initial value problem is f (), from which we get f (.4) 4.6. The percentage error is thus (4.6 4.)/4.6.9%. 54. Euler s Method gives an estimate f (.6).9. The solution to the initial value problem is f (), from which we get f (.6).96. The percentage error is thus (.96.9)/.96 %. (c) Writing to Learn Eplain wh ln C is a solution to the differential equation in the domain,,. d ln for all ecept. d (d) Show that the function ln C, { ln C, is a solution to the differential equation for an values of C and C. d d for all ecept. Etending the Ideas 67. Second-Order Differential Equations Find the general solution to each of the following second-order differential equations b first finding d/ and then finding. The general solution will have two unknown constants. (a) d 4 (b) d e sin (c) d 68. Second-Order Differential Equations Find the specific solution to each of the following second-order initial value problems b first finding d and then finding. (a) d 4. When, d and 5. (b) d cos sin. When, d 4 5 5 and (c) d e. When, d cos sin and. e 6 6 69. Differential Equation Potpourri For each of the following differential equations, find at least one particular solution. You will need to call on past eperience with functions ou have differentiated. For a greater challenge, find the general solution. (a) (b) (c) (d) (e) 7. Second-Order Potpourri For each of the following secondorder differential equations, find at least one particular solution. You will need to call on past eperience with functions ou have differentiated. For a significantl greater challenge, find the general solution (which will involve two unknown constants). (a) (b) (c) sin (d) (e) 55. At ever point (, ), (e () )(e () ) e () () e, so the slopes are negative reciprocals. The slope lines are therefore perpendicular. 6. False. For eample, f () is a solution of d, but f () is not a solution of d. 67. (a) C C (b) e sin C C 5 (c) C C 69. (a) C (b) C (d) Ce (e) Ce (c) Ce 7. (a) C 6 C (b) C 6 C (c) sin C C (d) C e C e (e) C sin C cos
Section 6. Antidifferentiation b Substitution 6. What ou ll learn about Indefinite Integrals Leibniz Notation and Antiderivatives Substitution in Indefinite Integrals Substitution in Definite Integrals... and wh Antidifferentiation techniques were historicall crucial for appling the results of calculus. Antidifferentiation b Substitution Indefinite Integrals If f () we can denote the derivative of f b either d or f(). What can we use to denote the antiderivative of f? We have seen that the general solution to the differential equation d/ f () actuall consists of an infinite famil of functions of the form F() C, where F() f (). Both the name for this famil of functions and the smbol we use to denote it are closel related to the definite integral because of the Fundamental Theorem of Calculus. DEFINITION Indefinite Integral The famil of all antiderivatives of a function f () is the indefinite integral of f with respect to and is denoted b f (). If F is an function such that F() f (), then f () F() C, where C is an arbitrar constant, called the constant of integration. As in Chapter 5, the smbol is an integral sign, the function f is the integrand of the integral, and is the variable of integration. Notice that an indefinite integral is not at all like a definite integral, despite the similarities in notation and name. A definite integral is a number, the limit of a sequence of Riemann sums. An indefinite integral is a famil of functions having a common derivative. If the Fundamental Theorem of Calculus had not provided such a dramatic link between antiderivatives and integration, we would surel be using a different name and smbol for the general antiderivative toda. EXAMPLE Evaluating an Indefinite Integral Evaluate ( sin ). Evaluating this definite integral is just like solving the differential equation d sin. Our past eperience with derivatives leads us to conclude that ( sin ) cos C (as ou can check b differentiating). Now tr Eercise. You have actuall been finding antiderivatives since Section 5., so Eample should hardl have seemed new. Indeed, each derivative formula in Chapter could be turned around to ield a corresponding indefinite integral formula. We list some of the most useful such indefinite integral formulas below. Be sure to familiarize ourself with these before moving on to the net section, in which function composition becomes an issue. (Incidentall, it is in anticipation of the net section that we give some of these formulas in terms of the variable u rather than.)
