Introductory Mathematical Analysis

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5 Introductory Mthemticl Anlysis For Business, Economics, nd the Life nd Socil Sciences Arb World Edition Ernest F. Heussler, Jr. The Pennsylvni Stte University Richrd S. Pul The Pennsylvni Stte University Richrd J. Wood Dlhousie University Sdi Khouyibb Americn University of Shrjh

6 Acquisitions Editor: Rsheed Roussn Senior Development Editor: Sophie Bulbrook Project Editor: Nicole Elliott Copy-editor: Alice Yew Proofreder: John King nd XXXX Design Mnger: Srh Fch Permissions Editor: XXXX Picture Resercher: XXXX Indeer: XXXX Mrketing Mnger: Sue Miney Senior Mnufcturing Controller: Christopher Crow Cover Designer: XXXX Typesetter: Integr Typefce: Tir Printed in Chin. Person Eduction Limited Edinburgh Gte Hrlow Esse CM JE Englnd nd Associted Compnies throughout the world c Person Eduction Limited Authorized for sle only in the Middle Est nd North Afric. The rights of Ernest Heussler, Richrd Pul, Richrd Wood, nd Sdi Khouyibb to be identified s uthors of this work hve been sserted by them in ccordnce with the Copyright, Designs nd Ptents Act 988. All rights reserved. No prt of this publiction my be reproduced, stored in retrievl system, or trnsmitted in ny form or by ny mens, electronic, mechnicl, photocopying, recording or otherwise, without either the prior written permission of the publisher or licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Sffron House, 6 Kirby Street, London ECN 8TS. All trdemrks used herein re the property of their respective owners. The use of ny trdemrk in this tet does not vest in the uthor or publisher ny trdemrk ownership rights in such trdemrks, nor does the use of such trdemrks imply ny ffilition with or endorsement of this book by such owners. Person Eduction is not responsible for the content of third prty internet sites. First published IMP ISBN:

7 About the Adpting Author Sdi Khouyibb, Ph. D., is n instructor of mthemtics in the Deprtment of Mthemtics nd Sttistics, Americn University of Shrjh, UAE. She received Mster s degree in Grph Theory from Montrel University, Cnd, nd Ph. D. degree in History of Mthemtics from Lvl University, Quebec, Cnd. Her reserch interests re relted to the history of mthemtics nd mthemticl eduction, though her first voction is teching mthemtics. At AUS since 6, she hs tught severl courses including Preclculus, Algebr, Clculus, nd Mthemtics for Business. Her dediction nd pssion for the teching profession mkes her ecellent instructor who lwys mnges to find the best wy to communicte her knowledge, cpture students interest, nd stimulte their curiosity. When not teching, Dr. Khouyibb enjoys the compny of her husbnd Guillume nd their two kids Yssine nd Skin, with whom she shres some very enjoyble nd rewrding moments. v

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9 Contents Foreword v Prefce vii Acknowledgments i CHAPTER Review of Bsic Algebr. Sets of Rel Numbers. Some Properties of Rel Numbers. Eponents nd Rdicls 9. Opertions with Algebric Epressions 6.5 Fctoring Polynomils.6 Rtionl Epressions Chpter Review 9 Importnt Terms nd Symbols 9 Review Problems 9 Chpter Test CHAPTER Equtions nd Inequlities. Equtions, in Prticulr Liner Equtions. Qudrtic Equtions. Applictions of Equtions 9. Liner Inequlities 58.5 Applictions of Inequlities 6.6 Absolute Vlue 66 Chpter Review 7 Importnt Terms nd Symbols 7 Summry 7 Review Problems 7 Chpter Test 7 Vrible-Qulity Recording 7 CHAPTER Functions, Grphs nd Lines 75. Functions 76. Specil Functions 85. Combintions of Functions 9. Inverse Functions 95.5 Grphs in Rectngulr Coordintes 98.6 Lines 7.7 Liner Functions nd Applictions 6.8 Qudrtic Functions nd Prbols Chpter Review Importnt Terms nd Symbols Summry Review Problems Chpter Test Mobile Phones 5 CHAPTER Eponentil nd Logrithmic Functions 7. Eponentil Functions 8. Logrithmic Functions 5. Properties of Logrithms 58. Logrithmic nd Eponentil Equtions 6 vii

10 viii Contents Chpter Review 69 Importnt Terms nd Symbols 69 Summry 69 Review Problems 7 Chpter Test 7 Drug Dosges 7 CHAPTER Mthemtics of Finnce 76. Summtion Nottion nd Sequences 76. Simple nd Compound Interest 9. Present Vlue 97. Interest Compounded Continuously.5 Annuities 5.6 Amortiztion of Lons.7 Perpetuities Chpter Review Importnt Terms nd Symbols Summry Review Problems 5 Chpter Test 5 Tresury Securities 7 CHAPTER 5 Mtri Algebr 9 5. Systems of Liner Equtions 5. Applictions of Systems of Liner Equtions 5. Mtrices 9 5. Mtri Addition nd Sclr Multipliction Mtri Multipliction Solving Systems of Liner Equtions by the Guss Jordn Method Inverses Leontief s Input Output Anlysis 99 Chpter 5 Review 5 Importnt Terms nd Symbols 5 Summry 6 Review Problems 6 Chpter Test 8 Insulin Requirements s Liner Process 9 CHAPTER 6 Liner Progrmming 6. Liner Inequlities in Two Vribles 6. Liner Progrmming: Grphicl Approch 7 6. The Simple Method: Mimiztion 9 6. The Simple Method: Nonstndrd Mimiztion Problems Minimiztion The Dul 66 Chpter 6 Review 76 Importnt Terms nd Symbols 76 Summry 76 Review Problems 77 Chpter Test 79 Drug nd Rdition Therpies 8 CHAPTER 7 Introduction to Probbility nd Sttistics 8 7. Bsic Counting Principle nd Permuttions 8 7. Combintions nd Other Counting Principles 9 7. Smple Spces nd Events 7. Probbility Conditionl Probbility nd Stochstic Processes

11 Contents i 7.6 Independent Events Byes s Formul 6 Chpter 7 Review 5 Importnt Terms nd Symbols 5 Summry 55 Review Problems 56 Chpter Test 58 Probbility nd Cellulr Automt 6 CHAPTER 8 Additionl Topics in Probbility 6 8. Discrete Rndom Vribles nd Epected Vlue 6 8. The Binomil Distribution 7 8. Mrkov Chins 78 Chpter 8 Review 88 Importnt Terms nd Symbols 88 Summry 88 Review Problems 89 Chpter Test 9 Mrkov Chins in Gme Theory 9 CHAPTER 9 Limits nd Continuity 9 9. Limits 9 9. One-Sided Limits nd Limits t Infinity 5 9. Continuity 5 9. Continuity Applied to Inequlities 57 Chpter 9 Review 5 Importnt Terms nd Symbols 5 Summry 5 Review Problems 5 Chpter Test 55 Public Debt 55 CHAPTER Differentition 57. The Derivtive 58. Rules for Differentition 56. The Derivtive s Rte of Chnge 5. The Product Rule nd the Quotient Rule The Chin Rule 565 Chpter Review 57 Importnt Terms nd Symbols 57 Summry 57 Review Problems 57 Chpter Test 576 Mrginl Propensity to Consume 577 CHAPTER Additionl Differentition Topics 579. Derivtives of Logrithmic Functions 58. Derivtives of Eponentil Functions 586. Elsticity of Demnd 59. Implicit Differentition Logrithmic Differentition 6.6 Higher-Order Derivtives 67 Chpter Review 6 Importnt Terms nd Symbols 6 Summry 6 Review Problems 6 Chpter Test 6 Economic Order Quntity 65

12 Contents CHAPTER Curve Sketching 67. Reltive Etrem 68. Absolute Etrem on Closed Intervl 6. Concvity 6. The Second-Derivtive Test 6.5 Asymptotes 6.6 Applied Mim nd Minim 65 Chpter Review 66 Importnt Terms nd Symbols 66 Summry 66 Review Problems 66 Chpter Test 667 Popultion Chnge over Time 668 CHAPTER Integrtion 67. Differentils 67. The Indefinite Integrl 676. Integrtion with Initil Conditions 68. More Integrtion Formuls Techniques of Integrtion 69.6 The Definite Integrl The Fundmentl Theorem of Integrl Clculus 75.8 Are between Curves 7.9 Consumers nd Producers Surplus 7 Chpter Review 76 Importnt Terms nd Symbols 76 Summry 77 Review Problems 78 Chpter Test 7 Delivered Price 7 CHAPTER Methods nd Applictions of Integrtion 7. Integrtion by Prts 75. Integrtion by Tbles 79. Averge Vlue of Function 76. Differentil Equtions 78.5 More Applictions of Differentil Equtions Improper Integrls 76 Chpter Review 765 Importnt Terms nd Symbols 765 Summry 765 Review Problems 766 Chpter Test 768 Dieting 769 CHAPTER 5 Continuous Rndom Vribles Continuous Rndom Vribles The Norml Distribution The Norml Approimtion to the Binomil Distribution 78 Chpter 5 Review 787 Importnt Terms nd Symbols 787 Summry 787 Review Problems 788 Chpter Test 788 Cumultive Distribution from Dt 789 CHAPTER 6 Multivrible Clculus Functions of Severl Vribles Prtil Derivtives 8

13 Contents i 6. Applictions of Prtil Derivtives Higher-Order Prtil Derivtives Mim nd Minim for Functions of Two Vribles Lgrnge Multipliers Lines of Regression 8 Chpter 6 Review 86 Importnt Terms nd Symbols 86 Summry 86 Review Problems 87 Chpter Test 88 Dt Anlysis to Model Cooling 8 APPENDIX A Compound Interest Tbles 8 APPENDIX B Tble of Selected Integrls 85 APPENDIX C Ares Under the Stndrd Norml Curve 85 English Arbic Glossry of Mthemticl Terms G- Answers to Odd-Numbered Problems AN- Inde I- Photo Credits P-

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15 Prefce The Arb World edition of Introductory Mthemticl Anlysis for Business, Economics, nd the Life nd Socil Sciences is built upon one of the finest books of its kind. This edition hs been dpted specificlly to meet the needs of students in the Arb world, nd provides mthemticl foundtion for students in vriety of fields nd mjors. It begins with preclculus nd finite mthemtics topics such s functions, equtions, mthemtics of finnce, mtri lgebr, liner progrmming, nd probbility. Then it progresses through both single vrible nd multivrible clculus, including continuous rndom vribles. Technicl proofs, conditions, nd the like re sufficiently described but re not overdone. Our guiding philosophy led us to include those proofs nd generl clcultions tht shed light on how the corresponding clcultions re done in pplied problems. Informl intuitive rguments re often given s well. Approch The Arb World Edition of Introductory Mthemticl Anlysis for Business, Economics, nd the Life nd Socil Sciences follows unique pproch to problem solving. As hs been the cse in erlier editions of this book, we estblish n emphsis on lgebric clcultions tht sets this tet prt from other introductory, pplied mthemtics books. The process of clculting with vribles builds skill in mthemticl modeling nd pves the wy for students to use clculus. The reder will not find definition-theorem-proof tretment, but there is sustined effort to imprt genuine mthemticl tretment of rel world problems. Emphsis on developing lgebric skills is etended to the eercises, in which mny, even those of the drill type, re given with generl coefficients. In ddition to the overll pproch to problem solving, we im to work through emples nd eplntions with just the right blend of rigor nd ccessibility. The tone of the book is not too forml, yet certinly not lcking precision. One might sy the book reds in reled tone without scrificing opportunities to bring students to higher level of understnding through strongly motivted pplictions. In ddition, the content of this edition is presented in more logicl wy for those teching nd lerning in the Arb region, in very mngeble portions for optiml teching nd lerning. Wht s New in the Arb World Edition? A number of dpttions nd new fetures hve been dded to the Arb World Edition. Additionl Emples nd Problems: Hundreds of rel life emples nd problems bout the Arb World hve been incorported. Additionl Applictions: Mny new Apply It fetures from cross the Arb region hve been dded to chpters to provide etr reinforcement of concepts, nd to provide the link between theory nd the rel world. Chpter test: This new feture hs been dded to every chpter to solidify the lerning process. These problems do not hve solutions provided t the end of the book, so cn be used s clss tests or homework. Biogrphies: These hve been included for prominent nd importnt mthemticins. This historicl ccount gives its rightful plce to both Arb nd interntionl contributors of this gret science. English-Arbic Glossry: Mthemticl, finncil nd economic terms with trnsltion to Arbic hs been dded to the end of the book. Any instructor with eperience in the Arb World knows how helpful this is for the students who studied in high school in Arbic. v

16 vi Prefce CAUTION Other Fetures nd Pedgogy Applictions: An bundnce nd vriety of new nd dditionl pplictions for the Arb udience pper throughout the book; students continully see how the mthemtics they re lerning cn be used in fmilir situtions, providing rel-world contet. These pplictions cover such diverse res s business, economics, biology, medicine, sociology, psychology, ecology, sttistics, erth science, nd rcheology. Mny of these rel-world situtions re drwn from literture nd re documented by references, sometimes from the Web. In some, the bckground nd contet re given in order to stimulte interest. However, the tet is self-contined, in the sense tht it ssumes no prior eposure to the concepts on which the pplictions re bsed. (See, for emple, pge XXX, Emple X in X.X) Apply It: The Apply It eercises provide students with further pplictions, with mny of these covering compnies nd trends from cross the region. Locted in the mrgins, these dditionl eercises give students rel-world pplictions nd more opportunities to see the chpter mteril put into prctice. An icon indictes Apply It problems tht cn be solved using grphing clcultor. Answers to Apply It problems pper t the end of the tet nd complete solutions to these problems re found in the Solutions Mnuls. (See, for emple, pge XXX, Apply It X in X.X) Now Work Problem N: Throughout the tet we hve retined the populr Now Work Problem N feture. The ide is tht fter worked emple, students re directed to n end of section problem (lbeled with blue eercise number) tht reinforces the ides of the worked emple. This gives students n opportunity to prctice wht they hve just lerned. Becuse the mjority of these keyed eercises re odd-numbered, students cn immeditely check their nswer in the bck of the book to ssess their level of understnding. The complete solutions to these eercises cn be found in the Student Solutions Mnul. (See, for emple, pge XXX, Emple X in XX.X) Cutions: Throughout the book, cutionry wrnings re presented in very much the sme wy n instructor would wrn students in clss of commonly-mde errors. These Cutions re indicted with n icon to help students prevent common misconceptions. (See, for emple, pge XXX, Emple X in XX.X) Definitions, key concepts, nd importnt rules nd formuls re clerly stted nd displyed s wy to mke the nvigtion of the book tht much esier for the student. (See, for emple, pge XXX, Definition of Derivtive in XX.X) Eplore & Etend Activities: Strtegiclly plced t the end of the chpter, these to bring together multiple mthemticl concepts studied in the previous sections within the contet of highly relevnt nd interesting ppliction. Where pproprite, these hve been dpted to the Arb World. These ctivities cn be completed in or out of clss either individully or within group. (See, for emple, pge XXX, in Chpter XX) Review Mteril: Ech chpter hs review section tht contins list of importnt terms nd symbols, chpter summry, nd numerous review problems. In ddition, key emples re referenced long with ech group of importnt terms nd symbols. (See, for emple, pge XXX, in Chpter XX) Bck-of-Book Answers: Answers to odd-numbered problems pper t the end of the book. For mny of the differentition problems, the nswers pper in both unsimplified nd simplified forms. This llows students to redily check their work. (See, for emple, pge AN-XX, in Answers for XX.X) Emples nd Eercises Most instructors nd students will gree tht the key to n effective tetbook is in the qulity nd quntity of the emples nd eercise sets. To tht end, hundreds emples re worked out in detil. Mny of these re new nd bout the Arb World, with rel regionl dt nd sttistics included wherever possible. These problems tke the reder from the popultion growth of Ciro, to the Infnt Mortlity rte in Tunisi, the life epectncy in Morocco, the

17 Prefce vii divorce rte in Algeri, the unemployment rte in Sudi Arbi, the eports nd imports of Kuwit, the oil production in Tunisi nd Sudi Arbi, Lbor Force in Morocco, the CPI of Liby, the GDC of Lebnon, the popultion of Bhrin in the ge group of 5 to 6, nd the number of doctors in Jordn. They lso include populr products from the region, nd locl compnies like Air Arbi, Royl Jordnin Airline, Emirtes, oil compnies such s Armco, postl compnies like Arme, telecommuniction providers such s Etislt or Mentel, the stocks of Emr. Regionl trends re lso covered in these problems, such s internet users in Yemen, mobile subscriptions in Syri, the emission of CO in Qtr, the number of shops in Dubi, the production of oil nd nturl gs in Omn, the production of electricity nd fresh ornge in Morocco, the prticiption to the Olympic gmes by the Arb ntions, nd the concept of Murbh in Islmic finnce. Some emples include strtegy bo designed to guide students through the generl steps of the solution before the specific solution is obtined (See pges XXX XXX, XX.X emple X). In ddition, n bundnt number of digrms nd eercises re included. In ech eercise set, grouped problems re given in incresing order of difficulty. In most eercise sets the problems progress from the bsic mechnicl drill-type to more interesting thought-provoking problems. The eercises lbeled with blue eercise number correlte to Now Work Problem N sttement nd emple in the section. A gret del of effort hs been put into producing proper blnce between the drilltype eercises nd the problems requiring the integrtion nd ppliction of the concepts lerned. (see pges XXX XXX, Eplore nd Etend for Chpter X; XXX, Eplore nd Etend for Chpter X; XXX XXX, Emple X in XX.X on Lines of Regression) Technology In order tht students pprecite the vlue of current technology, optionl grphing clcultor mteril ppers throughout the tet both in the eposition nd eercises. It ppers for vriety of resons: s mthemticl tool, to visulize concept, s computing id, nd to reinforce concepts. Although clcultor displys for TI-8 Plus ccompny the corresponding technology discussion, our pproch is generl enough so tht it cn be pplied to other grphing clcultors. In the eercise sets, grphing clcultor problems re indicted by n icon. To provide fleibility for n instructor in plnning ssignments, these problems re typiclly plced t the end of n eercise set. Course Plnning One of the obvious ssets of this book is tht considerble number of courses cn be served by it. Becuse instructors pln course outline to serve the individul needs of prticulr clss nd curriculum, we will not ttempt to provide detiled smple outlines. Introductory Mthemticl Anlysis is designed to meet the needs of students in Business, Economics, nd Life nd Socil Sciences. The mteril presented is sufficient for two semester course in Finite Mthemtics nd Clculus, or three semester course tht lso includes College Algebr nd Core Preclculus topics. The book consists of three importnt prts: Prt I: College Algebr The purpose of this prt is to provide students with the bsic skills of lgebr needed for ny subsequent work in Mthemtics. Most of the mteril covered in this prt hs been tught in high school. Prt II: Finite Mthemtics The second prt of this book provides the student with the tools he needs to solve rel-world problems relted to Business, Economic or Life nd Socil Sciences. Prt III: Applied Clculus In this lst prt the student will lern how to connect some Clculus topics to rel life problems.

