Section 4: Integration ECO4112F 2011


 Morgan Smith
 2 years ago
 Views:
Transcription
1
2 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic in nture, tht is optimising the vlue of function without ny reference to time. In sttic optimistion nd comprtive sttic nlysis we mke the ssumption tht the process of economic djustment leds to n equilirium, nd we then emine the effect of chnges of the eogenous vriles on the equilirium vlues of the endogenous vriles. With dynmic nlysis, time is eplicitly considered in the nlysis. While we re not covering dynmic nlysis t this point, certin mthemtic tools re required for dynmic nlysis, such s integrtion nd differentil equtions. Without these tools, it ecomes impossile to consider prolems which re not sttic in nture. We will e covering oth of these topics in minly mthemticl wy, leving economic prolems for lter dte. Integrtion is the reverse process of differentition. If function () integrl of f () will yield F (). The nottion to denote integrtion is s follows: F hs first derivtive f () then the f ( ) d, where the integrl sign is n elongted S. f () is referred to s the integrnd, nd the d sign reminds us tht we re integrting with respect to the vrile. We go through the following eption to determine where this nottion comes from. Suppose we re given n ritrry function f () nd sked to find the re of the curve etween points, for emple the re under the curve f() etween nd. Figure. With liner function, this equtes to finding the re of tringle nd rectngle s follows: Figure.
3 However with nonliner function the prolem ecomes slightly more comple. Wht we cn do, however, is ttempt to find the re under the curve using numer of pproimting rectngles s follows: Figure. We let ech of the rectngles hve equl width nd we cll this width Δ. Ech rectngle hs height equl to the function vlue, for emple the height of the lst rectngle where = is equl to f()=. Thus the re of the lst rectngle is equl to (Δ), s re of rectngle equls length times redth, nd here redth is Δ nd length is f()=. The re of ny of the rectngles is equl to length times redth, which equls f() times Δ, s ll rectngles hve equl redth equl to Δ. To find the re under the curve we dd up the re of ech of the rectngles. This gives us the epression: n A f ( ) Where n = numer of rectngles i = the vlue of t ech point = the sum of ll the res, strting from the first one (i = ) nd ending t the nth one (i = ). i Oviously this sum will not e very ccurte representtion of the re. But perhps if we mke our Δ smller, then this epression will ecome more ccurte representtion of the re under the curve, s there will e less overshooting y ech rectngle. If initilly we hd ten rectngles, the re given y the sum of these rectngles res would oviously e more of n overestimte (or mye underestimte) thn if we douled the numer of rectngles, nd then summed their re. The more rectngles we use in this pproimting process the etter our estimte for the re under the curve. For emple, imgine we wish to find the re under the curve f() = etween nd. i Figure.
4 We could tke four rectngles, ech with redth Δ equl to., nd tke the heights from the right hnd side of ech rectngle. Hence the height for ech rectngle will e: (.) (.) (.7) Therefore the entire re equls to.(.) +.(.) +.(.7) +.() = /=.687 If we doule the mount of rectngles from four to eight, we will use Δ of., nd the following right hnd heights (rememer the height of the rectngle is given y the function vlue f()). (.) (.) (.7) (.) (.6) (.7) (.87) The corresponding totl re is given y the sum of ech of the res which is Δ multiplied y ech function vlue: The finl vlue we get is.987 As cn e seen in figure., using right end points for the rectngles for n incresing function will give n overestimte, while using right end points for decresing function will yield n overestimte. Thus douling the numer of rectngles while trying to estimte the re under the grph f()= will egin to ring our estimte down to its true vlue. It ppers tht s the numer of rectngles increses, our estimtions ecome etter nd etter pproimtions of the re. If we let the numer of rectngles tend to infinity, we will otin perfectly ccurte estimte for the re under our grph. Our epression for the re under the curve now ecomes: A lim n n i f ( ) This gives us the epression for the definite integrl, which gives us wy of finding the re under the continuous function f() etween = nd =: An eption of the terminology: f ( ) d lim n n i i f ( ) The integrtion sign is n elongted S, nd ws so chosen ecuse n integrl is limit of sums., re the limits of integrtion, f () is known s the integrnd. d is the lower limit of integrtion is the upper limit of integrtion. hs no mening y itself, ut merely reminds us tht we re integrting with respect to the vrile. Fortuntely when we wnt to find the re under curve, we do not hve to go into the long process of finding n epression for the sum of the re of n rectngles: numer of theorems mke the process esier. i
5 Before we set out the properties of the definite integrl, some rules of integrtion re s follows: (see pge 9 nd onwrds in Ching for emples).. The Power Rule ( ) n n n d c n. The Eponentil Rule e d e c. The Logrithmic Rule d c ( ) Properties of the definite integrl:. cd c( ). [ f ( ) g( )] d f ( ) d g( ) d. cf ( ) d c f ( ) d. [ f ( ) g( )] d f ( ) d g( ) d c c. f ( ) d f ( ) d f ( ) d Property sttes tht the integrl of constnt function y=c is the constnt times the length of the intervl, s seen in figure. Figure. Property sys tht the integrl of sum is the sum of the integrls. The re under f+g is the re under f plus the re under g. This property follows from the property of limits nd sums.