Chapter 6 Differential Equations and Mathematical Modeling Properties of Indefinite Integrals k f () kf () for an constant k (f () g()) f () g() Power Formulas u n u du n C when n n u du du ln u C u (see Eample ) Trigonometric Formulas cos u du sin u C sec u du tan u C sec u tan u du sec u C sin u du cos u C csc u du cot u C csc u cot u du csc u C Eponential and Logarithmic Formulas e u du e u C a u du ln a u a C ln u du u ln u u C (See Eample ) log a u du l n u du u ln u u C ln a ln a A Note on Absolute Value Since the indefinite integral does not specif a domain, ou should alwas use the absolute value when finding /udu. The function ln u C is onl defined on positive u-intervals, while the function ln u C is defined on both the positive and negative intervals in the domain of /u (see Eample ). EXAMPLE Verifing Antiderivative Formulas Verif the antiderivative formulas: (a) u du u du ln u C (b) ln u du u ln u u C We can verif antiderivative formulas b differentiating. d d (a) For u, we have (ln u C) (ln u C) d u d u u u. d d For u, we have (ln u C) (ln(u) C) () d u d u u u. d Since (lnu C) in either case, ln u C is the general antiderivative of d u u the function on its entire domain. u d (b) (u ln u u C) ln u u d u ( u ) ln u ln u. Now tr Eercise.
Section 6. Antidifferentiation b Substitution Leibniz Notation and Antiderivatives b The appearance of the differential in the definite integral a f () is easil eplained b the fact that it is the limit of a Riemann sum of the form n f ( k ) Δ (see Section 5.). k The same almost seems unnecessar when we use the indefinite integral f () to represent the general antiderivative of f, but in fact it is quite useful for dealing with the effects of the Chain Rule when function composition is involved. Eploration will show ou wh this is an important consideration. EXPLORATION Are f (u) du and f (u) the Same Thing? Let u and let f (u) u.. Find f (u) du as a function of u.. Use our answer to question to write f (u) du as a function of.. Show that f (u) 6 and find f (u) as a function of. 4. Are the answers to questions and the same? Eploration shows that the notation f (u) is not sufficient to describe an antiderivative when u is a function of another variable. Just as du/du is different from du when differentiating, f (u) du is different from f (u) when antidifferentiating. We will use this fact to our advantage in the net section, where the importance of or du in the integral epression will become even more apparent. EXAMPLE Paing Attention to the Differential Let f () and let u. Find each of the following antiderivatives in terms of : (a)f () (b) f (u) du (c) f (u) (a) f () ( ) 4 4 C 4 (b) f (u) du (u ) du u u C ( C C 4 4 4 (c) f (u) (u ) (( ) ) ( 6 ) C 7 Substitution in Indefinite Integrals ) 4 8 7 Now tr Eercise 5. A change of variables can often turn an unfamiliar integral into one that we can evaluate. The important point to remember is that it is not sufficient to change an integral of the form f () into an integral of the form g(u). The differential matters. A complete substitution changes the integral f () into an integral of the form g(u) du. EXAMPLE 4 Evaluate sin e cos. Using Substitution continued
4 Chapter 6 Differential Equations and Mathematical Modeling Let u cos. Then du/ sin, from which we conclude that du sin. We rewrite the integral and proceed as follows: sin ecos (sin )ecos e cos (sin ) e u du Substitute u for cos and du for sin. e u C e cos C Re-substitute cos for u after antidifferentiating. Now tr Eercise 9. If ou differentiate e cos C, ou will find that a factor of sin appears when ou appl the Chain Rule. The technique of antidifferentiation b substitution reverses that effect b absorbing the sin into the differential du when ou change sin e cos into e u du. That is wh a u-substitution alwas involves a du-substitution to convert the integral into a form read for antidifferentiation. EXAMPLE 5 Evaluate 5. Using Substitution This integral invites the substitution u 5, du 6. 5. (5 ) 6 (5 ) 6 Set up the substitution with a factor of 6. 6 u du Substitute u for 5 and du for 6. 6 u C 9 (5 ) C Re-substitute after antidifferentiating. Now tr Eercise 7. EXAMPLE 6 Evaluate cot 7. Using Substitution We do not recall a function whose derivative is cot 7, but a basic trigonometric identit changes the integrand into a form that invites the substitution u sin 7, du 7cos 7. We rewrite the integrand as shown on the net page.