18 viii Prefce Supplements The Student Solutions Mnul includes worked solutions for ll odd-numbered problems nd ll Apply It problems. ISBN XXXXX XXXXX The Instructor s Solution Mnul hs worked solutions to ll problems, including those in the Apply It eercises nd in the Eplore & Etend ctivities. It is downlodble from the Instructor s Resource Center t XXXXX. TestGen ( enbles instructors to build, edit, nd print, nd dminister tests using computerized bnk of questions developed to cover ll the objectives of the tet. TestGen is lgorithmiclly bsed, llowing instructors to crete multiple but equivlent versions of the sme question or test with the click of button. Instructors cn lso modify test bnk questions or dd new questions. The softwre nd testbnk re vilble for downlod from Person Eduction s online ctlog nd from the Instructor s Resource Center t XXXXXX. MyMthLb, gretly pprecited by instructors nd students, is powerful online lerning nd ssessment tool with interctive eercises nd problems, uto-grding, nd ssignble sets of questions tht cn be ssigned to students by the click of mouse.

19 Acknowledgments We epress our pprecition to the following collegues who contributed comments nd suggestions tht were vluble to us in the evolution of this tet: Nizr Bu Fkhreeddine, Deprtment of Mthemtics nd Sttistics, Notre Dme University Zouk Mousbeh, Lebnon Dr. Mged Isknder, Fculty of Business Administrtion, Economics nd Politicl Science, British University in Egypt Dr. Fud A. Kittneh, Deprtment of Mthemtics, University of Jordn, Jordn Hithm S. Solh, Deprtment of Mthemtics, Americn University in Dubi, UAE Michel M. Zlzli, Deprtment of Mthemtics, UAE University, UAE Mny reviewers nd contributors hve provided vluble contributions nd suggestions for previous editions of Introductory Mthemticl Anlysis. Mny thnks to them for their insights, which hve informed our work on this dpttion. Sdi Khouyibb i

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21 Introductory Mthemticl Anlysis For Business, Economics, nd the Life nd Socil Sciences Arb World Edition

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23 Integrtion. Differentils. The Indefinite Integrl. Integrtion with Initil Conditions. More Integrtion Formuls.5 Techniques of Integrtion.6 The Definite Integrl.7 The Fundmentl Theorem of Integrl Clculus.8 Are between Curves.9 Consumers nd Producers Surplus Chpter Review Delivered Price Anyone who runs business knows the need for ccurte cost estimtes. When jobs re individully contrcted, determining how much job will cost is generlly the first step in deciding how much to bid. For emple, pinter must determine how much pint job will tke. Since gllon of pint will cover certin number of squre meters, the key is to determine the re of the surfces to be pinted. Normlly, even this requires only simple rithmetic wlls nd ceilings re rectngulr, nd so totl re is sum of products of bse nd height. But not ll re clcultions re s simple. Suppose, for instnce, tht the bridge shown below must be sndblsted to remove ccumulted soot. How would the contrctor who chrges for sndblsting by the squre meter clculte the re of the verticl fce on ech side of the bridge? A D The re could be estimted s perhps three-qurters of the re of the trpezoid formed by points A, B, C, nd D. But more ccurte clcultion which might be desirble if the bid were for dozens of bridges of the sme dimensions (s long stretch of rilrod) would require more refined pproch. If the shpe of the bridge s rch cn be described mthemticlly by function, the contrctor could use the method introduced in this chpter: integrtion. Integrtion hs mny pplictions, the simplest of which is finding res of regions bounded by curves. Other pplictions include clculting the totl deflection of bem due to bending stress, clculting the distnce trveled underwter by submrine, nd clculting the electricity bill for compny tht consumes power t differing rtes over the course of month. Chpters delt with differentil clculus. We differentited function nd obtined nother function, its derivtive. Integrl clculus is concerned with the reverse process: We re given the derivtive of function nd must find the originl function. The need for doing this rises in nturl wy. For emple, we might hve mrginl-revenue function nd wnt to find the revenue function from it. Integrl clculus lso involves concept tht llows us to tke the limit of specil kind of sum s the number of terms in the sum becomes infinite. This is the rel power of integrl clculus! With such notion, we cn find the re of region tht cnnot be found by ny other convenient method. C B 67

24 67 Chpter Integrtion Objective To define the differentil, interpret it geometriclly, nd use it in pproimtions. Also, to restte the reciprocl reltionship between d/dy nd dy/d.. Differentils We will soon give reson for using the symbol dy/d to denote the derivtive of y with respect to. To do this, we introduce the notion of the differentil of function. Definition Let y = f () be differentible function of, nd let denote chnge in, where cn be ny rel number. Then the differentil of y, denoted dy or d( f ()), is given by dy = f () Note tht dy depends on two vribles, nmely, nd. In fct, dy is function of two vribles. EXAMPLE Computing Differentil Find the differentil of y = +, nd evlute it when = nd =.. Solution: The differentil is dy = d d ( + ) = ( + ) When = nd =., Isc Newton Isc Newton (6 77) is considered to be one of the most influentil physicists ever. His groundbreking findings, published in 687 in Philosophie Nturlis Principi Mthemtic ( Mthemticl Principles of Nturl Philosophy ), form the foundtion of clssicl mechnics. He nd Leibniz independently developed wht could be clled the most importnt discovery in mthemtics: the differentil nd integrl clculus. dy = [() () + ](.) =.8 Now Work Problem If y =, then dy = d() = =. Hence, the differentil of is. We bbrevite d() by d. Thus, d =. From now on, it will be our prctice to write d for when finding differentil. For emple, d( + 5) = d d ( + 5) d = d Summrizing, we sy tht if y = f () defines differentible function of, then dy = f () d where d is ny rel number. Provided tht d, we cn divide both sides by d: dy d = f () Tht is, dy/d cn be viewed either s the quotient of two differentils, nmely, dy divided by d, or s one symbol for the derivtive of f t. It is for this reson tht we introduced the symbol dy/d to denote the derivtive. EXAMPLE Finding Differentil in Terms of d. If f () =, then d( ) = d d ( ) d = / d = d b. If u = ( + ) 5, then du = 5( + ) () d = ( + ) d. Now Work Problem

25 Section. Differentils 67 y y f( ) f ( d) Q f ( d) f( ) f( ) P R S L dy y d d FIGURE. Geometric interprettion of dy nd. The differentil cn be interpreted geometriclly. In Figure., the point P(, f ()) is on the curve y = f (). Suppose chnges by d, rel number, to the new vlue + d. Then the new function vlue is f ( + d), nd the corresponding point on the curve is Q( + d, f ( + d)). Pssing through P nd Q re horizontl nd verticl lines, respectively, tht intersect t S. A line L tngent to the curve t P intersects segment QS t R, forming the right tringle PRS. Observe tht the grph of f ner P is pproimted by the tngent line t P. The slope of L is f () but it is lso given by SR/PS so tht Since dy = f () d nd d = PS, f () = SR PS dy = f () d = SR PS = SR PS Thus, if d is chnge in t P, then dy is the corresponding verticl chnge long the tngent line t P. Note tht for the sme d, the verticl chnge long the curve is y = SQ = f (+d) f (). Do not confuse y with dy. However, from Figure., the following is pprent: When d is close to, dy is n pproimtion to y. Therefore, y dy APPLY IT. The number of personl computers in Kuwit from 995 to 5 cn be pproimted by N(t) = where t = corresponds to the yer 995. Use differentils to pproimte the chnge in the number of computers s t goes from 995 to 5. Source: Bsed on dt from the United Ntions Sttistics Division. This fct is useful in estimting y, chnge in y, s Emple shows. EXAMPLE Using the Differentil to Estimte Chnge in Quntity A governmentl helth gency in the Middle Est emined the records of group of individuls who were hospitlized with prticulr illness. It ws found tht the totl proportion P tht re dischrged t the end of t dys of hospitliztion is given by ( ) P = P(t) = + t Use differentils to pproimte the chnge in the proportion dischrged if t chnges from to 5. Solution: The chnge in t from to 5 is t = dt = 5 = 5. The chnge in P is P = P(5) P(). We pproimte P by dp: ( ) ( P dp = P (t) dt = ) dt = + t ( + t) ( + t) dt

26 67 Chpter Integrtion When t = nd dt = 5, dp = 6 5 = 5 6 = =. For comprison, the true vlue of P is P(5) P() = =.7 (to five deciml plces). Now Work Problem Formul () is used to pproimte function vlue, wheres the formul y dy is used to pproimte chnge in function vlues. so tht We sid tht if y = f (), then y dy if d is close to zero. Thus, y = f ( + d) f () dy f ( + d) f () + dy () This formul gives us wy of estimting function vlue f ( + d). For emple, suppose we estimte ln(.6). Letting y = f () = ln, we need to estimte f (.6). Since d(ln ) = (/) d, we hve, from Formul (), f ( + d) f () + dy ln ( + d) ln + d We know the ect vlue of ln, so we will let = nd d =.6. Then + d =.6, nd d is close to zero. Therefore, ln ( +.6) ln () + (.6) ln (.6) +.6 =.6 The true vlue of ln(.6) to five deciml plces is.587. EXAMPLE Using the Differentil to Estimte Function Vlue A shoe mnufcturer in Sudn estblished tht the demnd function for its sports shoes is given by p = f (q) = q where p is the price per pir of shoes in dollrs for q pirs. By using differentils, pproimte the price when 99 pirs of shoes re demnded. Solution: We wnt to pproimte f (99). By Formul (), where f (q + dq) f (q) + dp dp = q dq dp dq = q / We choose q = nd dq = becuse q + dq = 99, dq is smll, nd it is esy to compute f () = =. We thus hve f (99) = f [ + ( )] f () ( ) f (99) +.5 =.5 Hence, the price per pir of shoes when 99 pirs re demnded is pproimtely $.5. Now Work Problem 5 The eqution y = defines y s function of. We could write f () = However, the eqution lso defines implicitly s function of y. In fct,

27 Section. Differentils 675 if we restrict the domin of f to some set of rel numbers so tht y = f () is one-toone function, then in principle we could solve for in terms of y nd get = f (y). [Actully, no restriction of the domin is necessry here. Since f () = + >, for ll, we see tht f is strictly incresing on (, ) nd is thus one-to-one on (, ).] As we did in Section., we cn look t the derivtive of with respect to y, d/dy nd we hve seen tht it is given by d dy = dy d provided tht dy/d Since d/dy cn be considered quotient of differentils, we now see tht it is the reciprocl of the quotient of differentils dy/d. Thus d dy = + It is importnt to understnd tht it is not necessry to be ble to solve y = for in terms of y, nd the eqution d dy = holds for ll. + EXAMPLE 5 Find dp dq if q = 5 p. Solution: Finding dp/dq from dq/dp Strtegy There re number of wys to find dp/dq. One pproch is to solve the given eqution for p eplicitly in terms of q nd then differentite directly. Another pproch to find dp/dq is to use implicit differentition. However, since q is given eplicitly s function of p, we cn esily find dq/dp nd then use the preceding reciprocl reltion to find dp/dq. We will tke this pproch. We hve Hence, dq dp = (5 p ) / p ( p) = 5 p dp dq = dq dp = 5 p p Now Work Problem 5 PROBLEMS. In Problems 9, find the differentil of the function in terms of nd d.. y = + b. y =. f () = 9. f () = ( 5 + ) 5. u = 6. u = 7. p = ln ( + 7) 8. p = e y = ln + In Problems, find y nd dy for the given vlues of nd d.. y = 5 ; =, d =.. y = + b; for ny nd ny d. y = + 5 7; =, d =.. y = ( + ) ; =, d =.. y = ; =, d =.5 Round your nswer to three deciml plces. 5. Let f () = () Evlute f (). (b) Use differentils to estimte the vlue of f (.). In Problems 6, pproimte ech epression by using differentils (Hint: 7 = 89.)

28 676 Chpter Integrtion. ln.97. ln.. e.. e. In Problems 9, find d/dy or dp/dq.. y = y = 6. q = (p + 5) 7. q = p q = p 9. q = e p. If y = 7 6 +, find the vlue of d/dy when =.. If y = ln, find the vlue of d/dy when =. In Problems nd, find the rte of chnge of q with respect to p for the indicted vlue of q.. p = 5 q + ; q = 8. p = 6 q; q = 5. Profit Suppose tht the profit (in dollrs) of producing q units of product is P = 97q.q Using differentils, find the pproimte chnge in profit if the level of production chnges from q = 9 to q = 9. Find the true chnge. 5. Revenue Given the revenue function r = 5q + 5q q use differentils to find the pproimte chnge in revenue if the number of units increses from q = to q =. Find the true chnge. 6. Demnd The demnd eqution for product is p = q Using differentils, pproimte the price when units re demnded. 7. Demnd Given the demnd function p = q + 8 use differentils to estimte the price per unit when units re demnded. 8. If y = f (), then the proportionl chnge in y is defined to be y/y, which cn be pproimted with differentils by dy/y. Use this lst form to pproimte the proportionl chnge in the cost function c = f (q) = q + 5q + when q = nd dq =. Round your nswer to one deciml plce. 9. Sttus/Income Suppose tht S is numericl vlue of sttus bsed on person s nnul income I (in thousnds of dollrs). For certin popultion, suppose S = I. Use differentils to pproimte the chnge in S if nnul income decreses from $5, to $,5.. Biology The volume of sphericl cell is given by V = πr, where r is the rdius. Estimte the chnge in volume when the rdius chnges from 6.5 cm to 6.6 cm.. Muscle Contrction The eqution (P + )(v + b) = k is clled the fundmentl eqution of muscle contrction. Here P is the lod imposed on the muscle, v is the velocity of the shortening of the muscle fibers, nd, b, nd k re positive constnts. Find P in terms of v, nd then use the differentil to pproimte the chnge in P due to smll chnge in v.. Profit The demnd eqution for monopolist s product is nd the verge-cost function is p = q 66q + 7 c = 5 q + 8, q () Find the profit when units re demnded. (b) Use differentils nd the result of prt () to estimte the profit when units re demnded. Objective To define the ntiderivtive nd the indefinite integrl nd to pply bsic integrtion formuls.. The Indefinite Integrl Given function f, if F is function such tht F () = f () () then F is clled n ntiderivtive of f. Thus, An ntiderivtive of f is simply function whose derivtive is f. Multiplying both sides of Eqution () by the differentil d gives F () d = f () d. However, becuse F () d is the differentil of F, we hve df = f () d. Hence, we cn think of n ntiderivtive of f s function whose differentil is f () d. R. W. Stcy et l., Essentils of Biologicl nd Medicl Physics (New York: McGrw-Hill, 955).

29 Section. The Indefinite Integrl 677 Definition An ntiderivtive of function f is function F such tht Equivlently, in differentil nottion, F () = f () df = f () d For emple, becuse the derivtive of is, is n ntiderivtive of. However, it is not the only ntiderivtive of : Since d d ( + ) = nd d d ( 5) = both + nd 5 re lso ntiderivtives of. In fct, it is obvious tht becuse the derivtive of constnt is zero, +C is lso n ntiderivtive of for ny constnt C. Thus, hs infinitely mny ntiderivtives. More importntly, ll ntiderivtives of must be functions of the form + C, becuse of the following fct: Any two ntiderivtives of function differ only by constnt. Since + C describes ll ntiderivtives of, we cn refer to it s being the most generl ntiderivtive of, denoted by d, which is red the indefinite integrl of with respect to. Thus, we write d = + C The symbol is clled the integrl sign, is the integrnd, nd C is the constnt of integrtion. The d is prt of the integrl nottion nd indictes the vrible involved. Here is the vrible of integrtion. More generlly, the indefinite integrl of ny function f with respect to is written f () d nd denotes the most generl ntiderivtive of f. Since ll ntiderivtives of f differ only by constnt, if F is ny ntiderivtive of f, then f () d = F() + C, where C is constnt To integrte f mens to find f () d. In summry, f () d = F() + C if nd only if F () = f () Thus we hve ( d d ) f () d = f () nd d (F()) d = F() + C d which shows the etent to which differentition nd indefinite integrtion re inverse procedures.

30 678 Chpter Integrtion APPLY IT. Suppose tht the mrginl cost for Mhrn Co. is f (q) = 8., find 8. dq, which gives the form of the cost function. CAUTION A common mistke is to omit C, the constnt of integrtion. EXAMPLE 6 Find 5 d. Solution: Finding n Indefinite Integrl Strtegy First we must find (perhps better words re guess t) function whose derivtive is 5. Then we dd the constnt of integrtion. Since we know tht the derivtive of 5 is 5, 5 is n ntiderivtive of 5. Therefore, 5 d = 5 + C Now Work Problem Abu Ali Ibn l-hythm Abu Ali l-hsn ibn l-hythm (965 ), born in Irq, ws one of the most fmousarb scientists, who left importnt works in stronomy, mthemtics, medicine nd physics. More thn 6 yers before integrls were known; he developed in his mnuscript Kitāb l-mnāzir ( Book of Optics ) method to clculte wht were, in fct, integrls of fourth-degree polynomils. Tble. Elementry Integrtion Formuls. k d = k + C k is constnt. d = C. d = d =. e d = e + C kf () d = k f () d (f () ± g()) d = d = = ln + C for > f () d ± k is constnt g() d Using differentition formuls from Chpters nd, we hve compiled list of elementry integrtion formuls in Tble.. These formuls re esily verified. For emple, Formul () is true becuse the derivtive of + /( + ) is for =. (We must hve = becuse the denomintor is when =.) Formul () sttes tht the indefinite integrl of power of, other thn, is obtined by incresing the eponent of by, dividing by the new eponent, nd dding constnt of integrtion. The indefinite integrl of will be discussed in Section.. To verify Formul (5), we must show tht the derivtive of k f () d is kf (). Since the derivtive of k f () d is simply k times the derivtive of f () d, nd the derivtive of f () d is f (), Formul (5) is verified. The reder should verify the other formuls. Formul (6) cn be etended to ny number of terms. EXAMPLE 7. Find d. Indefinite Integrls of Constnt nd of Power of Solution: By Formul () with k = d = + C = + C Usully, we write d s d. Thus, d = + C.

