Math 107H Topics for the first exam. csc 2 x dx = cot x + C csc x cotx dx = csc x + C tan x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C


 Justina Betty Bradley
 4 years ago
 Views:
Transcription
1 Integrtion Mth 07H Topics for the first exm Bsic list: x n dx = xn+ + C (provided n ) n + sin(kx) dx = cos(kx) + C k sec x dx = tnx + C sec x tnx dx = sec x + C /x dx = ln x + C cos(kx) dx = sin(kx) + C k csc x dx = cot x + C csc x cotx dx = csc x + C tn x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C dx x + = Arctn( x ) + c Bsic integrtion rules: for k=constnt, k = k sec x dx = ln sec x + tnx + C csc x dx = ln csc x + cotx + C dx = Arcsin( x x ) + c dx x x = Arcsec( x ) + c (f(x) ± g(x) dx = ± g(x) dx The Fundmentl Theorem of Clculus x f(t) dt = F(x) is function of x. F(x) = the re under grph of f, from to x. FTC : If f is cts, then F (x) = f(x) (F is n ntideriv of f!) Since ny two ntiderivtives differ by constnt, nd F(b) = b f(t) dt, we get FTC : If f is cts, nd F is n ntideriv of f, then b = F(b) F() = F(x) b Integrtion by substitution. The ide: reverse the chin rule! g(x) = u, then d dx f(g(x))= d dx f(u) = f (u) du dx, so f (u) du dx dx = f (u) du = f(u)+c f(g(x))g (x) dx ; set u = g(x), then du = g (x) dx, so f(g(x))g (x) dx = f(u) du, where u = g(x) Exmple: x(x 3) 4 dx ; set u = x 3, so du=x dx. Then x(x 3) 4 dx = (x 3) 4 x dx = u 4 du u=x 3 = u c u=x 3 = (x 3) 5 + c 0 The three most importnt points:. Mke sure tht you clculte (nd then set side) your du before doing step!. Mke sure everything gets chnged from x s to u s 3. Don t push x s through the integrl sign! They re not constnts! We cn use usubstitution directly with definite integrl, provided we remember tht b relly mens x=b x=, nd we remember to chnge ll x s to u s! x(+x ) 6 dx; set u = +x, du = x dx. when x =, u = ; when x =, u = 5; so x( + x ) 6 dx = 5 u 6 du =...
2 Bsic integrtion formuls (AKA dirty tricks) complete the squre x + bx + c = (x + rx) + c = (x + r/) + (c (r/) ) x + x + dx = (x + ) + dx use trig identities sin x + cos x =, tn x + = sec x, sin(x) = sinxcos x, tn x =sin x, etc. sec x sin x cos cos x dx = x dx cos x pull frctions prt; put frctions together! x + dx = x + x 3 dx x 3 do polynomil long division x 3 x dx = x + x x dx dd zero, multiply byone sec x(tnx + sec x) sec x dx = dx sec x + tnx Integrtion by prts Product rule: d(uv) = (du)v + u(dv) reverse: u dv = uv v du x cosx dx : set u=x, dv=cos x dx du=dx, v = sin x (or ny other ntiderivtive) So: x cos x = x sinx sin x dx specil cse: ; u = f(x), dv=dx = xf(x) xf (x) dx x Arcsin x dx = x Arcsin x x The bsic ide: integrte prt of the function ( prt tht you cn), differentite the rest. Gol: rech n integrl tht is nicer. x 3 lnx dx = (x 4 /4) lnx (x 4 /4)(/x) dx Trig substitution Ide: get rid of squre roots, by turning the stuff inside into perfect squre! x : set x = sin u. dx = cosu du, x = cos u x x dx = cos u sin u cosu du x=sin u + x : set x = tnu. dx = sec u du, + x = sec u sec u dx = (x + 4) 3/ ( sec u) du x= 3 tn u x : set x = sec u. dx = secutnu du, x = tnu
3 sec u tnu x x dx = sec u tnu du x=sec u Undoing the usubstitution : use right tringles! (Drw right tringle!) x = sinu, then ngle u hs opposite = x, hypotenuse =, so djcent = x. So cos u = ( x )/, tnu = x/ x, etc. Trig integrls: Wht trig substitution usully leds to! sin n x cos m x dx If n is odd, keep one sinx nd turn the others, in pirs, into cosx (using sin x = cos x), then do usubstitution u = cos x. If m is odd, reverse the roles of sin x nd cos x. If both re even, turn the sinx into cos x (in pirs) nd use the double ngle formul cos x = ( + cos(x)) This will convert cos m x into bunch of lower powers of cos(x); odd powers cn be delt with by substitution, even powers by nother ppliction of the ngle doubling formul! sec n x tn m sin m x x dx = cos n+m x dx If n is even, set two of them side nd convert the rest to tnx using sec x = tn x +, nd use u = tnx. If m is odd, set one ech of sec x, tnx side, convert the rest of the tnx to sec x using tn x = sec x, nd use u = sec x. If n is odd nd m is even, convert ll of the tnx to sec x (in pirs), leving bunch of powers of sec x. Then use the reduction formul: sec n x dx = n secn x tnx + n sec n x dx n At the end, rech sec x dx = tnx + C or sec x dx = ln sec x + tnx + C A little trick worth knowing: the substitution u = π x, since sin(π x) = cos x nd cos(π x) = sin x, will reverse the roles of sin x nd cos x, so will turn cot x into tnu nd csc x into sec u. So, for exmple, the integrl cos 4 x sin 7 x dx = csc 3 x cot 4 x dx, which our techniques don t cover, becomes sec 3 u tn 4 u du, which our techniques do cover. 3
