《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

EXAMPLE3Finds (5r + 1)-5 d.THEOREM 5the Substitution Rulefag(r) isadifferemtiable fumctionwhcse range is an inlerval I,and Jis contimuous on /,thnWesubstiute=5t+ andd= 5dhThenSolutionNgt)g'(x)dr-[(n) dusee (5r + 1)-5 dt =seeudiLau-5el-3a tanr +CnvscSupposethat Fis theantiderivative off,then tan(5t + 1) +CShimtoSr + ISorn量Fg()=F(g(0)g/()Chain RalrThe Substitution Rale provides the following substitution method io evaluate the imtegnl= f(g(x)-g(x)P-1ng(t)g'(d dIf we make the substitutiom a = g(x), then[ng()g(),一/层目g(0),when f and g" are continuous functions:Substitute= gtx)and dr = (du/dt) dx= g(cx) drto obeain the integral=Fig(a) +Cteriad TheammHn=F(a)+C[o) dn.[Fr(a)dFuslarsttad Thoam2Inegrate withresp to-JrmaF-/3Replacbyt)inthreut4-31Some TipsEXAMPLE4Find cos (78 + 3) a8.We let u = 7e + 3 so that dv 7 do. The constamt factoe 7 is nSolutissmetr+ 1. Sometimes we observe that a power ofx appears in the integrandthe do term in themtegral.We for it by multiplying and dividing by 7uing thesmc procedure ain Exmple 2Ththat is onceless thanthe powerofx appearinginthe argument ofafunction we want to integrate.This observationimmediately suggestscos (70 + 3) d6 = +,eos (70 + 3)-7 d6we try a substitution for the higher power ofx.Wecanverifythissolutionby differentiating and[2ea=[e+xa-e.idaaduchecking that we obtainthe(/3)-originalfunctioncos(70+3)-fiw+cedar+Integrate wiahtespect tot= +sin(76 + 3) +:CThere is another apgrooch to ihis proelemWitha = 78 + 3 and dr = 7 as bce+cReplace wu by sfore, we solve forde to obtain dy = (1/7) dlk. Then the integral becomesThe successofthe substitutionoos(78 + 3) 0 -osa-famethoddepends onfinding a+sina + Csubstitutionthatwill changeanintegral we cannot evaluate号 sin(7 + 3) + Cdirectlyintothe onethatwe canSlide 4- 34Evaluating Indefinite IntegralsExercisesEXAMPLETV2x+idxEvaluateEvaluatc theindefinite integralsinExercises116by using the givensubstitutions to reduce the integrals to standard form.Method1 Let u-2x+1, then x=(1/2) (u-1), and dx=1/2 du, So0s(2x)+C8.rsin(2)de,w=2r2[rV2r +1d =1 / (u DVrdr 1 / (u 1)a/Par Sbme--/an-wy duMultiply term-cos+c10 (-c)sind,=1-c(sn-号wn)+c专bategrete9raw=16M-+11.V-r(2r+1)-(2x+p+CRoplaceuby2r +112.12(04+ 42+ 1PG + 20), = y*+4p2+1Method 2 Let u = 2x+1, then x= ,(u -I, and dx=u-du. So*+4y+y+CJx2x+dx=(a-1)-uud=,(-)d2-Vesin(rn = 1)d ==113,4111.s(2x+1)"_))+C:"+C:(2x+1)*"+C25101066()-sin(2 2)+CSlide 4- 35Slide 4-38

2016/11/15 6 Slide 4 - 31 Suppose that F is the antiderivative of f, then Slide 4 - 32 Slide 4 - 33 We can verify this solution by differentiating and checking that we obtain the original function cos(7θ+3) The success of the substitution method depends on finding a substitution that will change an integral we cannot evaluate directly into the one that we can. Slide 4 - 34 1. Sometimes we observe that a power of x appears in the integrand that is once less than the power of x appearing in the argument of a function we want to integrate. This observation immediately suggests we try a substitution for the higher power of x. Slide 4 - 35 Method 1 Let u=2x+1, then x=(1/2) (u-1), and dx=1/2 du. So Method 2 Let , then , and dx=u·du. So u x   2 1 1 2 ( 1) 2 x u   1 1 2 4 2 2 1 ( 1) ( ) 2 2 x x dx u u u du u u du           1 1 1 1 1 1 1 5 3 5 3 5/2 3/2 ( ) (2 1) (2 1) 2 5 3 10 6 10 6            u u C u u C x x C Slide 4 - 36 Exercises 1 2 cos(2 ) 4   x C 2 3 (1 cos ) 3 2 t  C 3    6 1 r C 4 2 3 ( 4 1) y y C    1 1 3/2 3/2 ( 1) sin(2 2) 3 6 x x C    

