《微积分》课程教学课件（Calculus）05. Integration

CALCULUSTHOMAS'CALCULUSARLYTRANSCENDENTAEChapter 5JAMESSTEWARTIntegrationERMStiheA-5.11.FindingAntiderivatives:Indefinite IntegralsIndefiniteIntegrals,DifferentialEquations,and ModelingWe have studied how to find the derivative of a function, However, many problems requirethat we recover a function from its known derivative (froin its known rate of change). For in-stance, we mayknow the velocity functionof an objctfaling from an initial height andneed to know its height at any time. More generally, we want to find a function F from itsderivative Jfsuch a function Fexists, it is.called an antiderivativeof fWe willsee in thenext chapter that antiderivatives are the link connecting thetwo major elements of calculusderivatives and definite intogralsC=2DEFINITIONA function F is an antiderivative of J on an interval I ifC-1F'(n) f(x) forall x in L*1. The process ofrecovering a function F(x) from is derivativefx) is called antidifferentiation, Weuse capital letters such as F torepresent an antiderivative ofafunetion f, G to represent anantiderivative ofg, and so forth.2. The most general antiderivative off onl is a family of functionsF(x)+C(C is an arbirary constant) whose graphs are verticalFIGURE 4.50 The curvesy=+Ctranslations ofone another.fillthe coordinate plane withoutovelapping,In Example 2, we identify tcurve y x2 as the one that passesAthrough the given point (1, 1),AEN

2016/11/15 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 5 Integration Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5.1 Indefinite Integrals, Differential Equations, and Modeling Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 1. Finding Antiderivatives: Indefinite Integrals Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1. The process of recovering a function F(x) from its derivative f(x) is called antidifferentiation. We use capital letters such as F to represent an antiderivative of a function f, G to represent an antiderivative of g, and so forth. 2. The most general antiderivative of f on I is a family of functions F(x)+C(C is an arbitrary constant) whose graphs are vertical translations of one another. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE1Find an antiderivative for eich of the following functions.DEFINITIONThe set of all antiderivatives of f is the indefinite integral of f(0)0)-2r( () ↓ + 2ewith respect to x, denoted by(b) g(x) = cos[rx) dSolutionThe symbol J is an integral signThe fiunction f is the integrand of the inte-aFu)=(o Hn) m]+e(b) G(x) = sinxgral, and x is the variable of integration.For example2rd-r+C,A国IntegralfomulasIndefinite integralReverse derivative formula-fra-+C,nxl, nnarional2.Integral Formulas, Properties2(0)-1Jax=+C,(speciaf case)of IndefiniteIntegrals andRules2fsinkrdt.-c(-cosk)=sinkrforIndefinite Integration3feoskrdink+c(cok(tan x) = sec* x4.Jsecxdr=tanx+C(-cot x)-ese'r5. fesc x dr =-t+C6.Jsectanxdt=secx+C(secx)= ecx tan xa7. jcxcotxdt=-cr+Ccscx)=csc.xcot.xProperties of Indefinite IntegralsProperties of Indefinite IntegralsProperty 2 (Linear property)Property 1(J((;(J(x) -(x),or j(x)-(r)t(2) [()±g(r)-()J2(x)tJ(xM=f(x)+C,or Jf(r)=f(x)+c2r+/+ldrExample. EvaluateDue to the inverse relation between integrals and differentials (or derivatives),Solution.we may obtain indefinite integrals of some simple functions from the[+Eα-[2+α+,acorrespondingexpressions of derivatives.x=2x+2/ +In|x/+C白E-11-12

2016/11/15 2 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For example Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 2. Integral Formulas, Properties of Indefinite Integrals and Rules for Indefinite Integration Integral formulas Indefinite integral Reverse derivative formula 1 1. , 1, 1 n n x x dx C n n rational n        dx x C special case   , ( )  1 ( ) 1 n d x n x dx n    ( ) 1 d x dx  2. sin coxkx kx dx C k     sin 3. cos kx kx dx C k    2 4. sec tan x dx x C    2 5. csc cot x dx x C     6. sec tan sec x x dx x C    7. csc cot csc x x dx x C     cos ( ) sin d kx kx dx k   sin ( ) cos d kx kx dx k  2 (tan ) sec d x x dx  2 ( cot ) csc d x x dx   (sec ) sec tan d x x x dx  ( csc ) csc cot d x x x dx   •11 Properties of Indefinite Integrals Property 1  f x dx f x ( ) ( )     , or d f x dx f x dx ( ) ( )   f x dx f x C ( ) ( )    , or df x f x C ( ) ( )    . Due to the inverse relation between integrals and differentials (or derivatives), we may obtain indefinite integrals of some simple functions from the corresponding expressions of derivatives. •12 Properties of Indefinite Integrals Property 2 (Linear property) (1) kf x dx k f x dx ( ) ( )    ; (2)  f x g x dx f x dx g x dx ( ) ( ) ( ) ( )        Example. Evaluate 2 1 x x dx x    . Solution. 1 2 2 1 1 2 2 2 ln | | x x dx dx x dx dx x x x x x C              