6 Property tells us tht constnt (ut only constnt) cn e tken in front of n integrl sign. This lso follows from the properties of limits nd sums. Property follows from property nd, using c=. Property tells us we cn find the re under the grph etween nd c, y splitting it up into two res, etween nd, nd etween nd c. We now find our rule for evluting the definite integrl: f ( ) d F( ) F( ) where the derivtive of F() is f(), i.e. F is ny ntiderivtive of f. For emple, if we differentite F( ) we otin Thus d F() F(). f ( ), so F() is n ntiderivtive of f(). Therefore the re under the curve f() = etween nd, is equl to third, or. recurring. Incidentlly this nswers our previous question which we ttempted using the sum of the res of n rectngles. The fundmentl theorem of clculus motivtes this use of the evlution theorem. In short, it sttes tht differentition nd integrtion re opposite processes. Thus, if we strt with function F(), nd differentite it to otin f(), [ i.e. F ( ) f ( ) ], if we then integrte the function f(), the result will e the initil function F(). Similrly if we integrte f() to otin F(), [ i.e. f ) d F( ) C ( where C is n ritrry constnt ]. Thus to find the integrl of function f(), we must find the function which when differentited yields f(). This theorem is very useful to us, s otherwise whenever we wish to find the vlue of the re tht lies underneth curve, we hve to go through the entire process of finding the limit of the sum of the res of n pproimting rectngles, which is time consuming process! Prior to the discovery of the fundmentl theorem, finding res, volumes nd other similr types of prolems were nigh on impossile. For completeness, the fundmentl theorem is presented elow: The fundmentl theorem of clculus: Suppose f() is continuous function on the closed intervl [,] d g ( f ( t) dt then g ( ) f ( ) i.e. g( ) f ( t) dt f ( ) d. If ). f ( ) d F( ) F( ) where the derivtive of F() is f(), i.e. F is ny ntiderivtive of f. Wht it sys, roughly speking, is tht if you integrte function nd then differentite the result, you retrieve the originl function. More out the ritrry constnt little lter
7 We now need to discuss two different types of integrls definite nd indefinite. A definite integrl involves finding the integrl of function etween two numer limits i.e. f ( ) d. The nswer to definite integrl is numer, s we know ccording to the evlution rule the nswer to this is just the ntiderivtive F() evluted etween nd, i.e. F()F(). An indefinite integrl yields function of s its nswer (if we re integrting with respect to ). An indefinite integrl is n integrl of the form f ( ) d (i.e without upper nd lower limits) nd the solution is f ( ) d F( ) C where C is n ritrry constnt which cn tke on ny vlue. The reson we include the ritrry constnt is illustrted in the following emple. Given the prolem: d potentil solution is we otin. However s this is n ntiderivtive of the cuic function (If we differentite is lso solution to this prolem, s is. This is ecuse when differentiting these epressions, the constnt differentited moves to zero. So it would pper tht the most generl form to give the nswer to this prolem would e s follows: d C, where C is n ritrry constnt. Just smll note on ritrry constnts when we dd two together, we otin third one which hs ggregted the first two, when multiplying, dividing dding or sutrcting numer y/to/from n ritrry constnt, the result is just the ritrry constnt. However the ritrry constnt when multiplied y function of, will sty s just tht: f ) g( ) d F( ) C G( ) C F( ) G( ) C ( where C nd C re two ritrry constnts, nd F() nd G() re two ntiderivtes of f() nd g() respectively. f ( ) d [ F( ) C] F( ) C F( ) C However: f ( ) d [ F( ) C] F( ) C We now turn to some rules of integrtion (definite nd indefinite) nd then some emples.. ( ) d c cf f ( ) d. ( f ( ) g( )) d f ( ) d g( ) d n n n. d C. d C (n cnnot equl ) 6
8 . e d e C sin d cos 6. d C 7. C 8. cos d sin C Rememer to check the nswer to ny integrtion sum just differentite it nd you should rrive ck t the originl function. Some emples:. d C. d C. d d. d d 7 ( ). [ sin ] d cos C cos C C 6. ( 6) d 6 (9) t t t t dt 8 () ( t 9 (t t t t t 6) dt 6t 8 8 c c t t 9 ) dt (t t t ) dt t t (Note the trick here) t 9 7