ExercisesSome TipsE1siar+c14.t，=-cos 2. An integrand may fequire some algebraic manipulafion before2x4the sutbstitution method can be applied.edx+cot(20)+C15,.ese 28 cot28 d8(a)Mulnigly by te'/e) -+ea2+1≥. Using u = cot 20b. Using cse 2Letw-en-e-c0r(20)+C,-一+1山-edcot(20)+Ccse*(20)+C,(a)-(6)-= tan'μ + Ctntegrale withrespectto= tan"i(e") + Ceplacea by e16.V5r+8h. Using i = V5r + 8seext=(sx)d=Se1-S0C1± Using r = 5r + 8omotb)(a)=/5x+8+c(6)=95x+8+ct stnsecx+tan x-/赠-(= In[a] + C = In[see x + tan | + C..Slide 4-37Some TipsSome Tips+3. We can use the subsitution method ofintegrafion as an 3. We can use the substitution method aof integrafion as anexplorationtool: Substitutefor the mosttroublesome part of theexploration tool:Substituteforthemosttroublesome partof theintegrand and seehow things work out.integrand and seehowthingswork out.2d2:dEXAMPLE8EvaluateEXAMPLE8EvaluateV+1W+ISolution 1: Substitute ur = +* + 1Solution 2: Substitute w = V + linstead2=dduLetu-+1-[-[W41-tV-2 +/u-l/ duloae fim /n-d=3/ud+℃-tategrate=3.号+CIndirgralc号u+C号(+1/+CReplaewbyt-+1s=号(2*+ 1)/ +CRepleeu liy 2+1Slide 4- 39Slide 4- 40ExercisesExercisesin(+cos(++1)+C21.+o29,Va+vsin(2#+1)31.L22,/Eos (3 + 4± =sin(3 +4)+Ccos(2+1)2cos(21+1)s(l--sin--n+cc(3x+2d=an(3+2)+C33.23.25-sine1+c34,cos(Vi+3d=2sin(F+3)+C35.26+cCosVo36.+(2. [----+c8sinyoVosiVaalgebraic-(1-)+(38.manpulationSlide 4- 41slide 4- 42

2016/11/15 7 Slide 4 - 37 Exercises 1 1 2 sin( ) 2 4 C x x     2 1 1 ( ) cot (2 ) 4 a C    2 2 1 ( ) csc (2 ) 4 b C    1 2 ( ) 5 8 5 a x C   2 2 ( ) 5 8 5 b x C   2 2 2 2 1 (1 cot (2 )) 4 1 1 cot (2 ) ( ) 4 4 C C          Slide 4 - 38 2. An integrand may require some algebraic manipulation before the substitution method can be applied. Slide 4 - 39 3. We can use the substitution method of integration as an exploration tool: Substitute for the most troublesome part of the integrand and see how things work out. Slide 4 - 40 3. We can use the substitution method of integration as an exploration tool: Substitute for the most troublesome part of the integrand and see how things work out. Slide 4 - 41 Exercises 2 1 C x     1 sin(3 4) 3    z C 1 tan(3 2) 3    x C 1 6 sin 2 3 x  C 1 8 tan 4 2 x  C 3 6 ( 1) 18 r   C Slide 4 - 42 Exercises 2 3/2 cos( 1) 3     x C 1 1 2 cos(2 1) C t     1 sin( 1) C t        2sin( 3) t C 1 2 cos 4 C    2 sin C     2 1 3/2 (1 ) 3 C x    algebraic manipulation

