6.5 Improper integrals


 Adrian Fields
 1 years ago
 Views:
Transcription
1 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y = f () long the intervl [, ]. By definition, these integrls n only e used to ompute res of ounded regions. In some situtions, however, we re interested in unounded regions these re regions tht etend towrds infinity in t lest one diretion. Yet, mny unounded regions still hve finite re. We strt with si emple of this phenomenon: Prolem 6.6: Wht is the re of the region ordered y the urve y =, the line =, nd the is? Solution for Prolem 6.6: We sketh piture of this region t right. Notie tht this region is unounded: the region etends towrds + s grows lrge. Even though this region is unounded, we n ttempt to determine its re. We ertinly n ompute the re of the portion of the region to the left of = (for ny > ) s the definite integrl y d. As grows lrger, we epet tht the re under the urve on [, ] pprohes the re of the entire region under the urve on [, +). Speifilly, this re is y =! d. The integrl is esy to evlute: d = = Thus, when we tke the it, we get tht the re of the region is. d =!! = =. Note the prdo here: even though the region is unounded, it hs finite re. Prolem 6.6 suggests logil definition: Definition: integrl Let f e ontinuous funtion nd R suh tht (, ) Dom( f ). We define the improper f () d = f () d,! provided the it is defined. If the it is defined nd is not ±, we sy tht the improper integrl onverges. Otherwise, we sy tht the improper integrl diverges. There s n oviously similr definition for improper integrls in the other diretion: 7 Copyrighted Mteril
2 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY Definition: integrl Let f e ontinuous funtion nd R suh tht (, ) Dom( f ). We define the improper f () d = f () d,! provided the it is defined. If the it is defined nd is not ±, we sy tht the improper integrl onverges. Otherwise, we sy tht the improper integrl diverges. Let s generlize Prolem 6.6: Prolem 6.7: Let r e rel numer. Compute r d. Solution for Prolem 6.7: hve: This equls By definition, we ompute the improper integrl y writing it. If r,, then we d =! r!! r. (r ) r. r If r >, then the term pprohes s pprohes. Thus, in this se, the improper integrl onverges r to r. If r <, then the term grows without ound s pprohes. Thus, the integrl diverges. We might lso r write d = if r <. r Our originl integrtion ws not vlid for r =, so we hve to do tht se seprtely:! d =! (log ) =! (log ). As goes towrds infinity, this grows without ound, so the integrl diverges. In summry: r d = 8 >< r if r >, >: diverges if r pple. The net prolem is nother ommon emple of n improper integrl: Prolem 6.8: Compute e d, where is rel numer. Solution for Prolem 6.8: We ompute, for, (we ll investigte = t the end):! e d =! (e ) =! (e ). 8 Copyrighted Mteril
3 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS If is positive, then e =, so the integrl diverges. If is negtive, then e =, so the integrl equls!!. (Note this is positive numer when is negtive, so this nswer mkes sense.) Finlly, if =, then the integrl is R d, whih lerly diverges. Thus, the integrl diverges for nonnegtive eponents, nd onverges for negtive eponents. The result of Prolem 6.8 is typilly written s follows: if r >, then e r d = r. Prolem 6.9: Suppose f nd g re ontinuous funtions on [, ) nd f () pple g() for ll. () Show tht, if R f nd R g oth onverge, then Show tht if oth funtions re positive, nd R () Show tht if oth funtions re positive, nd R Solution for Prolem 6.9: () For ny, we hve (g f )() for ll [, ], thus Therefore, f pple g. g onverges, then R f onverges. f diverges, then R g diverges. (g f )() d. f () d pple g() d, nd sine its preserve nonstrit inequlities, we onlude tht f () d = f () d pple g() d = g() d.!! Define funtion F() = f (t) dt. Note tht F is n inresing funtion (sine f () for ll ), nd tht R eists. Also, sine pple f () pple g() for ll, we hve f () d =! F(), if this it pple F() = f (t) dt pple g(t) dt pple g(t) dt. Thus F is inresing nd hs n upper ound (nmely, R g(t) dt, whih y ssumption onverges), so y the result of Prolem 6.5, the it F() = f (t) dt! eists, so the integrl onverges. 9 Copyrighted Mteril
