and that at t = 0 the object is at position 5. Find the position of the object at t = 2.


 Annabel Maxwell
 3 years ago
 Views:
Transcription
1 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 end up with mthemtics tht looks like the two emples, though of course the function involved will not lwys be so simple. Even better, we now see tht while the second problem did not pper to be menble to pproch one, it cn in fct be solved in the sme wy. The resoning is this: we know tht problem one cn be solved esily by finding function whose derivtive is 3t. We lso know tht mthemticlly the two problems re the sme, becuse both cn be solved by tking it of sum, nd the sums re identicl. Therefore, we don t relly need to compute the it of either sum becuse we know tht we will get the sme nswer by computing function with the derivtive 3t or, which is the sme thing, 3. It struethtthefirstproblemhdtheddedcomplictionofthe 0,ndwecertinly need to be ble to del with such minor vritions, but tht turns out to be quite simple. The lesson then is this: whenever we cn solve problem by tking the it of sum of certin form, we cn insted of computing the (often nsty) it find new function with certin derivtive. Eercises 7... Suppose n object moves in stright line so tht its speed t time t is given by v(t) = 2t+2, nd tht t t = the object is t position 5. Find the position of the object t t = Suppose n object moves in stright line so tht its speed t time t is given by v(t) = t 2 +2, nd tht t t = 0 the object is t position 5. Find the position of the object t t = By method similr to tht in emple 7.2, find the re under y = 2 between = 0 nd ny positive vlue for. 4. By method similr to tht in emple 7.2, find the re under y = 4 between = 0 nd ny positive vlue for. 5. By method similr to tht in emple 7.2, find the re under y = 4 between = 2 nd ny positive vlue for bigger thn By method similr to tht in emple 7.2, find the re under y = 4 between ny two positive vlues for, sy < b. 7. Let f() = Approimte the re under the curve between = 0 nd = 2 using 4 rectngles nd lso using 8 rectngles. 8. Let f() = Approimte the re under the curve between = nd = 3 using 4 rectngles. º¾ Ì ÙÒ Ñ ÒØ Ð Ì ÓÖ Ñ Ó ÐÙÐÙ Let s recst the first emple from the previous section. Suppose tht the speed of the object is 3t t time t. How fr does the object trvel between time t = nd time t = b? We re no longer ssuming tht we know where the object is t time t = 0 or t ny other Pge 5 of 6
2 50 Chpter 7 Integrtion time. Itiscertinlytruethtitissomewhere,solet ssupposethttt = 0thepositionisk. Thenjustsintheemple, weknowthtthepositionoftheobjecttnytimeis3t 2 /2+k. This mens tht t time t = the position is 3 2 /2+k nd t time t = b the position is 3b 2 /2+k. Therefore the chnge in position is 3b 2 /2+k (3 2 /2+k) = 3b 2 /2 3 2 /2. Notice tht the k drops out; this mens tht it doesn t mtter tht we don t know k, it doesn t even mtter if we use the wrong k, we get the correct nswer. In other words, to find the chnge in position between time nd time b we cn use ny ntiderivtive of the speed function 3t it need not be the one ntiderivtive tht ctully gives the loction of the object. Wht bout the second pproch to this problem, in the new form? We now wnt to pproimte the chnge in position between time nd time b. We tke the intervl of time between nd b, divide it into n subintervls, nd pproimte the distnce trveled during ech. The strting time of subintervl number i is now +(i )(b )/n, which we bbrevite s t i, so tht t 0 =, t = + (b )/n, nd so on. The speed of the object is f(t) = 3t, nd ech subintervl is (b )/n = t seconds long. The distnce trveled during subintervl number i is pproimtely f(t i ) t, nd the totl chnge in distnce is pproimtely f(t 0 ) t+f(t ) t+ +f(t ) t. The ect chnge in position is the it of this sum s n goes to infinity. We bbrevite this sum using sigm nottion: f(t i ) t = f(t 0 ) t+f(t ) t+ +f(t ) t. The nottion on the left side of the equl sign uses lrge cpitl sigm, Greek letter, nd the left side is n bbrevition for the right side. The nswer we seek is f(t i ) t. Since this must be the sme s the nswer we hve lredy obtined, we know tht f(t i ) t = 3b The significnce of 3t 2 /2, into which we substitute t = b nd t =, is of course tht it is function whose derivtive is f(t). As we hve discussed, by the time we know tht we Pge 6 of 6