Section 6. Antidifferentiation b Substitution 5 cot 7 c os 7 sin 7 7 7co s 7 sin 7 d u u Trigonometric identit Note that du 7 cos 7 when u sin 7 We multipl b 7 7, or. Substitute u for 7 sin and du for 7 cos 7. ln u C 7 Notice the absolute value! ln sin 7 C 7 Re-substitute sin 7 for u after antidifferentiating. Now tr Eercise 9. EXAMPLE 7 Setting Up a Substitution with a Trigonometric Identit Find the indefinite integrals. In each case ou can use a trigonometric identit to set up a substitution. (a) (b) cos d cot (c) cos d (a) co s sec sec sec u du Let u and du. tan u C tan C Re-substitute after antidifferentiating. (b)cot (csc ) (csc ) (csc u ) du Let u and du. (cot u u) C (cot ) C cot C Re-substitute after antidifferentiation. continued
6 Chapter 6 Differential Equations and Mathematical Modeling (c)cos (cos ) cos ( sin ) cos ( u ) du u u C sin sin C Let u sin and du cos. Re-substitute after antidifferentiating. Now tr Eercise 47. Substitution in Definite Integrals Antiderivatives pla an important role when we evaluate a definite integral b the Fundamental Theorem of Calculus, and so, consequentl, does substitution. In fact, if we make full use of our substitution of variables and change the interval of integration to match the u-substitution in the integrand, we can avoid the resubstitution step in the previous four eamples. EXAMPLE 8 Evaluating a Definite Integral b Substitution p/ Evaluate tan sec. Let u tan and du sec. Note also that u() tan and u(p/) tan(p/). So p/ tan sec u du u Substitute u-interval for -interval. Now tr Eercise 55. EXAMPLE 9 That Absolute Value Again Evaluate. 4 Let u 4 and du. Then u() 4 4 and u() 4. continued
Section 6. Antidifferentiation b Substitution 7 So 4 4 d u 4 u Substitute u-interval for -interval. ln u 4 (ln ln 4) ln 4 Notice that ln u would not have eisted over the interval of integration [4, ]. The absolute value in the antiderivative is important. Now tr Eercise 6. Finall, consider this historical note. The technique of u-substitution derived its importance from the fact that it was a powerful tool for antidifferentiation. Antidifferentiation derived its importance from the Fundamental Theorem, which established it as the wa to evaluate definite integrals. Definite integrals derived their importance from real-world applications. While the applications are no less important toda, the fact that the definite integrals can be easil evaluated b technolog has made the world less reliant on antidifferentiation, and hence less reliant on u-substitution. Consequentl, ou have seen in this book onl a sampling of the substitution tricks calculus students would have routinel studied in the past. You ma see more of them in a differential equations course. Quick Review 6. (For help, go to Sections.6 and.9.) In Eercises and, evaluate the definite integral.. 4 5. 5 In Eercises, find d.. t 4. t 6 5. 4 4( ) ( 4) 6. sin 4 5 8 sin (4 5) cos (4 5) 7. ln cos tan 8. ln sin cot 9. ln sec tan sec. ln csc cot csc Section 6. Eercises In Eercises 6, find the indefinite integral.. cos. C sin C. t t t t C 4. t tan t C 5. 4 sec 6. e sec tan (5) 5 tan C e sec () C 8. (csc u C)(csc u cot u) csc u cot u In Eercises 7, use differentiation to verif the antiderivative formula. 7. csc u du cot u C 8. csc u cot u csc u C (cot u C)(csc u) csc u 9. e e C. 5 5 C See page 4. ln 5 See page 4.. u du tan u C. du sin u u C See page 4. See page 4.
