Chapter 5. The Riemann Integral. 5.1 The Riemann integral Partitions and lower and upper integrals. Note: 1.5 lectures
|
|
- Reginald Bennett Allen
- 5 years ago
- Views:
Transcription
1 Chpter 5 The Riem Itegrl 5.1 The Riem itegrl Note: 1.5 lectures We ow get to the fudmetl cocept of itegrtio. There is ofte cofusio mog studets of clculus betwee itegrl d tiderivtive. The itegrl is (iformlly) the re uder the curve, othig else. Tht we c compute tiderivtive usig the itegrl is otrivil result we hve to prove. I this chpter we defie the Riem itegrl usig the Drboux itegrl, which is techiclly simpler th (but equivlet to) the trditiol defiitio s doe by Riem Prtitios d lower d upper itegrls We wt to itegrte bouded fuctio defied o itervl [,b]. We first defie two uxiliry itegrls tht c be defied for ll bouded fuctios. Oly the c we tlk bout the Riem itegrl d the Riem itegrble fuctios. Defiitio A prtitio P of the itervl [,b] is fiite set of umbers {x,x 1,x 2,...,x } such tht = x < x 1 < x 2 < < x 1 < x = b. We write x i := x i x i 1. Nmed fter the Germ mthemtici Georg Friedrich Berhrd Riem ( ). Nmed fter the Frech mthemtici Je-Gsto Drboux ( ). 147
2 148 CHAPTER 5. THE RIEMANN INTEGRAL Let f : [,b] R be bouded fuctio. Let P be prtitio of [,b]. Defie m i := if{ f (x) : x i 1 x x i }, M i := sup{ f (x) : x i 1 x x i }, L(P, f ) := U(P, f ) := m i x i, M i x i. We cll L(P, f ) the lower Drboux sum d U(P, f ) the upper Drboux sum. The geometric ide of Drboux sums is idicted i Figure 5.1. The lower sum is the re of the shded rectgles, d the upper sum is the re of the etire rectgles. The width of the ith rectgle is x i, the height of the shded rectgle is m i d the height of the etire rectgle is M i. Figure 5.1: Smple Drboux sums. Propositio Let f : [,b] R be bouded fuctio. Let m,m R be such tht for ll x we hve m f (x) M. For y prtitio P of [,b] we hve m(b ) L(P, f ) U(P, f ) M(b ). (5.1) Proof. Let P be prtitio. The ote tht m m i for ll i d M i M for ll i. Also m i M i for ll i. Filly x i = (b ). Therefore, m(b ) = m ( x i ) = m x i m i x i M i x i M x i = M ( x i ) = M(b ).
3 5.1. THE RIEMANN INTEGRAL 149 Hece we get (5.1). I other words, the set of lower d upper sums re bouded sets. Defiitio As the sets of lower d upper Drboux sums re bouded, we defie f (x) dx := sup{l(p, f ) : P prtitio of [,b]}, f (x) dx := if{u(p, f ) : P prtitio of [,b]}. We cll the lower Drboux itegrl d the upper Drboux itegrl. To void worryig bout the vrible of itegrtio, we ofte simply write f := f (x) dx d f := If itegrtio is to mke sese, the the lower d upper Drboux itegrls should be the sme umber, s we wt sigle umber to cll the itegrl. However, these two itegrls my i fct differ for some fuctios. Exmple 5.1.4: Tke the Dirichlet fuctio f : [,1] R, where f (x) := 1 if x Q d f (x) := if x / Q. The 1 f = d 1 f = 1. The reso is tht for every i we hve tht m i = if{ f (x) : x [x i 1,x i ]} = d M i = sup{ f (x) : x [x i 1,x i ]} = 1. Thus L(P, f ) = U(P, f ) = x i =, 1 x i = x i = 1. Remrk The sme defiitio of f d f is used whe f is defied o lrger set S such tht [,b] S. I tht cse, we use the restrictio of f to [,b] d we must esure tht the restrictio is bouded o [,b]. To compute the itegrl we ofte tke prtitio P d mke it fier. Tht is, we cut itervls i the prtitio ito yet smller pieces. Defiitio Let P = {x,x 1,...,x } d P = { x, x 1,..., x m } be prtitios of [,b]. We sy P is refiemet of P if s sets P P.