3 wnt to compute 7.2 The Fundmentl Theorem of Clculus 5 f(t i ) t, it no longer mtters wht f(t) stnds for it could be speed, or the height of curve, or something else entirely. We know tht the it cn be computed by finding ny function with derivtive f(t), substituting nd b, nd subtrcting. We summrize this in theorem. First, we introduce some new nottion nd terms. We write = f(t i ) t if the it eists. Tht is, the left hnd side mens, or is n bbrevition for, the right hnd side. The symbol is clled n integrl sign, nd the whole epression is red s the integrl of f(t) from to b. Wht we hve lerned is tht this integrl cn be computed by finding function, sy F(t), with the property tht F (t) = f(t), nd then computing F(b) F(). The function F(t) is clled n ntiderivtive of f(t). Now the theorem: THEOREM 7.3 Fundmentl Theorem of Clculus Suppose tht f() is continuous on the intervl [,b]. If F() is ny ntiderivtive of f(), then f()d = F(b) F(). Let s rewrite this slightly: = F() F(). We ve replced the vrible by t nd b by. These re just different nmes for quntities, so the substitution doesn t chnge the mening. It does mke it esier to think of the two sides of the eqution s functions. The epression is function: plug in vlue for, get out some other vlue. The epression F() F() is of course lso function, nd it hs nice property: d d (F() F()) = F () = f(), Pge 7 of 6
4 52 Chpter 7 Integrtion since F() is constnt nd hs derivtive zero. In other words, by shifting our point of view slightly, we see tht the odd looking function G() = hs derivtive, nd tht in fct G () = f(). This is relly just resttement of the Fundmentl Theorem of Clculus, nd indeed is often clled the Fundmentl Theorem of Clculus. To void confusion, some people cll the two versions of the theorem The Fundmentl Theorem of Clculus, prt I nd The Fundmentl Theorem of Clculus, prt II, lthough unfortuntely there is no universl greement s to which is prt I nd which prt II. Since it relly is the sme theorem, differently stted, some people simply cll them both The Fundmentl Theorem of Clculus. Suppose tht f() is con THEOREM 7.4 Fundmentl Theorem of Clculus tinuous on the intervl [,b] nd let G() =. Then G () = f(). We hve not relly proved the Fundmentl Theorem. In nutshell, we gve the following rgument to justify it: Suppose we wnt to know the vlue of = f(t i ) t. We cn interpret the right hnd side s the distnce trveled by n object whose speed is given by f(t). We know nother wy to compute the nswer to such problem: find the position of the object by finding n ntiderivtive of f(t), then substitute t = nd t = b nd subtrct to find the distnce trveled. This must be the nswer to the originl problem s well, even if f(t) does not represent speed. Wht s wrong with this? In some sense, nothing. As prcticl mtter it is very convincing rgument, becuse our understnding of the reltionship between speed nd distnce seems to be quite solid. From the point of view of mthemtics, however, it is unstisfctory to justify purely mthemticl reltionship by ppeling to our understnding of the physicl universe, which could, however unlikely it is in this cse, be wrong. A complete proof is bit too involved to include here, but we will indicte how it goes. First, if we cn prove the second version of the Fundmentl Theorem, theorem 7.4, then we cn prove the first version from tht: Pge 8 of 6
5 Proof of Theorem 7.3. We know from theorem 7.4 tht 7.2 The Fundmentl Theorem of Clculus 53 G() = is n ntiderivtive of f(), nd therefore ny ntiderivtive F() of f() is of the form F() = G()+k. Then F(b) F() = G(b)+k (G()+k) = G(b) G() =. It is not hrd to see tht = 0, so this mens tht F(b) F() = which is ectly wht theorem 7.3 sys., So the rel job is to prove theorem 7.4. We will sketch the proof, using some fcts tht we do not prove. First, the following identity is true of integrls: = c + c. This cn be proved directly from the definition of the integrl, tht is, using the its of sums. It is quite esy to see tht it must be true by thinking of either of the two pplictions of integrls tht we hve seen. It turns out tht the identity is true no mtter wht c is, but it is esiest to think bout the mening when c b. First, if f(t) represents speed, then we know tht the three integrls represent the distnce trveled between time nd time b; the distnce trveled between time nd time c; nd the distnce trveled between time c nd time b. Clerly the sum of the ltter two is equl to the first of these. Second, if f(t) represents the height of curve, the three integrls represent the re under the curve between nd b; the re under the curve between nd c; nd the re under the curve between c nd b. Agin it is cler from the geometry tht the first is equl to the sum of the second nd third. Pge 9 of 6