31 b. Find 5 d. Section. The Indefinite Integrl 679 Solution: By Formul () with n = 5, 5 d = C = C Now Work Problem APPLY IT. If the rte of chnge of Hossm Compny s revenues cn be modeled by dr =.t, then find.t dt, dt which gives the form of the compny s revenue function. CAUTION Only constnt fctor of the integrnd cn pss through n integrl sign. EXAMPLE 8 Find 7 d. Indefinite Integrl of Constnt Times Function Solution: By Formul (5) with k = 7 nd f () =, 7 d = 7 d Since is, by Formul () we hve d = C = + C where C is the constnt of integrtion. Therefore, ( ) 7 d = 7 d = 7 + C = 7 + 7C Since 7C is just n rbitrry constnt, we will replce it by C for simplicity. Thus, 7 d = 7 + C It is not necessry to write ll intermedite steps when integrting. More simply, we write 7 d = (7) + C = 7 + C Now Work Problem 5 EXAMPLE 9 Indefinite Integrl of Constnt Times Function Find 5 e d. Solution: 5 e d = e d Formul (5) 5 = 5 e + C Formul () Now Work Problem APPLY IT. Suppose tht due to new competition, the number of subscriptions toarb World mgzine is declining t rte of ds = 8 subscriptions per month, dt t where t is the number of months since the competition entered the mrket. Find the form of the eqution for the number of subscribers to the mgzine. EXAMPLE Finding Indefinite Integrls. Find dt. t Solution: Here t is the vrible of integrtion. We rewrite the integrnd so tht bsic formul cn be used. Since / t = t /, pplying Formul () gives dt = t / dt = t( /)+ t / t + C = + C = t + C +

32 68 Chpter Integrtion b. Find Solution: 6 d. 6 d = 6 d = ( ) C = + C = + C Now Work Problem 9 APPLY IT 5. The growth of the popultion of Riydh for the yers 995 to is estimted to follow the lw dn dt = t where t is the number of yers fter 995 nd N is in thousnds of individuls. Find n eqution tht describes the popultion of Riydh. Source: Bsed on dt from mongby.com/popultion_estimtes/full/ Riydh-Sudi_Arbi. html (ccessed November 8, ). When integrting n epression involving more thn one term, only one constnt of integrtion is needed. APPLY IT 6. The growth rte of pssengers flown by Royl Jordnin Airlines from to 8 cn be modeled by dn dt =.68t +.55t.95t +.76 where t is the time in yers nd N is the number of pssengers in millions. Find the form of the eqution describing the number of pssengers flown by Royl Jordnin. Source: Bsed on dt from the 9 Royl Jordnin Annul Report. EXAMPLE Indefinite Integrl of Sum Find ( + ) d. Solution: By Formul (6), ( + ) d = Now, nd Thus, d = d + d d = C = + C d = () C = + C ( + ) d = + + C + C For convenience, we will replce the constnt C + C by C. We then hve ( + ) d = + + C Omitting intermedite steps, we simply integrte term by term nd write ( + ) d = + () + C = + + C Now Work Problem EXAMPLE Indefinite Integrl of Sum nd Difference Find ( e ) d. Solution: ( e ) d = = () 9/5 9 5 /5 d 7 d + e d d Formuls (5) nd (6) (7) + e + C Formuls (), (), nd () = 9 9/5 7 + e + C Now Work Problem 5

33 Section. The Indefinite Integrl 68 Sometimes, in order to pply the bsic integrtion formuls, it is necessry first to perform lgebric mnipultions on the integrnd, s Emple shows. EXAMPLE Using Algebric Mnipultion to Find n Indefinite Integrl ( Find y y + ) dy. CAUTION In Emple, we first multiplied the fctors in the integrnd. The nswer could not hve been found simply in terms of y dy nd (y + ) dy. There is not formul for the integrl of generl product of functions. Solution: The integrnd does not fit fmilir integrtion form. However, by multiplying the integrnd we get ( y y + ) dy = (y + ) y dy ( ) = y y + + C = y + y 9 + C Now Work Problem 9 EXAMPLE ( )( + ). Find d. 6 Using Algebric Mnipultion to Find n Indefinite Integrl Solution: By fctoring out the constnt nd multiplying the binomils, we get 6 ( )( + ) d = ( + 5 ) d 6 6 = ) (() 6 + (5) + C Another lgebric pproch to prt (b) is d = ( ) d = ( ) d nd so on. b. Find d. = C Solution: We cn brek up the integrnd into frctions by dividing ech term in the numertor by the denomintor: ( d = ) d = ( ) d = + C = + + C Now Work Problem 7 PROBLEMS. In Problems 5, find the indefinite integrls.. 7 d. d. 8 d. 5 d z d 6. dz d 8. d 7 9/ 9. dt t7/.. ( + t) dt.. (5 w 6w ) dw. 5. (t t + 5) dt 6. (7r 5 + r + ) dr (y 5 5y) dy ( + t + t + t 6 ) dt ( + e) d

34 68 Chpter Integrtion 7. (5 ) d 8. ( ) d.. πe d.. (.7y + + y ) dy. d d 7. ( 8. ) d 9. ( w. ) dw. w u. du. 5. (u e + e u ) du dt 7. ( 7 ) d (e + + ) d ( ) d dz () d ( ) d 7e s ds ( ) e d ( ) d ( ) d ( ) 8. u + u du 9. ( + 5)( ) d (. ( ) d. + ) d. (z + ) dz. (u + ) du ( ). 5 d 5. ( + 5) d 6. (6e u u ( z u + )) du + z 7. dz z e + e d 9. d 5 e 5. ( + ) d 5. If F() nd G() re such tht F () = G (), is it true tht F() G() must be zero? 5. () Find function F such tht F() d = e + C. (b) Is there only one function F stisfying the eqution given in prt (), or re there mny such functions? 5. Find d d ( + ) d. Objective To find prticulr ntiderivtive of function tht stisfies certin conditions. This involves evluting constnts of integrtion.. Integrtion with Initil Conditions If we know the rte of chnge, f, of the function f, then the function f itself is n ntiderivtive of f (since the derivtive of f is f ). Of course, there re mny ntiderivtives of f, nd the most generl one is denoted by the indefinite integrl. For emple, if f () = then f () = f () d = d = + C. () Tht is, ny function of the form f () = +C hs its derivtive equl to. Becuse of the constnt of integrtion, notice tht we do not know f () specificlly. However, if f must ssume certin function vlue for prticulr vlue of, then we cn determine the vlue of C nd thus determine f () specificlly. For instnce, if f () =, then, from Eqution (), Thus, f () = + C = + C C = f () = + Tht is, we now know the prticulr function f () for which f () = nd f () =. The condition f () =, which gives function vlue of f for specific vlue of, is clled n initil condition.

35 Section. Integrtion with Initil Conditions 68 APPLY IT 7. The rte of growth of species of bcteri is estimted by dn = 8+e t, dt where N is the number of bcteri (in thousnds) fter t hours. If N(5) =,, find N(t). EXAMPLE 5 Initil-Condition Problem Suppose tht the mrginl profit of plstics fctory in Qtr is given by the function P () = where is the number (in thousnds) of items produced nd P represents the profit in thousnds of dollrs. Find the profit function, ssuming tht selling no items results in loss of $,. Solution: The profit function is P() = + 5 d = 5 5 d = d 5 d + 5 d d + 5 d = C = C () We determine the vlue of C by using the initil condition: substitute = nd P() = into Eqution () to get Hence, 75 () + 5() + C = C = P() = () Now Work Problem APPLY IT 8. The ccelertion of n object fter t seconds is given by y = 8t +, the velocity t 8 seconds is given by y (8) = 89 m, nd the position t seconds is given by y() = 85 m. Find y(t). EXAMPLE 6 Initil-Condition Problem Involving y Given tht y = 6, y () =, nd y() =, find y. Solution: Strtegy To go from y to y, two integrtions re needed: the first to tke us from y to y nd the other to tke us from y to y. Hence, there will be two constnts of integrtion, which we will denote by C nd C. Since y = d d (y ) = 6, y is n ntiderivtive of 6. Thus, y = ( 6) d = 6 + C () Now, y () = mens tht y = when = ; therefore, from Eqution (), we hve Hence, C =, so = 6() + C y = 6 +

36 68 Chpter Integrtion By integrtion, we cn find y: ( ) y = 6 + d ( ) = (6) + + C so y = + + C (5) Now, since y = when =, we hve, from Eqution (5), Thus, C =, so = () + () + C y = + Now Work Problem 5 Integrtion with initil conditions is pplicble to mny pplied situtions, s the net three emples illustrte. EXAMPLE 7 Income nd Eduction Suppose tht for prticulr Arb group, sociologists studied the current verge yerly income y (in dollrs) tht person cn epect to receive with yers of eduction before seeking regulr employment. They estimted tht the rte t which income chnges with respect to eduction is given by where y = 8,7 when = 9. Find y. dy d = / 6 Solution: Here y is n ntiderivtive of /. Thus, y = / d = / d = () 5/ 5 + C y = 5/ + C (6) The initil condition is tht y = 8,7 when = 9. By putting these vlues into Eqution (6), we cn determine the vlue of C: 8,7 = (9) 5/ + C = () + C 8,7 = 97 + C Therefore, C = 9,, nd y = 5/ + 9, Now Work Problem 7

37 Section. Integrtion with Initil Conditions 685 EXAMPLE 8 Finding Revenue from Mrginl Averge Revenue Suppose tht the mrginl verge revenue in dollrs of Ali Bb Museum resulting from the sle of tickets is given by R () = + If the verge revenue from the sle of tickets is $5, wht is the revenue when 5 tickets re sold? Solution: To find the revenue function, we first find the verge revenue. We hve R() = + d = + ln + C To find C, we use the initil condition R() = 5. This gives R() = + ln() + C = 5 C = 5 ln() Therefore R() = + ln() + nd hence R() = ( + ln() + ). So the revenue from the sle of 5 tickets is R(5) = 5(5 + ln(5) + ) 796 dollrs EXAMPLE 9 Finding the Demnd Function from Mrginl Revenue If the mrginl-revenue function for mnufcturer s product is find the demnd function. Solution: dr = q q dq Strtegy By integrting dr/dq nd using n initil condition, we cn find the revenue function r. But revenue is lso given by the generl reltionship r = pq, where p is the price per unit. Thus, p = r/q. Replcing r in this eqution by the revenue function yields the demnd function. Since dr/dq is the derivtive of totl revenue r, r = ( q q ) dq so tht = q () q ()q + C Revenue is when q is. r = q q q + C (7) We ssume tht when no units re sold, there is no revenue; tht is, r = when q =. This is our initil condition. Putting these vlues into Eqution (7) gives Although q = gives C =, this is not true in generl. It occurs in this section = () () + C becuse the revenue functions re polynomils. In lter sections, evluting Hence, C =, nd t q = my produce nonzero vlue for C. r = q q q

38 686 Chpter Integrtion To find the demnd function, we use the fct tht p = r/q nd substitute for r: p = r q = q q q q p = q q Now Work Problem EXAMPLE Finding Cost from Mrginl Cost Suppose tht Al Hllb Resturnt s fied costs per week re $. (Fied costs re costs, such s rent nd insurnce, tht remin constnt t ll levels of production during given time period.) If the mrginl-cost function is dc dq =.(.q 5q) +. where c is the totl cost (in dollrs) of producing q mels per week, find the cost of producing mels in week. Solution: Since dc/dq is the derivtive of the totl cost c, c(q) = [.(.q 5q) +.] dq =. (.q 5q) dq +. dq When q is, totl cost is equl to fied cost. Although q = gives C vlue equl to fied costs, this is not true in generl. It occurs in this section becuse the cost functions re polynomils. In lter sections, evluting t q = my produce vlue for C tht is different from fied cost. (.q ) c(q) =. 5q +.q + C Fied costs re constnt regrdless of output. Therefore, when q =, c =, which is our initil condition. Putting c() = in the lst eqution, we find tht C =, so (.q ) c(q) =. 5q +.q + (8) From Eqution (8), we hve c() = Thus, the totl cost for producing mels in week is $88.7. Now Work Problem 5 PROBLEMS. In Problems nd, find y subject to the given conditions.. dy/d = ; y( ) =. dy/d = ; y() = 9 In Problems, if y stisfies the given conditions, find y() for the given vlue of.. y = 9 8, y(6) = ; = 9 In Problems 7, find y subject to the given conditions.. y = + ; y () =, y() = 5 5. y = + ; y () =, y() = 6. y = ; y ( ) =, y () =, y() = 7. y = e + ; y () = 7, y () = 5, y() = In Problems 8, dr/dq is mrginl-revenue function. Find the demnd function. 8. dr/dq =.7 9. dr/dq = 6 q. dr/dq = 5, (q+q ). dr/dq = 75 q.q

39 Section. More Integrtion Formuls 687 In Problems 5, dc/dq is mrginl-cost function nd fied costs re indicted in brces. For Problems nd, find the totl-cost function. For Problems nd 5, find the totl cost for the indicted vlue of q.. dc/dq =.7; {59}. dc/dq = q + 75; {}. dc/dq =.q.6q + 6; {5,}; q = 5. dc/dq =.8q.6q + 6.5; {8}; q = 5 6. Winter Moth A study of the winter moth ws mde in Nov Scoti, Cnd. The prepupe of the moth fll onto the ground from host trees. It ws found tht the (pproimte) rte t which prepupl density y (the number of prepupe per squre foot of soil) chnges with respect to distnce (in feet) from the bse of host tree is If y = 59.6 when =, find y. dy d = Diet for Rts A group of biologists studied the nutritionl effects on rts tht were fed diet contining % protein. The protein consisted of yest nd corn flour. If G = 8 when P =, find G. dg dp = P 5 + P 8. Fluid Flow In the study of the flow of fluid in tube of constnt rdius R, such s blood flow in portions of the body, one cn think of the tube s consisting of concentric tubes of rdius r, where r R. The velocity v of the fluid is function of r nd is given by v = (P P )r lη where P nd P re pressures t the ends of the tube, η ( Greek letter red et ) is fluid viscosity, nd l is the length of the tube. If v = when r = R, show tht dr v = (P P )(R r ) lη 9. Averge Cost Amrn mnufctures jens nd hs determined tht the mrginl-cost function is dc dq =.q.q + where q is the number of pirs of jens produced. If mrginl cost is $7.5 when q = 5 nd fied costs re $5, wht is the verge cost of producing pirs of jens? Over period of time, the group found tht the (pproimte) rte of chnge of the verge weight gin G (in grms) of rt with respect to the percentge P of yest in the protein mi ws. If f () = + nd f () =, evlute f (965.55) f ( ) Objective To lern nd pply the formuls for u du, e u du, nd u du.. More Integrtion Formuls Power Rule for Integrtion The formul d = + n + + C if = which pplies to power of, cn be generlized to hndle power of function of. Let u be differentible function of. By the power rule for differentition, if =, then ( (u()) +) d d + = ( + )(u()) u () + = (u()) u () Adpted from D. G. Embree, The Popultion Dynmics of the Winter Moth in Nov Scoti, 95 96, Memoirs of the Entomologicl Society of Cnd, no. 6 (965). Adpted from R. Bressni, The Use of Yest in Humn Foods, in Single-Cell Protein, eds. R. I. Mteles nd S. R. Tnnenbum (Cmbridge, MA: MIT Press, 968). R. W. Stcy et l., Essentils of Biologicl nd Medicl Physics (New York: McGrw-Hill, 955).

40 688 Chpter Integrtion Thus, (u()) u () d = (u())+ + + C We cll this the power rule for integrtion. Note tht u ()d is the differentil of u, nmely du. In mthemticl shorthnd, we cn replce u() by u nd u () d by du: Power Rule for Integrtion If u is differentible, then u du = u+ + C + if () It is importnt to pprecite the difference between the power rule for integrtion nd the formul for d. In the power rule, u represents function, wheres in d, is vrible. EXAMPLE Applying the Power Rule for Integrtion. Find ( + ) d. Solution: Since the integrnd is power of the function +, we will set u = +. Then du = d, nd (+) d hs the form u du. By the power rule for integrtion, ( + ) d = u du = u + C = ( + ) + C Note tht we give our nswer not in terms of u, but eplicitly in terms of. b. Find ( + 7) d. Solution: We observe tht the integrnd contins power of the function + 7. Let u = + 7. Then du = d. Fortuntely, ppers s fctor in the integrnd nd we hve ( + 7) d = ( + 7) [ d] = u du After integrting, you my wonder wht hppened to. We note gin tht du = d. = u + C = ( + 7) + C Now Work Problem In order to pply the power rule for integrtion, sometimes n djustment must be mde to obtin du in the integrnd, s Emple illustrtes. APPLY IT 9. A shoe mnufcturer in Lebnon finds tht the mrginl cost of producing pirs of shoes is pproimted by C () = +9 if the fied costs re $.. Find the cost function EXAMPLE Adjusting for du Find + 5 d. Solution: We cn write this s ( + 5) / d. Notice tht the integrnd contins power of the function + 5. If u = + 5, then du = d. Since the constnt fctor in du does not pper in the integrnd, this integrl does not hve the

41 Section. More Integrtion Formuls 689 CAUTION The nswer to n integrtion problem must be epressed in terms of the originl vrible. CAUTION We cn djust for constnt fctors, but not vrible fctors. form u n du. However, from du = d we cn write d = du so tht the integrl becomes ( + 5) / d = ( + 5) / / du [ d] = u Moving the constnt fctor in front of the integrl sign, we hve ( + 5) / d = ( ) u / du = u / + C = u/ + C which in terms of (s is required) gives + 5 d = ( + 5) / + C Now Work Problem 5 In Emple, the integrnd + 5 missed being of the form (u()) / u () by the constnt fctor of. In generl, if we hve (u()) u () d, for k nonzero k constnt, then we cn write (u()) u () d = u du k k = u du k to simplify the integrl, but such djustments of the integrnd re not possible for vrible fctors. When using the form u du, do not neglect du. For emple, ( + ) ( + ) d + C The correct wy to do this problem is s follows. Let u = +, from which it follows tht du = d. Thus d = du nd ( + ) d = [ ] du u = u du = u + C = ( + ) + C EXAMPLE Applying the Power Rule for Integrtion. Find 6y dy. Solution: The integrnd is (6y) /, power of function. However, in this cse the obvious substitution u = 6y cn be voided. More simply, we hve 6y dy = 6 / y / dy = 6 y / y / dy = 6 + C = 6 y/ + C b. Find + ( + + 7) d. Solution: We cn write this s ( + + 7) ( + ) d. Let us try to use the power rule for integrtion. If u = + + 7, then du = ( + 6) d, which is two times the quntity ( + ) d in the integrl. Thus ( + ) d = du nd we gin illustrte the djustment technique: [ ] du ( + + 7) [( + ) d] = u = u du = u + C = 6u + C = 6( + + 7) + C Now Work Problem 5

