# LECTURE. INTEGRATION AND ANTIDERIVATIVE.

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1 ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development of the modern clculus Definite integrl nd the re under the grph. Computing the re of the shpe bounded, sy, by the grph of f(x) = x 2, the horizontl xis y = 0 nd the verticl line x = 1, cn be done by elementry methods: this mounts to computtion of the sums n 1 ( ) k 2 Sn 1 n ( ) k 2 = n n, 1 S+ n = n n k=0 for lrge n N. One cn prove by induction n explicit formul for the sum n 1 k2 = 1 n3 6n(n+1)(2n+1) = 3 + n2 2 + n 6 nd conclude tht S± n = 1 3 +o(1) s n. However, if the function f(x) = x 2 is replced by more generl polynomil of higher degree, then the summtion becomes more nd more sophisticted. Alterntive is required. But first of ll, wht is the re? more specificlly, wht is the re of the trpeze between the grph of positive function f : [, b] R +, the xis y = 0 nd two lines x = nd x = b? The definition is bsed on the ide of the inequlity vlid for polygons on R 2 : if P, Q re two polygons with P Q, then re(p ) re(q). Let P = (x 0,..., x n ) be n points on [, b], = x 0 < x 1 < < x n = b (it will be clled n n-prtition), nd f : [, b] R function on [, b]. Define the two uxiliry functions, f P (x) = f + P (x) = k=1 inf f(x), if x [x k 1, x k ], x [x k 1,x k ] sup f(x), if x [x k 1, x k ]. x [x k 1,x k ] The grph of ech of these functions is polygon on the (x, y)-plne, hence the re S ± P (f) cn be esily computed for ny prtition P. Since f P (x) f(x) f + P (x), we lwys hve the inequlity S P (f) S+ P (f). If the re Dte: December 22,

2 2 ROTHSCHILD CAESARIA COURSE, 2015/6 S(f) under the grph of f exists in ny sense, it must seprte the two sets, S P (f) S(f) S+ P (f) for ny two prtitions P, P. If there is more thn one such number, we re in trouble, but if this number is unique, there is no other choice but to cll it the re. Definition 1. A function f : [, b] R is clled (Riemnn) integrble (on this segment), if sup S P (f) = inf P P S+ P (f), the supremum/infimum being tken over ll n-prtitions P for ll n. The (unique) common vlue bove is clled the Riemnn integrl of f nd denoted f(x) dx. The corresponding sums re clled lower (resp., upper) integrl sums for f on [, b]. Remrk 1. An unbounded (on finite segment) function f does not dmit either lower or upper pproximtion f ± so one of integrl sums if ±. Thus by definition it cnnot be Riemnn integrble. Remrk 2. The nottion is both historiclly nd mthemticlly justified: it is for good reson tht insted of the function f we integrte the differentil f(x) dx. Lter we will lern tht this nottion is well dpted mnemoniclly to the chnge of the independent vrible x. By the wy, the nottion of the derivtive f () lso concels the choice of the independent vrible, so in sense the old-fshioned nottion df dx () is better! Obviously, the constnt function f(x) = c is integrble on ny (finite) intervl, nd c dx = c(b ). The functions f(x) = x2 nd f(x) = x 3 re lso integrble, s the bove computtions show (one hs to ensure tht there is no gp between the lower nd the upper sums). Theorem 1. Any function f : [, b] R continuous on [, b], is integrble. Some functions re non-integrble, however. Exmple 1. The Dirichlet function f(x) = 1 or 0 depending on whether x is rtionl or not, is non-integrble on ny segment. Indeed, since both rtionl nd irrtionl numbers re dense, ll upper sums will be b, while ll lower sums re zero. Exmple 2. If f is integrble on [, b] nd g differs from f only in finitely mny points, then g is lso integrble on [, b]. Exmple 3. The function f(x) = 1/x is non-integrble on the segment [0, 1] regrdless of the wy how one defines f(0). Remrk 3. Integrbility on infinite segments is out of question currently: ny finite prtition will necessrily involve the infinite difference x n x n 1, lthough sometimes the res of the infinite shpes re finite. Remrk 4. If f is integrble on [, b], then it is integrble on [, c] for ny c (, b). Prove it!