ExampleProperties of Indefinite IntegralsEvaluate the following infinite integrals:(o)jrd=+acEml Eaute Ju s(6)]=Jx"=24C=2/+CSolution(0 jsin2x dt=-02+c[awes(a=J)2=2mc=tanx-cotx+1/2(e)[2-2+5)由-Jz±-J2+Jsa-号-+5+CU/ (++2e)d = m +e2+ C.鱼国-13ExercisesFinding Indefinite Integrals“[()32+2.1+4x3+0+C=In Exercises 2570, find the most genenal antiderivative or indefinit1/2integraCheck youramwensbydiflerentiation37. [(元)4号-2元++C=4r-14c25[u+山-号+x+C /(a+号)=β+号+C1-1438[(-)=-++C=y+4y+c29 (+→+)dx =lmx/-5 tan+C-1/4392(1-)=F-2+C=X+2+C3 (-2-→)=[(--x*-)dx=-f-↓x+C--1---+C. Vga-Jo+m-2/-2/c33. J(e"+4)d --*++C2/(V+V=[(2/+x/)dx++=g+/++/+c45. [7sin号 d8 =[21singa()=-21cos0+(国47./(-3esex)dr=3cotx+C65fα+o)=jseod8=tano+c(Hinr: 1 + tan’ e = sec 8)3.Initial Value Problems67. cofFrd =[(csc x1) dx= cot xx+C(Hint: 1 + co'r= cs"s)69, cos (tamo + sec 0) =J(sinO+1) d =cosO+8+C70 /00-mg0=/-sim-0d8 = Jsec* o do=tang+C白人

2016/11/15 3 •13 Properties of Indefinite Integrals Example. Evaluate 2 2 sin cos dx x x  . Solution. 2 2 2 2 2 2 2 2 sin cos sec csc sin cos sin cos tan cot dx x x dx xdx xdx x x x x x x C            Evaluate the following infinite integrals: 5 ( ) a x dx  1 ( ) b dx x  ( ) sin 2 c x dx  ( ) cos 2 x d dx  6 6 x  C 1/2 1/2 x dx x C x C 2 2        cos 2 2 x   C 1 sin(1/ 2) cos( ) 2sin 2 1/ 2 2 x x      x dx C C  ()e ()f Example Exercises 29. 33. ( 4 ) x x e dx    3/2 1/2 3 1 2 2 1 1 2 4 2 3 / 2 1/ 2 3 x x         C x x C 3/4 3 2 2 4 8 4 2 4 3 / 4 3 y        y C y y C 1/4 1 4 1 1 4 7 1/ 4 7 y y C y y C          1 2 2 1 2 2 1 x x C x x C           1 3 1 1 2 2 2 2 ( ) 2 2 t t dt t t C          21sin ( ) 21cos( ) 3 3 3 d C           3cot x C 2    sec tan    d C  2       (csc 1) cot x dx x x C        (sin 1) cos     d C  2 2 1 sec tan 1 sin d d C               Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 3. Initial Value Problems

SomeApplications of Indefinite IntegralsExample. Find the curve whose slope at the point (x,y) is 3x* if the curveThe problem of finding a function y of x when we know its derivative andis required to pass through the point (1,1).its value, ataparticular point, is called an initial value problem WeSolution, (continue)solve such problems in two steps, as demonstrated in following example.Step 1: Solve the differential equsation:Finding a Curve fiom its Slope Function and a PoirntExample.Find the curve whose slope at the point (x,y) is 3x’ if the curveis required to pass through the point (1,1),[Ad-J3r'ddSolution, In mathematical language, we are asked to solve the initial valueJ+C,-+C,ptohlem that comsists ofthe fllowingCostantsefintegraionccalinedy=r+cgiving thegeaeralsolution=3,(becue'sopeis 32)The differential eqnation:drThe constant C can be found by the condition j(I)=1.J()--1The inifial conditionAAE-19-20EsercisesExample.Find the curve whose slope at the point (x,y) is 3x’ if the curveSohe the initial value probiems in Exercises 91112.岁一2—。 0)-0红会-+ ≥000-1is required to pass through the point (l,1).ySolution. (continue)0岁=95 岁r, (-1) -5-, (4) =.0Step 2: Evaluate C;2Vxy=x+c98.90 - - mo。 (0) 0 cost + sint/ 5(m) .1-1-()+CInitial condition101岁1ser tanf.C=-2(0) = 1102.-8+()--)The curve we want isydy105,= 2 6r, y(0) = 4,y(0) = 1drNote. The curves y=x +C fll the coordinate plane without overlapping制兴手107, . = I, r(1) = 1LdWe solve the differential equation by finding is109.= 6; y(0) = 8, (0) = 0, (0) = 5gencral solution, then solve the imital valucprobiem by finding the particular solution that11. y(= sini + cosr;SA-21Satisfies the inital condition y(r)=%y"(0) = 7, y (0) =y(0) = 1, y(0) = 0The Modeling ProcessStep 1 Observe real-world behaviorStep 2 Make assumptions to identify vaniables and theirrelationships, creating a model.4.Mathematical ModelingStep 3 Solve the model to obtain mathematical solutions.Step 4 Interpret the model andvenify it is consistent with real-worldobservation.Slide 4- 23Slide 4-24