4 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY () This is just the ontrpositive sttement to prt, so there is nothing dditionl to prove. WARNING!! j We n only use the omprison tests in prts nd () of Prolem 6.9 if oth funtions re positive. As trivil emple, if f () = nd g() =, then for ny R, R g =, so R g onverges, ut R f diverges. Thus fr in this setion, we hve looked t improper integrls tht ompute res of regions tht re unounded in the diretion. There is nother type of improper integrl tht ours when the region tht we re emining is unounded in the ydiretion, s in the following emple: Prolem 6.: Compute p d. Solution for Prolem 6.: Skething the grph will immeditely show the issue. We hve! + p =. So the re under y = p is potentilly infinite (nd in ft the funtion is not even defined t ). We n do essentilly the sme thing we did for improper integrls with it of integrtion of ±. We define y y = p p d =! + p d. Note the + sine we only re out the intervl (,], we only re out wht hppens to the right of. This integrl is now esy to ompute: p d = p = p. As! +, this pprohes. Hene p d =. One gin, seemingly infinite re turns out to e finite. We n generlize the definition from Prolem 6.: Definition: integrl Suppose f is funtion, ontinuous on (, ], suh tht f () = ±. We define the improper! + f () d = f () d,! + provided this it is defined. If the it is defined, we sy tht this improper integrl onverges, nd if it is undefined, we sy tht the improper integrl diverges. Of ourse, we n do the sme thing if the funtion hs it of ± t the end of [, ). (We will omit writing out the forml definition.) Copyrighted Mteril
5 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS Sidenote: Note tht the ove definition is onsistent with our usul (nonimproper) integrls. In prtiulr, if R f is defined, then y the Fundmentl Theorem of Clulus, the funtion g() = is di erentile, hene ontinuous, nd thus f (t) dt f (t) dt = g() = g() = f (t) dt.! +! + We know tht for regulr (not improper) integrls, we n rek them prt t ny point into two seprte integrls. Speifilly, if (, ), then f = f + f. This is lso how we evlute integrls tht re improper t oth ends, s in the following emple: Prolem 6.: Compute d for ll r > (or determine when it diverges). r Solution for Prolem 6.: The orret thing to do with n integrl tht is improper t oth ends is to split it somewhere in the middle. For emple, we n write r d = r d + r d. (We didn t hve to pik = s the point t whih to split them, ut it seems onvenient sine r is niely ehved t =.) We lredy know y Prolem 6.7 tht R d onverges if nd only if r >. The other integrl is r! d = d = = r!+ r!+ (r ) r r.!+ r If r >, then the frtion gets ritrrily lrge, so the it is infinite. Thus R d diverges for r >. r Hene our originl doulyimproper integrl is never onvergent: the integrl on (, ] diverges for r >, nd the integrl on [, ) diverges for r pple. Importnt: If (, ) Dom( f ) nd f (t) dt is improper t oth ends of (, ), then for ny (, ). f (t) dt = f (t) dt + f (t) dt,! +! As noted in the solution to Prolem 6., it doesn t mtter t whih point we rek up the doulyimproper integrl. Copyrighted Mteril
6 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY Conept: We n rek n integrl prt s f = f + f t ny (, ) tht we hoose. Thus, hoose to e s onvenient s possile. We will leve it s n eerise to prove this. Also, it is not orret to try to tke shortut nd del with oth ends of douleimproper integrl t one. In prtiulr: WARNING!! j f () d is not the sme s f () d.! The orret wy to evlute n integrl over ll of R is to hoose R, nd then ompute f () d = f () d + f () d = f () d + f () d.!!+ We will leve it s n eerise to eplore this further. We lso hve to e it utious when deling with funtions with domins tht re not ll of R. Integrls of suh funtions might e improper ut not immeditely pper so. For emple: 3 Prolem 6.: Compute d. Solution for Prolem 6.: If you weren t pying lose ttention, you might do this: Bogus Solution: 3 d = 3 = 3 + = 6. We n t do this, euse the funtion is not defined t! To e little more preise, the funtion does not hve n ntiderivtive on the intervl [, 3], euse it is not defined t =, so we nnot pply the Fundmentl Theorem of Clulus. In order to evlute the integrl, we need to rek it up into sum of two improper integrls t the point t whih the funtion is undefined: 3 d = 3 d + d. As we sw in Prolem 6., oth of these diverge. Thus, the originl integrl itself diverges. d More generlly, when omputing something like, it might e tempting to sy is n odd funtion, so the integrl from to will nel out the integrl from to, nd thus the overll integrl is. This is lso the result tht nive lultion will give: Bogus Solution: d = log = log() log() =. Copyrighted Mteril
7 Eerpt from "Clulus" 3 AoPS In. REVIEW PROBLEMS But this is not orret! The only wy leglly to evlute this integrl is to rek it up into its improper prts. Neither prt onverges, so the integrl diverges. Eerises 6.5. Compute the following improper integrls: () 6.5. () 3 d ( ) Compute + d. Compute Compute + d. e d. (log ) d d = d + d. () e d (d) 4 d Show tht it doesn t mtter t whih point we rek up doulyimproper integrl. Speifilly, show tht, for ny, d (, ), if R f nd R f onverge, then R d f nd R f lso onverge, nd d d f + f = f + f. d Hints: 3, ? () Show tht if f () d onverges, then f () d = f () d. Hints: 65! Show tht the onverse of prt () is not true; tht is, it is possile tht f () d diverges. Hints: 8, 5! f () d onverges ut tht Review Prolems 6.3 Compute the following: ()! log os! 3! () e 4 (Soure: Rie) 6.4 sin t tn t Suppose nd re nonzero rel numers. Find nd. Hints: 3 t! sin t t! tn t 6.5 Compute () e d 3 d () d Hints: Copyrighted Mteril
AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals
AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More information1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of 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 informationGreen s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e
Green s Theorem. Let be the boundry of the unit squre, y, oriented ounterlokwise, nd let F be the vetor field F, y e y +, 2 y. Find F d r. Solution. Let s write P, y e y + nd Q, y 2 y, so tht F P, Q. Let
More informationFor a, b, c, d positive if a b and. ac bd. Reciprocal relations for a and b positive. If a > b then a ab > b. then
Slrs7.2ADV.7 Improper Definite Integrls 27.. D.dox Pge of Improper Definite Integrls Before we strt the min topi we present relevnt lger nd it review. See Appendix J for more lger review. Inequlities:
More informationMath 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1)
Green s Theorem Mth 3B isussion Session Week 8 Notes Februry 8 nd Mrh, 7 Very shortly fter you lerned how to integrte singlevrible funtions, you lerned the Fundmentl Theorem of lulus the wy most integrtion
More informationNumbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point
GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply
More informationPart 4. Integration (with Proofs)
Prt 4. Integrtion (with Proofs) 4.1 Definition Definition A prtition P of [, b] is finite set of points {x 0, x 1,..., x n } with = x 0 < x 1
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 informationSection 4.4. Green s Theorem
The Clulus of Funtions of Severl Vriles Setion 4.4 Green s Theorem Green s theorem is n exmple from fmily of theorems whih onnet line integrls (nd their higherdimensionl nlogues) with the definite integrls
More informationMATH 409 Advanced Calculus I Lecture 22: Improper Riemann integrals.
MATH 409 Advned Clulus I Leture 22: Improper Riemnn integrls. Improper Riemnn integrl If funtion f : [,b] R is integrble on [,b], then the funtion F(x) = x f(t)dt is well defined nd ontinuous on [,b].
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 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 informationMAT 403 NOTES 4. f + f =
MAT 403 NOTES 4 1. Fundmentl Theorem o Clulus We will proo more generl version o the FTC thn the textook. But just like the textook, we strt with the ollowing proposition. Let R[, ] e the set o Riemnn
More informationThe Double Integral. The Riemann sum of a function f (x; y) over this partition of [a; b] [c; d] is. f (r j ; t k ) x j y k
The Double Integrl De nition of the Integrl Iterted integrls re used primrily s tool for omputing double integrls, where double integrl is n integrl of f (; y) over region : In this setion, we de ne double
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 informationCalculus Cheat Sheet. Integrals Definitions. where F( x ) is an antiderivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx
Clulus Chet Sheet Integrls Definitions Definite Integrl: Suppose f ( ) is ontinuous AntiDerivtive : An ntiderivtive of f ( ) on [, ]. Divide [, ] into n suintervls of is funtion, F( ), suh tht F = f.
More informationSECTION A STUDENT MATERIAL. Part 1. What and Why.?
SECTION A STUDENT MATERIAL Prt Wht nd Wh.? Student Mteril Prt Prolem n > 0 n > 0 Is the onverse true? Prolem If n is even then n is even. If n is even then n is even. Wht nd Wh? Eploring Pure Mths Are
More informationProject 6: Minigoals Towards Simplifying and Rewriting Expressions
MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy
More informationINTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable
INTEGRATION NOTE: These notes re supposed to supplement Chpter 4 of the online textbook. 1 Integrls of Complex Vlued funtions of REAL vrible If I is n intervl in R (for exmple I = [, b] or I = (, b)) nd
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 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 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 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 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 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 information(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.
Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time
More informationT b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.
Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene
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 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 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 informationUniversity of Sioux Falls. MAT204/205 Calculus I/II
University of Sioux Flls MAT204/205 Clulus I/II Conepts ddressed: Clulus Textook: Thoms Clulus, 11 th ed., Weir, Hss, Giordno 1. Use stndrd differentition nd integrtion tehniques. Differentition tehniques
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 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 informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More 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 informationMore Properties of the Riemann Integral
More Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologil Sienes nd Deprtment of Mthemtil Sienes Clemson University Februry 15, 2018 Outline More Riemnn Integrl Properties The Fundmentl
More informationIntroduction to Olympiad Inequalities
Introdution to Olympid Inequlities Edutionl Studies Progrm HSSP Msshusetts Institute of Tehnology Snj Simonovikj Spring 207 Contents Wrm up nd AmGm inequlity 2. Elementry inequlities......................
More informationType 2: Improper Integrals with Infinite Discontinuities
mth imroer integrls: tye 6 Tye : Imroer Integrls with Infinite Disontinuities A seond wy tht funtion n fil to be integrble in the ordinry sense is tht it my hve n infinite disontinuity (vertil symtote)
More informationThe RiemannStieltjes Integral
Chpter 6 The RiemnnStieltjes Integrl 6.1. Definition nd Eistene of the Integrl Definition 6.1. Let, b R nd < b. ( A prtition P of intervl [, b] is finite set of points P = { 0, 1,..., n } suh tht = 0
More information5.2 Exponent Properties Involving Quotients
5. Eponent Properties Involving Quotients Lerning Objectives Use the quotient of powers property. Use the power of quotient property. Simplify epressions involving quotient properties of eponents. Use
More information6.1 Definition of the Riemann Integral
6 The Riemnn Integrl 6. Deinition o the Riemnn Integrl Deinition 6.. Given n intervl [, b] with < b, prtition P o [, b] is inite set o points {x, x,..., x n } [, b], lled grid points, suh tht x =, x n
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More 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 informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More information10. AREAS BETWEEN CURVES
. AREAS BETWEEN CURVES.. Ares etween curves So res ove the xxis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in
More informationChapter 9 Definite Integrals
Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished
More 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 informationCALCULUS STUDY MATERIAL. B.Sc. MATHEMATICS III SEMESTER UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION
CALCULUS STUDY MATERIAL BS MATHEMATICS III SEMESTER UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION CALICUT UNIVERSITY PO MALAPPURAM, KERALA, INDIA  67 65 5 UNIVERSITY OF CALICUT SCHOOL OF DISTANCE
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More 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 informationLogarithms LOGARITHMS.
Logrithms LOGARITHMS www.mthletis.om.u Logrithms LOGARITHMS Logrithms re nother method to lulte nd work with eponents. Answer these questions, efore working through this unit. I used to think: In the
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 informationpadic Egyptian Fractions
padic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Setup 3 4 pgreedy Algorithm 5 5 pegyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction
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 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 informationSection 7.1 Area of a Region Between Two Curves
Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region
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 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 informationImproper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Improper Integrls MATH 2, Clculus II J. Robert Buchnn Deprtment of Mthemtics Spring 28 Definite Integrls Theorem (Fundmentl Theorem of Clculus (Prt I)) If f is continuous on [, b] then b f (x) dx = [F(x)]
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 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 informationEvaluating Definite Integrals. There are a few properties that you should remember in order to assist you in evaluating definite integrals.
Evluting Definite Integrls There re few properties tht you should rememer in order to ssist you in evluting definite integrls. f x dx= ; where k is ny rel constnt k f x dx= k f x dx ± = ± f x g x dx f
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More 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 informationParse trees, ambiguity, and Chomsky normal form
Prse trees, miguity, nd Chomsky norml form In this lecture we will discuss few importnt notions connected with contextfree grmmrs, including prse trees, miguity, nd specil form for contextfree grmmrs
More informationINTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable
INTEGRATION 1 Integrls of Complex Vlued funtions of REAL vrible If I is n intervl in R (for exmple I = [, b] or I = (, b)) nd h : I C writing h = u + iv where u, v : I C, we n extend ll lulus 1 onepts
More informationPart I: Study the theorem statement.