42 69 Chpter Integrtion In using the power rule for integrtion, tke cre when mking choice for u. In Emple (b), letting u = + does not led very fr. At times it my be necessry to try mny different choices. Sometimes wrong choice will provide hint s to wht does work. Skill t integrtion comes only fter mny hours of prctice nd conscientious study. EXAMPLE An Integrl to Which the Power Rule Does Not Apply Find ( + ) d. Solution: If we set u = +, then du = d. To get du in the integrl, we need n dditionl fctor of the vrible. However, we cn djust only for constnt fctors. Thus, we cnnot use the power rule. Insted, to find the integrl, we will first epnd ( + ) : ( + ) d = ( ) d = ( ) d ( ) = C Integrting Nturl Eponentil Functions Now Work Problem 65 We now turn our ttention to integrting eponentil functions. If u is differentible function of, then d d (eu ) = e u du d CAUTION Do not pply the power-rule formul for u du to e u du. Corresponding to this differentition formul is the integrtion formul e u du d d = eu + C But du d is the differentil of u, nmely, du. Thus, d e u du = e u + C () APPLY IT. When n object is moved from one environment to nother, its temperture T chnges t rte given by dt = kce kt, where t is the time (in dt hours) fter chnging environments, C is the temperture difference (originl minus new) between the environments, nd k is constnt. If the originl environment is 7, the new environment is 6, nd k =.5, find the generl form of T(t). EXAMPLE 5 Integrls Involving Eponentil Functions. Find e d. Solution: Let u =. Then du = d, nd, by Eqution (), e d = e [ d] = e u du = e u + C = e + C

43 b. Find ( + )e + d. Section. More Integrtion Formuls 69 Solution: If u = +, then du = ( + ) d = ( + ) d. If the integrnd contined fctor of, the integrl would hve the form e u du. Thus, we write ( + )e + d = e + [( + ) d] = e u du = eu + C = e + + C where in the second step we replced ( + ) d by du but wrote outside the integrl. Now Work Problem 9 Integrls Involving Logrithmic Functions As we know, the power-rule formul u du = u + /( + ) + C does not pply when =. To hndle tht sitution, nmely, u du = du, we first recll from u Section. tht d d ( ln u ) = du u d for u = which gives us the integrtion formul du = ln u + C u for u = () In prticulr, if u =, then du = d, nd d = ln + C for = () APPLY IT. If the rte of vocbulry memoriztion of the verge student in foreign lnguge is given by dv dt = 5 t +, where v is the number of vocbulry words memorized in t hours of study, find the generl form of v(t). EXAMPLE 6 7. Find d. Integrls Involving u du Solution: From Eqution (), 7 d = 7 d = 7 ln + C Using properties of logrithms, we cn write this nswer nother wy: 7 d = ln 7 + C b. Find + 5 d. Solution: Let u = + 5. Then du = d. From Eqution (), + 5 d = [ d] = + 5 u du = ln u + C = ln C Since + 5 is lwys positive, we cn omit the bsolute-vlue brs: + 5 d = ln ( + 5) + C Now Work Problem 9

44 69 Chpter Integrtion EXAMPLE 7 An Integrl Involving u du The mnger of Al Mdin Superstore determines tht the price of bottle of minerl wter is chnging t the rte of dp dq = q q + 5 where q is the number (in hundreds) of bottles demnded by customers t price p. If the quntity demnded is 5 when the price is $.5, t wht price would no bottles be sold? Solution: The price function is p(q) = q q + 5 dq Let u = q + 5; then du = q dq, so tht q du = du. Hence p(q) = q du q + 5 dq = u Rewriting u in terms of q, we get = du = ln u + C u p(q) = ln (q + 5) + C To find the vlue of C, we use the fct tht p(5) =.5. Thus, p(5) = ln ((5) + 5) + C =.5 C =.8 So p(q) = ln (q + 5) +.8, nd the price t which no bottles of wter would be sold is p() = ln (() + 5) +.8 =.78 Now Work Problem 9 EXAMPLE 8 An Integrl Involving Two Forms ( Find ( w) + ) dw. w Solution: ( ( w) + w ) dw = = ( w) dw + w dw ( w) [ dw] + w dw The first integrl hs the form u du, nd the second hs the form ( ( w) + ) ( w) dw = + ln w + C w dv. Thus, v = w + ln w + C

45 Section. More Integrtion Formuls 69 PROBLEMS. In Problems 78, find the indefinite integrls.. ( + 5) 7 d. 5( + ) d. ( + ) 5 d. ( + )( + + ) d 5. (y + 6y)(y + y + ) / dy 6. (5t 6t + )(5t t + t) 7 dt d 8. ( ) + d 9. d. (7 6) d 5. ( + 7) d. u(5u 9) du. + 5 d. ( 5) 9 d 5. (7 + 5 ) / d 6. e d 7. 5e t+7 dt 8. (t + )e t +t+ dt 9. w e w dw. e 7 d. e d. e d. 5 e 6 +7 d. + 5 d d d 8z 7. dz 8. (z 5) 7 (5v ) dv 9. d. + y dy s. ds. s d 5 7t. d. 5t 6 dt 5 5. d 6. () d b d 8. 9 d 9. y e y + dy. d. v e v dv. d + +. (e 5 + e ) d. y + dy (8 + )(7 5) d ye y dy d 8. (e + e e 5 6s ) d 9. s + s ds 5. (6t + t)(t + t + ) 6 dt 5. ( + ) d 5. (5w + 8w + )(w 5 + w + ) dw 5. ( 5 )( 6 ) d 5. (v )e v+v dv 55. ( + )( + ) d (e. ) d 57. (5 ) d 58. (e e ) d 59. ( + )e + d 6. (u ue 6 u ) du 6. (8 5 ) d ( ) 6. e d 6. d 6. d 65. ( + ) d e [ ( 6) ] d + 5 ( ) [ ] d 68. ( + ) + d ( ) [ ] ( 8 5 )( 6 ) 8 d [ ] 7. (r + 5) dr 7. + d + ( ) d ( + ) e 7. d 7. (e 5 e ) d + e 75. d 76. e t t + 9 dt ln ( + )d 78. e 8 d In Problems 79 8, find y subject to the given conditions. 79. y = ( ) ; y() = 8. y = + 6 ; y() = 8. y = ; y ( ) =, y() = 8. y = ( + ) / ; y (8) = 9, y() = Rel Estte The rte of chnge of the vlue of house in Djerb, Tunisi, tht cost $5, to build cn be modeled by dv = 8e.5t, where t is the time in yers since the house ws built dt nd V is the vlue (in thousnds of dollrs) of the house. Find V(t).

46 69 Chpter Integrtion 8. Oygen in Cpillry In discussion of the diffusion of oygen from cpillries, 5 concentric cylinders of rdius r re used s model for cpillry. The concentrtion C of oygen in the cpillry is given by ( Rr C = K + B ) dr r where R is the constnt rte t which oygen diffuses from the cpillry, nd K nd B re constnts. Find C. (Write the constnt of integrtion s B.) 85. Find f () if f ( ) = nd f () = e +. Objective To discuss techniques of hndling more chllenging integrtion problems, nmely, by lgebric mnipultion nd by fitting the integrnd to fmilir form. To integrte n eponentil function with bse different from e nd to find the consumption function, given the mrginl propensity to consume. Here we split up the integrnd. Here we used long division to rewrite the integrnd..5 Techniques of Integrtion We turn now to some more difficult integrtion problems. When integrting frctions, sometimes preliminry division is needed to get fmilir integrtion forms, s the net emple shows. EXAMPLE 9 +. Find d. Preliminry Division before Integrtion Solution: A fmilir integrtion form is not pprent. However, we cn brek up the integrnd into two frctions by dividing ech term in the numertor by the denomintor. We then hve ( + d = + ) ( d = + ) d = + ln + C b. Find d. + Solution: Here the integrnd is quotient of polynomils in which the degree of the numertor is greter thn or equl to tht of the denomintor. In such sitution we first use long division. Recll tht if f nd g re polynomils, with the degree of f greter thn or equl to the degree of g, then long division llows us to find (uniquely) polynomils q nd r, where either r is the zero polynomil or the degree of r is strictly less thn the degree of g, stisfying f g = q + r g Using n obvious, bbrevited nottion, we see tht ( f g = q + r ) r = q + g g Since integrting polynomil is esy, we see tht integrting rtionl functions reduces to the tsk of integrting proper rtionl functions those for which the degree of the numertor is strictly less thn the degree of the denomintor. In this cse we obtin + ( + + d = + + ) d + + = d = + + d( + ) + = + + ln + + C Now Work Problem 5 W. Simon, Mthemticl Techniques for Physiology nd Medicine (New York: Acdemic Press, Inc., 97).

47 Section.5 Techniques of Integrtion 695 EXAMPLE. Find Indefinite Integrls ( ) d. Solution: We cn write this integrl s ( ) d. Let us try the power rule Here the integrl is fit to the form to which the power rule for integrtion pplies. Here the integrl fits the fmilir form u du. Here the integrl is fit to the form to which the power rule for integrtion pplies. for integrtion with u =. Then du = ( ) d = b. Find ln d. = d, so tht d = du, nd ( [ ] d ) ( u u ) du = + C = u + C = ( ) + C Solution: If u = ln, then du = d, nd ( ) ln d = ln d = u du c. Find 5 dw. w(ln w) / = ln u + C = ln ln + C Solution: If u = ln w, then du = dw. Applying the power rule for integrtion, we w hve [ ] 5 dw = 5 (ln w) / w(ln w) / w dw = 5 u / du = 5 u / + C = + C = u/ (ln w) + C / Now Work Problem Integrting b u In Section., we integrted n eponentil function to the bse e: e u du = e u + C Now let us consider the integrl of n eponentil function with n rbitrry bse, b. b u du To find this integrl, we first convert to bse e using b u = e ( ln b)u () (s we did in mny differentition emples too). Emple will illustrte.

48 696 Chpter Integrtion APPLY IT. The rte of chnge of Syri s totl fertility rte (verge number of children born to ech womn) cn be pproimted by the function f (t) =.(.97) t where t = corresponds to. Find the totl fertility rte function f (t) if the totl fertility rte in ws.9. Source: Bsed on dt from the CIA World Fctbook. EXAMPLE An Integrl Involving b u Find d. Solution: Strtegy We wnt to integrte n eponentil function to the bse. To do this, we will first convert from bse to bse e by using Eqution (). d = e ( ln )( ) d The integrnd of the second integrl is of the form e u, where u = ( ln )( ). Since du = ln d, we cn solve for d nd write e ( ln )( ) d = e u du ln = ln eu + C = ln e( ln )( ) + C = ln + C Thus, d = ln + C Notice tht we epressed our nswer in terms of n eponentil function to the bse, the bse of the originl integrnd. Now Work Problem 7 Generlizing the procedure described in Emple, we cn obtin formul for integrting b u : b u du = e ( ln b)u du = e ( ln b)u d(( ln b)u) ln b is constnt ln b = ln b e( ln b)u + C Hence, we hve = ln b bu + C b u du = ln b bu + C Applying this formul to the integrl in Emple gives d b =, u = = d( ) d( ) = d = ln + C which is the sme result tht we obtined before.

49 Appliction of Integrtion Section.5 Techniques of Integrtion 697 We will now consider n ppliction of integrtion tht reltes consumption function to the mrginl propensity to consume. EXAMPLE Finding Consumption Function from Mrginl Propensity to Consume Suppose tht, the mrginl propensity to consume for Bhrin is given by dc di = I where consumption C is function of ntionl income I. Here I is epressed in lrge denomintions of money. Determine the consumption function for Bhrin if it is known tht consumption is (C = ) when I =. Solution: Since the mrginl propensity to consume is the derivtive of C, we hve ( C = C(I) = ) di = I di (I) / di = I (I) / di This is n emple of n initil-vlue problem. If we let u = I, then du = di = d(i), nd C = ( ) I (I) / d(i) = I (I) / + K 6 C = I I + K When I =, C =, so = () () + K = 9 + K Thus, K =, nd the consumption function is C = I I + Now Work Problem 59 PROBLEMS.5 In Problems 5, determine the indefinite integrls d. d. ( + ) + + d. + d 5. d e d e 7. 5 t dt 9. 7 d (7 e / ) d e + d e. ( + )( ) d. 6(e ) d d d d 5e 7e + d 5e / d d 5e s + e ds s

50 698 Chpter Integrtion 5( / + ). d. t(. t t).6 dt. + d ln d e r ln (r + ). dr 5. d r ln ( ) d ln 7. d 8. e + + b d 9. d c = c + d 8. d. (e e + e ) d ( + ) ln ( + ) + ln +. d. d + ln (. + ) d 5. ( + ) / e + d + ( ) 6. ln 7 d 7. d e e 8. d 9. d e + e. d. + ( + ) ln ( + ) d. e e + d 5. ( + )[ + ln ( + )] d (e + 5). d e [ ] d e (8 + e ) 6. ( + e) + e d 7. ln ( + ln ) d [Hint: d ( ln ) = + ln ] d 7 s 8. ( ln ) d π 9. e ds s 5. e ln ( +) d 5. d ln ( e ) d e f ()+ln (f ()) d ssuming f > In Problems 5 nd 55, dr/dq is mrginl-revenue function. Find the demnd function. dr 5. dq = dr 55. (q + ) dq = 9 (q + ) In Problems 56 nd 57, dc/dq is mrginl-cost function. Find the totl-cost function if fied costs in ech cse re. dc 56. dq = dc 57. q + 5 dq = e.5q In Problems 58 6, dc/di represents the mrginl propensity to consume. Find the consumption function subject to the given condition. dc 58. di = I ; C() = dc di = I ; C(9) = 8 dc di = 6 ; C(5) = I 6. Cost Function The mrginl-cost function for mnufcturer s product is given by dc = dq q + where c is the totl cost in dollrs when q units re produced. When units re produced, the verge cost is $5 per unit. To the nerest dollr, determine the mnufcturer s fied cost. 6. Cost Function Suppose the mrginl-cost function for mnufcturer s product is given by dc dq = q 998q + 6 q q + where c is the totl cost in dollrs when q units re produced. () Determine the mrginl cost when units re produced. (b) If fied costs re $,, find the totl cost of producing units. (c) Use the results of prts () nd (b) nd differentils to pproimte the totl cost of producing units. 6. Cost Function The mrginl-cost function for mnufcturer s product is given by dc dq = 9 q.q / + where c is the totl cost in dollrs when q units re produced. Fied costs re $6. () Determine the mrginl cost when 5 units re produced. (b) Find the totl cost of producing 5 units. (c) Use the results of prts () nd (b) nd differentils to pproimte the totl cost of producing units. 6. Vlue of Lnd Suppose it is estimted tht t yers from now the vlue V (in dollrs) of n cre of lnd ner the town of Dmmm, Sudi Arbi will be incresing t the rte of 8t dollrs per yer. If the lnd is currently worth.t + 8 $5 per cre, how much will it be worth in yers? Epress your nswer to the nerest dollr. 65. Svings A certin Arb country s mrginl propensity to sve is given by ds di = 5 (I + ) where S nd I represent totl ntionl svings nd income, respectively, nd re mesured in billions of dollrs. If totl ntionl consumption is $7.5 billion when totl ntionl income

51 Section.6 The Definite Integrl 699 is $8 billion, for wht vlue(s) of I is totl ntionl svings equl to zero? 66. Consumption Function The mrginl propensity to sve of certin country in North Afric is given by ds di = 5.6 I where S nd I represent totl ntionl svings nd income, respectively, nd re mesured in billions of dollrs. () Determine the mrginl propensity to consume when totl ntionl income is $6 billion. (b) Determine the consumption function, given tht svings re $ billion when totl ntionl income is $5 billion. (c) Use the result in prt (b) to show tht consumption is $ 8 = 6. billion when totl ntionl income is $6 billion 5 ( deficit sitution). (d) Use differentils nd the results in prts () nd (c) to pproimte consumption when totl ntionl income is $8 billion. Objective To motivte, by mens of the concept of re, the definite integrl s limit of specil sum; to evlute simple definite integrls by using limiting process..6 The Definite Integrl Figure. shows the region R bounded by the lines y = f () =, y = (the -is), nd =. The region is simply right tringle. If b nd h re the lengths of the bse nd the height, respectively, then, from geometry, the re of the tringle is A = bh = ()() = squre unit. (Henceforth, we will tret res s pure numbers nd write squre unit only if it seems necessry for emphsis.) We will now find this re by nother method, which, s we will see lter, pplies to more comple regions. This method involves the summtion of res of rectngles. Let us divide the intervl [, ] on the -is into four subintervls of equl length by mens of the eqully spced points =, =, =, =, nd = =. (See Figure..) Ech subintervl hs length =. These subintervls determine four subregions of R: R, R, R, nd R, s indicted. With ech subregion, we cn ssocite circumscribed rectngle (Figure.) tht is, rectngle whose bse is the corresponding subintervl nd whose height is the mimum vlue of f () on tht subintervl. Since f is n incresing function, the mimum vlue of f () on ech subintervl occurs when is the right-hnd endpoint. Thus, the res of the circumscribed rectngles ssocited with regions R, R, R, nd R re f ( ), f ( ), f ( ), nd f ( ), respectively. The re of ech rectngle is n pproimtion to the re of its corresponding subregion. Hence, the sum of the res of these rectngles, denoted by S (red S upper br sub or the fourth upper sum ), pproimtes the re A of the tringle. We hve S = f ( ) + f ( ) + f ( ) + f ( ) ( ( ) ( + ) ( + ) ( + )) = 5 = y y f y f() = f() f() R f f R R f R FIGURE. Region bounded by f () =, y =, nd =. R FIGURE. Four subregions of R. FIGURE. rectngles. Four circumscribed

52 7 Chpter Integrtion f f f f() y f() FIGURE.5 rectngles. f f f f f f y 6 6 S 6 f() FIGURE.6 rectngles. f f f f f f() y S 6 6 Four inscribed 6 f() FIGURE.7 rectngles Si circumscribed Si inscribed TO REVIEW summtion nottion, refer to Section.. You cn verify tht S = i= f ( i). The fct tht S is greter thn the ctul re of the tringle might hve been epected, since S includes res of shded regions tht re not in the tringle. (See Figure..) On the other hnd, with ech subregion we cn lso ssocite n inscribed rectngle (Figure.5) tht is, rectngle whose bse is the corresponding subintervl, but whose height is the minimum vlue of f () on tht subintervl. Since f is n incresing function, the minimum vlue of f () on ech subintervl will occur when is the left-hnd endpoint. Thus, the res of the four inscribed rectngles ssocited with R, R, R, nd R re f (), f ( ), f ( ), nd f ( ), respectively. Their sum, denoted S (red S lower br sub or the fourth lower sum ), is lso n pproimtion to the re A of the tringle. We hve S = f () + f ( ) + f ( ) + f ( ) ( ( () + ) ( + ) ( + )) = = Using summtion nottion, we cn write S = i= f ( i). Note tht S is less thn the re of the tringle, becuse the rectngles do not ccount for the portion of the tringle tht is not shded in Figure.5. Since = S A S = 5 we sy tht S is n pproimtion to A from below nd S is n pproimtion to A from bove. If [, ] is divided into more subintervls, we epect tht better pproimtions to A will occur. To test this, let us use si subintervls of equl length = 6. Then S 6, the totl re of si circumscribed rectngles (see Figure.6), nd S 6, the totl re of si inscribed rectngles (see Figure.7), re S 6 = 6 f ( ) f ( ) f ( ) f ( ) f ( ) f ( ) 6 6 = ( ( 6 ) ( 6 + ) ( 6 + ) ( 6 + ) ( ) ( )) = 7 6 nd S 6 = 6 f () + 6 f ( ) f ( ) f ( ) f ( ) f ( ) 5 6 ( ( = 6 () + ) ( 6 + ) ( 6 + ) ( 6 + ) ( )) = 5 6 Note tht S 6 A S 6, nd, with pproprite lbeling, both S 6 nd S 6 will be of the form f (). Clerly, using si subintervls gives better pproimtions to the re thn does four subintervls, s epected. More generlly, if we divide [, ] into n subintervls of equl length, then = /n, nd the endpoints of the subintervls re =, /n, /n,..., (n )/n, nd n/n =. (See Figure.8.) The endpoints of the kth subintervl, for k =,... n, re (k )/n nd k/n nd the mimum vlue of f occurs t the righthnd endpoint k/n. It follows tht the re of the kth circumscribed rectngle is /n f (k/n) = /n (k/n) = k/n, for k =,..., n. The totl re of ll n circumscribed rectngles is S n = n f (k/n) = k= = n k n k= n k= k n () by fctoring from ech term n = n(n + ) from Section. n = n + n (We recll tht n k= k = ++ +n is the sum of the first n positive integers nd the formul used bove ws derived in Section. in nticiption of its ppliction here.)

