Topic 1 Notes Jeremy Orloff


 Aubrey Payne
 2 years ago
 Views:
Transcription
1 Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble to derive the differentil eqution modeling phsicl or geometric sitution. 4. Be ble to solve seprble differentil eqution, including finding lost solutions. 5. Be ble to solve n initil vlue problem (IVP) b solving the differentil eqution nd using the initil condition to find the constnt of integrtion. 1.2 Differentil equtions nd solutions A differentil eqution (DE) is n eqution with derivtives! Emple 1.1. (DE s modeling phsicl processes, i.e., rte equtions) 1. Newton s lw of cooling: dt = k(t A), where T is the temperture of bod in n dt environment with mbient temperture A. 2. Grvit ner the erth s surfce: m d2 = mg, where is the height of mss m dt2 bove the surfce of the erth. 3. Hooke s lw: m d2 dt 2 = k, where is the displcement from equilibrium of spring with spring constnt k. Other emples: Below we will give some emples of differentil equtions modeling some geometric situtions. A solution to differentil eqution is n function tht stisfies the DE. Let s focus on wht this mens b contrsting it with solving n lgebric eqution. The unknown in n lgebric eqution, such s = 0 is the number. The eqution is solved b finding numericl vlue for tht stisfies the eqution. You cn check b substitution tht = 1 is solution to the eqution shown. The unknown in the differentil eqution d 2 d d d + = 0 1
2 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 2 is the function ().The eqution is solved b finding function () tht stisfies the eqution One solution to the eqution shown is () = e. You cn check this b substituting () = e into the eqution. Agin, note tht the solution is function. More often we will s tht the solution is fmil of functions, e.g. = Ce t. The prmeter C is like the constnt of integrtion in Ever vlue of C gives different function which solves the DE. 1.3 The most importnt differentil eqution in Here, in the ver first clss, we stte nd give solutions to our most importnt differentil equtions. In this cse we will check the solutions b substitution. As we proceed in the course we will lern methods tht help us discover solutions to equtions. The most importnt DE we will stud is d dt =, (1) where is constnt (in units of 1/time). In words the eqution ss tht the rte of chnge of is proportionl to. Becuse of its importnce we will write down some other ws ou might see it: = ; d dt = (t); = 0;. = 0. In the lst eqution we used the phsicist dot nottion to indicte the derivtive is with respect to time. You should recognize tht ll of these re the sme eqution. The solution to this eqution is (t) = Ce t, where C is n constnt Checking the solution b substitution The bove solution is esil checked b substitution. Becuse this eqution is so importnt we show the detils. Substituting (t) = Ce t into Eqution 1 we hve: Left side of (1): Right side of (1): = Ce t = Ce t Since fter substitution the left side equls the right, we hve shown tht (t) = Ce t is indeed solution of Eqution 1.
3 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS The phsicl model of the most importnt DE As phsicl model this eqution ss tht the quntit chnges t rte proportionl to. Becuse of the form the solution tkes we s tht Eqution 1 models eponentil growth or dec. In this course we will lern mn techniques for solving differentil equtions. We will test lmost ll of them on Eqution 1. You should of course understnd how to use these techniques to solve (1). However: whenever ou see this eqution ou should remind ourself tht it models eponentil growth or dec nd ou should know the solution without computtion. 1.4 The second most importnt differentil eqution Our second most importnt DE is m + k = 0, (2) where m nd k re constnts. You cn esil check tht, with ω = k/m, the function (t) = C 1 cos(ωt) + C 2 sin(ωt) is solution. Eqution 2 models simple hrmonic oscilltor. More prosicll, it models mss m oscillting t the end of spring with spring constnt k. 1.5 Solving differentil equtions b the method of optimism In our first nd second most importnt equtions bove we simpl told ou the solution. Once ou hve possible solution it is es to check it b substitution into the differentil eqution. We will cll this method, where ou guess solution nd check it b plugging our guess into the eqution, the method of optimism. In ll seriousness, this will be n importnt method for us. Of course, its utilit depends on lerning how to mke good guesses! 1.6 Generl form of differentil eqution We cn lws rerrnge differentil eqution so tht the right hnd side is 0. For emple, = cn be written s = 0. With this in mind the most generl form for differentil eqution is F (t,,,..., (n) ) = 0, where F is function. For emple, ( ) 2 + e sin(t) (4) = 0. The order of differentil eqution is the order of the highest derivtive tht occurs. So, the emple just bove shows DE of order 4.
