Math 31S. Rumbos Fall Solutions to Assignment #16


 Marjory Cook
 5 years ago
 Views:
Transcription
1 Mth 31S. Rumbos Fll Solutions to Assignment #16 1. Logistic Growth 1. Suppose tht the growth of certin niml popultion is governed by the differentil eqution 1000 dn N dt = 100 N, (1) where N(t) denote the number of individuls in the popultion t time t. () Suppose there re 200 individuls in the popultion t time t = 0. Sketch the grph of N = N(t). Solution: The eqution in (1) describes logistic growth in popultion with intrinsic growth rte r = 100/1000 nd crrying cpcity K = 100. A sketch of the solution with initil popultion N(0) = 200 is shown in Figure 1. N 100 t Figure 1: Sketch of Solution to (1) with N o = 200 (b) Will there ever be more thn 200 individuls in the popultion? Will there ever be fewer thn 100 individuls? Explin your nswer. Solution: The sketch of the solution to (1) subject to the initil condition N(0) = 200 shows tht the popultion size will never be bove 200 or below Spred of virl infection 2. Let I(t) denote the totl number of people infected with virus. Assume tht I(t) grows ccording to logistic model. Suppose 1 Adpted from Problem 6 on pge 521 in Hughes Hllett et l, Clculus, Third Edition, Wiley, Adpted from Problem 7 on pge 521 in Hughes Hllett et l, Clculus, Third Edition, Wiley, 2002
2 Mth 31S. Rumbos Fll tht 10 people hve the virus originlly nd tht, in the erly stges of the infection the number of infected people doubles every 3 dys. It is lso estimted tht, in the long run 5000 people in given re will become infected. () Solve n pproprite logistic model to find formul for computing I(t), where t is the time from the initil infection mesured in weeks. Sketch the grph of I(t). Solution: The function I solves the logistic eqution di dt = ri(k I), (2) where r is the intrinsic growth rte of infection nd K is the limiting number of people who will become infected in the long run. Thus, K = (3) In order to estimte r, we pproximte the spred of the infection with n exponentil model with doubling time of 3 dys or 3/7 weeks. Thus, r = ln 2 3/7 = , (4) in units of 1/week. The solution to (2) subject to the initil condition I(0) = I o is given by I(t) = I o K, for t R. (5) I o + (K I o )e rt Substituting the vlues of I o = 10, nd K nd r given in (3) nd (4), respectively, into (5) yields the solution I(t) = 50000, for t R. (6) 10 + (4990)e t A sketch of the grph of the function in (6) is pictured in Figure 2. (b) Estimte the time when the rte of infected people begins to decrese. Solution: The rte of infection will begin to decrese when the number of infected people is hlf of the limiting vlue; nmely, when I(t) = 2500,
3 Mth 31S. Rumbos Fll I K t Figure 2: Sketch of function in (6) or, ccording to (6), when Solving the eqution in (7) yields t = = (7) 10 + (4990)e t 1 ln(499) = 3.84 weeks Thus, the rte of infection will begin to decrese in bout 3 weeks nd 5 dys nd 21 hours. 3. Non Logistic Growth 3. There re mny clsses of orgnisms whose birth rte is not proportionl to the popultion size. For exmple, suppose tht ech member of the popultion requires prtner for reproduction, nd ech member relies on chnce encounters for meeting mte. Assume tht the expected number of encounters is proportionl to the product of numbers of femle nd mle members in the popultion, nd tht these re eqully distributed; hence, the number of encounters will be proportionl to the squre of the size of the popultion. Use conservtion principle to derive the popultion model dn dt = N 2 bn, (8) 3 Adpted from Problem 12 on pge 39 in Brun, Differentil Equtions nd their Applictions, Fourth Edition, Springer Verlg, 1993