4 15 CHAPTER 5. THE RIEMANN INTEGRAL Tht is, P is refiemet of prtitio if it cotis ll the umbers i P d perhps some other umbers i betwee. For exmple, {,.5,1,2} is prtitio of [,2] d {,.2,.5,1,1.5,1.75,2} is refiemet. The mi reso for itroducig refiemets is the followig propositio. Propositio Let f : [,b] R be bouded fuctio, d let P be prtitio of [,b]. Let P be refiemet of P. The L(P, f ) L( P, f ) d U( P, f ) U(P, f ). Proof. The tricky prt of this proof is to get the ottio correct. Let P := { x, x 1,..., x m } be refiemet of P := {x,x 1,...,x }. The x = x d x = x m. I fct, we c fid itegers k < k 1 < < k such tht x j = x for j =,1,2,...,. Let x j = x j 1 x j. We get tht x j = x p. p= 1 +1 Let m j be s before d correspod to the prtitio P. Let m j := if{ f (x) : x j 1 x x j }. Now, m j m p for 1 < p. Therefore, m j x j = m j x p = p= 1 +1 m j x p p= 1 +1 m p x p. p= 1 +1 So L(P, f ) = m j x j j=1 j=1 p= 1 +1 m p x p = m m j x j = L( P, f ). j=1 The proof of U( P, f ) U(P, f ) is left s exercise. Armed with refiemets we prove the followig. The key poit of this ext propositio is tht the lower Drboux itegrl is less th or equl to the upper Drboux itegrl. Propositio Let f : [,b] R be bouded fuctio. Let m,m R be such tht for ll x we hve m f (x) M. The m(b ) f Proof. By Propositio we hve for y prtitio P m(b ) L(P, f ) U(P, f ) M(b ). f M(b ). (5.2)
5 5.1. THE RIEMANN INTEGRAL 151 The iequlity m(b ) L(P, f ) implies m(b ) f. Also U(P, f ) M(b ) implies f M(b ). The key poit of this propositio is the middle iequlity i (5.2). Let P 1,P 2 be prtitios of [,b]. Defie P := P 1 P 2. The set P is prtitio of [,b]. Furthermore, P is refiemet of P 1 d it is lso refiemet of P 2. By Propositio we hve L(P 1, f ) L( P, f ) d U( P, f ) U(P 2, f ). Puttig it ll together we hve L(P 1, f ) L( P, f ) U( P, f ) U(P 2, f ). I other words, for two rbitrry prtitios P 1 d P 2 we hve L(P 1, f ) U(P 2, f ). Now we recll Propositio Tkig the supremum d ifimum over ll prtitios we get I other words f f Riem itegrl sup{l(p, f ) : P prtitio} if{u(p, f ) : P prtitio}. We c filly defie the Riem itegrl. However, the Riem itegrl is oly defied o certi clss of fuctios, clled the Riem itegrble fuctios. Defiitio Let f : [,b] R be bouded fuctio such tht f (x) dx = The f is sid to be Riem itegrble. The set of Riem itegrble fuctios o [,b] is deoted by R[,b]. Whe f R[,b] we defie As before, we ofte simply write f (x) dx := f := f (x) dx = The umber f is clled the Riem itegrl of f, or sometimes simply the itegrl of f. By defiitio, y Riem itegrble fuctio is bouded. By ppelig to Propositio we immeditely obti the followig propositio. Propositio Let f : [,b] R be Riem itegrble fuctio. Let m,m R be such tht m f (x) M. The m(b ) f M(b ).
6 152 CHAPTER 5. THE RIEMANN INTEGRAL Ofte we use weker form of this propositio. Tht is, if f (x) M for ll x [,b], the f M(b ). Exmple : We itegrte costt fuctios usig Propositio If f (x) := c for some costt c, the we tke m = M = c. I iequlity (5.2) ll the iequlities must be equlities. Thus f is itegrble o [,b] d f = c(b ). Exmple : Let f : [, 2] R be defied by 1 if x < 1, f (x) := 1/2 if x = 1, if x > 1. We clim tht f is Riem itegrble d tht 2 f = 1. Proof: Let < ε < 1 be rbitrry. Let P := {,1 ε,1+ε,2} be prtitio. We use the ottio from the defiitio of the Drboux sums. The m 1 = if{ f (x) : x [,1 ε]} = 1, M 1 = sup{ f (x) : x [,1 ε]} = 1, m 2 = if{ f (x) : x [1 ε,1 + ε]} =, M 2 = sup{ f (x) : x [1 ε,1 + ε]} = 1, m 3 = if{ f (x) : x [1 + ε,2]} =, M 3 = sup{ f (x) : x [1 + ε,2]} =. Furthermore, x 1 = 1 ε, x 2 = 2ε d x 3 = 1 ε. We compute L(P, f ) = U(P, f ) = 3 3 m i x i = 1 (1 ε) + 2ε + (1 ε) = 1 ε, M i x i = 1 (1 ε) + 1 2ε + (1 ε) = 1 + ε. Thus, 2 2 f f U(P, f ) L(P, f ) = (1 + ε) (1 ε) = 2ε. By Propositio we hve 2 f 2 f. As ε ws rbitrry we see tht 2 f = 2 f. So f is Riem itegrble. Filly, 1 ε = L(P, f ) 2 f U(P, f ) = 1 + ε. Hece, 2 f 1 ε. As ε ws rbitrry, we hve tht 2 f = 1.
