7. Indefinite Integrals


 Bernard Kelley
 3 years ago
 Views:
Transcription
1 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 function F stisfying F = f. Then, f f = F() F(). Such function F is clled primitive function (%/&$8 %*78&5), n ntiderivtive or n indefinite integrl (.*&2/! 9#)*!) of f. This chpter presents techniques for clculting indefinite integrls. Given function f, we denote its indefinite integrl y f. To denote tht we evlute primitive function of f t we write f. A more stndrd nottion is f (t).
2 00 Indefinite Integrls The stndrd nottion is useful when rther thn putting the nme of function in the integrnd we write its functionl form. For emple, for f () = 2 we my either write f or t 2. Recll tht n indefinite integrl is only determined up to n dditive constnt. Thus, when we write n identity like f = g we men tht ny primitive of f is lso primitive of g, nd when we write n identity like f = h+ g we men tht ny primitive of g plus h is primitive of f. Comment 7. It is not generlly true tht if F nd G re primitive functions of f then they differ y constnt. Consider the function f () = 2 R{0}. Then F() = nd G() = + < 0 +5 > 0 re oth primitive functions of f. Generlly, if function is defined on domin consisting of disconnected components, its indefinite integrl is unique up to different dditive constnt in ech connected component. The most elementry method for finding indefinite integrls is to differentite functions nd crete tle of functions nd their derivtive. Then, we cn red this tle ckwrd. This is not very prcticl men, ut still, we hve t the outset few functions whose indefinite integrls we know. Emple 7. Since we know tht sin = cos, it follows tht cost= sin. It is lso true tht cost= 5+sin, ut it is certinly not true tht sin = 5+sin. Recll tht the equlity relly mens tht sin elongs to the set of primitive functions of cos. In fct, we should hve written sin cos, ut we won t e doing it.
3 7.2 Elementry functions 0 We present elow tle of elementry integrls", which re otined y differentition: f e sin cos tn f() e cos sin cos 2 ln sin 2 cos 2 tn (+ 2 ) ln(sin) cot ln(cos) tn sinh cosh cosh sinh tnh tnh 2 sinh 2 + cosh 2 tnh ( 2 ) This tle is n strting point for clculting indefinite integrls of mny more functions. 7.2 Elementry functions The functions for which we hve nmes powers, trigonometric functions, hyperolic functions, the eponentil nd the logrithm, roots, inverse trigonometric functions nd inverse hyperolic functions re clled commonly elementry functions. So is ny function tht cn e otined y dding, multiplying, dividing nd composing elementry functions (note the recursive definitions). Emple 7.2 The function is n elementry function. f () = sin 2 e +sin (tnh 3 ) 3 5 +ln(tn ()) By the lws of derivtion we know how to differentite every elementry function, getting gin n elementry function. Note, however, tht there re mny more functions tht re not elementry. We simply don t hve nmes for them (sometimes we do). It turns out tht mny elementry functions don t hve elementry
4 02 Indefinite Integrls indefinite integrls. A theorem y Liouville in 835 provided the first proof tht nonelementry ntiderivtives of elementry functions eist. Emple 7.3 The following elementry functions don t hve elementry ntiderivtives: f () = 4 f () = ln(ln) f () = sin nd f () = e 2. It should e cler, however, tht these functions hve ntiderivtives. We simply cn t epress them s elementry functions. In fct, the function e t2 is sufficiently importnt in sttistics tht it hs nme the error function. 7.3 Integrtion y prts (.*8(" %*79#)*!) Recll the Leiniz rule, (uv) = u v+uv. Since uv is primitive function of (uv), we hve uv = u v+ uv, or u v = uv uv (7.) which in more stndrd nottion reds u (t)v(t) = u()v() u(t)v (t). This innocentlooking identity is useful strting point for evluting indefinite integrls. Emple 7.4 Consider the indefinite integrl I() = lnt. We evlute it s follows: set u() = nd v() = ln. Then, I() = lnt = ln u (t) v(t) u() v() t t u(t) v (t) = ln.
5 7.3 Integrtion y prts (.*8(" %*79#)*!) 03 Note tht we could hve s well tken u() = + 5, in which cse we would hve gotten I() = lnt = ( +5) ln u() v() u (t) v(t) (t +5) u(t) t v (t) = ln, so s epected, it doesn t mtter which ntiderivtive of u we tke. Emple 7.5 For n N consider the indefinite integrls I n () = t n lnt. Here we ll tke u () = n nd v() = ln, hence I n () = t n lnt= n+ n+ ln t n+ n+ n+ n+ = ln t n+ (n+) 2. Emple 7.6 For n N consider the indefinite integrls I n () = t n e t. Tking e = u () nd n = v() we get I n () = n e n t n e t = n e ni n (). This is recurrence reltion (%#*2 2(*). For n = 0, Then, It is esy to see tht I = e e = ( )e I 0 () = e. I 2 = 2 e 2( )e = ( )e I 3 = 3 e 3( )e = ( )e = I n () = n! n n! n (n )! + n 2 (n 2)! +( )n e = n!e n ( ) n k k. k! k=0 Emple 7.7 Consider the sequence of indefinite integrls I n () = sin n t.