53 Section.6 The Definite Integrl 7 f f f n n n n y n f() n FIGURE.8 rectngles. f n n f n y n FIGURE.9 n n f() n n n circumscribed n n n n n inscribed rectngles. For inscribed rectngles, we note tht the minimum vlue of f occurs t the lefthnd endpoint, (k )/n, of [(k )/n, k/n], so tht the re of the kth inscribed rectngle is /n f (k /n) = /n ((k )/n) = (k )/n, for k =,... n. The totl re determined of ll n inscribed rectngles (see Figure.9) is S n = n f ((k )/n) = k= = n k n k= = n k n k= = (n )n n = n n n k= (k ) n () by fctoring from ech term n djusting the summtion dpted from Section. From Equtions () nd (), we gin see tht both S n nd S n re sums of the form n ( ) k n ( ) k f (), nmely, Sn = f nd S n n = f. n k= From the nture of S n nd S n, it seems resonble nd it is indeed true tht S n A S n As n becomes lrger, S n nd S n become better pproimtions to A. In fct, let us tke the limits of S n nd S n s n pproches through positive integrl vlues: ( lim S n n = lim n = lim ) = n n n n ( lim S n + n = lim = lim + ) = n n n n n Since S n nd S n hve the sme limit, nmely, nd since k= lim S n = lim S n n n = () S n A S n we will tke this limit to be the re of the tringle. Thus A =, which grees with our prior finding. It is importnt to understnd tht here we developed definition of the notion of re tht is pplicble to mny different regions. We cll the common limit of S n nd S n, nmely,, the definite integrl of f () = on the intervl from = to =, nd we denote this quntity by writing d = () The reson for using the term definite integrl nd the symbolism in Eqution () will become pprent in the net section. The numbers nd ppering with the integrl sign in Eqution () re clled the limits of integrtion; is the lower limit nd is the upper limit. In generl, for function f defined on the intervl from = to = b, where < b, we cn form the sums S n nd S n, which re obtined by considering the mimum nd

54 7 Chpter Integrtion minimum vlues, respectively, on ech of n subintervls of equl length. 6 We cn now stte the following: The common limit of S n nd S n s n, if it eists, is clled the definite integrl of f over [, b] nd is written b f () d The numbers nd b re clled limits of integrtion; is the lower limit nd b is the upper limit. The symbol is clled the vrible of integrtion nd f () is the integrnd. The definite integrl is the limit of sums of the form f (). This definition will be useful in lter sections. APPLY IT. A compny in Mnm hs determined tht its mrginl-revenue function is given by R () = 6.5, where R is the revenue (in dollrs) received when units re sold. Find the totl revenue received for selling units by finding the re in the first qudrnt bounded by y = R () = 6.5 nd the lines y =, =, nd =. In generl, over [, b], we hve y = b n f() In terms of limiting process, we hve b f () f () d Two points must be mde bout the definite integrl. First, the definite integrl is the limit of sum of the form f (). In fct, we cn think of the integrl sign s n elongted S, the first letter of Summtion. Second, for n rbitrry function f defined on n intervl, we my be ble to clculte the sums S n nd S n nd determine their common limit if it eists. However, some terms in the sums my be negtive if f () is negtive t points in the intervl. These terms re not res of rectngles (n re is never negtive), so the common limit my not represent n re. Thus, the definite integrl is nothing more thn rel number; it my or my not represent n re. As we sw in Eqution (), lim n S n is equl to lim n S n. For n rbitrry function, this is not lwys true. However, for the functions tht we will consider, these limits will be equl, nd the definite integrl will lwys eist. To sve time, we will just use the right-hnd endpoint of ech subintervl in computing sum. For the functions in this section, this sum will be denoted S n. EXAMPLE Computing n Are by Using Right-Hnd Endpoints Find the re of the region in the first qudrnt bounded by f () = nd the lines = nd y =. Solution: A sketch of the region ppers in Figure.. The intervl over which vries in this region is seen to be [, ], which we divide into n subintervls of equl length. Since the length of [, ] is, we tke = /n. The endpoints of the subintervls re =, /n, (/n),..., (n )(/n), nd n(/n) =, which re shown in Figure.. The digrm lso shows the corresponding rectngles obtined by using the right-hnd endpoint of ech subintervl. The re of the kth rectngle, for k =,... n, is the product of its width, /n, nd its height, f (k(/n)) = (k/n), which is the function vlue t the right-hnd endpoint of its bse. Summing these res, we get S n = = n f k= n k= ( k ( n )) = ( ) 8 n 8k = n n k= ( n k= = 8 n n 8 n(n + )(n + ) n 6 ( ) (n + )(n + ) 8 n n k= ( k n 8k n ) ) = 8 n n n 8 n n k= k= k = 8 n FIGURE. Region of Emple. 6 Here we ssume tht the mimum nd minimum vlues eist.

55 Section.6 The Definite Integrl 7 y n f() n n The second line of the preceding computtions uses bsic summtion mnipultions s discussed in Section.. The third line uses two specific summtion formuls, lso from Section.: The sum of n copies of is n nd the sum of the first n squres is n(n + )(n + ). 6 Finlly, we tke the limit of the S n s n : ( )) (n + )(n + ) lim S n = lim n n ( 8 n ( n ) + n + = 8 lim n n = 8 ( lim + n + n ) n n (n ) n FIGURE. n subintervls nd corresponding rectngles for Emple. = 8 8 = 6 Hence, the re of the region is 6. Now Work Problem 7 EXAMPLE Evluting Definite Integrl No units re ttched to the nswer, since definite integrl is simply number. n n n FIGURE. subintervls. 5 n (n ) n y n n Dividing [, ] into n (n ) n n n f() 5 FIGURE. f () is negtive t ech right-hnd endpoint. Evlute ( ) d. Solution: We wnt to find the definite integrl of f () = over the intervl [, ]. Thus, we must compute lim n S n. But this limit is precisely the limit 6 found in Emple, so we conclude tht EXAMPLE 5 ( ) d = 6 Integrting Function over n Intervl Now Work Problem 9 Integrte f () = 5 from = to = ; tht is, evlute ( 5) d. Solution: We first divide [, ] into n subintervls of equl length = /n. The endpoints re, /n, (/n),..., (n )(/n), n(/n) =. (See Figure..) Using right-hnd endpoints, we form the sum nd simplify n ( S n = f k ) n n = k= n k= (( k ) ) n 5 = n = 9 n ( n(n + ) = 9 n + n Tking the limit, we obtin 5 = 9 lim S n = lim n n ) 5 n (n) n k= ( 9 n k 5 ) n ( + ) 5 n = 9 n n k= k 5 n ( ( 9 + ) ) 5 = 9 5 = n n k=

56 7 Chpter Integrtion Thus, y y = f() FIGURE. If f is continuous nd f () on [, b], then b f () d represents the re under the curve. b ( 5) d = Note tht the definite integrl here is negtive number. The reson is cler from the grph of f () = 5 over the intervl [, ]. (See Figure..) Since the vlue of f () is negtive t ech right-hnd endpoint, ech term in S n must lso be negtive. Hence, lim n S n, which is the definite integrl, is negtive. Geometriclly, ech term in S n is the negtive of the re of rectngle. (Refer gin to Figure..) Although the definite integrl is simply number, here we cn interpret it s representing the negtive of the re of the region bounded by f () = 5, =, =, nd the -is (y = ). Now Work Problem 7 In Emple 5, it ws shown tht the definite integrl does not hve to represent n re. In fct, there the definite integrl ws negtive. However, if f is continuous nd f () on [, b], then S n for ll vlues of n. Therefore, lim n S n, so b f () d. Furthermore, this definite integrl gives the re of the region bounded by y = f (), y =, =, nd = b. (See Figure..) Although the pproch tht we took to discuss the definite integrl is sufficient for our purposes, it is by no mens rigorous. The importnt thing to remember bout the definite integrl is tht it is the limit of specil sum. Here is progrm for the TI-8 Plus grphing clcultor tht will estimte the limit of S n s n for function f defined on [, b]. PROGRAM:RIGHTSUM Lbl Input SUBINTV,N (B A)/N H S A + H X I Lbl Y + S S X + H X I + I If I N Goto H S S Disp S Puse Goto RIGHTSUM will compute S n for given number n of subintervls. Before eecuting the progrm, store f (),, nd b s Y, A, nd B, respectively. Upon eecution of the progrm, you will be prompted to enter the number of subintervls. Then the progrm proceeds to disply the FIGURE.5 Vlues of S n for f () = 5 on [, ]. vlue of S n. Ech time ENTER is pressed, the progrm repets. In this wy, disply of vlues of S n for vrious numbers of subintervls my be obtined. Figure.5 shows vlues of S n (n =,, nd ) for the function f () = 5 on the intervl [, ]. As n, it ppers tht S n.5. Thus, we estimte tht Equivlently, lim S n.5 n ( 5) d.5 which grees with our result in Emple 5. It is interesting to note tht the time required for n older clcultor to compute S in Figure.5 ws in ecess of.5 minutes. The time required on TI-8 Plus is less thn minute.

57 Section.7 The Fundmentl Theorem of Integrl Clculus 75 PROBLEMS.6 In Problems, sketch the region in the first qudrnt tht is bounded by the given curves. Approimte the re of the region by the indicted sum. Use the right-hnd endpoint of ech subintervl.. f () = +, y =, =, = ; S. f () =, y =, = ; S 5. f () =, y =, = ; S. f () = +, y =, =, = ; S In Problems 5 nd 6, by dividing the indicted intervl into n subintervls of equl length, find S n for the given function. Use the right-hnd endpoint of ech subintervl. Do not find lim n S n. 5. f () = ; [, ] 6. f () = + ; [, ] In Problems 7 nd 8, () simplify S n nd (b) find lim n S n. 7. S n = [( ) ( ) ( n ) ] n n + + n n + [ ( 8. S n = ) ( + ) ( + + n ) ] n n n n In Problems 9, sketch the region in the first qudrnt tht is bounded by the given curves. Determine the ect re of the region by considering the limit of S n s n. Use the right-hnd endpoint of ech subintervl. 9. Region s described in Problem. Region s described in Problem. Region s described in Problem. y =, y =, =, =. f () =, y =, =. f () = 9, y =, = In Problems 5 9, evlute the given definite integrl by tking the limit of S n. Use the right-hnd endpoint of ech subintervl. Sketch the grph, over the given intervl, of the function to be integrted d 6. d 8. ( + ) d ( b d ( + ) d ) d without the use of limits.. Find d d. Find f () d without the use of limits, where f () = + if if if > In ech of Problems, use progrm, such s RIGHTSUM, to estimte the re of the region in the first qudrnt bounded by the given curves. Round your nswer to one deciml plce.. f () = +, y =, =, =.7. f () =, y =, =, = 9. f () = ln, y =, =, = In ech of Problems 5 8, use progrm, such s RIGHTSUM, to estimte the vlue of the definite integrl. Round your nswer to one deciml plce d 6. + ( + ) d 8. d ln d Objective To develop informlly the Fundmentl Theorem of Integrl Clculus nd to use it to compute definite integrls. y y f() FIGURE.6 On [, b], f is continuous nd f (). b.7 The Fundmentl Theorem of Integrl Clculus The Fundmentl Theorem Thus fr, the limiting processes of both the derivtive nd definite integrl hve been considered seprtely. We will now bring these fundmentl ides together nd develop the importnt reltionship tht eists between them. As result, we will be ble to evlute definite integrls more efficiently. The grph of function f is given in Figure.6. Assume tht f is continuous on the intervl [, b] nd tht its grph does not fll below the -is. Tht is, f (). From the preceding section, the re of the region below the grph nd bove the -is from = to = b is given by b f () d. We will now consider nother wy to determine this re. Suppose tht there is function A = A(), which we will refer to s n re function, tht gives the re of the region below the grph of f nd bove the -is from to, where b. This region is shded in Figure.7. Do not confuse A(), which is n re, with f (), which is the height of the grph t.

58 76 Chpter Integrtion y y f() A() FIGURE.7 function. y b A() is n re b h FIGURE.8 A( + h) gives the re of the shded region. y b h FIGURE.9 Are of shded region is A( + h) A(). f( h) y f() y FIGURE. Are of rectngle is the sme s re of shded region in Figure.9. h From its definition, we cn stte two properties of A immeditely:. A() =, since there is no re from to. A(b) is the re from to b; tht is, A(b) = b f () d If is incresed by h units, then A( + h) is the re of the shded region in Figure.8. Hence, A( + h) A() is the difference of the res in Figures.8 nd.7, nmely, the re of the shded region in Figure.9. For h sufficiently close to zero, the re of this region is the sme s the re of rectngle (Figure.) whose bse is h nd whose height is some vlue y between f () nd f ( + h). Here y is function of h. Thus, on the one hnd, the re of the rectngle is A( + h) A(), nd, on the other hnd, it is hy, so Equivlently, A( + h) A() = hy A( + h) A() = y dividing by h h Since y is between f () nd f ( + h), it follows tht s h, y pproches f (), so A( + h) A() lim = f () () h h But the left side is merely the derivtive of A. Thus, Eqution () becomes A () = f () We conclude tht the re function A hs the dditionl property tht its derivtive A is f. Tht is, A is n ntiderivtive of f. Now, suppose tht F is ny ntiderivtive of f. Then, since both A nd F re ntiderivtives of the sme function, they differ t most by constnt C: A() = F() + C. () Recll tht A() =. So, evluting both sides of Eqution () when = gives so tht Thus, Eqution () becomes If = b, then, from Eqution (), But recll tht From Equtions () nd (5), we get b = F() + C C = F() A() = F() F() () A(b) = F(b) F() () A(b) = b f () d = F(b) F() f () d (5) A reltionship between definite integrl nd ntidifferentition hs now become cler. To find b f () d, it suffices to find n ntiderivtive of f, sy, F, nd subtrct the vlue of F t the lower limit from its vlue t the upper limit b. We ssumed here tht f ws continuous nd f () so tht we could ppel to the concept of n re. However, our result is true for ny continuous function 7 nd is known s the Fundmentl Theorem of Integrl Clculus. 7 If f is continuous on [, b], it cn be shown tht b f () d does indeed eist.

59 Section.7 The Fundmentl Theorem of Integrl Clculus 77 Fundmentl Theorem of Integrl Clculus If f is continuous on the intervl [, b] nd F is ny ntiderivtive of f on [, b], then b f () d = F(b) F() The definite integrl is number, nd n indefinite integrl is function. It is importnt tht you understnd the difference between definite integrl nd n indefinite integrl. The definite integrl b f () d is number defined to be the limit of sum. The Fundmentl Theorem sttes tht the indefinite integrl f () d (the most generl ntiderivtive of f ), which is function of relted to the differentition process, cn be used to determine this limit. Suppose we pply the Fundmentl Theorem to evlute ( ) d. Here f () =, =, nd b =. Since n ntiderivtive of is F() = ( /), it follows tht ( ) d = F() F() = ( 8 8 ) () = 6 This confirms our result in Emple of Section.6. If we hd chosen F() to be ( /) + C, then we would hve [( F() F() = 8 8 ) ] + C [ + C] = 6 s before. Since the choice of the vlue of C is immteril, for convenience we will lwys choose it to be, s originlly done. Usully, F(b) F() is bbrevited by writing F(b) F() = F() b Since F in the Fundmentl Theorem of Clculus is ny ntiderivtive of f nd f () d is the most generl ntiderivtive of f, it showcses the nottion to write b ( ) b f () d = f () d Using the b nottion, we hve ) ( ) d = ( = ( 8 8 ) = 6 APPLY IT. The income (in dollrs) from fstfood chin in Beirut is incresing t rte of f (t) =,e.t, where t is in yers. Find 6,e.t dt, which gives the totl income for the chin between the third nd sith yers. EXAMPLE 6 Applying the Fundmentl Theorem The rte of production of electricity in Morocco for the yers to 6, mesured in billions of kilowtt hours per yer, cn be modeled by E (t) =.e.875t where t = corresponds to. Wht is the totl mount of electricity produced from to 6? Solution: The totl mount of electricity produced is given by summing (integrting) the (instntneous) production rte between yer nd yer 6: 6.e.875t dt =.(.875)e.875t =.5(e.875(6) e.875() )

60 78 Chpter Integrtion Therefore, pproimtely billion kilowtt hours of electricity were produced from nd 6 in Morocco. Source: Interntionl EnergyAgency, Energy Sttistics of Non-OECD Countries, Energy Blnces of Non-OECD Countries, Energy Sttistics of OECD Countries, nd Energy Blnces of OECD Countries. Now Work Problem Properties of the Definite Integrl For b f () d, we hve ssumed tht < b. We now define the cses in which > b or = b. First, If > b, then b f () d = b f () d. Tht is, interchnging the limits of integrtion chnges the integrl s sign. For emple, ( ) d = If the limits of integrtion re equl, we hve f () d = ( ) d Some properties of the definite integrl deserve mention. The first of the properties tht follow resttes more formlly our comment from the preceding section concerning re. Properties of the Definite Integrl. If f is continuous nd f () on [, b], then b f () d cn be interpreted s the re of the region bounded by the curve y = f (), the -is, nd the lines = nd = b.. b kf () d = k b f () d, where k is constnt. b [f () ± g()] d = b f () d ± b g() d Properties nd re similr to rules for indefinite integrls becuse definite integrl my be evluted by the Fundmentl Theorem in terms of n ntiderivtive. Two more properties of definite integrls re s follows.. b f () d = b f (t) dt The vrible of integrtion is dummy vrible in the sense tht ny other vrible produces the sme result tht is, the sme number. To illustrte property, you cn verify, for emple, tht d = t dt 5. If f is continuous on n intervl I nd, b, nd c re in I, then c f () d = b f () d + c b f () d