7 5.1. THE RIEMANN INTEGRAL 153 It my be worthwhile to sum up prt of the techique of the exmple i propositio. Propositio Let f : [,b] R be bouded fuctio. The f is Riem itegrble if for every ε >, there exists prtitio P such tht Proof. If for every ε >, P exists we hve: Therefore, f = f, d f is itegrble. U(P, f ) L(P, f ) < ε. f f U(P, f ) L(P, f ) < ε. 1 Exmple : Let us show tht 1+x is itegrble o [,b] for y b >. We will see lter tht ll cotiuous fuctios re itegrble, but let us demostrte how we c do it directly. Let ε > be give. Tke N d pick x j := ib/, to form the prtitio P := {x,x 1,...,x } of [,b]. We hve x j = b/ for ll j. For for y subitervl [x j 1,x j ] we obti { } 1 m j = if 1 + x : x [x j 1,x j ] = 1 { } 1, M j = sup 1 + x j 1 + x : x [x j 1,x j ] = 1. x j 1 The we hve U(P, f ) L(P, f ) = x j j=1(m j m j ) = = b ( 1 j= ) ( j 1)b/ 1 + ib/ = b ( ) = b2 b/ 1 + b/ (b + 1). The sum telescopes, the terms successively ccel ech other, somethig we hve see before. b Pickig to be such tht 2 (b+1) < ε the propositio is stisfied d the fuctio is itegrble More ottio Whe f : S R is defied o lrger set S d [,b] S, we sy tht f is Riem itegrble o [,b] if the restrictio of f to [,b] is Riem itegrble. I this cse, we sy f R[,b], d we write f to me the Riem itegrl of the restrictio of f to [,b]. It is useful to defie the itegrl f eve if b. Suppose tht b < d tht f R[b,], the defie f := f. b
8 154 CHAPTER 5. THE RIEMANN INTEGRAL For y fuctio f we defie f :=. At times, the vrible x my lredy hve some other meig. Whe we eed to write dow the vrible of itegrtio, we my simply use differet letter. For exmple, Exercises f (s) ds := Exercise 5.1.1: Let f : [,1] R be defied by f (x) := x 3 d let P := {,.1,.4,1}. Compute L(P, f ) d U(P, f ). Exercise 5.1.2: Let f : [,1] R be defied by f (x) := x. Show tht f R[,1] d compute 1 f usig the defiitio of the itegrl (but feel free to use the propositios of this sectio). Exercise 5.1.3: Let f : [,b] R be bouded fuctio. Suppose tht there exists sequece of prtitios {P k } of [,b] such tht ( lim U(Pk, f ) L(P k, f ) ) =. k Show tht f is Riem itegrble d tht Exercise 5.1.4: Fiish the proof of Propositio f = lim k U(P k, f ) = lim k L(P k, f ). Exercise 5.1.5: Suppose tht f : [ 1,1] R is defied s Prove tht f R[ 1,1] d compute 1 propositios of this sectio). 1 f (x) := { 1 if x >, if x. Exercise 5.1.6: Let c (,b) d let d R. Defie f : [,b] R s f usig the defiitio of the itegrl (but feel free to use the f (x) := { d if x = c, if x c. Prove tht f R[,b] d compute f usig the defiitio of the itegrl (but feel free to use the propositios of this sectio).
The Definite Integral
The Defiite Itegrl A Riem sum R S (f) is pproximtio to the re uder fuctio f. The true re uder the fuctio is obtied by tkig the it of better d better pproximtios to the re uder f. Here is the forml defiitio,
More informationRiemann Integration. Chapter 1
Mesure, Itegrtio & Rel Alysis. Prelimiry editio. 8 July 2018. 2018 Sheldo Axler 1 Chpter 1 Riem Itegrtio This chpter reviews Riem itegrtio. Riem itegrtio uses rectgles to pproximte res uder grphs. This
More informationReview of the Riemann Integral
Chpter 1 Review of the Riem Itegrl This chpter provides quick review of the bsic properties of the Riem itegrl. 1.0 Itegrls d Riem Sums Defiitio 1.0.1. Let [, b] be fiite, closed itervl. A prtitio P of
More information1.3 Continuous Functions and Riemann Sums
mth riem sums, prt 0 Cotiuous Fuctios d Riem Sums I Exmple we sw tht lim Lower() = lim Upper() for the fuctio!! f (x) = + x o [0, ] This is o ccidet It is exmple of the followig theorem THEOREM Let f be
More informationSequence and Series of Functions
6 Sequece d Series of Fuctios 6. Sequece of Fuctios 6.. Poitwise Covergece d Uiform Covergece Let J be itervl i R. Defiitio 6. For ech N, suppose fuctio f : J R is give. The we sy tht sequece (f ) of fuctios
More informationPOWER SERIES R. E. SHOWALTER
POWER SERIES R. E. SHOWALTER. sequeces We deote by lim = tht the limit of the sequece { } is the umber. By this we me tht for y ε > 0 there is iteger N such tht < ε for ll itegers N. This mkes precise
More informationEVALUATING DEFINITE INTEGRALS
Chpter 4 EVALUATING DEFINITE INTEGRALS If the defiite itegrl represets re betwee curve d the x-xis, d if you c fid the re by recogizig the shpe of the regio, the you c evlute the defiite itegrl. Those
More informationReview of Sections
Review of Sectios.-.6 Mrch 24, 204 Abstrct This is the set of otes tht reviews the mi ides from Chpter coverig sequeces d series. The specific sectios tht we covered re s follows:.: Sequces..2: Series,
More informationWe will begin by supplying the proof to (a).