ExercisesProperties of Indefinite Integrals18tan67.Table of fundamental integralb12+tnJee'xit--cotx+CJkdr =kr+C (K is a constant)tan x,followed byw u,then by w2+uaHh.M= tan'x.followedby y =2+6fr'dreex+C+C (a+1)see.xtan+Ce.μ=2+tanxa+12+tan'x[些=m|x]+Csee.x+Creotxa68V+ sin* (x 1)sin (x 1)cos (x = 1) dJa'dr-,_+C (a>0, a±1)aresinx+Ca. u = + 1, followed by ur = sinu, then by w = I + vVi-b. u= sin (x - 1),followed byv =1 +Jed=e+cT+1+c. u= 1 + sin (x - 1)Jcosxdk=sinx+Csinh xdx =coshx+C-=(1+sin'(x-1)32+CJsin xde--cosx+csinhx+CnshJsee' xdr =tanr+Clide 4- 431004-45.3The area under the graph ofa positive function, the distance traveled by a moving object thatdoesn't change direction, and the average value of a nongative function over an intervalcan all be approximated by finite sums. First we subdivide the interval into subintervals,treating the appropriate fiunction f as if it were constant over each particular subinterval.Estimating withFinite SumsThen we multiply the width of each subinterval by the valuc of f at some point within it,and add these products togethet. Ir the interval [a, b] is subdivided ino a subinterals arequal widths Ar (b a)/n, and if f(ca) is the value of f at the chosen point e in theth subinteral, this process givesafinitesumofefomf(ei) Ar + f(cg) dr + f(cg) Ar +*-*+ f(c,) Ar.Thechoicesforthec could maximize orminimize the value of finthe kth subinterval,orgive some value in between. The true value lies somewhere between the approximationsgiven by upper sums and lower suns. The finite sum approximations we looked at im-proved as we took more subintervals of thinner width.slide 4- 45Slide 4- 48Suppose we want to find the area of the shaded region R that lies above the x-axis, belowthe graph of y = 1 - x, and between the vertical lines x = 0 and = 1 (Figure 5.1).-1-r1.TheArea Under theGraph of a0.5Positive Function0.5FIGURE 5.1 The area of the regioR cannot be found by a simpleformula,determiming the eaictarea ofceptforapproxmatingSlide 4- 47

2016/11/15 8 Slide 4 - 43 Exercises 3 6 2 tan C x    1 2 3/2 (1 sin ( 1)) 3    x C •44Slide 4 - 44 Properties of Indefinite Integrals Table of fundamental integrals kdx kx C    (k is a constant) 2 csc cot xdx x C     1 1 x x dx C         (  1) sec tan sec x xdx x C    ln | | dx x C x    csc cot sec x xdx x C     ln x x a a dx C a    (a  0, a  1) 2 arcsin 1 dx x C x     x x e dx e C    2 arctan 1 dx x C x     cos sin xdx x C    sinh cosh xdx x C    sin cos xdx x C     cosh sinh xdx x C    2 sec tan xdx x C    Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 45 5.3 Estimating with Finite Sums Slide 4 - 46 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 47 1. The Area Under the Graph of a Positive Function Slide 4 - 48 No method for determining the exact area of R except for approximating

The upper sum is obtained by taking the heightThe lower sum is obtained by taking the heightofeach rectangle as the maximum value of f(x)ofeach rectangle as the minimum value off(x)y=l-y= 1 32for a point x in each base interval of thefor a point x in each base interval of therectangle.rectangle.The length of each.The sum of the areas of the n rectangles issubinterval is.xf(Gla...4The sum of the areas of the n rectangles is-Sorecf(c)s..24ZoIn each of our computed sums, the interval [a, b] over which the function f is definedThe lkength ofeachwas subdivided into n subintervals of equal width (also ealled length) Js (b - a)/n,and / was evaluatod at a point in cach subinterval: , in the first subinterval, cz in the see-subnterval isAxond subinterval, and so on. The finite sums then all take the form(ei)x + f(e) + f(e) x +*++ f(e) r.By taking more and more rectangles, with each rectangle thinner than before, it appearsthat these finite sums give better and better approximations to the true area of the region Rlide 4- 49Slide 4-50When we don' know an antiderivative for the velocity function (c), we can apply thesame principle of approximating the distance traveled with finitesums ina way similartour estimates forarea discussed before, Wesubdivide the intenal[a, ] into short time intervals on each of which the velocity is considered to be fairly constant. Then we approxi-mate the distance traveled on each time subinterval with the usual distance formuladistance = velocity × timeand add the results across [a, b].2.The Distance Traveled byaSuppose the subdivided intenal looks likeMoving Object Without Changing-ar--4r--r-Direction+ (sec).with the subintervals all of equal length r, Pick a number tj in the first interval. If r isso small that the velocity barely changes over a short time interval of duration ,then thdistance traveled in the first time interval is about u(r) t, If ta is a mumber in the secondinterval, the distance traveled in the second time interval is about u(t) At. The sum of thedistances traveled over all the time intervals isD (h) Ar + o(r) A +*+ () r,where x is the total mumber of subintervals.Slide 4- 51EXAMPLE2The velocity function of a projectile fired straight into the air is(b)Thieebinasflengh,ihuluteghtndpointsgvingowerf(r) =160 -9.&r m/sec.Use the summation technique just described to estimate howfar the projectile rises during the first 3 sec, How close-do the sums come to the exact.value of 435.9 m?-ar-jSolutionWeexploretheresults for differentnumbers of intervals anddifferent choicesofvaluationoints.Notice that isdceasing, sochoosing tenointsgives anWith f evalunted at t = 1, 2, and 3, we haveper sum estimate; choosing right endpoints gives a lower sum estimate.D Jn) r + F(n) r + f(a) r(a)Thee subintenuls of length 1,with fevauatedar lefi endpointsghing anupper sam:= [160 9.8(1)](1) + [160 9.8(2)](1) + [160 9.8(3)](1)格= 421.2.X02[-ar-With evaluated at =0, 1,and 2, we hveD () + () + f) = [160 9.8(0)](1) + [160 9.8(1)](1) + [160 9.8(2)](1) 450.6.Slide 4- 53Slide 4- 54