61 Section.7 The Fundmentl Theorem of Integrl Clculus 79 Property 5 mens tht the definite integrl over n intervl cn be epressed in terms of definite integrls over subintervls. Thus, ( ) d = ( ) d + ( ) d We will look t some emples of definite integrtion now nd compute some res in Section.8. EXAMPLE 7 Using the Fundmentl Theorem CAUTION In Emple 7, the vlue of the ntiderivtive ( + ) / t the lower limit is ()/. Do not ssume tht n evlution t the limit zero will yield. Find + d. Solution: To find n ntiderivtive of the integrnd, we will pply the power rule for integrtion: d = ( + ) / d + = = ( + ) / ( + ) / d( + ) = = ( ( () / () /) ) ( + ) / = ( ) Now Work Problem EXAMPLE 8 Evluting Definite Integrls. Find [t / + t(t + ) ] dt. Solution: [t / + t(t + ) ] dt = b. Find e t dt. = () t/ t / dt + (t + ) d(t + ) ( ) (t + ) + = ( / ) + 8 (5 ) = / = Solution: e t dt = e t d(t) ( ) = e t = (e e ) = (e ) Now Work Problem 5

62 7 Chpter Integrtion y EXAMPLE 9 Finding nd Interpreting Definite Integrl FIGURE. intervl [, ]. y Grph of y = on the CAUTION Remember tht b f () d is limit of sum. In some cses this limit represents n re. In others it does not. When f () on [, b], the integrl represents the re between the grph of f nd the -is from = to = b. Evlute d. Solution: d = = ( ) = 6 = 5 The reson the result is negtive is cler from the grph of y = on the intervl [, ]. (See Figure..) For <, f () is negtive. Since definite integrl is limit of sum of the form f (), it follows tht d is not only negtive number, but lso the negtive of the re of the shded region in the third qudrnt. On the other hnd, d is the re of the shded region in the first qudrnt, since f () on [, ]. The definite integrl over the entire intervl [, ] is the lgebric sum of these numbers, becuse, from property 5, d = d + d Thus, d does not represent the re between the curve nd the -is. However, if re is desired, it cn be given by d + d Now Work Problem 5 The Definite Integrl of Derivtive Since function f is n ntiderivtive of f, by the Fundmentl Theorem we hve b f () d = f (b) f () (6) But f () is the rte of chnge of f with respect to. Hence, if we know the rte of chnge of f nd wnt to find the difference in function vlues f (b) f (), it suffices to evlute b f () d. EXAMPLE Totl Sles During n dvertising cmpign, Al Atll Nuts found tht the rte of chnge of sles is given by S (t) = 8. + e.t where S is the mount of sles in thousnds of dollrs nd t is the number of weeks tht the cmpign hs been running. Wht re the totl sles t the end of the sith week of the dvertising cmpign? Solution: The totl sles mount t the end of 6 weeks is given by 6 S (t) dt = = = e.t dt 6 8. dt + 8. dt +. e.t dt 6 e.t d(.t)

63 Section.7 The Fundmentl Theorem of Integrl Clculus 7 = 8.t e.t 6 = ( 8.(6) e.(6)) ( 8.() e.()).6 Tht is, bout $,6. APPLY IT 5. A mngeril service determines tht the rte of increse in mintennce costs (in dollrs per yer) for prticulr prtment comple in downtown Khrtoum is given by M () = 9 + 5, where is the ge of the prtment comple in yers nd M() is the totl (ccumulted) cost of mintennce for yers. Find the totl cost for the first five yers. EXAMPLE Finding Chnge in Function Vlues by Definite Integrtion A mnufcturer s mrginl-cost function is dc dq =.6q + If production is presently set t q = 8 units per week, how much more would it cost to increse production to units per week? Solution: The totl-cost function is c = c(q), nd we wnt to find the difference c() c(8). The rte of chnge of c is dc/dq, so, by Eqution (6), dc c() c(8) = 8 dq dq = (.6q + )dq 8 [.6q ] = + q = [.q + q] 8 = [.() + ()] [.(8) + (8)] = 8 = If c is in dollrs, then the cost of incresing production from 8 units to units is $. Now Work Problem 55 8 Mny grphing clcultors hve the cpbility to estimte the vlue of definite integrl. On TI-8 Plus, to estimte 8 (.6q + ) dq we use the fnint( commnd, s indicted in Figure.. The four prmeters tht must be entered with this commnd re function to vrible of lower upper be integrted integrtion limit limit We see tht the vlue of the definite integrl is pproimtely, which grees with the result in Emple. Similrly, to estimte d we enter FIGURE. Estimting 8 (.6q + ) dq. fnint(x, X,, ) or, lterntively, if we first store s Y, we cn enter fnint(y, X,, ) In ech cse we obtin.75, which grees with the result in Emple 9.

64 7 Chpter Integrtion PROBLEMS.7 In Problems, evlute the definite integrl / 8 5 d. 5 d. ( ) d 6. (y + y + ) dy 8. e π e+ / / b e 95 (w w ) dw. t dt. d. d 6. ( / / ) d 8. ( + ) d. d. d (e + e) d 5 d ( 9y) dy (t t ) dt dt / / 8 d ( + + ) d (z + ) dz ( ) d y dy e 5 d ( + )( + ) d 5 e d 6. ( + ) d + 5 d 8. p dp q q + dq. ( ) d. ( + ) / (m + ny) dy. ( + ) d ( 6 ) d 7. ln e d / d 7 + d d 8 d ( + )e + d.. e d (Hint: Multiply the integrnd by.) + e e f () d, where f () = { if < if (. Evlute d). Suppose f () =. Evlute 5. If 6. If f () d. 7 7 d. t dt. Evlute f () d. e d + d. f () d = 5 nd f () d = 6, 7. Suppose tht f () = e f () d =, find f () d = 5, nd e t e t dt e t + e where t e f () d. f () d =, find > e. Find f (). 8. Severity Inde In discussing trffic sfety, Shonle 8 considers how much ccelertion person cn tolerte in crsh so tht there is no mjor injury. The severity inde is defined s S.I. = T α 5/ dt where α ( Greek letter red lph ) is considered constnt involved with weighted verge ccelertion, nd T is the durtion of the crsh. Find the severity inde. 9. Sttistics In sttistics, the men µ ( Greek letter red mu ) of the continuous probbility density function f defined on the intervl [, b] is given by µ = b f () d nd the vrince σ (σ is Greek letter red sigm ) is given by σ = b ( µ) f () d Compute µ nd then σ if =, b =, nd f () = 6( ). 5. Distribution of Incomes The economist Preto 9 hs stted n empiricl lw of distribution of higher incomes tht gives the 8 J. I. Shonle, Environmentl Applictions of Generl Physics (Reding, MA: Addison-Wesley Publishing Compny, Inc., 975). 9 G. Tintner, Methodology of Mthemticl Economics nd Econometrics (Chicgo: University of Chicgo Press, 967), p. 6.

65 Section.7 The Fundmentl Theorem of Integrl Clculus 7 number N of persons receiving or more dollrs. If dn d = A B where A nd B re constnts, set up definite integrl tht gives the totl number of persons with incomes between nd b, where < b. 5. Biology In discussion of gene muttion, the following integrl occurs: Evlute this integrl. / d 5. Biology In biology, problems frequently rise involving the trnsfer of substnce between comprtments. An emple is trnsfer from the bloodstrem to tissue. Evlute the following integrl, which occurs in two-comprtment diffusion problem: t (e τ e bτ )dτ Here, τ (red tu ) is Greek letter; nd b re constnts. 5. Demogrphy For certin smll Arb popultion, suppose l is function such tht l() is the number of persons who rech the ge of in ny yer of time. This function is clled life tble function. Under pproprite conditions, the integrl b l(t) dt gives the epected number of people in the popultion between the ect ges of nd b, inclusive. If l() = for determine the number of people between the ect ges of nd 9, inclusive. Give your nswer to the nerest integer, since frctionl nswers mke no sense. Wht is the size of the popultion? 5. Minerl Consumption If C is the yerly consumption of minerl t time t =, then, under continuous consumption, the totl mount of the minerl used in the intervl [, t] is t Ce kτ dτ where k is the rte of consumption. For rre-erth minerl, it hs been determined tht C = units nd k =.5. Evlute the integrl for these dt. 55. Mrginl Cost A mnufcturer s mrginl-cost function is dc dq =.q + 8 If c is in dollrs, determine the cost involved to increse production from 65 to 75 units. 56. Mrginl Cost Repet Problem 55 if dc dq =.q.5q + 5 nd production increses from 9 to 8 units. 57. Mrginl Revenue A mnufcturer s mrginl-revenue function is dr dq = q If r is in dollrs, find the chnge in the mnufcturer s totl revenue if production is incresed from 5 to 8 units. 58. Mrginl Revenue Repet Problem 57 if dr = + 5q q dq nd production is incresed from 5 to units. 59. Crime Rte A sociologist is studying the crime rte in certin city. She estimtes tht t months fter the beginning of net yer, the totl number of crimes committed will increse t the rte of 8t + crimes per month. Determine the totl number of crimes tht cn be epected to be committed net yer. How mny crimes cn be epected to be committed during the lst si months of tht yer? 6. Production Imgine one-dimensionl country of length R. (See Figure.. ) Suppose the production of goods for this country is continuously distributed from border to border. If the mount produced ech yer per unit of distnce is f (), then the country s totl yerly production is given by G = R R f () d Evlute G if f () = i, where i is constnt. R One-dimensionl country Border FIGURE. R Border 6. Eports For the one-dimensionl country of Problem 6, under certin conditions the mount of the country s eports is given by E = R R i [e k(r ) + e k(r+) ] d where i nd k re constnts (k ). Evlute E. 6. Averge Delivered Price In discussion of delivered price of good from mill to customer, DeCnio clims tht the verge delivered price pid by consumers is given by A = R (m + )[ (m + )] d R [ (m + )] d W. J. Ewens, Popultion Genetics (London: Methuen & Compny Ltd., 969). W. Simon, Mthemticl Techniques for Physiology nd Medicine (New York: Acdemic Press, Inc., 97). R. Tgeper, Why the Trde/GNP Rtio Decreses with Country Size, Socil Science Reserch, 5 (976), 85.

66 7 Chpter Integrtion where m is mill price, nd is the mimum distnce to the point of sle. DeCnio determines tht Verify this. A = m + R m mr R m R In Problems 6 65, use the Fundmentl Theorem of Integrl Clculus to determine the vlue of the definite integrl ( + + ) d 6. e t dt ( + ) d Round your nswer to two deciml plces. In Problems 66 69, estimte the vlue of the definite integrl. Round your nswer to two deciml plces d t + dt 69. ln d 6 q + q + dq Objective To find the re of region bounded by curves using integrtion over both verticl nd horizontl strips. y d d FIGURE. for Emple. y Digrm.8 Are between Curves In Sections.6 nd.7 we sw tht the re of region bounded by the lines =, = b, y =, nd curve y = f () with f () for b cn be found by evluting the definite integrl b f () d. Similrly, for function f () on n intervl [, b], the re of the region bounded by =, = b, y =, nd y = f () is given by b f () d = b f () d. Most of the functions f we hve encountered, nd will encounter, re continuous nd hve finite number of roots of f () =. For such functions, the roots of f () = prtition the domin of f into finite number of intervls on ech of which we hve either f () or f (). For such function we cn determine the re bounded by y = f (), y = nd ny pir of verticl lines = nd = b, with nd b in the domin of f. We hve only to find ll the roots c < c < < c k with < c nd c k c b < b; clculte the integrls c f () d, f () d,, f () d; ttch to ech integrl the correct sign to correspond to n c c k re; nd dd the results. Emple will provide modest emple of this ide. For such n re determintion, rough sketch of the region involved is etremely vluble. To set up the integrls needed, smple rectngle should be included in the sketch for ech individul integrl s in Figure.. The re of the region is limit of sums of res of rectngles. A sketch helps to understnd the integrtion process nd it is indispensble when setting up integrls to find res of complicted regions. Such rectngle (see Figure.) is clled verticl strip. In the digrm, the width of the verticl strip is. We know from our work on differentils in Section. tht we cn consistently write = d, for the independent vrible. The height of the verticl strip is the y-vlue of the curve. Hence, the rectngle hs re y = f () d. The re of the entire region is found by summing the res of ll such verticl strips between = nd = b nd finding the limit of this sum, which is the definite integrl. Symboliclly, we hve y b f () d For f () it is helpful to think of d s length differentil nd f ()d s n re differentil da. Then, s we sw in Section.7, we hve da = f () for some re d function A nd b f () d = b da = A(b) A()

67 Section.8 Are between Curves 75 [If our re function A mesures re strting t the line =, s it did in Section.7, then A() = nd the re under f (nd over ) from to b is just A(b).] It is importnt to understnd here tht we need f () in order to think of f () s length nd hence f ()d s differentil re. But if f () then f () so tht f () becomes length nd f ()d becomes differentil re. EXAMPLE An Are Requiring Two Definite Integrls CAUTION It is wrong to write hstily tht the re is y d, for the following reson: For the left rectngle, the height is y. However, for the rectngle on the right, y is negtive, so its height is the positive number y. This points out the importnce of sketching the region. Find the re of the region bounded by the curve y = nd the line y = (the -is) from = to =. Solution: A sketch of the region is given in Figure.. Notice tht the -intercepts re (, ) nd (, ). On the intervl [, ], the re of the verticl strip is yd = ( )d On the intervl [, ], the re of the verticl strip is Thus, re = ( = = ( y)d = ( )d ( ) d + ) [( + ) ( ) d ) ( ( 8 )] + [( 8 ) ( )] + = 9 Now Work Problem Before embrking on more complicted re problems, we motivte the further study of re by seeing the use of re s probbility in sttistics. EXAMPLE Sttistics Appliction In sttistics, (probbility) density function f of vrible, where ssumes ll vlues in the intervl [, b], hs the following properties: (i) f () (ii) b f () d = The probbility tht ssumes vlue between c nd d, which is written P(c d), where c d b, is represented by the re of the region bounded by the grph of f nd the -is between = c nd = d. Hence (see Figure.5), P(c d) = d c f () d [In the terminology of Chpters 7 nd 8, the condition c d defines n event nd P(c d) is consistent with the nottion of the erlier chpters. Note too tht the hypothesis (ii) bove ensures tht b is the certin event.]

68 76 Chpter Integrtion y y f () c d b P (c d) d f () d c FIGURE.5 Probbility s n re. For the density function f () = 6( ), where, find ech of the following probbilities.. P( ) Solution: Here [, b] is [, ], c is, nd d is. We hve b. P ( ) / P ( ) / = 6( ) d = 6 ( ) d ( ) = 6 / / = ( ) ( ( ) ( ) ) = = 5 Solution: Since the domin of f is, to sy tht mens tht. Thus, ( P ) = 6( ) d = 6 ( ) d / / ( ) = 6 = ( ) = / / Now Work Problem 7 EXAMPLE Witing Time While witing for bnk representtive, Sihm finds tht the witing time follows n eponentil probbility density function p() =.e., where the number of minutes before customer is served. Wht is the probbility tht Sihm will wit t most five minutes? Solution: At most five minutes mens t 5. The probbility tht Sihm wits t most five minutes is given by 5 5.e.t dt =.. e.t (.) dt = ( e.t )

69 Verticl Strips Section.8 Are between Curves 77 We will now find the re of region enclosed by severl curves. As before, our procedure will be to drw smple strip of re nd use the definite integrl to dd together the res of ll such strips. For emple, consider the re of the region in Figure.6 tht is bounded on the top nd bottom by the curves y = f () nd y = g() nd on the sides by the lines = nd = b. The width of the indicted verticl strip is d, nd the height is the y-vlue of the upper curve minus the y-vlue of the lower curve, which we will write s y upper y lower. Thus, the re of the strip is (y upper y lower ) d y (, y upper ) y f() y g() d (, y lower ) b which is FIGURE.6 Region between curves. ( f () g()) d Summing the res of ll such strips from = to = b by the definite integrl gives the re of the region: b ( f () g()) d ( f () g()) d = re We remrk tht there is nother wy to view this re problem. In Figure.6 both f nd g re bove y = nd it is cler tht the re we seek is lso the re under f minus the re under g. Tht pproch tells us tht the required re is b f () d b g() d = b ( f () g()) d However, our first pproch does not require tht either f or g lie bove. Our usge of y upper nd y lower is relly just wy of sying tht f g on [, b]. This is equivlent to sying tht f g on [, b] so tht ech differentil (f () g()) d is meningful s n re. y EXAMPLE 5 Finding n Are between Two Curves (, y upper ) (, ) y y (, y lower ) (, ) d FIGURE.7 Digrm for Emple 5. Find the re of the region bounded by the curves y = nd y =. Solution: A sketch of the region ppers in Figure.7. To determine where the curves intersect, we solve the system formed by the equtions y = nd y =. Eliminting y by substitution, we obtin = = = = ( ) = or = squring both sides

70 78 Chpter Integrtion It should be obvious tht knowing the points of intersection is importnt in determining the limits of integrtion. Since we squred both sides, we must check the solutions found with respect to the originl eqution. It is esily determined tht both = nd = re solutions of =. If =, then y = ; if =, then y =. Thus, the curves intersect t (, ) nd (, ). The width of the indicted strip of re is d. The height is the y-vlue on the upper curve minus the y-vlue on the lower curve: y upper y lower = Hence, the re of the strip is ( ) d. Summing the res of ll such strips from = to = by the definite integrl, we get the re of the entire region: re = = = ( ) d ( / ) d = ( ( / ) ( ) = 6 ) Now Work Problem 5 (, ) FIGURE.8 Emple 6. y 9 y 9 y y 8 d (, 8) y = Digrm for d d y EXAMPLE 6 Finding n Are between Two Curves Find the re of the region bounded by the curves y = + 8 nd y =. Solution: A sketch of the region ppers in Figure.8. To find where the curves intersect, we solve the system of equtions y = + 8 nd y = : + 8 =, =, =, ( + )( ) = fctoring = or = When =, then y = ; when =, then y = 8. Thus, the curves intersect t (, ) nd (, 8). The width of the indicted strip is d. The height is the y-vlue on the upper curve minus the y-vlue on the lower curve: Therefore, the re of the strip is y upper y lower = ( + 8) ( ) [( + 8) ( )] d = ( ) d Summing ll such res from = to =, we hve EXAMPLE 7 re = ( ) d = Now Work Problem 9 Are of Region Hving Two Different Upper Curves (, 5) FIGURE.9 y upper is 9 on [, ] nd is + on [, ]. Find the re of the region between the curves y = 9 nd y = + from = to =. Solution: The region is sketched in Figure.9. The curves intersect when 9 = + 8 = = = ± two solutions