(The solutios of problem re mostly from Jeffrey Mudrock s HWs) Problem 1. There re three sttemet from Exmple 5.4 i the textbook for which we will supply proofs. The sttemets re the followig: () The spce
More informationWeek 13 Notes: 1) Riemann Sum. Aim: Compute Area Under a Graph. Suppose we want to find out the area of a graph, like the one on the right:
Week 1 Notes: 1) Riem Sum Aim: Compute Are Uder Grph Suppose we wt to fid out the re of grph, like the oe o the right: We wt to kow the re of the red re. Here re some wys to pproximte the re: We cut the
More information1. (25 points) Use the limit definition of the definite integral and the sum formulas to compute. [1 x + x2
Mth 3, Clculus II Fil Exm Solutios. (5 poits) Use the limit defiitio of the defiite itegrl d the sum formuls to compute 3 x + x. Check your swer by usig the Fudmetl Theorem of Clculus. Solutio: The limit
More informationMAS221 Analysis, Semester 2 Exercises
MAS22 Alysis, Semester 2 Exercises Srh Whitehouse (Exercises lbelled * my be more demdig.) Chpter Problems: Revisio Questio () Stte the defiitio of covergece of sequece of rel umbers, ( ), to limit. (b)
More informationB. Examples 1. Finite Sums finite sums are an example of Riemann Sums in which each subinterval has the same length and the same x i
Mth 06 Clculus Sec. 5.: The Defiite Itegrl I. Riem Sums A. Def : Give y=f(x):. Let f e defied o closed itervl[,].. Prtitio [,] ito suitervls[x (i-),x i ] of legth Δx i = x i -x (i-). Let P deote the prtitio
More information1 Tangent Line Problem
October 9, 018 MAT18 Week Justi Ko 1 Tget Lie Problem Questio: Give the grph of fuctio f, wht is the slope of the curve t the poit, f? Our strteg is to pproimte the slope b limit of sect lies betwee poits,
More informationRiemann Integral and Bounded function. Ng Tze Beng
Riem Itegrl d Bouded fuctio. Ng Tze Beg I geerlistio of re uder grph of fuctio, it is ormlly ssumed tht the fuctio uder cosidertio e ouded. For ouded fuctio, the rge of the fuctio is ouded d hece y suset
More informationConvergence rates of approximate sums of Riemann integrals
Jourl of Approximtio Theory 6 (9 477 49 www.elsevier.com/locte/jt Covergece rtes of pproximte sums of Riem itegrls Hiroyuki Tski Grdute School of Pure d Applied Sciece, Uiversity of Tsukub, Tsukub Ibrki
More informationFOURIER SERIES PART I: DEFINITIONS AND EXAMPLES. To a 2π-periodic function f(x) we will associate a trigonometric series. a n cos(nx) + b n sin(nx),
FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES To -periodic fuctio f() we will ssocite trigoometric series + cos() + b si(), or i terms of the epoetil e i, series of the form c e i. Z For most of the
More informationINFINITE SERIES. ,... having infinite number of terms is called infinite sequence and its indicated sum, i.e., a 1
Appedix A.. Itroductio As discussed i the Chpter 9 o Sequeces d Series, sequece,,...,,... hvig ifiite umber of terms is clled ifiite sequece d its idicted sum, i.e., + + +... + +... is clled ifite series
More informationChapter 7 Infinite Series
MA Ifiite Series Asst.Prof.Dr.Supree Liswdi Chpter 7 Ifiite Series Sectio 7. Sequece A sequece c be thought of s list of umbers writte i defiite order:,,...,,... 2 The umber is clled the first term, 2
More informationMA123, Chapter 9: Computing some integrals (pp )
MA13, Chpter 9: Computig some itegrls (pp. 189-05) Dte: Chpter Gols: Uderstd how to use bsic summtio formuls to evlute more complex sums. Uderstd how to compute its of rtiol fuctios t ifiity. Uderstd how
More informationMath 104: Final exam solutions
Mth 14: Fil exm solutios 1. Suppose tht (s ) is icresig sequece with coverget subsequece. Prove tht (s ) is coverget sequece. Aswer: Let the coverget subsequece be (s k ) tht coverges to limit s. The there
More informationINTEGRATION IN THEORY
CHATER 5 INTEGRATION IN THEORY 5.1 AREA AROXIMATION 5.1.1 SUMMATION NOTATION Fibocci Sequece First, exmple of fmous sequece of umbers. This is commoly ttributed to the mthemtici Fibocci of is, lthough
More informationRiemann Integral Oct 31, such that
Riem Itegrl Ot 31, 2007 Itegrtio of Step Futios A prtitio P of [, ] is olletio {x k } k=0 suh tht = x 0 < x 1 < < x 1 < x =. More suitly, prtitio is fiite suset of [, ] otiig d. It is helpful to thik of
More informationTest Info. Test may change slightly.
9. 9.6 Test Ifo Test my chge slightly. Short swer (0 questios 6 poits ech) o Must choose your ow test o Tests my oly be used oce o Tests/types you re resposible for: Geometric (kow sum) Telescopig (kow
More informationDefinite Integral. The Left and Right Sums
Clculus Li Vs Defiite Itegrl. The Left d Right Sums The defiite itegrl rises from the questio of fidig the re betwee give curve d x-xis o itervl. The re uder curve c be esily clculted if the curve is give
More information0 otherwise. sin( nx)sin( kx) 0 otherwise. cos( nx) sin( kx) dx 0 for all integers n, k.
. Computtio of Fourier Series I this sectio, we compute the Fourier coefficiets, f ( x) cos( x) b si( x) d b, i the Fourier series To do this, we eed the followig result o the orthogolity of the trigoometric
More informationUNIVERSITY OF BRISTOL. Examination for the Degrees of B.Sc. and M.Sci. (Level C/4) ANALYSIS 1B, SOLUTIONS MATH (Paper Code MATH-10006)
UNIVERSITY OF BRISTOL Exmitio for the Degrees of B.Sc. d M.Sci. (Level C/4) ANALYSIS B, SOLUTIONS MATH 6 (Pper Code MATH-6) My/Jue 25, hours 3 miutes This pper cotis two sectios, A d B. Plese use seprte
More informationCourse 121, , Test III (JF Hilary Term)
Course 2, 989 9, Test III (JF Hilry Term) Fridy 2d Februry 99, 3. 4.3pm Aswer y THREE questios. Let f: R R d g: R R be differetible fuctios o R. Stte the Product Rule d the Quotiet Rule for differetitig
More information10.5 Test Info. Test may change slightly.
0.5 Test Ifo Test my chge slightly. Short swer (0 questios 6 poits ech) o Must choose your ow test o Tests my oly be used oce o Tests/types you re resposible for: Geometric (kow sum) Telescopig (kow sum)
More informationInfinite Series Sequences: terms nth term Listing Terms of a Sequence 2 n recursively defined n+1 Pattern Recognition for Sequences Ex:
Ifiite Series Sequeces: A sequece i defied s fuctio whose domi is the set of positive itegers. Usully it s esier to deote sequece i subscript form rther th fuctio ottio.,, 3, re the terms of the sequece
More informationlecture 16: Introduction to Least Squares Approximation
97 lecture 16: Itroductio to Lest Squres Approximtio.4 Lest squres pproximtio The miimx criterio is ituitive objective for pproximtig fuctio. However, i my cses it is more ppelig (for both computtio d
More information( a n ) converges or diverges.