4 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS Constructing differentil eqution to model phsicl sitution We use rte equtions, i.e. differentil equtions, to model sstems tht undergo chnge. The following rgument using t should be somewht fmilir from clculus. Emple 1.2. Suppose popultion P (t) hs constnt birth nd deth rtes: β = 2%/er, δ = 1%/er Build differentil eqution tht models this sitution. nswer: In the intervl [t, t + t], the chnge in P is given b P = number of births  number of deths. Over smll time intervl t the popultion is roughl constnt so: Births in the time intervl P (t) β t Deths in the time intervl P (t) δ t Combining these we hve P (β δ)p (t). t Finll, letting t go to 0 we hve derived the differentil eqution dp dt = (β δ)p = kp. Notice tht if β > δ then the popultion is incresing. Of course, this DE is our most importnt DE (1): the eqution of eponentil growth or dec. We know the solution is P = P 0 e kt. Note: If β nd δ re more complicted nd depend on t, s β = P + 2t nd δ = P/t. The derivtion of the DE is the sme, but becuse β nd δ re no longer constnts this is not sitution of eponentil growth nd the solution will be more complicted (nd probbl hrder to find). Emple 1.3. Bcteri growth. Suppose popultion of bcteri is modeled b the eponentil growth eqution P = kp. Suppose tht the popultion doubles ever 3 hours. Find the growth constnt k. nswer: The eqution P = kp hs solution P (t) = Ce kt. From the initil condition we hve tht P (0) = C. Since the popultion doubles ever 3 hours we hve P (3) = Ce 3k = 2C. Solving for k we get k = 1 3 ln 2 (in units of 1/hours.) 1.8 Initil vlue problems An initil vlue problem (IVP) is just differentil eqution where one vlue of the solution function is specified. We illustrte with some simple emples. Emple 1.4. Initil vlue problem. Solve the IVP ẏ = 3, (0) = 7.
5 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 5 nswer: We recognize this s n eponentil growth eqution, so (t) = Ce 3t. Using the initil condition we hve (0) = 7 = C. Therefore (t) = 7e 3t. Emple 1.5. Initil vlue problem. Solve the IVP = 2, (2) = 7. nswer: Note, the use of indictes tht the independent vrible in this problem is. This is rell n problem: integrting we get = 3 /3+C. Using the initil condition we find C = 7 8/ Seprble Equtions Now it s time to lern our first technique for solving differentil equtions. A first order DE is clled seprble if the vribles cn be seprted from ech other. We illustrte with series of emples. Emple 1.6. Eponentil growth. Use seprtion of vribles to solve the eponentil growth eqution = k. nswer: We rewrite the eqution s d = 4. Net we seprte the vribles b getting dt ll the s on one side nd the t s on the other. d = 4 dt. Now we integrte both sides: d = 4 dt ln = 4t + C. Now we solve for b eponentiting both sides: = e C e 4t or = ±e C e 4t. Since ±e C is just constnt we renme it simpl K. We now hve the solution we knew we d get: = Ke 4t. Emple 1.7. Here is stndrd emple where the solution goes to infinit in finite time (i.e. the solutions blow up ). One of the fun fetures of differentil equtions is how ver simple equtions cn hve ver surprising behvior. Solve the initil vlue problem d dt = 2 ; (0) = 1. nswer: We cn seprte the vribles b moving ll the s to one side nd the t s to the other d 2 = dt Integrting both sides we get: 1 = t + C
6 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 6 Think: The constnt of integrtion is importnt, but we onl need it on one side. Solving for we get the solution: = 1 t + C. Finll, we use initil condition (0) = 1 to find tht C = 1. (t) = 1 1 t. So the solution is: We grph this below. Note tht the grph of the solution hs verticl smptote t t = 1. = 1 1 t t Grph of solution Technicl definition of solution Looking t the previous emple we see the domin of consists of two intervls: (, 1) nd (1, ). For technicl resons we will require tht the domin consist of ectl one intervl. So the bove grph rell shows two solutions: Solution 1: (t) = 1/(1 t), where is in the intervl (, 1) Solution 2: (t) = 1/(1 t), where is in the intervl (1, ) In the emple problem, since our IVP hd (0) = 1 the solution must hve t = 0 in its domin. Therefore, solution 1 is the solution to the emple s IVP Lost solutions We hve to cover one more detil of seprble equtions. Sometimes solutions get lost nd hve to be recovered. This is smll detil, but ou wnt to p ttention since it s worth 1 es point on ems nd psets. Emple 1.8. In the emple = 2 we found the solution = 1. But it is es to + C check b substitution tht (t) = 0 is lso solution. Since this solution cn not be written s = 1/( + C) we cll it lost solution. The simple eplntion is tht it got lost when we divided b 2. After ll if = 0 it ws not legitimte to divide b 2. Generl ide of lost solutions for seprble DE s
7 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 7 Suppose we hve the differentil eqution = f()g() If g( 0 ) = 0 then ou cn check b substitution the () = 0 is solution to the DE. It m get lost in when we seprte vribles becuse dividing b b g() would then men dividing b 0. Emple 1.9. Find ll the (possible) lost solutions of = ( 2)( 3). nswer: In this cse g() = ( 2)( 3). The lost solutions re found b finding ll the roots of g(). Tht is, the lost solutions re () = 2 nd () = Implicit solutions Sometimes solving for s function of is too hrd, so we don t! Emple Implicit solutions. Solve = nswer: This is seprble nd fter seprtting vribles nd integrting we hve = C. This is too hrd to solve for s function of so we leve our nswer in this implicit form More emples Emple Solve d d =. nswer: Seprting vribles: d = d. Therefore d = d, which implies ln = C. Finll fter eponentition nd replcing ec b K we hve = Ke 2 /2. Think: There is lost solution tht ws found b some slopp lgebr. Cn ou spot the solution nd the slopp lgebr? Emple Solve d d = 3 2. nswer: Seprting vribles nd integrting gives: 1 = 4 4 There is lso lost solution: () = 0. Emple Solve + p() = 0. 4 = 4 + 4C. + C. Solving for we hve nswer: We first rewrite this so tht it s clerl seprble: d = p() d. After the usul seprtion nd integrtion we hve log( ) = p() d + C Therefore, () = e C e p() d nd () = 0 is lost solution.