71 Section.8 Are between Curves 79 When = ±, then y = 5, so the points of intersection re (±, 5). Becuse we re interested in the region from = to =, the intersection point tht is of concern to us is (, 5). Notice in Figure.9 tht in the region to the left of the intersection point (, 5), strip hs y upper = 9 nd y lower = + but for strip to the right of (, 5) the reverse is true, nmely, y upper = + nd y lower = 9 Thus, from = to =, the re of strip is but from = to =, it is (y upper y lower ) d = [(9 ) ( + ] d = (8 ) d (y upper y lower ) d = [( + ) (9 )] d = ( 8) d Therefore, to find the re of the entire region, we need two integrls: re = (8 ) d + ( 8) d ) = (8 ( ) + 8 [( = 6 6 ) ] [ ( )] 6 + (8 ) 6 = 6 Now Work Problem y y d FIGURE. y 9 9, Verticl strip of re. Horizontl Strips Sometimes re cn more esily be determined by summing res of horizontl strips rther thn verticl strips. In the following emple, n re will be found by both methods. In ech cse, the strip of re determines the form of the integrl. EXAMPLE 8 Verticl Strips nd Horizontl Strips Find the re of the region bounded by the curve y = nd the lines y = nd = (the y-is). Solution: The region is sketched in Figure.. When the curves y = nd y = intersect, 9 =, so = 9. Thus, the intersection point is ( 9, ). Since the width of the verticl strip is d, we integrte with respect to the vrible. Accordingly, y upper nd y lower must be epressed s functions of. For the lower curve, y =, we hve y = ±. But y for the portion of this curve tht bounds the region, so we use y =. The upper curve is y =. Hence, the height of the strip is y upper y lower = Therefore, the strip hs n re of ( ), nd we wish to sum ll such res from = to = 9. We hve 9/ re = ( ) ) d = ( / 9/

72 7 Chpter Integrtion CAUTION With horizontl strips, the width is dy. = [ ( ) 9 ( ) ] 9 / () y y dy FIGURE. strip of re. y 9 Horizontl 9, = 7 [ (9 ) / ] = 7 ( ) = 9 Let us now pproch this problem from the point of view of horizontl strip s shown in Figure.. The width of the strip is dy. The length of the strip is the -vlue on the rightmost curve minus the -vlue on the leftmost curve. Thus, the re of the strip is ( right left ) dy We wish to sum ll such res from y = to y = : (right left ) dy ( right left ) dy Since the vrible of integrtion is y, we must epress right nd left s functions of y. The rightmost curve is y = so tht = y /. The left curve is =. Thus, re = = ( right left ) dy ( y ) dy = y = 9 Note tht for this region, horizontl strips mke the definite integrl esier to evlute (nd set up) thn n integrl with verticl strips. In ny cse, remember tht the limits of integrtion re limits for the vrible of integrtion. Now Work Problem 5 EXAMPLE 9 Advntge of Horizontl Elements Find the re of the region bounded by the grphs of y = nd y =. Solution: The region is sketched in Figure.. The curves intersect when y y =. Thus, y y = ; equivlently, (y + )(y ) =, from which it follows tht y = or y =. This gives the intersection points (, ) nd (, ). Let us try verticl strips of re. [See Figure.().] Solving y = for y gives y = ±. As seen in Figure.(), to the left of =, the upper end of the strip lies on y = nd the lower end lies on y =. To the right of =, the upper curve is y = nd the lower curve is y = (or y = ). Thus, with verticl strips, two integrls re needed to evlute the re: re = ( ( )) d + ( ( )) d y y y y (, ) d (, ) () (, ) y (, ) dy y (b) FIGURE. strips. Region of Emple 9 with verticl nd horizontl

73 Section.8 Are between Curves 7 Perhps the use of horizontl strips cn simplify our work. In Figure.(b), the width of the strip is y. The rightmost curve is lwys y = (or = y + ), nd the leftmost curve is lwys y = (or = y ). Therefore, the re of the horizontl strip is [(y + ) y ] y, so the totl re is re = (y + y ) dy = 9 Clerly, the use of horizontl strips is the most desirble pproch to solving the problem. Only single integrl is needed, nd it is much simpler to compute. Now Work Problem 55 PROBLEMS.8 In Problems, use definite integrl to find the re of the region bounded by the given curve, the -is, nd the given lines. In ech cse, first sketch the region. Wtch out for res of regions tht re below the -is.. y = 5 +, =, =. y = + 5, =, =. y =, =, =. y =, =, = 5. y = + +, = 6. y =, =, = 7. y = 8. y =, =, = 9. y =, =, =. y = e, =, =. y = ( ), =, =. y =, =, = e. y = + 9, = 9, =. y =, =, = 6 5. y =, =, = 5 6. y = +, =, = 7. y =, = 8. y = e +, =, = 9. y =, =, =. y = +, =, =. y =, =, =. y =, =, = 6. y = +, =, =. Given tht { f () = if < 6 if determine the re of the region bounded by the grph of y = f (), the -is, nd the line =. Include sketch of the region. 5. Under conditions of continuous uniform distribution ( topic in sttistics), the proportion of persons with incomes between nd t, where t b, is the re of the region between the curve y = /(b ) nd the -is from = to = t. Sketch the grph of the curve nd determine the re of the given region. 6. Suppose f () = ( ), where. If f is density function (refer to Emple ), find ech of the following. () P ( ) (b) P ( ) 5 (c) P ( ) (d) P ( ) using your result from prt (c) 7. Suppose f () = /8, where. If f is density function (refer to Emple ), find ech of the following. () P( ) (b) P( ) (c) P( ) 8. Suppose f () = /, where e e. If f is density function (refer to Emple ), find ech of the following. () P( 7) (b) P( 5) (c) P( ) (d) Verify tht P(e e ) =. 9. () Let r be rel number, where r >. Evlute r d (b) Your nswer to prt () cn be interpreted s the re of certin region of the plne. Sketch this region. ( r ) (c) Evlute lim r d. (d) Your nswer to prt (c) cn be interpreted s the re of certin region of the plne. Sketch this region. In Problems, use definite integrtion to estimte the re of the region bounded by the given curve, the -is, nd the given lines. Round your nswer to two deciml plces.. y = +, =, =

74 7 Chpter Integrtion. y = + 5, =, = 7 7. See Figure.6.. y =, =, =. y = + y y In Problems 7, epress the re of the shded region in terms of n integrl (or integrls). Do not evlute your epression.. See Figure.. y 8 y y y y FIGURE. 5. See Figure.. y y y ( ) FIGURE.6 8. Epress, in terms of single integrl, the totl re of the region to the right of the line = tht is between the curves y = 5 nd y = 7. Do not evlute the integrl. In Problems 9 5, find the re of the region bounded by the grphs of the given equtions. Be sure to find ny needed points of intersection. Consider whether the use of horizontl strips mkes the integrl simpler thn when verticl strips re used. 9. y =, y =. y =, y =. y =, y = +, y =. y = +, =. = 8 + y, =, y =, y =. y = 6, y = 5. y =, y = 6. y =, y = + 6, =. (Hint: The only rel root of 6 = is.) 7. y =, y = 8. y =, y = 9. y = 8, y =, =, = 5. y = +, y =, =, = 5. y =, y = 5. y =, y = 5. y =, y = 5, y =, y = FIGURE y + 7 =, y = 55. Find the re of the region tht is between the curves 6. See Figure.5. y = nd y = 5 y y from = to =. 56. Find the re of the region tht is between the curves y = + nd y = y FIGURE.5 y from = to =. 57. Lorenz Curve A Lorenz curve is used in studying income distributions. If is the cumultive percentge of income recipients, rnked from poorest to richest, nd y is the cumultive percentge of income, then equlity of income distribution is given by the line y = in Figure.7, where nd y re epressed s decimls. For emple, % of the people receive % of totl income, % of the people receive % of the

75 Section.9 Consumers nd Producers Surplus 7 income, nd so on. Suppose the ctul distribution is given by the Lorenz curve defined by Cumultive percentge of income. y. y = y y 5 5 Lorentz curve Cumultive percentge of income recipients FIGURE.7 Note, for emple, tht % of the people receive only.% of totl income. The degree of devition from equlity is mesured by the coefficient of inequlity for Lorenz curve. This coefficient is defined to be the re between the curve nd the digonl, divided by the re under the digonl: re between curve nd digonl re under digonl For emple, when ll incomes re equl, the coefficient of inequlity is zero. Find the coefficient of inequlity for the Lorenz curve just defined. 58. Lorenz curve Find the coefficient of inequlity s in Problem 57 for the Lorenz curve defined by y = Find the re of the region bounded by the grphs of the equtions y = nd y = m, where m is positive constnt. 6. () Find the re of the region bounded by the grphs of y = nd y = +. (b) Wht percentge of the re in prt () lies bove the -is? 6. The region bounded by the curve y = nd the line y = is divided into two prts of equl re by the line y = k, where k is constnt. Find the vlue of k. In Problems 6 66, estimte the re of the region bounded by the grphs of the given equtions. Round your nswer to two deciml plces. 6. y = +, y = 6 6. y = 5, y = 7 6. y = 8 +, y = y = 5 +, y = 66. y = , y = + Objective To develop the economic concepts of consumers surplus nd producers surplus, which re represented by res. p p p p Demnd curve q q dq FIGURE.8 demnd curves. Supply curve Supply nd q.9 Consumers nd Producers Surplus Determining the re of region hs pplictions in economics. Figure.8 shows supply curve for product. The curve indictes the price p per unit t which the mnufcturer will sell (or supply) q units. The digrm lso shows demnd curve for the product. This curve indictes the price p per unit t which consumers will purchse (or demnd) q units. The point (q, p ) where the two curves intersect is clled the point of equilibrium. Here p is the price per unit t which consumers will purchse the sme quntity q of product tht producers wish to sell t tht price. In short, p is the price t which stbility in the producer consumer reltionship occurs. Let us ssume tht the mrket is t equilibrium nd the price per unit of the product is p. According to the demnd curve, there re consumers who would be willing to py more thn p. For emple, t the price per unit of p, consumers would buy q units. These consumers re benefiting from the lower equilibrium price p. The verticl strip in Figure.8 hs re p dq. This epression cn lso be thought of s the totl mount of money tht consumers would spend by buying dq units of the product if the price per unit were p. Since the price is ctully p, these consumers spend only p dq for the dq units nd thus benefit by the mount pdq p dq. This epression cn be written (p p ) dq, which is the re of rectngle of width dq nd height p p. (See Figure.9.) Summing the res of ll such rectngles from q = to q = q by definite integrtion, we hve q ( p p ) dq G. Stigler, The Theory of Price, rd ed. (New York: The Mcmilln Compny, 966), pp. 9 9.

76 7 Chpter Integrtion p p p Demnd curve dq q Supply curve FIGURE.9 Benefit to consumers for dq units. p p CS p f(q) FIGURE. p p PS q Demnd curve q q Consumers surplus. Supply curve p g(q) This integrl, under certin conditions, represents the totl gin to consumers who re willing to py more thn the equilibrium price. This totl gin is clled consumers surplus, bbrevited CS. If the demnd function is given by p = f (q), then CS = q [ f (q) p ] dq Geometriclly (see Figure.), consumers surplus is represented by the re between the line p = p nd the demnd curve p = f (q) from q = to q = q. Some of the producers lso benefit from the equilibrium price, since they re willing to supply the product t prices less thn p. Under certin conditions, the totl gin to the producers is represented geometriclly in Figure. by the re between the line p = p nd the supply curve p = g(q) from q = to q = q. This gin, clled producers surplus nd bbrevited PS, is given by EXAMPLE 5 PS = q Producers Surplus [ p g(q)] dq The mnger of Ibn Al-Hithm Sunglsses store hs found tht the demnd eqution for the most populr glsses is given by p = 5 + q, where p is the price in hundreds of dollrs, t which q pirs of glsses re sold. If the mrket price is $, wht is the producers surplus? Solution: When p = we hve = 5 + q, from which we find tht q = 57. The producers surplus is given by PS = q (p p) dq = = = ( ) 5 + q dq dq 57 dq 5 + q ) dq 57 (5 + q) d(5 + q) q FIGURE. Producers surplus. q = q 9 (5 + q) 6 Therefore the producers surplus is pproimtely $,6. EXAMPLE 5 Finding Consumers Surplus nd Producers Surplus The demnd function for product is p = f (q) =.5q where p is the price per unit (in dollrs) for q units. The supply function is p = g(q) = +.q Determine consumers surplus nd producers surplus under mrket equilibrium. Solution: First we must find the equilibrium point (p, q ) by solving the system formed by the functions p =.5q nd p = +.q. We thus equte the two epressions for p nd solve: +.q =.5q.5q = 9 q = 6

77 Section.9 Consumers nd Producers Surplus 75 When q = 6 then p = +.(6) = 7. Hence, q = 6 nd p = 7. Consumers surplus is CS = q [ f (q) p ] dq = 6 ) = (q.5 q 6 = 9 Producers surplus is PS = q [p g(q)] dq = 6 ) = (6q. q 6 = 8, (.5q 7) dq [7 ( +.q)] dq Therefore, consumers surplus is $9 nd producers surplus is $8,. Now Work Problem EXAMPLE 5 Using Horizontl Strips to Find Consumers Surplus nd Producers Surplus The demnd eqution for product is q = f (p) = 9 p nd the supply eqution is q = g(p) = p. Determine consumers surplus nd producers surplus when mrket equilibrium hs been estblished. Solution: Determining the equilibrium point, we hve p = 9 p p + p 9 = (p + )(p 9) = Thus, p = 9, so q = 9 = 8. (See Figure..) Note tht the demnd eqution epresses q s function of p. Since consumers surplus cn be considered n re, this re cn be determined by mens of horizontl strips of width dp nd length q = f (p). The res of these strips re summed from p = 9 to p = 5 by integrting with respect to p: p 5 q p dp 9 9 CS q PS p 8 q FIGURE. Digrm for Emple 5.

78 76 Chpter Integrtion CS = 5 9 ( ) 9 p = 9 ln dp = (9 ln p p) Using horizontl strips for producers surplus, we hve 9 (p ) PS = (p ) dp = 9 = 5 9 Now Work Problem 5 PROBLEMS.9 In Problems 5, the first eqution is demnd eqution nd the second is supply eqution of product. In ech cse, determine consumers surplus nd producers surplus under mrket equilibrium.. p =.8q p = 6 +.q. p = 9 q p = q + 5. q = ( p) q = 5(p ) 6. The demnd eqution for product is. p = q p = + q. q = p q = p q = p Clculte consumers surplus under mrket equilibrium, which occurs t price of $8. 7. The demnd eqution for product is nd the supply eqution is q = p p = q Find producers surplus nd consumers surplus under mrket equilibrium. 8. The demnd eqution for product is p = q, nd the supply eqution is p = q+, where p is the price per unit (in hundreds of dollrs) when q units re demnded or supplied. Determine, to the nerest thousnd dollrs, consumers surplus under mrket equilibrium. 9. The demnd eqution for product is nd the supply eqution is p = 6 5q q + 6 p = ln (q + ) 6 Determine consumers surplus nd producers surplus under mrket equilibrium. Round your nswers to the nerest integer.. Producers Surplus The supply function for product is given by the following tble, where p is the price per unit (in dollrs) t which q units re supplied to the mrket: q 5 p Estimte the producers surplus if the selling price is $8. Chpter Review Importnt Terms nd Symbols Section. Section. Section. Section. Section.5 Section.6 Emples Differentils differentil, dy, d E., p. 67 The Indefinite Integrl ntiderivtive indefinite integrl f () d integrl sign E. 6, p. 678 integrnd vrible of integrtion constnt of integrtion E. 7, p. 678 Integrtion with Initil Conditions initil condition E. 5, p. 68 More Integrtion Formuls power rule for integrtion E., p. 688 Techniques of Integrtion preliminry division E. 9, p. 69 The Definite Integrl definite integrl f () d limits of integrtion E., p. 7 b

79 Chpter Review 77 Section.7 Section.8 Section.9 The Fundmentl Theorem of Integrl Clculus Fundmentl Theorem of Integrl Clculus F() b E. 6, p. 77 Are between Curves verticl strip E., p. 75 horizontl strip E. 8, p. 79 Consumers nd Producers Surplus consumers surplus producers surplus E. 5, p. 7 Summry If y = f () is differentible function of, we define the differentil dy by dy = f () d where d = is chnge in nd cn be ny rel number. (Thus dy is function of two vribles, nmely nd d.) If d is close to zero, then dy is n pproimtion to y = f ( + d) f (). y dy Moreover, dy cn be used to pproimte function vlue using f ( + d) f () + dy An ntiderivtive of function f is function F such tht F () = f (). Any two ntiderivtives of f differ t most by constnt. The most generl ntiderivtive of f is clled the indefinite integrl of f nd is denoted f () d. Thus, f () d = F() + C where C is clled the constnt of integrtion, if nd only if F = f. Some elementry integrtion formuls re s follows: k d = k + C k constnt d = C d = ln + C for > e d = e + C kf () d = k f () d k constnt [f () ± g()] d = f () d ± g() d Another formul is the power rule for integrtion: u du = u+ + + C, if Here u represents differentible function of, nd du is its differentil. In pplying the power rule to given integrl, it is importnt tht the integrl be written in form tht precisely mtches the power rule. Other integrtion formuls re nd e u du = e u + C du = ln u + C u u If the rte of chnge of function f is known tht is, if f is known then f is n ntiderivtive of f. In ddition, if we know tht f stisfies n initil condition, then we cn find the prticulr ntiderivtive. For emple, if mrginl-cost function dc/dq is given to us, then by integrtion, we cn find the most generl form of c. Tht form involves constnt of integrtion. However, if we re lso given fied costs (tht is, costs involved when q = ), then we cn determine the vlue of the constnt of integrtion nd thus find the prticulr cost function c. Similrly, if we re given mrginl-revenue function dr/dq, then by integrtion nd by using the fct tht r = when q =, we cn determine the prticulr revenue function r. Once r is known, the corresponding demnd eqution cn be found by using the eqution p = r/q. It is helpful t this point to review summtion nottion from Section.. This nottion is especilly useful in determining res. For continuous f, to find the re of the region bounded by y = f (), y =, =, nd = b, we divide the intervl [, b] into n subintervls of equl length d = (b )/n. If i is the right-hnd endpoint of n rbitrry subintervl, then the product f ( i ) d is the re of rectngle. Denoting the sum of ll such res of rectngles for the n subintervls by S n, we define the limit of S n s n s the re of the entire region: lim S n = lim n n n f ( i ) d = re i= If the restriction tht f () is omitted, this limit is defined s the definite integrl of f over [, b]: lim n n f ( i ) d = i= b f () d Insted of evluting definite integrls by using limits, we my be ble to employ the Fundmentl Theorem of Integrl Clculus. Mthemticlly, b b f () d = F() = F(b) F() where F is ny ntiderivtive of f.