8 Chapter 6 Differential Equations and Mathematical Modeling 4. ln sec () C ln cos () C In Eercises 6, verif that f (u) du f (u). f (u) u and u ( ) See page 4. 4. f (u) u and u 5 See page 4. 5. f (u) e u and u 7 See page 4. 6. f (u) sin u and u 4 See page 4. In Eercises 7 4, use the indicated substitution to evaluate the integral. Confirm our answer b differentiation. 7. sin, u cos C 8. cos, u 4 sin ( ) C 9. sec tan, u sec C. 87, u 7 (7 ) 4 C (/) tan (/)., u C 6 r C 9. 9r dr, u r r. ( cos t t sin ), u cos t 4. 8 4 4 d, u 4 4 ( 4 4 ) C In Eercises 5 46, use substitution to evaluate the integral. 5. C 6. sec cos t C tan ( ) C 4. co t 44. 5 8 45. sec (Hint: Multipl the integrand b s ec tan ln sec tan C sec tan and then use a substitution to integrate the result.) 46. csc (Hint: Multipl the integrand b c sc cot ln csc cot C csc cot and then use a substitution to integrate the result.) In Eercises 47 5, use the given trigonometric identit to set up a u-substitution and then evaluate the indefinite integral. 47. sin, sin cos co s co s C 48. sec 4, sec tan tan tan C 49. sin, cos sin sin C 5. 4 cos, cos cos sin C 5. tan 4, tan sec tan tan C 5. (cos 4 sin 4 ), cos cos sin sin C 5 8 C 5 7. tan sec (tan ) C 8. sec ( u p ) tan ( u p ) du sec p C 9. tan(4 ). (sin ) cot C In Eercises 5 66, make a u-substitution and integrate from ua to ub. 5. d 4/ 54. r r dr / 55. 4 tan sec / 56. 5r 4 r dr. cos z 4 dz. cot csc (cot ) C sin (z 4) C. ln6 57. 7 (ln )7 C 4. tan 7 sec u du ( u /) / 58. cos 4 sin 4 tan8 C 5. s cos s 4 8 ds 6. sin 59. 6 t 5 t5t 4 6. cos u sin u du /4 cot () C 7 5 4 sin (s 4 8) C sin t 6 cos t 7. 8. c os t sin t 6 6..54 6..97 C sin t (/) sec (t ) C p/4 9. ln(ln ) C 4. tan sec 6. () tan.69 64. ln t C cot p/4 4 4. 4. 8 tan 5 5 C 65. e.85 66. e.954 () ln( ) C 9. (4) ln cos (4 ) C or (4) ln sec (4 ) C
7. True. Using the substitution u f (), du f (), we have b f ( ) f (b) d u ln u f (b) ( b) ln( f (b)) ln( f (a)) ln a f ( ) f (a) u f (a) f f ( a). Two Routes to the Integral In Eercises 67 and 68, make a substitution u (an epression in ), du. Then (a) integrate with respect to u from ua to ub. (b) find an antiderivative with respect to u, replace u b the epression in, then evaluate from a to b. 67. (a).8 4 9 68. cos sin (a) / (b) / 6 69. Show that ln c os cos Note that d tan 5 and () 5. is the solution to the initial value problem d tan, f 5. (See the discussion following Eample 4, Section 5.4.) 7. Show that ln sin s in Note that d cot 6 and () 6. is the solution to the initial value problem d cot, f 6. Standardized Test Questions You should solve the following problems without using a graphing calculator. 7. True or False B u-substitution, p/4 tan sec p/4 u du. Justif our answer. See below. 7. True or False If f is positive and differentiable on [a, b], then b f ()d a f ( ) ln ( b) f f ( a). Justif our answer. See above. 7. Multiple Choice tan D (A) tan C (B) ln cot C (C) ln cos C (D) ln cos C (E) ln cot C (b).8 74. Multiple Choice e E (A) e 4 (B) e 4 (C) e 4 (D) e 4 (E) e4 75. Multiple Choice If 5 f ( a) 7 where a is a constant, then 5a f () a B (A) 7 a (B) 7 (C) 7 a (D) a 7 (E) 7 Section 6. Antidifferentiation b Substitution 9 76. Multiple Choice If the differential equation d/ f () leads to the slope field shown below, which of the following could be f ()? A (A) sin C (B) cos C (C) sin C (D) cos C (E) sin C Eplorations 77. Constant of Integration Consider the integral. (a) Show that / C. (b) Writing to Learn Eplain wh See page 4. t and t are antiderivatives of. (c) Use a table of values for to find the value of C for which C. See page 4. (d) Writing to Learn Give a convincing argument that C See page 4.. 78. Group Activit Making Connections Suppose that f F C. (a) Eplain how ou can use the derivative of F C to confirm the integration is correct. (b) Eplain how ou can use a slope field of f and the graph of F to support our evaluation of the integral. (c) Eplain how ou can use the graphs of F and f t to support our evaluation of the integral. (d) Eplain how ou can use a table of values for, and defined as in part (c), to support our evaluation of the integral. (e) Eplain how ou can use graphs of f and NDER of F to support our evaluation of the integral. (f) Illustrate parts (a) (e) for f. 7. False. The interval of integration should change from [, p4] to [, ], resulting in a different numerical answer.