9 A few more useful properties:.. f ( ) d f ( ) d f ( ) d We re going to e looking t two very useful techniques used in integrtion: use of sustitution, nd integrtion y prts. There re whole host of other techniques which cn e useful, however it is these two which re most useful to us in economics. Integrtion using Sustitution We use sustitution, when the integrnd contins function nd its own derivtive. i.e: f ( ) f ( ) d For emple: ( )( ) d If this is the cse, we cn mke use of the following sustitution: We let u equl to the function whose derivtive we cn spot (or crete, using constnt: more out this lter). Let u = Then we know tht du ( ) d When we then sustitute the vlues of u nd du into the integrl, we otin the following integrl: udu which hs the nswer u udu C ut: u = Therefore our finl nswer is ( ) ( )( ) d C Thus the generl rule solution for the prolem is s follows: [ f ( )] f ( ) f ( ) d C or more simply f ( f. f ) d C However the sustitution rule cn e used for more complicted emples, when our function f() whose derivtive we cn spot occurs inside nother function: 8
10 For emple: f g ( f ) d or (6 8) ( 8 ) d In these cses the procedure does not chnge t ll we still mke the sustitution s follows: Let u = f() Therefore du=f`()d And proceed s usul: Some emples:. (6 8) ( 8 ) d Let u ( 8 ) Therefore du ( 6 8) d Therefore our trnsformed integrl is s follows: u u ( 8 ) du u du C C. Sometimes we cn find function whose derivtive we cn crete s follows: However only if we introduce constnt function. For emple: e ( ) d If we let u = (+), we know du = d. However while we hve the d, we do not hve. This is esily solved however through the following mnipultion: ( e ) d This integrl now contins function nd its derivtive, thus sustitution cn e used: Therefore let u = (+), nd du = d. The integrl ecomes: u e du e u C e () C Rememer, the sustitution of u into this function is device tht we employ. Therefore our finl nswer ought not to contin u, s the originl prolem does not contin it. Alwys rememer to sustitute ck for u. Note: Unlike differentition, there eists no generl formul giving the integrl of product of two function i.t.o. the seprte integrls of those functions. There is lso no generl formul giving the integrl of quotient of functions in terms of their seprte integrls. As result, integrtion is trickier thn differentition, on the whole. 9
11 Integrtion y Prts We use this technique when we hve to integrte product: Eg. f ( ) g'( ) d When we re given this type of emple, we mke use of the following formul: ( ) g'( ) d f ( ) g( ) f f '( ) g( ) d For emple: e d We pick f s the function which is esy to differentite, nd Often picking squred or cuic term for your f is good ide, s g ' s the function which is esy to integrte. f ' will hve power tht is then one f ', g', f nd g, to lower, nd hence simpler. It is very good ide to mke yourself mini tle with keep things stright. Also, note tht when finding g, we do not other with the ritrry constnt. e d Therefore: f f ' g' e g e e d e () e d e e C Thus we hve mnged to use the formul to integrte our originl question. ] [Cn lso use the lterntive nottion used in Ching: where vdu uv udv
12 Another emple: sin d Therefore: f f ' g' sin g cos sin d cos cos d cos sin C An emple with trick: d Oviously we do not know the integrl of, tht is why we re using this method. So we mke equl to f, nd we cn then find its derivtive. But then wht will our g ' e? Simple, mke it. f f ' g' g d d C This is hndy trick which cn lso e used to find the integrls of some of the trigonometric functions. Some more emples: e d Therefore: f f ' g' e e g
13 C e C e e d e e d e Another emple: d ' ' g f g f C d d d Another emple: d g f g f ' '
14 d d d C The lst line uses result tht we proved few emples go. Now let s try definite integrl using integrtion y prts: te t dt f t f ' g' e t g e t We first clculte the indefinite integrl, then go ck nd sustitute in the limits. te t dt te te t t e e t t dt C Therefore: te t dt te t e e e e e e t You should now mke sure you cn do the integrtion prctice questions. An emple of n economic ppliction of integrls: One simple ppliction is to find totl quntity from mrginl quntity. Suppose firm hs C '( ) e where denotes output. Then totl cost is: mrginl cost C( ) ( e ) d 6e B, where B is the constnt of integrtion.