80 78 Chpter Integrtion nd Some properties of the definite integrl re b b kf () d = k b [f () ± g()] d = c f () d = b f () d b f () d ± f () d + c k constnt b b f () d g() d If f () is continuous on [, b], then the definite integrl cn be used to find the re of the region bounded by y = f (), the -is, =, nd = b. The definite integrl cn lso be used to find res of more complicted regions. In these situtions, strip of re should be drwn in the region. This llows us to set up the proper definite integrl. In this regrd, both verticl strips nd horizontl strips hve their uses. One ppliction of finding res involves consumers surplus nd producers surplus. Suppose the mrket for product is t equilibrium nd (q, p ) is the equilibrium point (the point of intersection of the supply curve nd the demnd curve for the product). Then consumers surplus, CS, corresponds to the re from q = to q = q, bounded bove by the demnd curve nd below by the line p = p. Thus, CS = q (f (q) p ) dq where f is the demnd function. Producers surplus, PS, corresponds to the re from q = to q = q, bounded bove by the line p = p nd below by the supply curve. Therefore, PS = q where g is the supply function. (p g(q)) dq Review Problems In Problems, determine the integrls.. ( + 7) d. d (9 + ) d. 6 ( + 5) d 6. 6 d t + 8 dt y(y + ) dy. 9 5 d (y 6) dy e 5 d. d 7 8 d 7 t t (.5.) dt. d t. t dt 6. d + t + d 8. (6 + )( + ) / d (e y e y ) dy. ( + ) d. (y + y + y + y) dy. 5 5 d d e d + e d ( + )( + ) d [ ] 8 7. d 8. ( + ) d ( + ) / t z 9. dt. t z dz + ( + ). d. d + e + e 5. d. d e e ln d 6. d e + ( + e ) c 7. d 8. e e b ( + e b ) d n 9. for n = nd b = d d In Problems nd, find y, subject to the given condition.. y = e +, y() =. y = + 5, y() = In Problems 5, determine the re of the region bounded by the given curve, the -is, nd the given lines.. y =, =, =. y = e, =, = 5. y = +, = 6. y = 6, =, = 7. y = 5 8. y =, = 8, = 6 9. y = +, =, = 5. y =, =

81 Chpter Review 79 In Problems 5 58, find the re of the region bounded by the given curves. 5. y =, =, y = 5. y = 5, =, y = 5. y = ( ), y = for < 5. y =, y = y =, y = 56. y =, =, y = 57. y = ln, =, y =, y = 58. y =, y =, y =, y = 59. Mrginl Revenue If mrginl revenue is given by dr dq = q determine the corresponding demnd eqution. 6. Mrginl Cost If mrginl cost is given by dc dq = q + 7q + 6 nd fied costs re 5, determine the totl cost of producing si units. Assume tht costs re in dollrs. 6. Mrginl Revenue A mnufcturer s mrginl-revenue function is dr = 5 q.q dq If r is in dollrs, find the increse in the mnufcturer s totl revenue if production is incresed from 5 to 5 units. 6. Mrginl Cost A mnufcturer s mrginl-cost function is dc dq = q + 7 If c is in dollrs, determine the cost involved to increse production from to units. 6. Hospitl Dischrges For group of hospitlized individuls, suppose the dischrge rte is given by f (t) =.7e.7t where f (t) is the proportion dischrged per dy t the end of t dys of hospitliztion. Wht proportion of the group is dischrged t the end of dys? 6. Business Epenses The totl ependitures (in dollrs) of business over the net five yers re given by Evlute the ependitures. 5 e.5t dt 65. Find the re of the region between the curves y = 9 nd y = from = to =. 66. Find the re of the region between the curves y = nd y = 5 from = to =. 67. Consumers nd Producers Surplus The demnd eqution for product is nd the supply eqution is p =.q.q + p =.q + 8 Determine consumers surplus nd producers surplus when mrket equilibrium hs been estblished. 68. Consumers Surplus The demnd eqution for product is nd the supply eqution is p = (q ) p = q + q + 7 where p (in thousnds of dollrs) is the price per units when q hundred units re demnded or supplied. Determine consumers surplus under mrket equilibrium. 69. Biology In discussion of gene muttion, the eqution qn dq n = (u + v) dt q q q occurs, where u nd v re gene muttion rtes, the q s re gene frequencies, nd n is the number of genertions. Assume tht ll letters represent constnts, ecept q nd t. Integrte both sides nd then use your result to show tht n = u + v ln q q q n q 7. Fluid Flow In studying the flow of fluid in tube of constnt rdius R, such s blood flow in portions of the body, we cn think of the tube s consisting of concentric tubes of rdius r, where r R. The velocity v of the fluid is function of r nd is given by 5 v = (P P )(R r ) ηl where P nd P re pressures t the ends of the tube, η ( Greek letter red et ) is the fluid viscosity, nd l is the length of the tube. The volume rte of flow through the tube, Q, is given by Q = R πrv dr Show tht Q = πr (P P ). Note tht R occurs s fctor to 8ηl the fourth power. Thus, doubling the rdius of the tube hs the effect of incresing the flow by fctor of 6. The formul tht you derived for the volume rte of flow is clled Poiseuille s lw, fter the French physiologist Jen Poiseuille. 7. Inventory In discussion of inventory, Brbos nd Friedmn 6 refer to the function g() = k / ku r du where k nd r re constnts, k > nd r >, nd >. Verify the clim tht W. B. Mther, Principles of Quntittive Genetics (Minnepolis: Burgess Publishing Compny, 96). 5 R. W. Stcy et l., Essentils of Biologicl nd Medicl Physics (New York: McGrw-Hill, 955). 6 L. C. Brbos nd M. Friedmn, Deterministic Inventory Lot Size Models Generl Root Lw, Mngement Science,, no. 8 (978), 89 6.

82 7 Chpter Integrtion g () = r+ (Hint: Consider two cses: when r nd when r =.) In Problems 7 7, estimte the re of the region bounded by the given curves. Round your nswer to two deciml plces. 7. y = + 9 +, y = 7. y = + + +, y = y = + 5, y = The demnd eqution for product is nd the supply eqution is p = q + p = ln (q + ) + 5 Determine consumers surplus nd producers surplus under mrket equilibrium. Round your nswers to the nerest integer. Chpter Test. Find the differentil of the following function in terms of nd d: y = (9 + )e +. For the following function, find y nd dy for the given vlues of nd d:. Let f () =. y = ln ; =, d =.. Demnd The demnd, q, for monopolist s product is relted to the price per unit, p, ccording to the eqution + q = p () Verify tht units will be demnded when the price per unit is $. (b) Show tht dq =.5 when the price per unit is $. dp (c) Use differentils nd the results of prts () nd (b) to pproimte the number of units tht will be demnded if the price per unit is reduced to $ Find the indefinite integrl d ) 6. Find the indefinite integrl (y y + ey dy 6 7. If y stisfies the following conditions, find y() for the given vlue of : y = +, y() = ; = 8. Elsticity of Demnd The sole producer of product hs determined tht the mrginl-revenue function is dr = q dq Determine the point elsticity of demnd for the product when q = 5. (Hint: First find the demnd function.) 9. Find d ( 7) + +. Find + + d. Life Spn If the rte of chnge of the epected life spn l t birth of people born in Egypt cn be modeled by dl dt =, where t is the number of yers fter 9 nd t + 5 the epected life spn ws 6 yers in 9, find the epected life spn for people born in 998. ( + ). Find d. Find (8) / + d ln. Find d 5. Revenue Function The mrginl-revenue function for mnufcturer s product is of the form dr dq = e q + b for constnts nd b, where r is the totl revenue received (in dollrs) when q units re produced nd sold. Find the demnd function, nd epress it in the form p = f (q). (Hint: Rewrite dr/dq by multiplying both numertor nd denomintor by e q.) 6. Evlute the following definite integrl by tking the limit of S n. Use the right-hnd endpoint of ech subintervl. Sketch the grph, over the given intervl, of the function to be integrted. 7. Find 8. Evlute ( + ) d f () d without the use of limits, where if < f () = if < 5 if 6 9 ( ) d e e 9. Evlute d ( d ). Evlute e d d d (Hint: It is not necessry to find e d.)

83 Chpter Eplore & Etend 7. Continuous Income Flow The present vlue (in dollrs) of continuous flow of income of $ yer for five yers t 6% compounded continuously is given by 5 e.6t dt Evlute the present vlue to the nerest dollr.. Hospitl Dischrges For group of hospitlized individuls, suppose the dischrge rte is given by f (t) = 8 6 ( + t) where f (t) is the proportion of the group dischrged per dy t the end of t dys. Wht proportion hs been dischrged by the end of 7 dys?. Use definite integrl to find the re of the region bounded by the given curve, the -is, nd the given lines. First sketch the region; wtch out for res of regions tht re below the -is. y =, =, =. Epress, in terms of single integrl, the totl re of the region in the first qudrnt bounded by the -is nd the grphs of y = nd y =. Do not evlute the integrl. 5. The first eqution given below is demnd eqution nd the second is supply eqution of certin product. In ech cse, determine consumers surplus nd producers surplus under mrket equilibrium. p = 5 q + 5 p = q The demnd eqution for product is nd the supply eqution is (p + )(q + ) = q p + = () Verify, by substitution, tht mrket equilibrium occurs when p = nd q =. (b) Determine consumers surplus under mrket equilibrium. Delivered Price Suppose tht you re mnufcturer of product whose sles occur within R kilometers of your mill. Assume tht you chrge customers for shipping t the rte s, in dollrs per kilometer, for ech unit of product sold. If m is the unit price (in dollrs) t the mill, then the delivered unit price p to customer kilometers from the mill is the mill price plus the shipping chrge s: p = m + s R () The problem is to determine the verge delivered price of the units sold. of f nd bove the t-is from t = to t = represents the totl number of units Q sold to customers within km of the mill. [See Figure.().] You cn refer to f s the distribution of demnd. Becuse Q is function of nd is represented by re, f(t) Q() = f (t) dt Number of units sold within km R t () f(t) Totl number of units sold within mrket re R t Suppose tht there is function f such tht f (t) on the intervl [, R] nd such tht the re under the grph FIGURE. (b) Number of units sold s n re.

84 7 Chpter Integrtion In prticulr, the totl number of units sold within the mrket re is Q(R) = R f (t) dt [see Figure.(b)]. For emple, if f (t) = nd R =, then the totl number of units sold within the mrket re is Q() = dt = t = = The verge delivered price A is given by A = totl revenue totl number of units sold Becuse the denomintor is Q(R), A cn be determined once the totl revenue is found. To find the totl revenue, first consider the number of units sold over n intervl. If t < t [see Figure.()], then the re under the grph of f nd bove the t-is from t = to t = t represents the number of units sold within t kilometers of the mill. Similrly, the re under the grph of f nd bove the t-is from t = to t = t represents the FIGURE. f(t ) R t t f(t ) () (t, f (t )) dt R (b) Number of units sold over n intervl. number of units sold within t kilometers of the mill. Thus the difference in these res is geometriclly the re of the shded region in Figure.() nd represents the number of units sold between t nd t kilometers of the mill, which is Q(t ) Q(t ). Thus Q(t ) Q(t ) = t t t t f (t) dt For emple, if f (t) =, then the number of units sold to customers locted between nd 6 kilometers of the mill is 6 6 Q(6) Q() = dt = t = 6 = The re of the shded region in Figure.() cn be pproimted by the re of rectngle [see Figure.(b)] whose height is f (t) nd whose width is dt, where dt = t t. Thus the number of units sold over the intervl of length dt is pproimtely f (t) dt. Becuse the price of ech of these units is [from Eqution ()] pproimtely m + st, the revenue received is pproimtely (m + st)f (t) dt The sum of ll such products from t = to t = R pproimtes the totl revenue. Definite integrtion gives Thus, R (m + st)f (t) dt (m + st)f (t) dt totl revenue = R (m + st)f (t) dt Consequently, the verge delivered price A is given by Equivlently, A = A = R R (m + st)f (t) dt Q(R) (m + st)f (t) dt R f (t) dt For emple, if f (t) =, m =, s =.5, nd R =, then R (m + st)f (t) dt = From before, R = = f (t) dt = ( +.5t) dt ( +.5t) dt ( t + t 8 ) [( =, +, 8 =,5 Thus, the verge delivered price is,5/ = $.5. dt = ) ]

85 Chpter Eplore & Etend 7 Problems. If f (t) = t, determine the number of units sold to customers locted () within 5 km of the mill, nd (b) between nd 5 km of the mill.. If f (t) =.5t, m = 5, s =., nd R = 8, determine () the totl revenue, (b) the totl number of units sold, nd (c) the verge delivered price.. If f (t) = 9 t, m =, s =, nd R =, determine () the totl revenue, (b) the totl number of units sold, nd (c) the verge delivered price. Use grphing clcultor if you like.. How do rel-world sellers of such things s books nd clothing generlly determine shipping chrges for n order? (Visit n online retiler to find out.) How would you clculte verge delivered price for their products? Is the procedure fundmentlly different from the one discussed in this Eplore & Etend?

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87 English Arbic Glossry of Mthemticl Terms Absolute etrem We refer to either bsolute mimum or bsolute minimum s bsolute etremum (plurl: bsolute etrem). Absolute mimum A function f hs n bsolute mimum t if f () f () for ll in the domin of f Absolute minimum f hs n bsolute minimum t if f () f () for ll in the domin of f Absolute Vlue On the rel numbers line the bsolute vlue is the distnce of number from lw of mutully eclusive events If E nd F re mutully eclusive events then P(EUF)=P(E)+P(F) Amortiztion schedules A tble giving n nlysis of how ech pyment in the lon is hndled Amortizing A lon, such s mortgge, is mortized when prt of ech instllment pyment is used to py interest nd the remining prt is used to reduce the principl Annuity A finite sequence of pyments mde t fied periods of time over given intervl Annuity due is n nnuity in which ech pyment is mde t the beginning of pyment period Antiderivtive An ntiderivtive of f is function whose derivtive is f Are between the curves Arithmetic sequence An rithmetic sequence is sequence (bk) defined recursively by b = nd, for ech positive integer k, bk+ = d + bk for fied rel numbers nd d. Artificil objective function A function of the form W = Z Mt where z is the originl objective function, t is the rtificil vrible nd the constnt M is very lrge positive number. Artificil Vribles A nonnegtive vrible dded to the left side of the eqution in which the coefficient of the slck vrible is Asymptote An symptote is line tht curve pproches rbitrrily closely without crossing it Augmented mtri Obtined by dding n dditionl column for the constnts to the right of the coefficient mtri Averge totl cost Totl cost divided by the quntity produced Averge rte of chnge The slope of the secnt line connecting two given points Averge vlue of function over n intervl [, b] is denoted by f nd is given by f = b b f ()d Ais of symmetry prbol is symmetric bout verticl line, clled the is of symmetry of the prbol Bsic Counting Principle Suppose tht procedure involves sequence of k stges. Let n be the number of wys the first cn occur nd n be the number of wys the second cn occur. Continuing in this wy, let nk be the number of wys the kth stge cn occur. Then the totl number of different wys the procedure cn occur is n n nk Byes s formul If E is n event nd F, F,..., Fn is prtition then, to find the conditionl probbility of event Fi, given E, when prior nd conditionl probbilities re known, we cn use Byes s formul Byes s probbility tree Used to solve Byes-type problem Bernoulli trils Whenever we hve n independent trils of n eperiment in which ech tril hs only two possible outcomes (success nd filure) nd the probbility of success in ech tril remins the sme, the trils re clled Bernoulli trils. Binomil Distribution In Binomil eperiment, if X is the number of successes in n trils, then the distribution f of X is clled binomil distribution Binomil eperiment If there re only two possible outcomes (success nd filure) for ech independent tril, nd the probbilities of success nd filure do not chnge from tril to tril, then the eperiment is clled binomil eperiment. G-

88 G- English Arbic Glossry of Mthemticl Terms Bounded fesible region A fesible region tht cn be contined within circle Brek even point The point t which totl revenue equls totl cost Certin event The smple spce itself. It will then occur with probbility Chnge-of-bse formul log_b m=(log_ m)/(log_ b) Closed intervl The set of ll rel numbers for which b nd includes the numbers nd b, Coefficient mtri Mtri formed by the coefficients of the vribles in ech eqution. The first colomn consists on the coefficients of the first vrible, the second one to the second vrible nd so on Combintion A selection of r objects, without regrd to order nd without repetition, selected from n distinct objects Common Logrithm Logrithm to the bse. Competitive products A nd B re competitive products or substitutes if n increse in the price of B cuses n increse in the demnd for A, if it is ssumed tht the price of A does not chnge. Similrly, n increse in the price of A cuses n increse in the demnd for B when the price of B is held fied. Complement The event consisting of ll smple points in the smple spce tht re not in the event Complementry products A nd B re Complementry products if n increse in the price of B cuses decrese in the demnd for A if the price of A does not chnge. Similrly, n increse in the price of A cuses decrese in the demnd for B when the price of B is held fied. Composite function Given two functions f nd g, the opertion of pplying g nd then pplying f to the result is clled composition nd the resulting function, denoted f g, is clled the composite of f with g. Compound mount the originl principl invested or borrowed plus ll ccrued interest Compound interest The difference between the compound mount nd the originl principl Compounded continuously If the number of interest periods per yer, is incresed indefinitely then the length of ech period pproches nd we sy tht interest is compounded continuously. Concve down f is sid to be concve up down on (, b) if f is decresing on (, b). Concve up f is sid to be concve up on (, b) if f is incresing on (, b). Conditionl probbility The probbility of n event when it is known tht some other event hs occurred. Constnt function A function of the form h() = c, where c is constnt Constrint Restrictions on the domin Consumers surplus The totl gin to consumers who re willing to py more thn the equilibrium price Consumption function The function relting consumption to income Continuous t point A function f is continuous t if nd only if the following three conditions re met:. f () eists. lim f () eists. lim f () = f () Continuous function Continuous t ech point of the domin Continuous rndom vrible A rndom vrible continuous rndom vrible if it ssumes ll vlues in some intervl or intervls Convergent The improper integrl is convergent if its vlue is finite number Corner point A liner function defined on nonempty, bounded fesible region hs mimum (minimum) vlue, nd this vlue cn be found t corner point. Criticl point If is criticl vlue for function f then (, f()) is criticl point Criticl vlue For in the domin of f, if either f() = or f() does not eist, then is clled criticl vlue for f. Current ssets Are ssets such s csh, merchndise inventory, nd ccounts receivble to its current libilities

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