Chpter Ifiite Series Pge of Sectio E Rtio Test Chpter : Ifiite Series By the ed of this sectio you will be ble to uderstd the proof of the rtio test test series for covergece by pplyig the rtio test pprecite
More informationn 2 + 3n + 1 4n = n2 + 3n + 1 n n 2 = n + 1
Ifiite Series Some Tests for Divergece d Covergece Divergece Test: If lim u or if the limit does ot exist, the series diverget. + 3 + 4 + 3 EXAMPLE: Show tht the series diverges. = u = + 3 + 4 + 3 + 3
More informationMATH 104 FINAL SOLUTIONS. 1. (2 points each) Mark each of the following as True or False. No justification is required. y n = x 1 + x x n n
MATH 04 FINAL SOLUTIONS. ( poits ech) Mrk ech of the followig s True or Flse. No justifictio is required. ) A ubouded sequece c hve o Cuchy subsequece. Flse b) A ifiite uio of Dedekid cuts is Dedekid cut.
More informationMATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS. and x i = a + i. i + n(n + 1)(2n + 1) + 2a. (b a)3 6n 2
MATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS 6.9: Let f(x) { x 2 if x Q [, b], 0 if x (R \ Q) [, b], where > 0. Prove tht b. Solutio. Let P { x 0 < x 1 < < x b} be regulr prtitio
More information3.7 The Lebesgue integral
3 Mesure d Itegrtio The f is simple fuctio d positive wheever f is positive (the ltter follows from the fct tht i this cse f 1 [B,k ] = for ll k, ). Moreover, f (x) f (x). Ideed, if x, the there exists
More informationGeneral properties of definite integrals
Roerto s Notes o Itegrl Clculus Chpter 4: Defiite itegrls d the FTC Sectio Geerl properties of defiite itegrls Wht you eed to kow lredy: Wht defiite Riem itegrl is. Wht you c ler here: Some key properties
More informationThe Basic Properties of the Integral
The Bsic Properties of the Itegrl Whe we compute the derivtive of complicted fuctio, like x + six, we usully use differetitio rules, like d [f(x)+g(x)] d f(x)+ d g(x), to reduce the computtio dx dx dx
More informationApproximate Integration
Study Sheet (7.7) Approimte Itegrtio I this sectio, we will ler: How to fid pproimte vlues of defiite itegrls. There re two situtios i which it is impossile to fid the ect vlue of defiite itegrl. Situtio:
More informationis infinite. The converse is proved similarly, and the last statement of the theorem is clear too.
12. No-stdrd lysis October 2, 2011 I this sectio we give brief itroductio to o-stdrd lysis. This is firly well-developed field of mthemtics bsed o model theory. It dels ot just with the rels, fuctios o
More informationMATH 104: INTRODUCTORY ANALYSIS SPRING 2008/09 PROBLEM SET 10 SOLUTIONS. f m. and. f m = 0. and x i = a + i. a + i. a + n 2. n(n + 1) = a(b a) +
MATH 04: INTRODUCTORY ANALYSIS SPRING 008/09 PROBLEM SET 0 SOLUTIONS Throughout this problem set, B[, b] will deote the set of ll rel-vlued futios bouded o [, b], C[, b] the set of ll rel-vlued futios
More informationChapter 2 Infinite Series Page 1 of 9
Chpter Ifiite eries Pge of 9 Chpter : Ifiite eries ectio A Itroductio to Ifiite eries By the ed of this sectio you will be ble to uderstd wht is met by covergece d divergece of ifiite series recogise geometric
More informationContent: Essential Calculus, Early Transcendentals, James Stewart, 2007 Chapter 1: Functions and Limits., in a set B.
Review Sheet: Chpter Cotet: Essetil Clculus, Erly Trscedetls, Jmes Stewrt, 007 Chpter : Fuctios d Limits Cocepts, Defiitios, Lws, Theorems: A fuctio, f, is rule tht ssigs to ech elemet i set A ectly oe
More informationTHE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING
OLLSCOIL NA héireann, CORCAIGH THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING MATHEMATICS MA008 Clculus d Lier
More informationf(bx) dx = f dx = dx l dx f(0) log b x a + l log b a 2ɛ log b a.