4 Chapter 6 Differential Equations and Mathematical Modeling 79. Different Solutions? Consider the integral sin cos. (a) Evaluate the integral using the substitution u sin. (b) Evaluate the integral using the substitution u cos. (c) Writing to Learn Eplain wh the different-looking answers in parts (a) and (b) are actuall equivalent. 8. Different Solutions? Consider the integral sec tan. (a) Evaluate the integral using the substitution u tan. (b) Evaluate the integral using the substitution u sec. (c) Writing to Learn Eplain wh the different-looking answers in parts (a) and (b) are actuall equivalent. Etending the Ideas 8. Trigonometric Substitution Suppose u sin. Then cos u. (a) Use the substitution sin u, cos u du to show that du. (b) Evaluate du to show that sin C. 8. Trigonometric Substitution Suppose u tan. (a) Use the substitution tan u, sec u du to show that du. (b) Evaluate du to show that tan C. 8. Trigonometric Substitution Suppose sin. (a) Use the substitution sin, sin cos d to show that p/4 sin d. / (b) Use the identit given in Eercise 49 to evaluate the definite integral without a calculator. 84. Trigonometric Substitution Suppose u tan. (a) Use the substitution tan u, sec u du to show that p/ sec u du. (b) Use the hint in Eercise 45 to evaluate the definite integral without a calculator. Answers: 9. e C e e. 5 ln 5 C 5 ln 5 5 ln 5. (tan u C) u. (sin u C) u.f (u) du udu ()u / C (/) C f (u) u (/) C 4.f (u) du u du (/)u C (/) 5 C f (u) u (/) C 5.f (u) du e u du e u C e 7 C f (u) e u du e 7 ( 7)e 7 C 6.f (u)du sin u du cos u C cos 4 C f (u) sin u sin 4 (4) cos 4 C d 77. (a) d ( )/ C (b) Because d / and d / (c) 4 (d) C 79. (a) sin cos u du u C sin C (b) sin cos u du u C cos C (c) Since sin (cos ), the two answers differ b a constant (accounted for in the constant of integration). 8. (a) sec tan u du u C tan C (b) sec tan u du u C sec C (c) Since sec tan, the two answers differ b a constant (accounted for in the constant of integration). 8. (a) cos u d in u s u cos du du. co us u (Note cos u, so cos u cos u cos u.) (b) du u C sin C sec 8. (a) u du tan se c u du u sec du u (b) du u C tan C 8. (a) / / sin sin cos d sin sin sin p/4 sin cos d p/4 sin d cos (b) / p/4 sin d p/4 ( cos ) d [ (/)sin ] p/4 (p )4 84. (a) () sec tan u du p/ se c u du p/ sec u du tan () n ta u sec u (b) p/ sec u du ln sec u tan u p ln ( ) ln ( ) ln ( )
Section 6. Antidifferentiation b Parts 4 6. What ou ll learn about Product Rule in Integral Form Solving for the Unknown Integral Tabular Integration Inverse Trigonometric and Logarithmic Functions... and wh The Product Rule relates to derivatives as the technique of parts relates to antiderivatives. Antidifferentiation b Parts Product Rule in Integral Form When u and v are differentiable functions of, the Product Rule for differentiation tells us that d uv u d v v d u. d Integrating both sides with respect to and rearranging leads to the integral equation ( u d v d ) ( d d uv) ( v d u d ) uv ( v d u d ). When this equation is written in the simpler differential notation we obtain the following formula. Integration b Parts Formula u dv uv vdu This formula epresses one integral, u dv, in terms of a second integral, vdu. With a proper choice of u and v, the second integral ma be easier to evaluate than the first. This is the reason for the importance of the formula. When faced with an integral that we cannot handle analticall, we can replace it b one with which we might have more success. LIPET If ou are wondering what to choose for u, here is what we usuall do. Our first choice is a natural logarithm (L), if there is one. If there isn t, we look for an inverse trigonometric function (I). If there isn t one of these either, look for a polnomial (P). Barring that, look for an eponential (E) or a trigonometric function (T). That s the preference order: L IPET. In general, we want u to be something that simplifies when differentiated, and dv to be something that remains manageable when integrated. EXAMPLE Evaluate cos. Using Integration b Parts We use the formula u dv uv vduwith u, dv cos. To complete the formula, we take the differential of u and find the simplest antiderivative of cos. du v sin Then, cos sin sin sin cos C. Now tr Eercise. Let s eamine the choices available for u and v in Eample.
