# Math 360: A primitive integral and elementary functions

Size: px
Start display at page:

Transcription

1 Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth C: Integrl/functions 1 / 32

2 Setup for the integrl prtitions Definition: (prtitions of n intervl) Let < b, then sequence of numbers = x 0 < x 1 < x 2 < < x n = b is clled prtition of the intervl [, b]. There is no requirement tht the x i s be evenly spced, only tht they be strictly incresing nd tht x 0 = nd x n = b. Sometimes we refer to prtition by P. For instnce, we sy P is refinement of the prtition P (nd write P > P) if ll of the x i s in P re lso in P. The norm of the prtition P (denoted P ) is the mximum of x i x i 1 for i = 1,..., n. (The textbook clls this the gp of P). Given two prtitions P 1 nd P 2 of [, b] it is esy to see tht they hve common refinement just use ll the x s from both prtitions in new P 3. D. DeTurck Mth C: Integrl/functions 2 / 32

3 Setup for the integrl monotonic functions At first, we re only going to consider the integrls of monotonic functions. Let f : [, b] R be monotonic function. For the moment, we ll ssume tht f is incresing (i.e., f (x) f (y) if x y), but everything will work for decresing functions with the obvious djustments. f does not hve to be continuous (nd in fct it cn be llowed to be discontinuous t [countbly ] infinitely mny plces), s long s it is defined for ll x [, b] nd is monotonic. D. DeTurck Mth C: Integrl/functions 3 / 32

4 Setup for the integrl upper nd lower sums For f monotoniclly incresing on [, b], nd P prtition of [, b]. Definition: (upper nd lower sums) The upper sum of f corresponding to the prtition P is U(f, P) = n f (x i )(x i x i 1 ) i=1 nd the lower sum of f corresponding to the prtition P is L(f, P) = n f (x i 1 )(x i x i 1 ) i=1 Since f is incresing, we hve f (x i ) is the mximum vlue of f on [x i 1, x i ] nd f (x i 1 ) is the minimum. Therefore U(f, P) L(f, P) D. DeTurck Mth C: Integrl/functions 4 / 32

5 Setup for the integrl definition Since there s lwys common refinement P for ny prtitions P 1 nd P 2, we hve L(f, P 1 ) L(f, P) U(f, P) U(f, P 2 ) where P is common refinement for P 1 nd P 2. So for ny pir of prtitions P 1 nd P 2 we hve L(f, P 1 ) U(f, P 2 ). Therefore sup L(f, P) over ll prtitions P is less thn or equl to inf U(f, P). If these re equl then their common vlue is clled the integrl: f (x) dx D. DeTurck Mth C: Integrl/functions 5 / 32

6 Proving existence regulr prtitions To show tht the integrl exists, it is sufficient to find, for ny ε > 0, prtition P such tht U(f, P) < L(f, P) + ε. For monotonic functions, we cn do this by using sufficiently fine regulr prtitions these re prtitions hving the x i s evenly spced (so x i x i 1 = b for ll i = 1,..., n). n Theorem If f : [, b] R is monotonic, then f (x) dx exists. Proof: Given ε > 0, choose n so lrge tht the regulr prtition P f (b) f () with n steps, will hve U(f, P) L(f, P) = < ε. n D. DeTurck Mth C: Integrl/functions 6 / 32

7 Functions of bounded vrition Proposition (Greter generlity) If f (x) cn be written s the sum of two monotonic functions p(x) nd q(x) with p incresing nd q decresing on [, b] (such function is clled function of bounded vrition), then f (x) dx exists nd is equl to p(x) dx + q(x) dx. The set of functions of bounded vrition on closed bounded intervl is ctully quite generl, so even though we strted out with somewht restrictive definition we hve creted pretty powerful form of the integrl. D. DeTurck Mth C: Integrl/functions 7 / 32

8 Bsic properties Becuse the upper nd lower sums hve these properties, it follows tht the integrl does: 1 Linerity: αf (x) + βg(x) dx = α f (x) dx + β g(x) dx for constnts α, β nd f nd g re functions of bounded vrition on [, b]. 2 Monotonicity: If f (x) g(x) for ll x [, b] then b f (x) dx b g(x) dx. 3 If f (x) is of bounded vrition on [, b], then so is f (x) nd f (x) dx f (x) dx. D. DeTurck Mth C: Integrl/functions 8 / 32

