# Chapter 6 Notes, Larson/Hostetler 3e

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1 Contents 6. Antiderivtives nd the Rules of Integrtion Are nd the Definite Integrl Are Reimnn Sums nd Definite Integrls Reimnn Sums Properties of the definite integrls The Fundmentl Theorem of Clculus Integrtion by Substitution

2 6. Antiderivtives nd the Rules of Integrtion Consider the following derivtives: F () = F () = + 7 F () = F () = F () = F () = If we wnt to work bckwrds nd sk the question, Wht function hs derivtive F () = we cn see tht there re mny possible nswers. When we sk this question we re solving for n ntiderivtive. This is more often clled n integrl. So if f() =, then F () = + C is the ntiderivtive (or integrl) of f(). NOTE: Since the ntiderivtive is generl epression you MUST include the +C in every ntiderivtive. Nottion: We would like to be ble to epress these integrls (ntiderivtives) mthemticlly without hving to write out the words so we hve the integrl sign:. In the previous emple we would write d = + C A word bout nottion nd the use of d: You will remember tht we cn write the derivtive of y = f() mny equivlent wys: y = f () = dy d With this we cn get n ide of why the integrl hs the d: dy d = f () dy = f ()d If we consider dy to be the derivtive of y then the integrl (ntiderivtive) of dy is y. Similrly f () is the derivtive of f() then the integrl (ntiderivtive) of f () is f(). Now integrte: dy = f ()d y = f() + C

3 Formuls. Integrl of constt:. The Power Rule. Constnt Multiple Rule k d = k + C n d = n+ n + + C; k f()d = k f()d n R, n. The Sum Rule f() ± g()d = f()d ± g() d. Eponentil Function e d = e + C 6. Integrl of f() = d = ln + C Emple 6... Integrte the following:. d = d = d = d = / d = ( + + )d = ( + + e ) d = Initil Vlue Problems If you know something obut the eqution you cn solve for C. This piece of informtion is clled n initil vlue. Emple 6... f () = +, f() = 9

4 6. Are nd the Definite Integrl 6.. Are Consider the problem of finding the re under the curve on the function y = + over the domin [, ]. We cn pproimte this re by using fmilir shpe, the rectngle. If we divide the domin intervl into severl pieces, then drw rectngles hving the width of the pieces, nd the height of the curve, we cn get rough ide of the totl re. For emple suppose we divide the intervl [, ] into equl subintervls of length = b n, i.e., ech of width. The intervls re [,.], [.,.8], [.8,.], [.,.6], [.6,.]. The tble below shows the vlues obtined when y () is evluted t the corresponding points. y = Plotting these points yields the following grph. y y = +

5 If we find the minimum vlue in the subintervl, nd use this s our height for tht rectngle, we hve wht is known s n inscribed rectngle. See the grph below. y y = + Now ech of the bove rectngles hs the ect sme width, nmely /. For this function the height of ech rectngle is given by clculting the vlue of the function t the right hnd endpoint of ech subintervl. The re under the curve, cn then be pproimted by dding the res of ll the rectngles together. Notice tht when using the minimum vlues, i.e. using inscribed rectngles, we rrive t n estimte tht is lower thn the ctul re under the curve. Hence, this method results in wht is known s the lower sum or n underestimte. Let s clculte this estimte using the right endpoints R = F ( i ) i i= ( ) ( ) ( ) ( ) 6 8 = f + f + f + f + f () = [ ( ) ( ) ( ) ( ) ] 6 8 f + f + f + f + f () = [ ] = [6.] = 6.8

6 You cn lso clculte n estimte using the mimum vlue in the subintervl nd using it s the height of the rectngles. These rectngles re known s circumscribed rectngles. The resulting re pproimtion will be greter thn the re under the curve. Consequently, we cll this type of sum n upper sum or n oversetimte. y y = + Clculuting the sum with the left endpoints we get: L = F ( i ) i i= = f () ( ) ( ) ( ) ( ) f + f + f + f = [ ( ) ( ) ( ) ( )] 6 8 f () + f + f + f + f = [ ] = [.] = 8.8 From the two clcultions bove we cn conclude tht the re of the curve lies some where between the two pproimtions, i.e. 6.8 < re of region < 8.8 6

7 Another method tht cn yield better pproimtion is known s the midpoint rule. In the midpoint rule, you choose the vlue ectly in the middle of the subintervl to use in clculting the height of the rectngle; resulting in some rectngles being both inscribed nd circumscribed. y y = + Let s clculte the bove estimte: The Midpoint Sum. M = F ( i ) i= ( ) ( ) ( ) ( ) ( ) 7 9 = f + f + f + f + f = [ ( ) ( ) ( ) ( ) ( )] 7 9 f + f + f + f + f = [ ] = [8.] = 7.6 NOTE: The smller the subintervls, the better the pproimtion will be. This is becuse, the function s vlues re chnging less in the subintervl, i.e. the vlue of the function is firly constnt in ech subintervl. Consequently, we re not pproimting by such rough mount ech time. For emple, here is the sme region divided into rectngle insted of. Note tht the error is minute compred with the previous work. y y = + Ech of the bove processes (lower sum, upper sum, midpoint sum) re just pproimtions. They re not ect. 7