4 Chapter 6 Differential Equations and Mathematical Modeling EXPLORATION Choosing the Right u and dv Not ever choice of u and dv leads to success in antidifferentiation b parts. There is alwas a trade-off when we replace u dv with v du, and we gain nothing if v du is no easier to find than the integral we started with. Let us look at the other choices we might have made in Eample to find cos.. Appl the parts formula to cos, letting u and dv cos. Analze the result to eplain wh the choice of u is never a good one.. Appl the parts formula to cos, letting u cos and dv. Analze the result to eplain wh this is not a good choice for this integral.. Appl the parts formula to cos, letting u cos and dv. Analze the result to eplain wh this is not a good choice for this integral. 4. What makes a good choice for u and cos a good choice for dv? The goal of integration b parts is to go from an integral u dvthat we don t see how to evaluate to an integral vduthat we can evaluate. Keep in mind that integration b parts does not alwas work. Sometimes we have to use integration b parts more than once to evaluate an integral. EXAMPLE Evaluate e. Repeated Use of Integration b Parts With u, dv e, du,and v e, we have e e e. The new integral is less complicated than the original because the eponent on is reduced b one. To evaluate the integral on the right, we integrate b parts again with u, dv e. Then du, v e, and Hence, e e e e e C. e e e e e e C. The technique of Eample works for an integral n e in which n is a positive integer, because differentiating n will eventuall lead to zero and integrating e is eas. We will sa more on this later in this section when we discuss tabular integration. Now tr Eercise 5. EXAMPLE Solving an Initial Value Problem Solve the differential equation d/ ln() subject to the initial condition when. Confirm the solution graphicall b showing that it conforms to the slope field. continued
Section 6. Antidifferentiation b Parts 4 We find the antiderivative of ln() b using parts. It is usuall a better idea to differentiate ln() than to antidifferentiate it (do ou see wh?), so we let u ln() and dv. ln() ln() ln() Using the initial condition, ln() C 4 ln() C 4 C 4 Thus C 4. ln() 4 4. Figure 6.9 shows a graph of this function superimposed on a slope field for d/ ln(), to which it conforms nicel. Now tr Eercise. [, ] b [.5,.5] Figure 6.9 The solution to the initial value problem in Eample conforms nicel to a slope field of the differential equation. (Eample ) Solving for the Unknown Integral Integrals like the one in the net eample occur in electrical engineering. Their evaluation requires two integrations b parts, followed b solving for the unknown integral. EXAMPLE 4 Evaluate e cos. Solving for the Unknown Integral Let u e, dv cos. Then du e, v sin, and e cos e sin e sin. The second integral is like the first, ecept it has sin in place of cos. To evaluate it, we use integration b parts with u e, dv sin, v cos, du e. Then e cos e sin ( e cos cos e ) e sin e cos e cos. continued
44 Chapter 6 Differential Equations and Mathematical Modeling The unknown integral now appears on both sides of the equation. Adding the integral to both sides gives e cos e sin e cos C. Dividing b and renaming the constant of integration gives e e cos sin e cos C. Now tr Eercise 7. When making repeated use of integration b parts in circumstances like Eample 4, once a choice for u and dv is made, it is usuall not a good idea to switch choices in the second stage of the problem. Doing so will result in undoing the work. For eample, if we had switched to the substitution u sin, dv e in the second integration, we would have obtained e cos e sin ( e sin e cos ) undoing the first integration b parts. Tabular Integration e cos, We have seen that integrals of the form f g, in which f can be differentiated repeatedl to become zero and g can be integrated repeatedl without difficult, are natural candidates for integration b parts. However, if man repetitions are required, the calculations can be cumbersome. In situations like this, there is a wa to organize the calculations that saves a great deal of work. It is tabular integration, as shown in Eamples 5 and 6. EXAMPLE 5 Using Tabular Integration Evaluate e. With f and g e, we list: f and its derivatives g and its integrals e e e e We combine the products of the functions connected b the arrows according to the operation signs above the arrows to obtain e e e e C. Compare this with the result in Eample. Now tr Eercise.
Section 6. Antidifferentiation b Parts 45 EXAMPLE 6 Evaluate sin. Using Tabular Integration With f and g sin, we list: f and its derivatives g and its integrals sin cos 6 sin 6 cos sin Again we combine the products of the functions connected b the arrows according to the operation signs above the arrows to obtain sin cos sin 6 cos 6 sin C. Now tr Eercise. Inverse Trigonometric and Logarithmic Functions The method of parts is onl useful when the integrand can be written as a product of two functions (u and dv). In fact, an integrand f () satisfies that requirement, since we can let u f () and dv. There are not man antiderivatives of the form f () that ou would want to find b parts, but there are some, most notabl the antiderivatives of logarithmic and inverse trigonometric functions. EXAMPLE 7 Antidifferentiating ln Find ln. If we want to use parts, we have little choice but to let u ln and dv. ln (ln )() () udv uv vdu ln ln C EXAMPLE 8 Antidifferentiating sin Find the solution to the differential equation d/ sin if the graph of the solution passes through the point (, ). We find sin, letting u sin, dv. continued