3 ANALYSIS FOR HIGH SCHOOL TEACHERS 3 2. The Newton Leibniz formul: the mgic key Theorem 2. Assume tht f : [, b] R is continuous function, nd F (c) = c f(x) dx is the integrl of f over [, c] s function of the right endpoint c [, b]. Then F is differentible function on [, b], F () = 0 nd F (c) = f(c) t ny point c [, b]. Definition 2. A function F is clled n ntiderivtive of f, if F is differentible nd DF = f. Corollry 3 (inversion). If F is n ntiderivtive of f, then f(x) dx = F (b) F (). Computtion of derivtives is n explicit process, in prticulr, derivtive of n elementry function is gin n elementry function. Thus there is Google wy to compute integrls: sk computers to differentite ll possible functions nd serch mong the results to find out whether your specific function is somewhere in the dtbse. There re esy rules which follow from the corresponding rules of the differentition: linerity (ntiderivtive of liner combintion is liner combintion of ntiderivtives); nti-leibniz rule: if f = uv nd ntiderivtive of v is known, V = v, then one cn express the nswer in terms of uv nd the ntiderivtive of the function g = u V. There re no priori resons why the ltter ntiderivtive is esier to compute, but sometimes it indeed is. This trick is clled integrtion by prts. However, the thumb rule is tht you should be quite lucky to find n explicit ( elementry ) ntiderivtive for your function. Exmple 4. If u = x, v = e x, then V = e x, so the ntiderivtive for xe x cn be expressed through the function uv = xe x nd the ntiderivtive of u V = e x which is gin the exponent e x. Repeting the sme trick, one cn reduce computtion of the ntiderivtive of x 2 e x to tht of xe x, lredy known by the previous trick. It looks like trechery, but surprisingly it works nd, moreover, llows to define the vlue of the fctoril F (n) = n! from nturl to ll rel (nd even complex) vlues of the rgument (Euler). Mysteriously, ( 1 2 )! = 1 2 π... Exmple 5. All monomils x n, n Z, re ntiderivtives of the monomils x n+1 n+1, except for n = 1: there is no lgebric function f stisfying the eqution f = 1 x. Antiderivtive of x 1 is the logrithm ln x. This is how the logrithmic nd exponentil functions enter the lgebric world. Exmple 6. Antiderivtives of the elementry trigonometric functions (when they cn be computed) re usully trigonometric.

4 4 ROTHSCHILD CAESARIA COURSE, 2015/6 The bridge between the trigonometric functions nd lgebric functions is the identity (rctn x) = x 2. This hs mzing consequences! Trigonometric constnts (e.g., π, the most mysterious number in mth) cn be expressed through the rtionl series. Upon the second thought, π/2 is the re between the curve x 2 +y 2 = 1 (the unit circle) nd the upper hlf-plne {y 0}... Integrtion is source of mny mircles Chnge of vribles. Assume tht f is integrble on [, b], x = ϕ(t) is monotonously growing differentible function, ϕ(a) =, ϕ(b) = b. Theorem 4. f(x) dx = B A f(ϕ(t)) dϕ (t) dt = dt First proof. Use the chin rule for derivtion B A (f ϕ) dϕ. d df (ϕ(t)) (F (ϕ(t)) = dϕ(t) dt dx dt pply it to the cse where F = f nd integrte both sides from A to B. Second proof. Consider prtition P nd its ϕ-preimge Q = {A = t 0 < t 1 < < t n = B}, ϕ(t i ) = x i. Then the vlues of the functions f ± (ϕ(t)) on [t k 1, t k ] coincide with the vlues of f ± on [x k 1, x k ], but x k x k 1 by the Lgrnge theorem is ϕ (ξ k )(t k t k 1 ) for some intermedite points ξ k (t k 1, t k ). If the derivtive ϕ is continuous, the upper nd lower sums S ± P (f ϕ) converge to the integrl of (f ϕ) ϕ The Stieltjes integrl. One cn slightly generlize the notion of the Riemnn integrl. Let f, g be two functions on [, b], with f continuous nd g monotonous (growing). Insted of the Riemnn sums S ± P (f) = k f ± (x)(x k x k 1 ) one cn consider the Stieltjes sums S ± P (f, g) = f ± (x)(g(x k ) g(x k 1 )). k These sums converge in the sense tht sup S P (f, g) = inf P P S+ P (f, g). The common vlue is the Stieltjes integrl denoted by f(x) dg(x). Exmple 7. If g(x) hs continuous derivtive g, then the Stieltjes integrl is equl to the Riemnn integrl, f dg = f(x)g (x) dx.