8 Emple 6... Let f() = nd compute the upper nd lower sums of f over the intervl [, ], using. two subintervls of equl length (n = ). five subintervls of equl length (n = ) 8

9 6. Reimnn Sums nd Definite Integrls 6.. Reimnn Sums Definition 6.. Let f() be continuous function defined on [, b] where [, b] is prtitioned into n subintervls ech of width = b, then the definite integrl of f from to b is denoted by n b f()d = lim n [f( ) + f( ) + + f( n ) ], where = b n, i = + i. Emple 6... Set up definite integrl tht yields the re of the region shown for f() = y y = 6 Are = 6 d Emple 6... Set up definite integrl tht yields the re of the region f(y) = y y Are = y dy 9

10 6.. Properties of the definite integrls. c. e. f. b b f()d = b. cd = c(b ); c = constnt function. d. c f() ± g()d = f()d + b b f()d = f()d ± b c b f()d g() d b f()d = b b c f()d = c b f()d = f()d Emple 6... Given tht evlute the following: f()d = 7, f()d =, g()d =, nd g()d = 8, ) f()d = ) f()d = ) 6g()d = ) [f() + g()]d = ) d =

11 6. The Fundmentl Theorem of Clculus Fundmentl Theorem : If f() is continuous on [, b], then b f()d = F (b) F () where F () is ny ntiderivtive of f() on [, b], tht is, function such tht F () = f(). Emple 6... Evlute the following integrls ) ( + )d = ) (y + e y )dy = ) (8 )d = ) t dt =

12 Emple 6... Find the re of the region under the grph of the function f on the intervl [, b].. f() = + ; [, ]. f() = ; [, ]

13 6. Integrtion by Substitution Recll: The Chin Rule Suppose we wnt to find the derivtive of y = f() = ( + 7). To do this we need to use the chin rule becuse we hve n inside function u = + 7 nd n outside function f(u) = u. dy d = ( + 7) 9 () = ( + 7) 9 () When we integrte we cn use the chin rule in reverse. ( + 7) 9 ()d = ( + 7) + C The trick is to recognize the integrl. To do this we use trick known s u - substitution. We will solve the integrl gin but this time without knowing the nswer. ( + 7) 9 ()d Here we hve function inside the prentheses nd we will cll tht function u: u = + 7 Emple 6... ( + ) d

14 Emple 6... ( ) 9 d Emple 6... e d

15 Emple 6... (e + e ) d Emple d

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### The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O 1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

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### Chapter 0. What is the Lebesgue integral about? Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

### 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:

### 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

### 1 Functions Defined in Terms of Integrals November 5, 8 MAT86 Week 3 Justin Ko Functions Defined in Terms of Integrls Integrls llow us to define new functions in terms of the bsic functions introduced in Week. Given continuous function f(), consider

### a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1 Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the

### 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

### 5.1 How do we Measure Distance Traveled given Velocity? Student Notes . How do we Mesure Distnce Trveled given Velocity? Student Notes EX ) The tle contins velocities of moving cr in ft/sec for time t in seconds: time (sec) 3 velocity (ft/sec) 3 A) Lel the x-xis & y-xis

### Calculus - Activity 1 Rate of change of a function at a point. Nme: Clss: p 77 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus - Activit Rte of chnge of function t point. ) Strt Mths Helper Plus, then lod the file: Clculus

### APPM 1360 Exam 2 Spring 2016 APPM 6 Em Spring 6. 8 pts, 7 pts ech For ech of the following prts, let f + nd g 4. For prts, b, nd c, set up, but do not evlute, the integrl needed to find the requested informtion. The volume of the

### 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

### 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  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)

### ROB EBY Blinn College Mathematics Department ROB EBY Blinn College Mthemtics Deprtment Mthemtics Deprtment 5.1, 5.2 Are, Definite Integrls MATH 2413 Rob Eby-Fll 26 Weknowthtwhengiventhedistncefunction, wecnfindthevelocitytnypointbyfindingthederivtiveorinstntneous

### 1 The fundamental theorems of calculus. The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Tody we provide the connection

### Lecture 20: Numerical Integration III cs4: introduction to numericl nlysis /8/0 Lecture 0: Numericl Integrtion III Instructor: Professor Amos Ron Scribes: Mrk Cowlishw, Yunpeng Li, Nthnel Fillmore For the lst few lectures we hve discussed

### Bob Brown Math 251 Calculus 1 Chapter 5, Section 4 1 CCBC Dundalk Bo Brown Mth Clculus Chpter, Section CCBC Dundlk The Fundmentl Theorem of Clculus Informlly, the Fundmentl Theorem of Clculus (FTC) sttes tht differentition nd definite integrtion re inverse opertions

### Chapter 8: Methods of Integration Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln

### 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

### 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

### 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

### 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.

### 13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce

### Objectives. Materials Techer Notes Activity 17 Fundmentl Theorem of Clculus Objectives Explore the connections between n ccumultion function, one defined by definite integrl, nd the integrnd Discover tht the derivtive of the

### 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 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 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