6 04 Indefinite Integrls We tret it s follows: I n () = sin n t sint = sin n cos +(n ) = sin n cos +(n ) sin n 2 t cos 2 td sin n 2 t( sin 2 t)d Hence, we get recurrence reltion This time we need oth I 0 nd I : = sin n cos +(n )I n 2 () (n )I n (). I n () = n n I n 2() n sinn cos. I 0 () = nd I () = cos. Then, I 2 () = 2 I 0() 2 sin cos I 3 () = 2 3 I () 3 sin2 cos I 4 () = 3 4 I 2() 4 sin3 cos, nd so on. 7.4 Sustitution method (%"7%% ;)*:) Recll the chin rule, from which follows tht (F g) = (F g)g F g = (F g)g For F = f (t), g() f = ( f g)g (7.2) nd it more stndrd nottion, f (g(t))g (t) = g() f (t). As we will see, this identity is useful for finding mny indefinite integrls.
7 7.4 Sustitution method (%"7%% ;)*:) Sustitution of liner functions Suppose tht we know primitive function of f nd we wnt to evlute I() = f (t +), where is constnt. Set g() = +, nd use the sustitution formul (7.2), I() = f (g(t))g (t) = g() f (t) = + f (t). Comment 7.2 The physicists" wy of doing the sme procedure goes s follows: they define, sy, u = +, nd then write Then, du = du d = d. d f ( +)d = f (u)du, nd it is understood tht the right hnd side is evluted t u = +. Emple 7.8 cos(t +) = sin( +). Emple 7.9 Find I() = 2 t 2. First note tht I() = (t) 2. Setting g() =, we use the sustitution formul (7.2) to get I() = g 2 (t) = g (t) g 2 (t) = g() t 2 = sin (g()) = sin.
8 06 Indefinite Integrls Emple 7.0 As similr emple, consider I() = 8+2t t 2. The trick here is to first complete the squre under the root, I() = = 9 (t ) 2 3. ((t )3) 2 Set then Then, I() = 3 g 2 (t) = g() = 3. g (t) g 2 (t) = g() = t 2 sin Other emples Emple 7. Consider gin the integrls I n () = sin n t, this time only for odd vlues of n = 2m +. We ve lredy solved these integrls using integrtion y prts. This time we will use the method of sustitution. Strt with I n () = sint(sin 2 t) m d = sint( cos 2 t) m d. Setting we get f () = ( 2 ) m nd g() = cos, I n () = f (g(t))g cos (t) = ( t 2 ) m. Tke for emple n = 3. Then, I 3 () = cos cos3 3 Emple 7.2 Consider integrls of the form I() = r(t) t 2,
9 7.5 Integrls of rtionl functions 07 where r() is rtionl function. Setting g() = sin or equivlently = sing(), we get I() = r(sin(g(t)))g (t) = sin r(sin(t)). Tke for emple r() = 2. Then, Thus, r(sin(t)) = sin 2 t= 2 ( cos2t) = 2 4 sin2 = ( sin cos). 2 t 2 t 2 = 2 sin Integrls of rtionl functions Consider integrls of the form p(t) q(t), where p nd q re polynomils ( function tht is the rtio of two polynomils is clled rtionl function). If deg p degq then we cn divide the polynomils to get p() s() = r()+ q() q(), where r nd s re polynomils, nd degs < degq. Since we know how to integrte polynomils, it remins to lern how to integrte rtionl functions with deg p < degq Liner denomintors When q() is liner function, we end with n integrl of the form = ln + t + Emple 7.3 Consider the following indefinite integrl I() = t 3 t +5.