Eercise 5 For y < A < B, we hve B A f fb B d = = A B A f d f d For y ɛ >, there re N > δ >, such tht d The for y < A < δ d B > N, we hve ba f d f A bb f d l By ba A A B A bb ba fb d f d = ba < m{, b}δ
More informationf(t)dt 2δ f(x) f(t)dt 0 and b f(t)dt = 0 gives F (b) = 0. Since F is increasing, this means that
Uiversity of Illiois t Ur-Chmpig Fll 6 Mth 444 Group E3 Itegrtio : correctio of the exercises.. ( Assume tht f : [, ] R is cotiuous fuctio such tht f(x for ll x (,, d f(tdt =. Show tht f(x = for ll x [,
More informationy udv uv y v du 7.1 INTEGRATION BY PARTS
7. INTEGRATION BY PARTS Ever differetitio rule hs correspodig itegrtio rule. For istce, the Substitutio Rule for itegrtio correspods to the Chi Rule for differetitio. The rule tht correspods to the Product
More informationThe Weierstrass Approximation Theorem
The Weierstrss Approximtio Theorem Jmes K. Peterso Deprtmet of Biologicl Scieces d Deprtmet of Mthemticl Scieces Clemso Uiversity Februry 26, 2018 Outlie The Wierstrss Approximtio Theorem MtLb Implemettio
More informationBRAIN TEASURES INDEFINITE INTEGRATION+DEFINITE INTEGRATION EXERCISE I
EXERCISE I t Q. d Q. 6 6 cos si Q. Q.6 d d Q. d Q. Itegrte cos t d by the substitutio z = + e d e Q.7 cos. l cos si d d Q. cos si si si si b cos Q.9 d Q. si b cos Q. si( ) si( ) d ( ) Q. d cot d d Q. (si
More informationApproximations of Definite Integrals
Approximtios of Defiite Itegrls So fr we hve relied o tiderivtives to evlute res uder curves, work doe by vrible force, volumes of revolutio, etc. More precisely, wheever we hve hd to evlute defiite itegrl
More informationThe Reimann Integral is a formal limit definition of a definite integral
MATH 136 The Reim Itegrl The Reim Itegrl is forml limit defiitio of defiite itegrl cotiuous fuctio f. The costructio is s follows: f ( x) dx for Reim Itegrl: Prtitio [, ] ito suitervls ech hvig the equl
More informationLimit of a function:
- Limit of fuctio: We sy tht f ( ) eists d is equl with (rel) umer L if f( ) gets s close s we wt to L if is close eough to (This defiitio c e geerlized for L y syig tht f( ) ecomes s lrge (or s lrge egtive
More information n. A Very Interesting Example + + = d. + x3. + 5x4. math 131 power series, part ii 7. One of the first power series we examined was. 2!
mth power series, prt ii 7 A Very Iterestig Emple Oe of the first power series we emied ws! + +! + + +!! + I Emple 58 we used the rtio test to show tht the itervl of covergece ws (, ) Sice the series coverges
More informationANALYSIS HW 3. f(x + y) = f(x) + f(y) for all real x, y. Demonstration: Let f be such a function. Since f is smooth, f exists.
ANALYSIS HW 3 CLAY SHONKWILER () Fid ll smooth fuctios f : R R with the property f(x + y) = f(x) + f(y) for ll rel x, y. Demostrtio: Let f be such fuctio. Sice f is smooth, f exists. The The f f(x + h)
More informationConvergence rates of approximate sums of Riemann integrals
Covergece rtes of pproximte sums of Riem itegrls Hiroyuki Tski Grdute School of Pure d Applied Sciece, Uiversity of Tsuku Tsuku Irki 5-857 Jp tski@mth.tsuku.c.jp Keywords : covergece rte; Riem sum; Riem
More information4. When is the particle speeding up? Why? 5. When is the particle slowing down? Why?
AB CALCULUS: 5.3 Positio vs Distce Velocity vs. Speed Accelertio All the questios which follow refer to the grph t the right.. Whe is the prticle movig t costt speed?. Whe is the prticle movig to the right?
More informationCertain sufficient conditions on N, p n, q n k summability of orthogonal series
Avilble olie t www.tjs.com J. Nolier Sci. Appl. 7 014, 7 77 Reserch Article Certi sufficiet coditios o N, p, k summbility of orthogol series Xhevt Z. Krsiqi Deprtmet of Mthemtics d Iformtics, Fculty of
More informationLimits and an Introduction to Calculus
Nme Chpter Limits d Itroductio to Clculus Sectio. Itroductio to Limits Objective: I this lesso ou lered how to estimte limits d use properties d opertios of limits. I. The Limit Cocept d Defiitio of Limit
More information10.5 Power Series. In this section, we are going to start talking about power series. A power series is a series of the form
0.5 Power Series I the lst three sectios, we ve spet most of tht time tlkig bout how to determie if series is coverget or ot. Now it is time to strt lookig t some specific kids of series d we will evetully
More informationSection 6.3: Geometric Sequences
40 Chpter 6 Sectio 6.: Geometric Sequeces My jobs offer ul cost-of-livig icrese to keep slries cosistet with ifltio. Suppose, for exmple, recet college grdute fids positio s sles mger erig ul slry of $6,000.
More informationProbability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J.
Probbility d Stochstic Processes: A Friedly Itroductio for Electricl d Computer Egieers Roy D. Ytes d Dvid J. Goodm Problem Solutios : Ytes d Goodm,4..4 4..4 4..7 4.4. 4.4. 4..6 4.6.8 4.6.9 4.7.4 4.7.
More informationb a 2 ((g(x))2 (f(x)) 2 dx
Clc II Fll 005 MATH Nme: T3 Istructios: Write swers to problems o seprte pper. You my NOT use clcultors or y electroic devices or otes of y kid. Ech st rred problem is extr credit d ech is worth 5 poits.
More informationThe Definite Riemann Integral
These otes closely follow the presettio of the mteril give i Jmes Stewrt s textook Clculus, Cocepts d Cotexts (d editio). These otes re iteded primrily for i-clss presettio d should ot e regrded s sustitute
More information( ) dx ; f ( x ) is height and Δx is
Mth : 6.3 Defiite Itegrls from Riem Sums We just sw tht the exct re ouded y cotiuous fuctio f d the x xis o the itervl x, ws give s A = lim A exct RAM, where is the umer of rectgles i the Rectgulr Approximtio
More informationCrushed Notes on MATH132: Calculus
Mth 13, Fll 011 Siyg Yg s Outlie Crushed Notes o MATH13: Clculus The otes elow re crushed d my ot e ect This is oly my ow cocise overview of the clss mterils The otes I put elow should ot e used to justify
More informationTaylor Polynomials. The Tangent Line. (a, f (a)) and has the same slope as the curve y = f (x) at that point. It is the best
Tylor Polyomils Let f () = e d let p() = 1 + + 1 + 1 6 3 Without usig clcultor, evlute f (1) d p(1) Ok, I m still witig With little effort it is possible to evlute p(1) = 1 + 1 + 1 (144) + 6 1 (178) =
More informationMTH 146 Class 16 Notes
MTH 46 Clss 6 Notes 0.4- Cotiued Motivtio: We ow cosider the rc legth of polr curve. Suppose we wish to fid the legth of polr curve curve i terms of prmetric equtios s: r f where b. We c view the cos si
More informationMATRIX ALGEBRA, Systems Linear Equations
MATRIX ALGEBRA, Systes Lier Equtios Now we chge to the LINEAR ALGEBRA perspective o vectors d trices to reforulte systes of lier equtios. If you fid the discussio i ters of geerl d gets lost i geerlity,
More informationRemarks: (a) The Dirac delta is the function zero on the domain R {0}.