4 Mth 31S. Rumbos Fll where nd b re positive constnts. Explin your resoning. Solution: Begin with the conservtion principle dn dt In this cse we hve nd = Rte of individuls in Rte of individuls out. (9) Rte of individuls in = N 2, (10) Rte of individuls outbn, (11) where nd b re positive constnts of proportionlity. The eqution in (8) follows from (9) fter substituting (10) nd (11). 4. For the eqution in (8), () find the vlues of N for which the popultion size is not chnging; Solution: Rewrite the eqution in (8) s ( dn dt = N N b ). (12) We see from (12) tht dn dt = 0 when N = 0 or N = b. (b) find the rnge of positive vlues of N for which the popultion size is incresing, nd those for which it is decresing; Solution: We see from (12) tht dn dt > 0 for N > b, nd dn dt < 0 for N < b. This, the popultion size increses for N > b, nd decreses for N < b. (c) find rnges of positive vlues of N for which the grph of N = N(t) is concve up, nd those for which it is concve down; Solution: Differentite on both sides of (8) with respect to t to obtin d 2 N dt 2 = 2N dn dt bdn dt, (13)
5 Mth 31S. Rumbos Fll where we hve pplied the Chin Rule. The eqution in (13) cn be rewritten s ( d 2 N dt = 2 N b ) dn 2 2 dt. (14) Substituting the expression for dn dt d 2 N dt 2 = 22 N in (12) into (14) then yields ( N b ) ( N b ). (15) 2 In view of (15) we see tht, for positive vlues of N, the sign of d2 N is dt 2 determined by the signs of the two right most fctors in (15). The signs of these two fctors re displyed in Tble 1. The concvity of of the grph N b 2 N b b/2 b/ N (t) + + grph of N(t) concve up concve down concve up Tble 1: Concvity of the grph of N = N(t) of N = N(t) is lso displyed in Tble 1. From tht tble we get tht the grph of N = N(t) is concve up for nd concve down for 0 < N < b 2 b 2 < N < b. or N > b, (d) Sketch possible solutions. Solution: Putting together the informtion on concvity in Tble 1 nd the fct tht N(t) increses for N > b/ nd decreses for 0 < N < b/,
6 Mth 31S. Rumbos Fll N b t Figure 3: Possible Solutions to Logistic eqution we obtin the sketches of possible solutions to the eqution in (8) displyed in Figure For the eqution in (8), () use seprtion of vribles nd prtil frctions to find solution stisfying the initil condition N(0) = N o, for N o > 0. Solution: Seprte vrible in the eqution in (12) to obtin 1 N(N b/) dn = dt. (16) Use prtil frctions in the integrnd on the left hnd side to (16) nd integrte on the right hnd side to get to get { 1 } b N + 1 dn = t + c 1, (17) N b/ for some constnt c 1. Evlute the integrl on the left hnd side of (17) nd simplify to get ( ) N b/ ln = bt + c 2, (18) N for some constnt c 2. Next, tke the exponentil function on both sides of (18) to get N b/ = c 3 e bt, (19) N
7 Mth 31S. Rumbos Fll where we hve set c 3 = e c 2. Using the continuity of N nd of the exponentil function we deduce from (19) tht N b/ = c e 2t/b, (20) N for some constnt c. The eqution in (20) cn now be solved for N s function of t to get N(t) = b/. (21) 1 c ebt Next, use the initil condition N(0) = N o to obtin from (20) tht Substituting the vlue of c in (22) into (21) yields N(t) = c = N o b/ N o. (22) N o b/. (23) N o + (b/ N o ) ebt (b) Wht hppens to N(t) s t if N o > b/? Wht hppens if N o < b/? Why is b/ clled threshold vlue? Solution: We first consider the cse in which 0 < N o < b/. In this cse, the function in (23) is defined for ll vlues of t nd lim N(t) = 0, t since b > 0. On the other hnd, is N o > b/, then the function in (23) ceses to exist when (N o b/) e bt = N o. As t pproches tht time, N(t). Thus, depending on whether N o < b/ or N o > b/, the popultion will eventully go extinct or it will hve unlimited growth in finite time. Thus, b/ is the threshold popultion vlue which determines growth or extinction.