9 More bsic properties 4 If < b < c then ˆ c f (x) dx = ˆ c f (x) dx + f (x) dx. b 5 If m inf{f (x) x [, b]} nd M sup{f (x) x [, b]} then (b )m f (x) dx (b )M. 6 Men vlue theorem for integrls: If f is continuous nd of bounded vrition on [, b] then there is c with < c < b such tht f (c) = 1 f (x) dx. b D. DeTurck Mth C: Integrl/functions 9 / 32

10 Fundmentl Theorem integrls of derivtives Fundmentl theorem of clculus I: Integrls of derivtives Let F (x) be differentible function on [, b] with derivtive F (x), nd suppose F is function of bounded vrition on [, b]. Then F (x) dx = F (b) F (). We cn prove this for functions with monotonic derivtives nd then use linerity. If F (x) is monotoniclly incresing, then for ny intervl (x i 1, x i ) prtition P we hve F (x i 1 ) F (x i) F (x i 1 ) x i x i 1 F (x i ) by the men vlue theorem (for derivtives). D. DeTurck Mth C: Integrl/functions 10 / 32

11 Fundmentl Theorem proof conclusion But then L(f, P) = n F (x i 1 )(x i x i 1 ) i=1 n F (x i ) F (x i 1 ) i=1 n F (x i )(x i x i 1 ) = U(f, P). i=1 But the middle sum telescopes to F (x n ) F (x 0 ) = F (b) F (). Now F (b) F () is trpped between sup L(f, P) nd inf U(f, P), both of which re equl to the integrl P f (x) dx. P D. DeTurck Mth C: Integrl/functions 11 / 32

12 Second fundmentl theorem derivtives of integrls Fundmentl theorem of clculus II: Derivtives of integrls Let f (x) be continuous function of bounded vrition on [, b]. Define the function F (x) vi F (x) = ˆ x f (t) dt. Then F is differentible nd F (x) = f (x). First, if h > 0 we hve F (x + h) F (x) = ˆ x+h x f (t) dt = hf (c) for some c between x nd x + h by properties 4 nd 6 (men vlue theorem for integrls). D. DeTurck Mth C: Integrl/functions 12 / 32

13 Second fundmentl theorem proof conclusion Therefore the difference quotient F (x + h) F (x) h = f (c) where x < c < x + h But since f is continuous, f (c) f (x) s h 0 +. ˆ x For h < 0 we use tht F (x + h) F (x) = x+h f (t) dt = hf (c) for some c between x + h nd x, nd the proof goes through s for the h > 0 cse. D. DeTurck Mth C: Integrl/functions 13 / 32

14 Using FTC I to compute integrls Of course, we cn use the first FTC to clculte integrls, once we hve nti-derivtive formuls obtined by turning round derivtive formuls. Right now, though, we don t hve much more thn derivtives of rtionl powers of x. So we hve tht since if r is rtionl number, then d dx (x r ) = rx r 1 x r 1 dx = 1 r (br r ) provided r 0. Of course we usully replce r by r + 1 nd sy x r dx = 1 r + 1 (br+1 r+1 ) provided r 1. D. DeTurck Mth C: Integrl/functions 14 / 32

15 Wht bout r = 1? 1 So now we hve question: wht is dx? Becuse x f (x) = 1/x is monotoniclly decresing on ny intervl [, b] where > 0 the integrl should exist. So, provisionlly, let s define the function L(x) vi: L(x) = ˆ x 1 1 t dt nd see if we cn uncover some of its properties. To begin, we know only tht L(1) = 0 nd L (x) = 1 x. D. DeTurck Mth C: Integrl/functions 15 / 32

16 Multiplictive property of L(x) Consider the function L(x) for positive constnt. By wht we know bout L nd the chin rule we hve d dl L(x) = dx dx = x x = 1 x. So the derivtive of L(x) is the sme s the derivtive of L(x), therefore the two differ by constnt. Wht constnt? Well, putting x = 1 gives tht L(x) t 1 is L(). So L(x) = L() + L(x). Look fmilir? D. DeTurck Mth C: Integrl/functions 16 / 32

17 Power property of L How bout L(x r ) if r Q? we hve d dx L(x r ) = 1 d x r dx x r rx r 1 = x r = r x = r dl(x) dx. So the derivtive of L(x r ) is r times the derivtive of L(x). Moreover, t x = 1, we hve L(x r ) x=1 = 0. This shows tht L(x r ) = rl(x). Agin, look fmilir? So fr, we hve defined L(x) only for x 1. But we could define it for 0 < x < 1 either by sying tht L(x) = L(1/x) or else by ˆ 1 1 sying tht L(x) = dx. It s n esy exercise to show tht x x these definitions gree. Likewise, if we define L(x) = L( x) for x < 0 we will hve L (x) = 1/x there s well. D. DeTurck Mth C: Integrl/functions 17 / 32