6 54 Chpter 7 Integrtion Proof sketch for Theorem 7.4. We wnt to compute G (), so we strt with the definition of the derivtive in terms of it: G G(+ ) G() () = 0 ( + = 0 ( = + 0 = 0 + Now we need to know something bout. + ) ) + when is smll; in fct, it is very close to f(), but we will not prove this. Once gin, it is esy to believe this is true by thinking of our two pplictions: The integrl + cn be interpreted s the distnce trveled by n object over very short intervl of time. Over sufficiently short period of time, the speed of the object will not chnge very much, so the distnce trveled will be pproimtely the length of time multiplied by the speed t the beginning of the intervl, nmely, f(). Alterntely, the integrl my be interpreted s the re under the curve between nd +. When is very smll, this will be very close to the re of the rectngle with bse nd height f(); gin this is f(). If we ccept this, we my proceed: 0 which is wht we wnted to show. + f() = 0 = f(), It is still true tht we re depending on n interprettion of the integrl to justify the rgument, but we hve isolted this prt of the rgument into two fcts tht re not too hrd to prove. Once the lst reference to interprettion hs been removed from the proofs of these fcts, we will hve rel proof of the Fundmentl Theorem. Pge 0 of 6
7 7.2 The Fundmentl Theorem of Clculus 55 Now we know tht to solve certin kinds of problems, those tht led to sum of certin form, we merely find n ntiderivtive nd substitute two vlues nd subtrct. Unfortuntely, finding ntiderivtives cn be quite difficult. While there re smll number of rules tht llow us to compute the derivtive of ny common function, there re no such rules for ntiderivtives. There re some techniques tht frequently prove useful, but we will never be ble to reduce the problem to completely mechnicl process. Becuse of the close reltionship between n integrl nd n ntiderivtive, the integrl sign is lso used to men ntiderivtive. You cn tell which is intended by whether the its of integrtion re included: 2 2 d is n ordinry integrl, lso clled definite integrl, becuse it hs definite vlue, nmely We use 2 2 d = = d to denote the ntiderivtive of 2, lso clled n indefinite integrl. So this is evluted s 2 d = 3 3 +C. It is customry to include the constnt C to indicte tht there re relly n infinite number of ntiderivtives. We do not need this C to compute definite integrls, but in other circumstnces we will need to remember tht the C is there, so it is best to get into the hbit of writing the C. When we compute definite integrl, we first find n ntiderivtive nd then substitute. It is convenient to first disply the ntiderivtive nd then do the substitution; we need nottion indicting tht the substitution is yet to be done. A typicl solution would look like this: 2 2 d = = = 7 3. The verticl line with subscript nd superscript is used to indicte the opertion substitute nd subtrct tht is needed to finish the evlution. Pge of 6
Integration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More information(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35
7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know
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 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 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 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 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 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 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 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 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 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 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 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 informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More 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 informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors PreChpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More 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 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 informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More 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 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 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 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 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 informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
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 informationNow, given the derivative, can we find the function back? Can we antidifferenitate it?