Sectio Objective(s): The Dirc s Delt. Mi Properties. Applictios. The Impulse Respose Fuctio. 4.4.. The Dirc Delt. 4.4. Geerlized Sources Defiitio 4.4.. The Dirc delt geerlized fuctio is the limit δ(t)
More informationOptions: Calculus. O C.1 PG #2, 3b, 4, 5ace O C.2 PG.24 #1 O D PG.28 #2, 3, 4, 5, 7 O E PG #1, 3, 4, 5 O F PG.
O C. PG.-3 #, 3b, 4, 5ce O C. PG.4 # Optios: Clculus O D PG.8 #, 3, 4, 5, 7 O E PG.3-33 #, 3, 4, 5 O F PG.36-37 #, 3 O G. PG.4 #c, 3c O G. PG.43 #, O H PG.49 #, 4, 5, 6, 7, 8, 9, 0 O I. PG.53-54 #5, 8
More informationApplication: Volume. 6.1 Overture. Cylinders
Applictio: Volume 6 Overture I this chpter we preset other pplictio of the defiite itegrl, this time to fid volumes of certi solids As importt s this prticulr pplictio is, more importt is to recogize ptter
More informationMathacle. PSet Stats, Concepts In Statistics Level Number Name: Date:
APPENDEX I. THE RAW ALGEBRA IN STATISTICS A I-1. THE INEQUALITY Exmple A I-1.1. Solve ech iequlity. Write the solutio i the itervl ottio..) 2 p - 6 p -8.) 2x- 3 < 5 Solutio:.). - 4 p -8 p³ 2 or pî[2, +
More information1.1 The FTC and Riemann Sums. An Application of Definite Integrals: Net Distance Travelled
mth 3 more o the fudmetl theorem of clculus The FTC d Riem Sums A Applictio of Defiite Itegrls: Net Distce Trvelled I the ext few sectios (d the ext few chpters) we will see severl importt pplictios of
More informationProbability for mathematicians INDEPENDENCE TAU
Probbility for mthemticis INDEPENDENCE TAU 2013 21 Cotets 2 Cetrl limit theorem 21 2 Itroductio............................ 21 2b Covolutio............................ 22 2c The iitil distributio does
More informationMath 140B - Notes. Neil Donaldson. September 2, 2009
Mth 40B - Notes Neil Doldso September 2, 2009 Itroductio This clss cotiues from 40A. The mi purpose of the clss is to mke bsic clculus rigorous.. Nottio We will observe the followig ottio throughout this
More informationTheorem 5.3 (Continued) The Fundamental Theorem of Calculus, Part 2: ab,, then. f x dx F x F b F a. a a. f x dx= f x x
Chpter 6 Applictios Itegrtio Sectio 6. Regio Betwee Curves Recll: Theorem 5.3 (Cotiued) The Fudmetl Theorem of Clculus, Prt :,,, the If f is cotiuous t ever poit of [ ] d F is tiderivtive of f o [ ] (
More informationSUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11
UTCLIFFE NOTE: CALCULU WOKOWKI CHAPTER Ifiite eries Coverget or Diverget eries Cosider the sequece If we form the ifiite sum 0, 00, 000, 0 00 000, we hve wht is clled ifiite series We wt to fid the sum
More information18.01 Calculus Jason Starr Fall 2005
18.01 Clculus Jso Strr Lecture 14. October 14, 005 Homework. Problem Set 4 Prt II: Problem. Prctice Problems. Course Reder: 3B 1, 3B 3, 3B 4, 3B 5. 1. The problem of res. The ciet Greeks computed the res
More informationSimilar idea to multiplication in N, C. Divide and conquer approach provides unexpected improvements. Naïve matrix multiplication
Next. Covered bsics of simple desig techique (Divided-coquer) Ch. of the text.. Next, Strsse s lgorithm. Lter: more desig d coquer lgorithms: MergeSort. Solvig recurreces d the Mster Theorem. Similr ide
More information5.1 - Areas and Distances
Mth 3B Midterm Review Writte by Victori Kl vtkl@mth.ucsb.edu SH 63u Office Hours: R 9:5 - :5m The midterm will cover the sectios for which you hve received homework d feedbck Sectios.9-6.5 i your book.
More informationExploring the Rate of Convergence of Approximations to the Riemann Integral
Explorig the Rte of Covergece of Approximtios to the Riem Itegrl Lus Owes My 7, 24 Cotets Itroductio 2. Prelimiries............................................. 2.2 Motivtio..............................................