8 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS Geometric Applictions of DEs Since the slope of curve is given b its derivtives we cn often use differentil equtions to describe curves. Emple An hev object is drgged through the snd b rope. Suppose the object strts t (0, ) with the puller t the origin, so the rope hs length. The puller moves long the is so tht the rope is lws tut nd tngent to the curve followed b the object. This curve is clled trctri. Find n eqution for it. nswer: The digrm below shows tht d d = 2 2 (, ) 2 2 The trctri Thus 2 2 d = d. Integrting (detils below) we get ( + ) ln = + C. ( + ) The initil position (, ) = (0, ) implies C = 0. Therefore = ln To finish the problem we show tht the integrl is wht we climed it ws: Let I = 2 2 d. Now use the trig. substitution: = sin u: cos u cos 2 u I = cos u du = sin u sin u du 1 sin 2 u = du = csc u sin u du sin u = ln(csc u + cot u) cos u Bck substituting we get I = ( ) + ln 2 2, which is wht we climed bove.
9 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS 9 Emple Suppose = () is curve in the first qudrnt nd tht the prt of the curve s tngent line tht lies in the first qudrnt is bisected b the point of tngenc. Find nd solve the DE for this curve. nswer: The figure shows the piece of the tngent bisected b the point (, ) on the curve. Thus, the slope of tngent = d d =. This differentil eqution is seprble nd is esil solved: = C/. 2 (, )
Improper 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 informationES.182A Topic 30 Notes Jeremy Orloff
ES82A opic 3 Notes Jerem Orloff 3 Nonindependent vribles: chin rule Recll the chin rule: If w = f, ; nd = r, t, = r, t then = + r t r t r t = + t t t r nfortuntel, sometimes there re more complicted reltions
More informationES.182A Topic 32 Notes Jeremy Orloff
ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In
More information13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes
The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce
More information13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes
The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce
More informationTopics Covered AP Calculus AB
Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.
More 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 informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
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 informationLogarithms. Logarithm is another word for an index or power. POWER. 2 is the power to which the base 10 must be raised to give 100.
Logrithms. Logrithm is nother word for n inde or power. THIS IS A POWER STATEMENT BASE POWER FOR EXAMPLE : We lred know tht; = NUMBER 10² = 100 This is the POWER Sttement OR 2 is the power to which the
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 informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More 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 informationThe practical version
Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht
More 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 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 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 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 informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More 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 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
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 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 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 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 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 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 informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
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 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 informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
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 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 AB. For a function f(x), the derivative would be f '(
lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:
More 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 informationSTEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.
STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t
More informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationIntroduction to Algebra  Part 2
Alger Module A Introduction to Alger  Prt Copright This puliction The Northern Alert Institute of Technolog 00. All Rights Reserved. LAST REVISED Oct., 008 Introduction to Alger  Prt Sttement of Prerequisite
More information1 Part II: Numerical Integration
Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
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 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 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 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 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 informationCalculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.
Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite
More informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in
More 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 informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
More informationAppendix 3, Rises and runs, slopes and sums: tools from calculus
Appendi 3, Rises nd runs, slopes nd sums: tools from clculus Sometimes we will wnt to eplore how quntity chnges s condition is vried. Clculus ws invented to do just this. We certinly do not need the full
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More information5.1 Estimating with Finite Sums Calculus
5.1 ESTIMATING WITH FINITE SUMS Emple: Suppose from the nd to 4 th hour of our rod trip, ou trvel with the cruise control set to ectl 70 miles per hour for tht two hour stretch. How fr hve ou trveled during
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 informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationDisclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.