18 The rnge of ln(x) We hve defined the function L(x) s the integrl of 1/x with L(1) = 0 (nd L( 1) = 0). So L(x) grees with the nturl logrithm function ln(x). Since 1 1 x 1 2 for 1 < x < 2, we hve tht 1 > ln(2) > 1 2. Therefore ln(2 n ) > n 2 which shows tht ln(x) s x. Likewise, ln(1/2) < 1/2, so ln(1/2 n ) < n/2, showing tht ln(x) s x 0 +. Therefore the rnge of ln(x) is ll of R, nd ln(x) is strictly monotoniclly incresing function, since its derivtive is positive. Therefore it hs n inverse function tht will mp R to (0, ). Let s cll the inverse function E(x) nd study its properties. D. DeTurck Mth C: Integrl/functions 18 / 32

19 The inverse of ln x nd rbitrry powers The function E(x) is defined by E(ln(x)) = x for x > 0 nd ln(e(x)) = x for ll x R. Becuse ln(1) = 0 we hve E(0) = 1. And since ln(e()e(b)) = ln(e()) + ln(e(b)) = + b, we hve E( + b) = E()E(b). Likewise, since for R nd r Q, we hve E(r) = E() r. ln(e() r ) = r ln(e()) = r, Therefore E(x) behves like rising some number to the xth power, nd it gives us wy to define x y for ll rel y, nmely x y = E(y ln x). D. DeTurck Mth C: Integrl/functions 19 / 32

20 The number e We hve E(x) is like rising some number to the xth power. Wht number? Well, tht would be E(1), which we ll cll e. So e is the number such tht ln e = 1. Becuse ln 2 = ˆ 2 1 ˆ dt < 1 nd ln 4 = t 1 t dt > > 1 we know tht 2 < e < 4. We ll get better estimtes lter. But for now, we ll write e x for E(x). Derivtive: If y = e x then x = ln y so 1 = 1 y in other words d dx ex = e x. dy dy, nd so dx dx = y, D. DeTurck Mth C: Integrl/functions 20 / 32

21 An importnt differentil eqution So e x is solution of the differentil eqution y = y with initil vlue y(0) = 1. More generlly Importnt! The unique solution of the initil-vlue problem y = ky, y(0) = C is y = Ce kx. It is esy to check tht Ce kx stisfies the differentil eqution nd the initil condition. If f (x) were nother solution, then f = kf nd f (0) = C nd we could clculte: d dx ( ) f (x) Ce kx = 1C (e kx f (x) ke kx f (x)) = e kx (kf kf ) = 0 C so e kx f (x) is constnt. Wht constnt? Evlute t zero to get tht it s 1, so f (x) = Ce kx fter ll. D. DeTurck Mth C: Integrl/functions 21 / 32

22 Better estimtes of e Now tht we know tht e x stisfies y = y, we know tht ll the derivtives of e x re e x. In prticulr, ll of them re equl to 1 t x = 0. Therefore, by the Tylor estimte: e = e 1 = ! + 1 2! + 1 3! n! + M (n + 1)! where M is the vlue of the (n + 1)st derivtive of e x evluted somewhere between 0 nd 1. From our erlier primitive estimte we know tht 1 < M < 4. So for ny n > 0 we cn write 1 (n + 1)! < e [ ! + 1 2! + 1 3! n! ] < 4 (n + 1)! D. DeTurck Mth C: Integrl/functions 22 / 32

23 e is irrtionl Use the lst inequlity to show tht e is not rtionl number. If it were, then e = p/q for some positive integers p nd q nd we would hve 1 (n + 1)! < p q Multiply by n! nd get 1 (n + 1) < n!p q n! [ ! + 1 2! + 1 3! ] < n! [ ! + 1 2! + 1 3! ] < n! 4 (n + 1)! 4 (n + 1) If n q then the number in the middle is n integer, but if n > 4 then we ve trpped n integer between two positive rtionl numbers both of which re less thn 1. A contrdiction! D. DeTurck Mth C: Integrl/functions 23 / 32

24 Deciml pproximtion of e Strt gin from the Tylor estimte e = e 1 = ! + 1 2! + 1 3! n! + where we know tht 1 < M < 4. If we evlute this for n = 19 we get tht M (n + 1)! < e < Probbly close enough for most purposes. D. DeTurck Mth C: Integrl/functions 24 / 32