A. Limits  L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c
More informationMath 42 Chapter 7 Practice Problems Set B
Mth 42 Chpter 7 Prctice Problems Set B 1. Which of the following functions is solution of the differentil eqution dy dx = 4xy? () y = e 4x (c) y = e 2x2 (e) y = e 2x (g) y = 4e2x2 (b) y = 4x (d) y = 4x
More informationMath 116 Final Exam April 26, 2013
Mth 6 Finl Exm April 26, 23 Nme: EXAM SOLUTIONS Instructor: Section:. Do not open this exm until you re told to do so. 2. This exm hs 5 pges including this cover. There re problems. Note tht the problems
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationMATH SS124 Sec 39 Concepts summary with examples
This note is mde for students in MTH124 Section 39 to review most(not ll) topics I think we covered in this semester, nd there s exmples fter these concepts, go over this note nd try to solve those exmples
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationA. Limits  L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. 1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by
More information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a nonconstant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More information1. Find the derivative of the following functions. a) f(x) = 2 + 3x b) f(x) = (5 2x) 8 c) f(x) = e2x
I. Dierentition. ) Rules. *product rule, quotient rule, chin rule MATH 34B FINAL REVIEW. Find the derivtive of the following functions. ) f(x) = 2 + 3x x 3 b) f(x) = (5 2x) 8 c) f(x) = e2x 4x 7 +x+2 d)
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More informationThe Fundamental Theorem of Calculus, Particle Motion, and Average Value
The Fundmentl Theorem of Clculus, Prticle Motion, nd Averge Vlue b Three Things to Alwys Keep In Mind: (1) v( dt p( b) p( ), where v( represents the velocity nd p( represents the position. b (2) v ( dt
More informationSection 7.1 Integration by Substitution
Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationConservation Law. Chapter Goal. 5.2 Theory
Chpter 5 Conservtion Lw 5.1 Gol Our long term gol is to understnd how mny mthemticl models re derived. We study how certin quntity chnges with time in given region (sptil domin). We first derive the very
More informationCalculus II: Integrations and Series
Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More informationAQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions
Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationChapter 3 Exponential and Logarithmic Functions Section 3.1
Chpter 3 Eponentil nd Logrithmic Functions Section 3. EXPONENTIAL FUNCTIONS AND THEIR GRAPHS Eponentil Functions Eponentil functions re nonlgebric functions. The re clled trnscendentl functions. The eponentil
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
More informationf(a+h) f(a) x a h 0. This is the rate at which
M408S Concept Inventory smple nswers These questions re openended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnkoutnnswer problems! (There re plenty of those in the
More information3.1 Exponential Functions and Their Graphs
. Eponentil Functions nd Their Grphs Sllbus Objective: 9. The student will sketch the grph of eponentil, logistic, or logrithmic function. 9. The student will evlute eponentil or logrithmic epressions.
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More informationAPPROXIMATE INTEGRATION
APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose ntiderivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be
More informationAB Calculus Review Sheet
AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is
More informationdu = C dy = 1 dy = dy W is invertible with inverse U, so that y = W(t) is exactly the same thing as t = U(y),
29. Differentil equtions. The conceptul bsis of llometr Did it occur to ou in Lecture 3 wh Fiboncci would even cre how rpidl rbbit popultion grows? Mbe he wnted to et the rbbits. If so, then he would be
More informationDistance And Velocity
Unit #8  The Integrl Some problems nd solutions selected or dpted from HughesHllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl
More information1.1. Linear Constant Coefficient Equations. Remark: A differential equation is an equation
1 1.1. Liner Constnt Coefficient Equtions Section Objective(s): Overview of Differentil Equtions. Liner Differentil Equtions. Solving Liner Differentil Equtions. The Initil Vlue Problem. 1.1.1. Overview
More information4.1 OnetoOne Functions; Inverse Functions. EX) Find the inverse of the following functions. State if the inverse also forms a function or not.
4.1 OnetoOne Functions; Inverse Functions Finding Inverses of Functions To find the inverse of function simply switch nd y vlues. Input becomes Output nd Output becomes Input. EX) Find the inverse of
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More information( ) as a fraction. Determine location of the highest
AB Clculus Exm Review Sheet  Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationFirst midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009
Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No
More information( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More information1 Techniques of Integration
November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.