2016/11/15 9 Slide 4 - 49 1 2 ( ) ( ) ( ) n f c x f c x f c x    The sum of the areas of the n rectangles is C1 C2C3 Cn The length of each subinterval is ⊿x The lower sum is obtained by taking the height of each rectangle as the minimum value of f(x) for a point x in each base interval of the rectangle. Slide 4 - 50 1 2 ( ) ( ) ( ) n f c x f c x f c x    The sum of the areas of the n rectangles is C1 C2 C3 Cn The length of each subinterval is ⊿x The upper sum is obtained by taking the height of each rectangle as the maximum value of f(x) for a point x in each base interval of the rectangle. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 51 2. The Distance Traveled by a Moving Object Without Changing Direction Slide 4 - 52 Slide 4 - 53 Slide 4 - 54

(c) Wath six subinrervals oflength 1/2, we getIt wouldbereasonableto concludefrom the table last entries thuat the projectilemosea232about 436 m during itsfist3 sec oflight..→!TABLES.2Travel-distanceestimatesHHNumbererLength ofeachUpperLomerThese estimates give an upper sum using left endpoints: D 443.25; and a lowersubesternahsubinteralsumsumsum using right endpoints: D 428.55.These six-interval estimates are somewhat180450.6421.21closer than the three-interval estimates. The results improve as the subintervals get443.25428.55shorter.439.58432.23As wecansee in Table 52, the leff-endpointuppersums approachthe true valuc437.74434.06435.9 from above, whereas the right-endpoint lower sums approach it from below. The true436.82434.98value liesbetwentheseupperandlowersums.Themagnitdeoftheerorintheclosest435.44436.36entries is 0.23, a small percentage of the true valuc1/6443613435,67Error magnitude = [true value calculated valuelBe closer and closer to [435.9 435.67] = 0.23.Lhe exact value of435.90.23Ero percentge0.05%slide 4-55Slide 4-58Q: The average value ofa collction ofn numbers X, X-.,X, isobtained by adding them together and dividing by n., But what isthe average value ofa continuous function on an interval [a,b?When a function is constant, this question is easy to answer. A function with constantluonan il[] hsagealuWhnositvesgphor []givesrectangleofheightc.Theawernge valhueofthefinctioncam thenbeinterpeedgeometrically as theareaofthisreetangle divided by its widhb (Figure 5.6a).3.TheAverage Value ofNonnegative Continuous Function=2-OveranIntervaloof(h)aHGuRE.6(a)hmegevaleofft)mlabjiaeafngedgiarea beneath its graphi divided by b a.Slide 4- s7Side 4- saEXAMPLE4Estimate the average value of the function f(x) = sin.x on the interval[0,=]We donot haveasimpleway todeerminethearea, oweapproximateitwithfinie=(.38+ .71+92+1+1+92+7#+ 38)号= (6.02)-量=2.365sumsToget anupoersum aoproximationweadd theareas ofiht rectanlesofequalwidth w/8 that together contain the region beneath the graph of y = sinx and above theToestimate the avenge value ofsinx we divide theestimated areaby wand obtaintheapx-axis on [0, w]. We choose the heights of the rectangles to be the largest value of sin x onproximation 2.365/ 0.753each subinterval. Over aparticular subinterval, this larngest value may occur at the lftend.poin, the right endpoint, or somewhere berween them, We evaluate sin x at this point toget the height of the rectangle for an unper sum. The sum oftherectangle areasthenestimates the total area (Figure 5.7):x) sin?0AFIGURE5.7Appeuximating thearea under f(x) = sinx betweenO and w to compute the averagevalue of sin x over [0, ], usingStide 4- 59Slide 4- 60eight rectangles (Example 4)

2016/11/15 10 Slide 4 - 55 Slide 4 - 56 Be closer and closer to the exact value of 435.9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 57 3. The Average Value of Nonnegative Continuous Function Over an Interval Slide 4 - 58 Q: The average value of a collection of n numbers x1 , x2 , ., xn is obtained by adding them together and dividing by n. But what is the average value of a continuous function on an interval [a,b]? Slide 4 - 59 Slide 4 - 60