10 08 Indefinite Integrls Since it follows tht 3 +5 = ( +5) = ( +5)2 5( +5) = ( +5)2 5( +5)+25( +5) = , I() = ln Qudrtic denomintors When q() is qudrtic function there re three possiilities:. q hs two distinct relvlued roots, in which cse we otin n integrl of the form t + (t c)(t d), 2. The two roots of q coincide, in which cse we otin n integrl of the form t + (t c) 2 3. q is irreducile, i.e., hs two complevlued root, then the integrl cn e rought to the form t + (t c) 2 +d 2 Cse : Two distinct relvlued roots In this cse we split the the integrnd s follows: t + (t c)(t d) = A t c + Mtching coefficients we otin B t d A(t d)+b(t c) =. (t c)(t d) A+B = nd Ad Bc =. This is liner system tht lwys hs solution provided c d: A = c+ c d nd B = d + d c
11 7.5 Integrls of rtionl functions 09 Then, (t +) (t c)(t d) = c+ d + ln c+ ln d c d d c Emple 7.4 Consider the cse = 0, =, c = nd d = 2, (t +)(t +2) = ln + ln +2 Cse 2: Two equl roots We split the integrnd s follows, t + (t c)+(+c) = (t c) 2 (t c) 2. Thus, nmely, (t +) (t c) 2 = t c +(c+) (t +) (t c) 2 = ln c c+ c (t c) 2, Emple 7.5 2t +5 t 2 +4t +4 = 2t +5 = 2 ln +2 (t +2) Cse 3: Two complevlued roots Remins the cse where q is irreducile (with rel numers). In this cse we cn ring q y squre completion to the form q(t) = (t c) 2 +d 2. Now, Hence, t + (t c)+(c+) (t c) 2 = +d2 (t c) 2 +d 2. (t +) (t c) 2 +d 2 = (t c) (t c) 2 +d 2 +(c+) ( c) 2 +d 2. The first integrl equls 2 ln(t c)2 +d 2,
12 0 Indefinite Integrls wheres the second integrl equls, using the method of sustitution To conclude, c+ d 2 [(t c)d] 2 + = c+ d tn c d. (t +) (t c) 2 +d 2 = 2 ln( c)2 +d 2 + c+ d tn c d Denomintors of higher degree Let s consider n emple. Other follow similr route: Emple 7.6 (t +5) t 3 t 2 t + To do something we must e le to decompose the denomintor. In the present cse t 3 t 2 t + = (t +)(t ) 2. Since we hve simple root, ( ), nd doule root,, we look for decomposition, Mtching the numertors from which we get tht Then, t +5 t 3 t 2 t + = A t + B (t ) 2 + C t +. t +5 = A(t 2 )+B(t +)+C(t 2 2t +), A = C B 2C = nd A+B+C = 5. A = B = 3 nd C =, from which we conclude tht (t +5) 3 t 3 t 2 = ln +ln +. t Definite integrls We now turn to evlute definite integrls. By the NewtonLeiniz theorem it suffices to find primitive function F of f, in which cse f = F. In this section we will see how to pply the integrtion methods for the clcultion of definite integrls.
13 7.6 Definite integrls 7.6. Integrtion y prts Since it follows tht uv = (uv) = u v+ uv, u ()v()d = u()v() u()v ()d Emple 7.7 Clculte 6 5 ln Method of sustitution Let F = f. Since If follows tht i.e., F g = Emple 7.8 Clculte Letting g(t) = t 2, we hve I = We proceed y writing (F g) = f (g(t))g (t), f (g(t))g (t) = F g() g(), f (g(t))g (t) = I = g (t) (+g(t))(2+g(t)) = g(2) 2 g(0) otining A = B =. Hence I = g() g() t (+t 2 )(2+t 2 ). f (t) (+t)(2+t) = (+t)(2+t) = A +t + B 2+t = A(2+t)+B(+t), (+t)(2+t) 0 4 +t 0 4 (+t)(2+t). 2+t = 2 (ln5 ln ln6+ln2) = 2 ln 5 3.
14 2 Indefinite Integrls 7.7 Improper integrls Improper integrls (.**;*/!!.*9#)*!) re definite integrls in which something infinite" tkes plce: either the domin of integrtion is infinite or the integrnd diverges within the domin of integrtion. The first instnce comprises of integrls of the form f = lim f f lim y y f = lim f = lim Emple 7.9 Compre the two cses t 2 nd Note tht the Riemnn sums tht correspond to Dt = re the respectively convergent nd divergent series, nd f. + = p2 6 t =. The second instnce comprises the emples which we interpret s 0 lim 0 + Emple 7.20 Consider the integrl Since sin is ntisymmetric, for every nd t t nd lim sint. sint= 0. 0 t, 0 + It my seem therefore tht the integrl of sin over the whole line is zero, ut this is not the cse s sint= cosy cos does not hve limit s y nd y t.
Chapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More informationChapter 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
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationMA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations
LESSON 0 Chpter 7.2 Trigonometric Integrls. Bsic trig integrls you should know. sin = cos + C cos = sin + C sec 2 = tn + C sec tn = sec + C csc 2 = cot + C csc cot = csc + C MA 6200 Em 2 Study Guide, Fll
More informationRAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IITJEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties
M.Sc. (Mths), B.Ed, M.Phil (Mths) MATHEMATICS Mob. : 947084408 9546359990 M.Sc. (Mths), B.Ed, M.Phil (Mths) RAM RAJYA MORE, SIWAN XI th, XII th, TARGET IITJEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII
More informationSection 4: Integration ECO4112F 2011
Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic
More informationAP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals
AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationThe practical version
Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationAQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions
Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationMath& 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
More informationHOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016
HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationSYDE 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
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationOverview 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,
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More informationAppendix 3, Rises and runs, slopes and sums: tools from calculus
Appendi 3, Rises nd runs, slopes nd sums: tools from clculus Sometimes we will wnt to eplore how quntity chnges s condition is vried. Clculus ws invented to do just this. We certinly do not need the full
More information10 Vector Integral Calculus
Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve
More informationPolynomials 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
More information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationUnit #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
More informationSection 7.1 Integration by Substitution
Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Theorem Suppose f is continuous
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationMATH 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
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
More informationDefinition 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)
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More informationNow, given the derivative, can we find the function back? Can we antidifferenitate it?
Fundmentl Theorem of Clculus. Prt I Connection between integrtion nd differentition. Tody we will discuss reltionship between two mjor concepts of Clculus: integrtion nd differentition. We will show tht
More informationProperties 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
More informationFormulae For. Standard Formulae Of Integrals: x dx k, n 1. log. a dx a k. cosec x.cot xdx cosec. e dx e k. sec. ax dx ax k. 1 1 a x.
Forule For Stndrd Forule Of Integrls: u Integrl Clculus By OP Gupt [Indir Awrd Winner, +9965 35 48] A B C D n n k, n n log k k log e e k k E sin cos k F cos sin G tn log sec k OR log cos k H cot log sin
More information11 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
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationTotal Score Maximum
Lst Nme: Mth 8: Honours Clculus II Dr. J. Bowmn 9: : April 5, 7 Finl Em First Nme: Student ID: Question 4 5 6 7 Totl Score Mimum 6 4 8 9 4 No clcultors or formul sheets. Check tht you hve 6 pges.. Find
More information1 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
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More information63. 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 =
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationReview 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
More informationMATH1013 Tutorial 12. Indefinite Integrals
MATH Tutoril Indefinite Integrls The indefinite integrl f() d is to look for fmily of functions F () + C, where C is n rbitrry constnt, with the sme derivtive f(). Tble of Indefinite Integrls cf() d c
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationF (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
More informationReview on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.
Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5.  5.3) Remrks on the course. Slide Review: Sec. 5.5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description
More informationUnit 5. Integration techniques
18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationdifferent 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
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More information. Doubleangle 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?
More informationP 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
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationx = 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
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
More informationMath 3B Final Review
Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:4510:45m SH 1607 Mth Lb Hours: TR 12pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems
More informationFirst midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009
Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No
More informationTaylor Polynomial Inequalities
Tylor Polynomil Inequlities Ben Glin September 17, 24 Abstrct There re instnces where we my wish to pproximte the vlue of complicted function round given point by constructing simpler function such s polynomil
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More information1 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
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl WonKwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
More informationThe Evaluation Theorem
These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for inclss presenttion nd should not
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationPhysics 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
More information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a nonconstant 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
More informationCalculus AB. For a function f(x), the derivative would be f '(
lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationpadic Egyptian Fractions
padic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Setup 3 4 pgreedy Algorithm 5 5 pegyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 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
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors PreChpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More informationIf 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 ntiderivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible
More informationTopics Covered AP Calculus AB
Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationBernoulli Numbers Jeff Morton
Bernoulli Numbers Jeff Morton. We re interested in the opertor e t k d k t k, which is to sy k tk. Applying this to some function f E to get e t f d k k tk d k f f + d k k tk dk f, we note tht since f
More information378 Relations Solutions for Chapter 16. Section 16.1 Exercises. 3. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses on A.
378 Reltions 16.7 Solutions for Chpter 16 Section 16.1 Exercises 1. Let A = {0,1,2,3,4,5}. Write out the reltion R tht expresses > on A. Then illustrte it with digrm. 2 1 R = { (5,4),(5,3),(5,2),(5,1),(5,0),(4,3),(4,2),(4,1),
More informationChapter 1: Logarithmic functions and indices
Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4
More informationChapter 3 Single Random Variables and Probability Distributions (Part 2)
Chpter 3 Single Rndom Vriles nd Proilit Distriutions (Prt ) Contents Wht is Rndom Vrile? Proilit Distriution Functions Cumultive Distriution Function Proilit Densit Function Common Rndom Vriles nd their
More informationODE: 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() =
More information38 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 xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationIndefinite 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 information5.4. The Fundamental Theorem of Calculus. 356 Chapter 5: Integration. Mean Value Theorem for Definite Integrals
56 Chter 5: Integrtion 5.4 The Fundmentl Theorem of Clculus HISTORICA BIOGRAPHY Sir Isc Newton (64 77) In this section we resent the Fundmentl Theorem of Clculus, which is the centrl theorem of integrl
More information