Chapter 9 Definite Integrals
Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationTopics Covered AP Calculus AB
Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.
More informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in
More information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationThe practical version
Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht
More information7. Indefinite Integrals
7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find
More informationCalculus AB. For a function f(x), the derivative would be f '(
lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationCalculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.
Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite
More informationMat 210 Updated on April 28, 2013
Mt Brief Clculus Mt Updted on April 8, Alger: m n / / m n m n / mn n m n m n n ( ) ( )( ) n terms n n n n n n ( )( ) Common denomintor: ( ) ( )( ) ( )( ) ( )( ) ( )( ) Prctice prolems: Simplify using common
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More information2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationBridging the gap: GCSE AS Level
Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationMAT137 Calculus! Lecture 28
officil wesite http://uoft.me/mat137 MAT137 Clculus! Lecture 28 Tody: Antiderivtives Fundmentl Theorem of Clculus Net: More FTC (review v. 8.58.7) 5.7 Sustitution (v. 9.19.4) Properties of the Definite
More informationEvaluating Definite Integrals. There are a few properties that you should remember in order to assist you in evaluating definite integrals.
Evluting Definite Integrls There re few properties tht you should rememer in order to ssist you in evluting definite integrls. f x dx= ; where k is ny rel constnt k f x dx= k f x dx ± = ± f x g x dx f
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationAP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals
AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationNow, given the derivative, can we find the function back? Can we antidifferenitate it?
Fundmentl Theorem of Clculus. Prt I Connection between integrtion nd differentition. Tody we will discuss reltionship between two mjor concepts of Clculus: integrtion nd differentition. We will show tht
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors PreChpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More information10 Vector Integral Calculus
Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationMathematics Number: Logarithms
plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement
More informationReview of Gaussian Quadrature method
Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More informationAQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions
Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic
More information5.4. The Fundamental Theorem of Calculus. 356 Chapter 5: Integration. Mean Value Theorem for Definite Integrals
56 Chter 5: Integrtion 5.4 The Fundmentl Theorem of Clculus HISTORICA BIOGRAPHY Sir Isc Newton (64 77) In this section we resent the Fundmentl Theorem of Clculus, which is the centrl theorem of integrl
More informationChapter 7: Applications of Integrals
Chpter 7: Applictions of Integrls 78 Chpter 7 Overview: Applictions of Integrls Clculus, like most mthemticl fields, egn with tring to solve everd prolems. The theor nd opertions were formlized lter. As
More informationMath 131. Numerical Integration Larson Section 4.6
Mth. Numericl Integrtion Lrson Section. This section looks t couple of methods for pproimting definite integrls numericlly. The gol is to get good pproimtion of the definite integrl in problems where n
More informationMath 259 Winter Solutions to Homework #9
Mth 59 Winter 9 Solutions to Homework #9 Prolems from Pges 658659 (Section.8). Given f(, y, z) = + y + z nd the constrint g(, y, z) = + y + z =, the three equtions tht we get y setting up the Lgrnge multiplier
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More information4.6 Numerical Integration
.6 Numericl Integrtion 5.6 Numericl Integrtion Approimte definite integrl using the Trpezoidl Rule. Approimte definite integrl using Simpson s Rule. Anlze the pproimte errors in the Trpezoidl Rule nd Simpson
More information1 Part II: Numerical Integration
Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble
More informationBob Brown Math 251 Calculus 1 Chapter 5, Section 4 1 CCBC Dundalk
Bo Brown Mth Clculus Chpter, Section CCBC Dundlk The Fundmentl Theorem of Clculus Informlly, the Fundmentl Theorem of Clculus (FTC) sttes tht differentition nd definite integrtion re inverse opertions
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More information( ) Same as above but m = f x = f x  symmetric to yaxis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.
AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find
More informationMT Integral equations
MT58  Integrl equtions Introduction Integrl equtions occur in vriety of pplictions, often eing otined from differentil eqution. The reson for doing this is tht it my mke solution of the prolem esier or,
More informationAn Overview of Integration
An Overview of Integrtion S. F. Ellermeyer July 26, 2 The Definite Integrl of Function f Over n Intervl, Suppose tht f is continuous function defined on n intervl,. The definite integrl of f from to is
More informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
More information( ) as a fraction. Determine location of the highest
AB Clculus Exm Review Sheet  Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More informationLecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.
Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one
More information( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationThe Evaluation Theorem
These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for inclss presenttion nd should not
More informationFinal Exam  Review MATH Spring 2017
Finl Exm  Review MATH 5  Spring 7 Chpter, 3, nd Sections 5.5.5, 5.7 Finl Exm: Tuesdy 5/9, :37:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationAB Calculus Review Sheet
AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More informationLecture 20: Numerical Integration III
cs4: introduction to numericl nlysis /8/0 Lecture 0: Numericl Integrtion III Instructor: Professor Amos Ron Scribes: Mrk Cowlishw, Yunpeng Li, Nthnel Fillmore For the lst few lectures we hve discussed
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationLinear Inequalities. Work Sheet 1
Work Sheet 1 Liner Inequlities RentHep, cr rentl compny,chrges $ 15 per week plus $ 0.0 per mile to rent one of their crs. Suppose you re limited y how much money you cn spend for the week : You cn spend
More informationIntroduction to Algebra  Part 2
Alger Module A Introduction to Alger  Prt Copright This puliction The Northern Alert Institute of Technolog 00. All Rights Reserved. LAST REVISED Oct., 008 Introduction to Alger  Prt Sttement of Prerequisite
More informationLINEAR ALGEBRA APPLIED
5.5 Applictions of Inner Product Spces 5.5 Applictions of Inner Product Spces 7 Find the cross product of two vectors in R. Find the liner or qudrtic lest squres pproimtion of function. Find the nthorder
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationTime in Seconds Speed in ft/sec (a) Sketch a possible graph for this function.
4. Are under Curve A cr is trveling so tht its speed is never decresing during 1second intervl. The speed t vrious moments in time is listed in the tle elow. Time in Seconds 3 6 9 1 Speed in t/sec 3 37
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationSYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus
SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is
More information2. VECTORS AND MATRICES IN 3 DIMENSIONS
2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2dimensionl Vectors x A point in 3dimensionl spce cn e represented y column vector of the form y z zxis yxis z x y xxis Most of the
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationList all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.
Mth Anlysis CP WS 4.X Section 4.4.4 Review Complete ech question without the use of grphing clcultor.. Compre the mening of the words: roots, zeros nd fctors.. Determine whether  is root of 0. Show
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationFINALTERM EXAMINATION 9 (Session  ) Clculus & Anlyticl GeometryI Question No: ( Mrs: )  Plese choose one f ( x) x According to PowerRule of differentition, if d [ x n ] n x n n x n n x + ( n ) x n+
More information1 Error Analysis of Simple Rules for Numerical Integration
cs41: introduction to numericl nlysis 11/16/10 Lecture 19: Numericl Integrtion II Instructor: Professor Amos Ron Scries: Mrk Cowlishw, Nthnel Fillmore 1 Error Anlysis of Simple Rules for Numericl Integrtion
More informationChapters Five Notes SN AA U1C5
Chpters Five Notes SN AA U1C5 Nme Period Section 5: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles
More information