5 ANALYSIS FOR HIGH SCHOOL TEACHERS 5 Exmple 8. Assume tht g(x) is step function, g(x) = c i if x [x i 1, x i ) for some prtition P, c 1 < c 2 < < c n. Then the Stieltjes integrl of continuous function f reduces to finite sum, n 1 f dg = f(x i )(c i+1 c i ). i=1 Thus the Stieltjes integrl interpoltes between the genuine integrtion nd the finite sums (if g hs jumps). Theorem 5 (chnge of vribles in the Stieltjes integrl). If ϕ : [A, B] [, b], t ϕ(t) is monotone differentible chnge of vribles, f, g : [, b] R re two functions such tht the Stieltjes integrl f dg exists, then f dg = B A F dg, F = f ϕ, G = g ϕ. Proof. This is tutologicl sttement relted to the upper/lower sums. Contempltion of the Stieltjes integrl nd its trnsformtion clrifies the reson, why we integrte not function f, but rther differentil df = f(x) dx. The Newton Leibnitz formul then becomes the obvious identity df = F (b) F () Integrbility of discontinuous functions. Integrtion improves the regulrity of functions. Exmple 9. If f(x) = sign(x) = ±1 for ±x > 0, then (regrdless of the choice for f(0)) this function is integrble on ny segment, nd 0 signx dx = x. Note tht the function F (x) = x is not exctly ntiderivtive of f: F is non-differentible t = 0. Yet F is continuous nd F () = f() for ll 0. This exmple cn simplified even more: let f be continuous nd F = f. Consider the function f which differs from f t one point only. Then f is discontinuous t this point, yet integrble, nd F is n lmost ntiderivtive of f s well. Proposition 6. Any bounded function hving only finitely mny points of discontinuity, is integrble. In fct, stronger sttement cn be proved. Definition 3. We sy tht subset C R is ε-smll for some ε > 0, if it cn be covered by finitely mny open intervls of totl length less thn ε. The set C hs zero length, if it is ε-smll for ny ε > 0. Of course, ll finite sets hve zero length. One cn construct infinite sets of zero length, e.g., C = { 1 n : n N} (prove it!).

6 6 ROTHSCHILD CAESARIA COURSE, 2015/6 Theorem 7. A bounded function continuous on the complement to set of length zero, is integrble. This is lredy very close to the necessry nd sufficient condition of integrbility (Lebesgue, erly 20th century). Definition 4. A subset C R is sid to hve zero mesure, if for ny ε > 0 it cn be covered by countbly mny open intervls of totl length less thn ε > 0. Exmple 10. Any countble subset of R, e.g., C = Q, hs mesure zero. Theorem 8. A bounded function is integrble if nd only if it is continuous on the complement to set of mesure zero. 3. Conclusions The problem of computing res, volumes etc., cn be solved by computing definite (Riemnn) integrl. By the Newton Leibniz fundmentl theorem, insted of clculting the limits of integrl sums, one cn serch for n ntiderivtive. Unlike derivtion which is lwys explicitly computble (hence cn be trusted to computers mnging symbolic computtions), ntiderivtions cn be relly new functions not dmitting explicit expression. Integrtion regulrizes functions (mkes them behving better). The Riemnn pproch to integrtion is not the only one possible: there exist other constructions (most notbly the Lebesgue integrl) which cn be pplied to broder clss of functions. One cn work out some wys to define (nd compute) integrls of unbounded functions or integrls over infinite intervls. For instnce, the re under the grph of the function f(x) = 1 on the whole rel line R, is 1+x 2 equl to π. In fct, integrls over the entire rel line R re sometimes esier to compute thn integrls over finite segments (the resons for this lie in the complex domin).

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### SYDE 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

### 1.9 C 2 inner variations 46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for

### Calculus I-II Review Sheet Clculus I-II Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing

### Integrals - 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 non-liner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but

### The 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

### Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1 Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution

### 38 Riemann sums and existence of the definite integral. 38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

### Numerical Analysis: Trapezoidal and Simpson s Rule nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =

### Main topics for the Second Midterm Min topics for the Second Midterm The Midterm will cover Sections 5.4-5.9, Sections 6.1-6.3, nd Sections 7.1-7.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the

### The 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

### Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 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

### NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by. NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

### Improper 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

### Summer MTH142 College Calculus 2. Section J. Lecture Notes. Yin Su University at Buffalo Summer 6 MTH4 College Clculus Section J Lecture Notes Yin Su University t Bufflo yinsu@bufflo.edu Contents Bsic techniques of integrtion 3. Antiderivtive nd indefinite integrls..............................................

### Calculus II: Integrations and Series Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]

### P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0) 1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this

### W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)

### f(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

### The Henstock-Kurzweil integral fculteit Wiskunde en Ntuurwetenschppen The Henstock-Kurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft

### MA123, 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. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

### 7.2 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

### Indefinite Integral. Chapter Integration - reverse of differentiation Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the

### Math 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 x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

### different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s). Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different

### Chapter 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

### arxiv: v1 [math.ca] 11 Jul 2011 rxiv:1107.1996v1 [mth.ca] 11 Jul 2011 Existence nd computtion of Riemnn Stieltjes integrls through Riemnn integrls July, 2011 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde

### 1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q. Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the

### Week 10: Riemann integral and its properties Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the

### x = 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

### THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS. THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

### Big idea in Calculus: approximation Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:

### F (x) dx = F (x)+c = u + C = du, 35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil

### Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

### New Expansion and Infinite Series Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06-073 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University

### Improper Integrals, and Differential Equations Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

### Mapping the delta function and other Radon measures Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support

### 4.4 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

### MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1 MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further

### 1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above. 1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt SECTION : Fourier Series. MATH4. In section 4, we will study method clled Seprtion of Vribles for finding exct solutions to certin clss of prtil differentil equtions (PDEs. To do this, it will be necessry 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