25 Trig? How should we pproch trigonometric functions? Let s strt with the function ˆ x 1 A(x) = 1 + t 2 dt Clerly A(x) is n incresing function nd since A(x) < ˆ x 1 1 t 2 dt = 2 1 x, we hve tht A(x) is defined for ll x R (we hve A(x) is n odd function of x) nd A(x) < 2 for ll x. Since A(x) is incresing nd bounded bove, it must pproch limit s x. Let s cll this limit π/2: Definition of π π = 2 lim x A(x). D. DeTurck Mth C: Integrl/functions 25 / 32

26 Inverse functions Since A(x) is monotoniclly incresing function from R to the intervl ( 1 2 π, 1 2π), it hs n inverse function tht we will cll T (x) from ( 1 2 π, 1 2π) to R. Since d dx A(x) = 1, if x = A(y) (or y = T (x)) we hve 1 + x 2 1 = y 2 dy dx Therefore d dx T (x) = 1 + T (x)2. From our definition of π/2, we see tht lim T (x) = x π/2 D. DeTurck Mth C: Integrl/functions 26 / 32

27 S(x) nd C(x) Next. let S(x) nd C(x) be the functions stisfying S(x) = T (x) 1 + T (x) 2 nd C(x) = T (x) 2. It s immeditely pprent tht S(x) 2 + C(x) 2 = 1. Moreover d dx S(x) = = 1 + T 2 T T TT 1+T T 2 = (1 + T 2 ) 2 T 2 (1 + T 2 ) (1 + T 2 ) 3/2 1 (1 + T 2 = C(x) ) 1/2 nd likewise d C(x) = S(x). dx D. DeTurck Mth C: Integrl/functions 27 / 32

28 The differentil eqution for S(x) nd C(x) Both S(x) nd C(x) stisfy the second-order DE: y + y = 0. Additionlly S(0) = 0 nd S (0) = 1 nd C(0) = 1 nd C (0) = 0. A new kind of initil-vlue problem! Proposition There is t most one solution of the initil-vlue problem y + y = 0 y(0) = y (0) = b. If there were two, cll them y nd z, then their difference u = y z would stisfy u + u = 0 with u(0) = u (0) = 0. But then d dx (u2 + u 2 ) = 2uu + 2u u = 2u (u + u ) = 0, so this quntity is constnt, nd in fct is zero becuse it is zero when x = 0. But tht mens u 2 = 0, so u = 0 nd so y = z. D. DeTurck Mth C: Integrl/functions 28 / 32

29 Extensions, periodicity At this point, becuse T (x) is defined only for 1 2 π < x < 1 2π, the functions S(x) nd C(x) hve these sme restrictions. However, since T (x) s x 1 2 π, you cn see tht S(x) 1 nd C(x) 0 s x 1 2 π. It follows tht C (x) 1 nd S (x) 0 there, nd becuse we hve tht S(π x) nd C(π x) re solutions of y + y = 0, we see we cn extend S nd C to be defined up to π. Keep repeting this trick to get tht S nd C re defined for ll x nd re periodic with period 2π. D. DeTurck Mth C: Integrl/functions 29 / 32

30 More trig identities Addition formuls: Wht is sin( + b)? Well, the function y = sin( + x) stisfies y + y = 0 nd y(0) = sin nd y (0) = cos. Therefore we hve y = sin( + x) = sin cos x + cos sin x Get ddition formuls for other trig functions in similr wy. D. DeTurck Mth C: Integrl/functions 30 / 32

31 Integrtion by substitution We puse to do the proof of fmilir integrtion technique, which is the chin rule in reverse: Integrtion by subsitution Suppose f : [, b] R nd g : [c, d] R re of bounded vrition, nd tht the derivtive of g exists nd is of bounded vrition on (c, d). Further, ssume tht the imge of g : (c, d) R is contined in the intervl (, b). Then ˆ d c f (g(x))g (x) dx = Proof: Apply FTC to the function H(x) = ˆ x c f (g(t)) dt ˆ g(d) g(c) ˆ g(x) g(c) f (x) dx. f (t) dt. D. DeTurck Mth C: Integrl/functions 31 / 32

32 Wht is π? We ve defined π/2 to be the first positive zero of cos x, but now we cn give it geometric significnce. To do so, we ll use the substitution g(x) = sin x on the following integrl: ˆ x 2 dx = ˆ π/2 = = 0 ˆ π/2 0 1 sin 2 θ cos θ dθ cos 2 θ dθ = ( ) θ sin 2θ π/ ˆ π/2 0 = π 4 1 (1 + cos 2θ) dθ 2 This shows tht π is the re of the unit circle. At lst. D. DeTurck Mth C: Integrl/functions 32 / 32