Nme 1 Nme 2 Nme 3 A STUDY OF PYTHAGORAS THEOREM Instrutions: Together in groups of 2 or 3, fill out the following worksheet. You my lift nswers from the reding, or nswer on your own. Turn in one pket for
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 informationLinear Inequalities. Work Sheet 1
Work Sheet 1 Liner Inequlities RentHep, cr rentl compny,chrges $ 15 per week plus $ 0.0 per mile to rent one of their crs. Suppose you re limited y how much money you cn spend for the week : You cn spend
More informationApplications of Definite Integral
Chpter 5 Applitions of Definite Integrl 5.1 Are Between Two Curves In this setion we use integrls to find res of regions tht lie between the grphs of two funtions. Consider the region tht lies between
More informationTrigonometry Revision Sheet Q5 of Paper 2
Trigonometry Revision Sheet Q of Pper The Bsis  The Trigonometry setion is ll out tringles. We will normlly e given some of the sides or ngles of tringle nd we use formule nd rules to find the others.
More informationSolutions to Assignment 1
MTHE 237 Fll 2015 Solutions to Assignment 1 Problem 1 Find the order of the differentil eqution: t d3 y dt 3 +t2 y = os(t. Is the differentil eqution liner? Is the eqution homogeneous? b Repet the bove
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 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 informationLecture Solution of a System of Linear Equation
ChE Lecture Notes, Dept. of Chemicl Engineering, Univ. of TN, Knoville  D. Keffer, 5/9/98 (updted /) Lecture 8  Solution of System of Liner Eqution 8. Why is it importnt to e le to solve system of liner
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 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 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 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 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 informationArrow s Impossibility Theorem
Rep Voting Prdoxes Properties Arrow s Theorem Arrow s Impossiility Theorem Leture 12 Arrow s Impossiility Theorem Leture 12, Slide 1 Rep Voting Prdoxes Properties Arrow s Theorem Leture Overview 1 Rep
More informationCS 491G Combinatorial Optimization Lecture Notes
CS 491G Comintoril Optimiztion Leture Notes Dvi Owen July 30, August 1 1 Mthings Figure 1: two possile mthings in simple grph. Definition 1 Given grph G = V, E, mthing is olletion of eges M suh tht e i,
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 informationDiscrete Structures Lecture 11
Introdution Good morning. In this setion we study funtions. A funtion is mpping from one set to nother set or, perhps, from one set to itself. We study the properties of funtions. A mpping my not e funtion.
More informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
More 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 = f(b) f(a). a b f (x)dx is the limit of sums
Green s Theorem If f is funtion of one vrible x with derivtive f x) or df dx to the Fundmentl Theorem of lulus, nd [, b] is given intervl then, ording This is not trivil result, onsidering tht b b f x)dx
More informationNONDETERMINISTIC FSA
Tw o types of nondeterminism: NONDETERMINISTIC FS () Multiple strtsttes; strtsttes S Q. The lnguge L(M) ={x:x tkes M from some strtstte to some finlstte nd ll of x is proessed}. The string x = is
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 informationIntegration. antidifferentiation
9 Integrtion 9A Antidifferentition 9B Integrtion of e, sin ( ) nd os ( ) 9C Integrtion reognition 9D Approimting res enlosed funtions 9E The fundmentl theorem of integrl lulus 9F Signed res 9G Further
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
2 The Prllel Circuit Electric Circuits: Figure 2 elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More 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 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 informationMAT137 Calculus! Lecture 28
officil wesite http://uoft.me/mat137 MAT137 Clculus! Lecture 28 Tody: Antiderivtives Fundmentl Theorem of Clculus Net: More FTC (review v. 8.58.7) 5.7 Sustitution (v. 9.19.4) Properties of the Definite
More information] dx (3) = [15x] 2 0
Leture 6. Double Integrls nd Volume on etngle Welome to Cl IV!!!! These notes re designed to be redble nd desribe the w I will eplin the mteril in lss. Hopefull the re thorough, but it s good ide to hve
More informationAP CALCULUS Test #6: Unit #6 Basic Integration and Applications
AP CALCULUS Test #6: Unit #6 Bsi Integrtion nd Applitions A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS IN THIS PART OF THE EXAMINATION. () The ext numeril vlue of the orret
More informationMath 3B Final Review
Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:4510:45m SH 1607 Mth Lb Hours: TR 12pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems
More information