More information(200 terms) equals Let f (x) = 1 + x + x 2 + +x 100 = x101 1
SECTION 5. PGE 78.. DMS: CLCULUS.... 5. 6. CHPTE 5. Sectio 5. pge 78 i + + + INTEGTION Sums d Sigm Nottio j j + + + + + i + + + + i j i i + + + j j + 5 + + j + + 9 + + 7. 5 + 6 + 7 + 8 + 9 9 i i5 8. +
More information9.1 Sequences & Series: Convergence & Divergence
Notes 9.: Cov & Div of Seq & Ser 9. Sequeces & Series: Covergece & Divergece A sequece is simply list of thigs geerted by rule More formlly, sequece is fuctio whose domi is the set of positive itegers,
More information( ) k ( ) 1 T n 1 x = xk. Geometric series obtained directly from the definition. = 1 1 x. See also Scalars 9.1 ADV-1: lim n.
Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From 207.04.07 A.docx Pge of Algebric tricks ivolvig x. You c use lgebric tricks to simplify workig with the Tylor polyomils of certi fuctios..
More informationREVIEW OF CHAPTER 5 MATH 114 (SECTION C1): ELEMENTARY CALCULUS
REVIEW OF CHAPTER 5 MATH 4 (SECTION C): EEMENTARY CACUUS.. Are.. Are d Estimtig with Fiite Sums Emple. Approimte the re of the shded regio R tht is lies ove the -is, elow the grph of =, d etwee the verticl
More informationChapter Real Numbers
Chpter. - Rel Numbers Itegers: coutig umbers, zero, d the egtive of the coutig umbers. ex: {,-3, -, -,,,, 3, } Rtiol Numbers: quotiets of two itegers with ozero deomitor; termitig or repetig decimls. ex:
More informationFall 2004 Math Integrals 6.1 Sigma Notation Mon, 15/Nov c 2004, Art Belmonte
Fll Mth 6 Itegrls 6. Sigm Nottio Mo, /Nov c, Art Belmote Summr Sigm ottio For itegers m d rel umbers m, m+,...,, we write k = m + m+ + +. k=m The left-hd side is shorthd for the fiite sum o right. The
More information4.1 Bounded integral.
33 4. LEBESGUE INTEGRAL: BOUNDED CASE Hvig developed the cocept of the Lebesgue mesure, we will ow move to defiig the Lebesgue itegrl. It would be wste to restrict ttetio to itegrls with respect to the
More informationINSTRUCTOR: CEZAR LUPU. Problem 1. a) Let f(x) be a continuous function on [1, 2]. Prove that. nx 2 lim
WORKSHEET FOR THE PRELIMINARY EXAMINATION-REAL ANALYSIS (RIEMANN AND RIEMANN-STIELTJES INTEGRAL OF A SINGLE VARIABLE, INTEGRAL CALCULUS, FOURIER SERIES AND SPECIAL FUNCTIONS INSTRUCTOR: CEZAR LUPU Problem.
More informationMath 3B Midterm Review
Mth 3B Midterm Review Writte by Victori Kl vtkl@mth.ucsb.edu SH 643u Office Hours: R 11:00 m - 1:00 pm Lst updted /15/015 Here re some short otes o Sectios 7.1-7.8 i your ebook. The best idictio of wht
More informationLinford 1. Kyle Linford. Math 211. Honors Project. Theorems to Analyze: Theorem 2.4 The Limit of a Function Involving a Radical (A4)
Liford 1 Kyle Liford Mth 211 Hoors Project Theorems to Alyze: Theorem 2.4 The Limit of Fuctio Ivolvig Rdicl (A4) Theorem 2.8 The Squeeze Theorem (A5) Theorem 2.9 The Limit of Si(x)/x = 1 (p. 85) Theorem
More informationPROGRESSIONS AND SERIES
PROGRESSIONS AND SERIES A sequece is lso clled progressio. We ow study three importt types of sequeces: () The Arithmetic Progressio, () The Geometric Progressio, () The Hrmoic Progressio. Arithmetic Progressio.
More informationSection IV.6: The Master Method and Applications
Sectio IV.6: The Mster Method d Applictios Defiitio IV.6.1: A fuctio f is symptoticlly positive if d oly if there exists rel umer such tht f(x) > for ll x >. A cosequece of this defiitio is tht fuctio
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationis an ordered list of numbers. Each number in a sequence is a term of a sequence. n-1 term
Mthemticl Ptters. Arithmetic Sequeces. Arithmetic Series. To idetify mthemticl ptters foud sequece. To use formul to fid the th term of sequece. To defie, idetify, d pply rithmetic sequeces. To defie rithmetic
More informationNumbers (Part I) -- Solutions
Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets Numbers Prt I) -- Solutios. The equtio b c 008 hs solutio i which, b, c re distict
More informationSolutions to Problem Set 7
8.0 Clculus Jso Strr Due by :00pm shrp Fll 005 Fridy, Dec., 005 Solutios to Problem Set 7 Lte homework policy. Lte work will be ccepted oly with medicl ote or for other Istitute pproved reso. Coopertio
More informationSUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11
SUTCLIFFE S NOTES: CALCULUS SWOKOWSKI S CHAPTER Ifiite Series.5 Altertig Series d Absolute Covergece Next, let us cosider series with both positive d egtive terms. The simplest d most useful is ltertig
More informationINTEGRATION TECHNIQUES (TRIG, LOG, EXP FUNCTIONS)
Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of MK HOME TUITION Mthemtics Revisio Guides Level: AS / A Level AQA : C Edecel: C OCR: C OCR MEI: C INTEGRATION TECHNIQUES (TRIG, LOG, EXP FUNCTIONS)
More information8.1 Arc Length. What is the length of a curve? How can we approximate it? We could do it following the pattern we ve used before
8.1 Arc Legth Wht is the legth of curve? How c we pproximte it? We could do it followig the ptter we ve used efore Use sequece of icresigly short segmets to pproximte the curve: As the segmets get smller
More information