Fundmentl Theorem of Clculus. Prt I Connection between integrtion nd differentition. Tody we will discuss reltionship between two mjor concepts of Clculus: integrtion nd differentition. We will show tht
More 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 information5.7 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 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 informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
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 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 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 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 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 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 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 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 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 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 informationInfinite Geometric Series
Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to
More information7. Indefinite Integrals
7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find
More informationReversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b
Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite
More information5.3 The Fundamental Theorem of Calculus
CHAPTER 5. THE DEFINITE INTEGRAL 35 5.3 The Funmentl Theorem of Clculus Emple. Let f(t) t +. () Fin the re of the region below f(t), bove the tis, n between t n t. (You my wnt to look up the re formul
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: How to identify the leding coefficients nd degrees of polynomils How to dd nd subtrct polynomils How to multiply polynomils
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 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 informationRiemann Integrals and the Fundamental Theorem of Calculus
Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums
More informationx = b a N. (131) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is
Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According
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 5. Numerical Integration
Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationIntegrals  Motivation
Integrls  Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is nonliner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but
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 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 informationDuality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.
Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we
More informationHow do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?
XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk bout solving systems of liner equtions. These re problems tht give couple of equtions with couple of unknowns, like: 6 2 3 7 4
More informationMAT 168: Calculus II with Analytic Geometry. James V. Lambers
MAT 68: Clculus II with Anlytic Geometry Jmes V. Lmbers Februry 7, Contents Integrls 5. Introduction............................ 5.. Differentil Clculus nd Quotient Formuls...... 5.. Integrl Clculus nd
More informationSYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus
SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is
More informationMath 131. Numerical Integration Larson Section 4.6
Mth. Numericl Integrtion Lrson Section. This section looks t couple of methods for pproimting definite integrls numericlly. The gol is to get good pproimtion of the definite integrl in problems where n
More 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 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 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 informationThe Fundamental Theorem of Calculus
The Fundmentl Theorem of Clculus Professor Richrd Blecksmith richrd@mth.niu.edu Dept. of Mthemticl Sciences Northern Illinois University http://mth.niu.edu/ richrd/mth229. The Definite Integrl We define
More informationBob Brown Math 251 Calculus 1 Chapter 5, Section 4 1 CCBC Dundalk
Bo Brown Mth Clculus Chpter, Section CCBC Dundlk The Fundmentl Theorem of Clculus Informlly, the Fundmentl Theorem of Clculus (FTC) sttes tht differentition nd definite integrtion re inverse opertions
More 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 informationWeek 10: Line Integrals
Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.
More informationLinear Approximation and the Fundamental Theorem of Calculus
Mth 3A Discussion Session Week 9 Notes Mrch nd 3, 26 Liner Approimtion nd the Fundmentl Theorem of Clculus We hve three primry ols in tody s discussion of the fundmentl theorem of clculus. By the end of
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 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 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 informationLine and Surface Integrals: An Intuitive Understanding
Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
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 informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
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 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 Evaluation Theorem
These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for inclss presenttion nd should not
More 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 informationdt. However, we might also be curious about dy
Section 0. The Clculus of Prmetric Curves Even though curve defined prmetricly my not be function, we cn still consider concepts such s rtes of chnge. However, the concepts will need specil tretment. For
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationHandout: Natural deduction for first order logic
MATH 457 Introduction to Mthemticl Logic Spring 2016 Dr Json Rute Hndout: Nturl deduction for first order logic We will extend our nturl deduction rules for sententil logic to first order logic These notes
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 information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More information5.2 Volumes: Disks and Washers
4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict
More informationMathematics Number: Logarithms
plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement
More 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 informationLesson 1: Quadratic Equations
Lesson 1: Qudrtic Equtions Qudrtic Eqution: The qudrtic eqution in form is. In this section, we will review 4 methods of qudrtic equtions, nd when it is most to use ech method. 1. 3.. 4. Method 1: Fctoring
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 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 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 informationBernoulli Numbers Jeff Morton
Bernoulli Numbers Jeff Morton. We re interested in the opertor e t k d k t k, which is to sy k tk. Applying this to some function f E to get e t f d k k tk d k f f + d k k tk dk f, we note tht since f
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 informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More information4.5 THE FUNDAMENTAL THEOREM OF CALCULUS
4.5 The Funmentl Theorem of Clculus Contemporry Clculus 4.5 THE FUNDAMENTAL THEOREM OF CALCULUS This section contins the most importnt n most use theorem of clculus, THE Funmentl Theorem of Clculus. Discovere
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 information