2016/11/15 4 •19 Some Applications of Indefinite Integrals The problem of finding a function y of x when we know its derivative and its value 0 y at a particular point 0 x is called an initial value problem. We solve such problems in two steps, as demonstrated in following example. Example. Find the curve whose slope at the point ( , ) xy is 2 3x if the curve is required to pass through the point (1, 1)  . Finding a Curve from its Slope Function and a Point Solution. In mathematical language, we are asked to solve the initial value problem that consists of the following The differential equation: 2 3 dy x dx  , (the curve’s slope is 2 3x ) The initial condition: y(1) 1  •20 Solution. (continue) Step 1: Solve the differential equation: 2 3 dy x dx  2 3 dy dx x dx dx    3 1 2 y C x C    3 y x C   The constant C can be found by the condition y(1) 1  . Example. Find the curve whose slope at the point ( , ) xy is 2 3x if the curve is required to pass through the point (1, 1)  . Constants of integration combined, giving the general solution. •21 Solution. (continue) Step 2: Evaluate C: 3 y x C  3    1 (1) C C 2 The curve we want is 3 y x   2. Note. The curves 3 y x C   fill the coordinate plane without overlapping. Example. Find the curve whose slope at the point ( , ) xy is 2 3x if the curve is required to pass through the point (1, 1)  . Initial condition We solve the differential equation by finding its general solution, then solve the initial value problem by finding the particular solution that satisfies the initial condition . 0 0 y x y ( )  Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 23 4. Mathematical Modeling Slide 4 - 24 The Modeling Process Step 1 Observe real-world behavior. Step 2 Make assumptions to identify variables and their relationships, creating a model. Step 3 Solve the model to obtain mathematical solutions. Step 4 Interpret the model and verify it is consistent with real-world observation

EXAMPLE5A hot-air alloon aicemding at the rate of 12 f/sec is at a height 80Solutioni(continue) Solving this differential equation getsabove the ground when a package is dropped. How long does it take the package to reach[-J(-321+12)=$=-16r+121+Cthe ground?Solution Let v(t) denote the velocity ofthe4s(0) =80 = C = 80package at time t, and let s(t) denote its height( = 12above the ground. The acceleration of gravity nearThe package's height above ground at time tis s=-16t2 +12+80the surface ofthe earth is 32f/sec2,Assuming noLet s= -162 +121+80=0,other forces act on the dropped package, thenthen 4(2-3t-20=0, t=2.64, t =-1.89(has no physical meaning)会-2The package his the ground about 2.64sec afler it is droppedIntegating both idesatains Ja-f-2afrom the balloon.-diY=321+C(0)=12=C=12 Sov=-32+12And --2+12,(0)=804FIGURE4.52Apacluge droppedfrom a risling hot-air balloonlide 4- 25Slide 4-28(Example5)5.21.TheIntegralsofsin?xand cos?xIntegration by SubstitutionSlide 4- 27Slide 4- 28We can sometimes use trigonometric identities to transformintegrals we do not know how to evaluate into integrals we doknowhow to evaluate.Example Integrating sin?x and cos2x(a2)22.Substitution:Running the Chain1:2+c=-2+cRuleBackwards2422()[±=Slide 4- 29Slide 4-30

2016/11/15 5 Slide 4 - 25 Solution: Let v(t) denote the velocity of the package at time t, and let s(t) denote its height above the ground. The acceleration of gravity near the surface of the earth is 32ft/sec2 . Assuming no other forces act on the dropped package, then 32 dv dt  Integrating both sides attains 32 dv dt dt dt         v t C 32 v C (0) 12 12    So v=-32t+12 And 32 12, (0) 80 ds t s dt     Slide 4 - 26 Solution:(continue) Solving this differential equation gets 2 ( 32 12) 16 12 ds dt t dt s t t C dt           s C (0) 80 80    The package’s height above ground at time t is s= -16t2 +12t+80 Let s= -16t2 +12t+80=0, then 4t2 -3t-20=0, t ≈2.64, t ≈-1.89(has no physical meaning) The package hits the ground about 2.64sec after it is dropped from the balloon. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 27 5.2 Integration by Substitution Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 28 1. The Integrals of sin2x and cos2x Slide 4 - 29 We can sometimes use trigonometric identities to transform integrals we do not know how to evaluate into integrals we do know how to evaluate. Example Integrating sin2x and cos2x 2 ( ) sin a x dx  2 ( ) cos b x dx  1 cos 2 1 1 1 (1 cos 2 ) cos 2 2 2 2 2 x dx x dx dx x dx           1 1 sin 2 sin 2 2 2 2 2 4 x x x        x C C 1 cos 2 sin 2 2 2 4 x x x dx C       Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 30 2. Substitution: Running the Chain Rule Backwards