Disclimer: This is ment to help you strt studying. It is not necessrily complete list of everything you need to know. The MTH 33 finl exm minly consists of stndrd response questions where students must
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
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 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 informationSUMMER KNOWHOW STUDY AND LEARNING CENTRE
SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Statistical Physics I Spring Term Solutions to Problem Set #1
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Deprtment 8.044 Sttisticl Physics I Spring Term 03 Problem : Doping Semiconductor Solutions to Problem Set # ) Mentlly integrte the function p(x) given in
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 informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More 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 informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More informationA. 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 informationLoudoun Valley High School Calculus Summertime Fun Packet
Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember!
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 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 informationChapter 6: Transcendental functions: Table of Contents: 6.3 The natural exponential function. 6.2 Inverse functions and their derivatives
Chpter 6: Trnscendentl functions: In this chpter we will lern differentition nd integrtion formuls for few new functions, which include the nturl nd generl eponentil nd the nturl nd generl logrithmic function
More information1.9 C 2 inner variations
46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for
More informationMath 31S. Rumbos Fall Solutions to Assignment #16
Mth 31S. Rumbos Fll 2016 1 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)
More informationStudent Session Topic: Particle Motion
Student Session Topic: Prticle Motion Prticle motion nd similr problems re on the AP Clculus exms lmost every yer. The prticle my be prticle, person, cr, etc. The position, velocity or ccelertion my be
More informationAnswers for Ch. 5 Review: The Integral
Answers for Ch. 5 Review: The Integrl. So, I m to sketch grph of. I sketch things with softwre. Four subregions, Δ ½. The region R is onl in the first qudrnt, nd the intercepts re nd, so the verticl seprtors
More informationSCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics. Basic Algebra
SCHOOL OF ENGINEERING & BUILT ENVIRONMENT Mthemtics Bsic Algebr Opertions nd Epressions Common Mistkes Division of Algebric Epressions Eponentil Functions nd Logrithms Opertions nd their Inverses Mnipulting
More informationReview Exercises for Chapter 4
_R.qd // : PM Pge CHAPTER Integrtion Review Eercises for Chpter In Eercises nd, use the grph of to sketch grph of f. To print n enlrged cop of the grph, go to the wesite www.mthgrphs.com... In Eercises
More informationMAT187H1F Lec0101 Burbulla
Chpter 6 Lecture Notes Review nd Two New Sections Sprint 17 Net Distnce nd Totl Distnce Trvelled Suppose s is the position of prticle t time t for t [, b]. Then v dt = s (t) dt = s(b) s(). s(b) s() is
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationThe Fundamental Theorem of Calculus Part 2, The Evaluation Part
AP Clculus AB 6.4 Funmentl Theorem of Clculus The Funmentl Theorem of Clculus hs two prts. These two prts tie together the concept of integrtion n ifferentition n is regre by some to by the most importnt
More informationSection 4: Integration ECO4112F 2011
Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic
More information2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
More information(i) b b. (ii) (iii) (vi) b. P a g e Exponential Functions 1. Properties of Exponents: Ex1. Solve the following equation
P g e 30 4.2 Eponentil Functions 1. Properties of Eponents: (i) (iii) (iv) (v) (vi) 1 If 1, 0 1, nd 1, then E1. Solve the following eqution 4 3. 1 2 89 8(2 ) 7 Definition: The eponentil function with se
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationMath 231E, Lecture 33. Parametric Calculus
Mth 31E, Lecture 33. Prmetric Clculus 1 Derivtives 1.1 First derivtive Now, let us sy tht we wnt the slope t point on prmetric curve. Recll the chin rule: which exists s long s /. = / / Exmple 1.1. Reconsider
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More 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 informationWhat Is Calculus? 42 CHAPTER 1 Limits and Their Properties
60_00.qd //0 : PM Pge CHAPTER Limits nd Their Properties The Mistress Fellows, Girton College, Cmridge Section. STUDY TIP As ou progress through this course, rememer tht lerning clculus is just one of
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More 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 informationThe Periodically Forced Harmonic Oscillator
The Periodiclly Forced Hrmonic Oscilltor S. F. Ellermeyer Kennesw Stte University July 15, 003 Abstrct We study the differentil eqution dt + pdy + qy = A cos (t θ) dt which models periodiclly forced hrmonic
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More information1 Probability Density Functions
Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our
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 informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More information