More information( ) where f ( x ) is a. AB/BC Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB/ Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 Find the intersection of f ( x) nd g( x). A3 Show tht f ( x) is even. A4 Show tht
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More information6.5 Numerical Approximations of Definite Integrals
Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 6.5 Numericl Approximtions of Definite Integrls Sometimes the integrl of function cnnot be expressed with elementry functions, i.e., polynomil,
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationMathematics Extension 1
04 Bored of Studies Tril Emintions Mthemtics Etension Written by Crrotsticks & Trebl. Generl Instructions Totl Mrks 70 Reding time 5 minutes. Working time hours. Write using blck or blue pen. Blck pen
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More information5 Accumulated Change: The Definite Integral
5 Accumulted Chnge: The Definite Integrl 5.1 Distnce nd Accumulted Chnge * How To Mesure Distnce Trveled nd Visulize Distnce on the Velocity Grph Distnce = Velocity Time Exmple 1 Suppose tht you trvel
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationMath 113 Exam 2 Practice
Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationROB EBY Blinn College Mathematics Department
ROB EBY Blinn College Mthemtics Deprtment Mthemtics Deprtment 5.1, 5.2 Are, Definite Integrls MATH 2413 Rob EbyFll 26 Weknowthtwhengiventhedistncefunction, wecnfindthevelocitytnypointbyfindingthederivtiveorinstntneous
More informationMath 1431 Section 6.1. f x dx, find f. Question 22: If. a. 5 b. π c. π5 d. 0 e. 5. Question 33: Choose the correct statement given that
Mth 43 Section 6 Question : If f d nd f d, find f 4 d π c π d e  Question 33: Choose the correct sttement given tht 7 f d 8 nd 7 f d3 7 c d f d3 f d f d f d e None of these Mth 43 Section 6 Are Under
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06073 HIKARI Ltd, www.mhikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Theorem Suppose f is continuous
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationTaylor Polynomial Inequalities
Tylor Polynomil Inequlities Ben Glin September 17, 24 Abstrct There re instnces where we my wish to pproximte the vlue of complicted function round given point by constructing simpler function such s polynomil
More informationHOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016
HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M
More informationMATH 115 FINAL EXAM. April 25, 2005
MATH 115 FINAL EXAM April 25, 2005 NAME: Solution Key INSTRUCTOR: SECTION NO: 1. Do not open this exm until you re told to begin. 2. This exm hs 9 pges including this cover. There re 9 questions. 3. Do
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationAP Calculus. Fundamental Theorem of Calculus
AP Clculus Fundmentl Theorem of Clculus Student Hndout 16 17 EDITION Click on the following link or scn the QR code to complete the evlution for the Study Session https://www.surveymonkey.com/r/s_sss Copyright
More informationMath 116 Calculus II
Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More information1 Functions Defined in Terms of Integrals
November 5, 8 MAT86 Week 3 Justin Ko Functions Defined in Terms of Integrls Integrls llow us to define new functions in terms of the bsic functions introduced in Week. Given continuous function f(), consider
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationRead section 3.3, 3.4 Announcements:
Dte: 3/1/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: 1. f x = 3x 6, find the inverse, f 1 x., Using your grphing clcultor, Grph 1. f x,f
More information13.4. Integration by Parts. Introduction. Prerequisites. Learning Outcomes
Integrtion by Prts 13.4 Introduction Integrtion by Prts is technique for integrting products of functions. In this Section you will lern to recognise when it is pproprite to use the technique nd hve the
More informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
More informationPractice Final. Name: Problem 1. Show all of your work, label your answers clearly, and do not use a calculator.
Nme: MATH 2250 Clculus Eric Perkerson Dte: December 11, 2015 Prctice Finl Show ll of your work, lbel your nswers clerly, nd do not use clcultor. Problem 1 Compute the following limits, showing pproprite
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationUNIT 1 FUNCTIONS AND THEIR INVERSES Lesson 1.4: Logarithmic Functions as Inverses Instruction
Lesson : Logrithmic Functions s Inverses Prerequisite Skills This lesson requires the use of the following skills: determining the dependent nd independent vribles in n exponentil function bsed on dt from
More information4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve
Dte: 3/14/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: Use your clcultor to solve 4 7x =250; 5 3x =500; HW Requests: Properties of Log Equtions
More informationUnit 5. Integration techniques
18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationSections 1.3, 7.1, and 9.2: Properties of Exponents and Radical Notation
Sections., 7., nd 9.: Properties of Eponents nd Rdicl Nottion Let p nd q be rtionl numbers. For ll rel numbers nd b for which the epressions re rel numbers, the following properties hold. i = + p q p q
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationFirst Semester Review Calculus BC
First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewiseliner function f, for 4, is shown below.
More information