### The 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

### Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when

### Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

### 7.2 Riemann Integrable Functions

7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous

### 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)

### Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivtives/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

### Main topics for the First Midterm

Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

### Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

### 1 The Riemann Integral

The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

### 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

### LECTURE. INTEGRATION AND ANTIDERIVATIVE.

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

### UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

### Math 554 Integration

Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

### f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the

### Polynomial 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

### Lecture 1: Introduction to integration theory and bounded variation

Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You

### Euler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )

Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s

### Chapter 6. Riemann Integral

Introduction to Riemnn integrl Chpter 6. Riemnn Integrl Won-Kwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl

### Chapter 6 Notes, Larson/Hostetler 3e

Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

### The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

### Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

### 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]

### Lecture 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

### Topics 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.

### 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

### Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

### ODE: Existence and Uniqueness of a Solution

Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =

### Definite integral. Mathematics FRDIS MENDELU

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

### Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015

Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n

### Math Solutions to homework 1

Mth 75 - Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht

### 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

### Lecture 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

### 11 An introduction to Riemann Integration

11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in

### 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

### Math 61CM - Solutions to homework 9

Mth 61CM - Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ

### 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

### 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

### Math& 152 Section Integration by Parts

Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

### Overview of Calculus I

Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

### Review of Riemann Integral

1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

### Disclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.

Disclimer: This is ment to help you strt studying. It is not necessrily complete list of everything you need to know. The MTH 33 finl exm minly consists of stndrd response questions where students must

### 1 Probability Density Functions

Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our

### Principles of Real Analysis I Fall VI. Riemann Integration

21-355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will

### Best Approximation. Chapter The General Case

Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given

### UNIFORM 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,

### 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

### MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

### Properties of the Riemann Integral

Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2

### Review of basic calculus

Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

### MATH , Calculus 2, Fall 2018

MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly

### 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

### IMPORTANT THEOREMS CHEAT SHEET

IMPORTANT THEOREMS CHEAT SHEET BY DOUGLAS DANE Howdy, I m Bronson s dog Dougls. Bronson is still complining bout the textbook so I thought if I kept list of the importnt results for you, he might stop.

### 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

### Review of Calculus, cont d

Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

### 63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

### Riemann 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

### 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

### Math 3B Final Review

Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:45-10:45m SH 1607 Mth Lb Hours: TR 1-2pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems

### 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

### Math 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

### Riemann 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

### APPROXIMATE INTEGRATION

APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

### Presentation Problems 5

Presenttion Problems 5 21-355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).

### 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.

### Math 100 Review Sheet

Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s

### 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

### The Riemann Integral

Deprtment of Mthemtics King Sud University 2017-2018 Tble of contents 1 Anti-derivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Anti-derivtive Function Definition Let f : I R be function

### Math 1B, lecture 4: Error bounds for numerical methods

Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the

### MA 124 January 18, Derivatives are. Integrals are.

MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

### A 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

### If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du

Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find nti-derivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible

### n 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

### 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

### 5.5 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 nti-derivtive is not esily recognizble, then we re in

### Z b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...

Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.

### Section 4.8. D v(t j 1 ) t. (4.8.1) j=1

Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions

### 1 Techniques of Integration

November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.

### MAT187H1F Lec0101 Burbulla

Chpter 6 Lecture Notes Review nd Two New Sections Sprint 17 Net Distnce nd Totl Distnce Trvelled Suppose s is the position of prticle t time t for t [, b]. Then v dt = s (t) dt = s(b) s(). s(b) s() is

### Stuff You Need to Know From Calculus

Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you

### 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

### 7. 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

### MATH 144: Business Calculus Final Review

MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives

### Chapters 4 & 5 Integrals & Applications

Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions

### Calculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties

Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ fuehr@mth.rwth-chen.de Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009 Overview 1 Motivtion:

### approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below

. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.

### 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

### 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

### Riemann is the Mann! (But Lebesgue may besgue to differ.)

Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >

### 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

### 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

### Goals: 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

### 7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series

7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts

### . Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =

Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos( - 1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin( - 1 ) = -π 2 6 2 6 Cn you do similr problems?

### 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

### 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

### f(a+h) f(a) x a h 0. This is the rate at which

M408S Concept Inventory smple nswers These questions re open-ended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnk-out-n-nswer problems! (There re plenty of those in the

### Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner

### 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