4 Prtil frctions rtionl function = quotient of polynomils Ide: integrte by writing function s sum of simpler functions Procedure: f(x) = p(x)/q(x) (0): rrnge for degree(p)<degree(q); do long division if it isn t (): fctor q(x) into liner nd irreducible qudrtic fctors (): group common fctors together s powers (3): for ech group (x ) n dd together: (3b): for ech group (x + bx + c) n dd together: x + b x + + n (x ) n n x + b n (x + bx + c) n x + bx + c + + (4) set f(x) = sum of ll sums; solve for the undetermined coefficients put sum over common denomentor (=q(x)); set numertors equl. lwys works: multiply out, group common powers, set coeffs of the two polys equl x + 3 = (x ) + b(x ) = ( + b)x + ( b); = + b, 3 = b liner term (x ) n : set x =, will llow you to solve for coefficient if n, tke derivtives of both sides! set x=, gives nother coeff. Improper integrls x (x ) (x + ) = A x + Fund Thm of Clc: b B (x ) + Cx + D x + = A(x )(x + ) + B(x + ) + (Cx + D)(x ) (x ) (x + ) = F(b) F(), where F (x) = f(x) Problems: =, b = ; f blows up t or b or somewhere in between integrl is improper ; usul technique doesn t work. Solution to this: = lim b b b b (blow up t ) = lim r + (similrly for blowup t b (or both!)) b = lim s b s r = lim ɛ 0 + b b = lim = lim ɛ 0 + b ɛ r (blows up t c (b/w nd b)) = lim r c The integrl converges if (ll of the) limit(s) re finite Comprison: 0 f(x) g(x) for ll x; if if b g(x) dx converges, so does diverges, so does 4 +ɛ b b + lim s c + s g(x) dx
5 Applictions of integrtion Volume by slicing. To clculte volume, prroximte region by objects whose volume we cn clculte. Volume (volumes of cylinders ) = (re of bse)(height) = (re of crosssection) x i. So volume = right (re of cross section) dx Solids of revolution: disks nd wshers. Solid of revolution: tke region in the plne nd revolve it round n xis in the plne. tke crosssections perpendiculr to xis of revolution; crosssection = disk (re=πr ) or wsher (re=πr πr ) rotte round xxis: write r (nd R) s functions of x, integrte dx rotte round yxis: write r (nd R) s functions of y, integrte dy Otherwise, everything is s before: volume = right A(x) dx or volume = top region bottom rotte A(y) dy The sme is true if xis is prllel to x or y xis; r nd R just chnge (we dd constnt). Cylindricl shells. Different picture, sme volume! Solid of revolution; use cylinders centered on the xis of revolution. The intersection is cylinder, with re = (circumference)(height) = πrh volume = right (re of cylinder) dx or top bottom (re of cylinder) dy!) region revolve round verticl line: integrte dx revolve round horizontl line: integrte dy r h rotte 5
6 region in plne between y = 4x, y = x, revolved round yxis =0, right=4, r = x, h = (4x x ) volume = Arclength nd surfce re 4 0 πx(4x x ) dx rotte Arclength. Ide: pproximte curve by lots of short line segments; length of curve sum of lengths of line segments. Line segment between (c i, f(c i )) nd (c i+, f(c i+ )): + ( f(c i+) f(c i ) ) c i+ c (c i+ c i ) + (f (c i )) x i i right So length of curve = + (f (x)) dx The problem: integrting + (f (x))! Sometimes, + (f (x)) turns out to be perfect squre... Surfce re. Ide: find the re of surfce (of revolution) by pproximting the surfce by things whose re we cn figure out. Frustrum of cone! re of frustrum = π (f(c i+ ) + f(c i ) ) ( f(ci+ ) f(c i ) ) + (ci+ c i ) c i+ c i πf(c i + (f (c i )) x i. So re of surfce = right πf(x) + (f (x)) dx The problem: sme problem! But sometimes it s possible to do... for f(x) = r x, the thing to integrte simplifies to: πr! 6
Antiderivatives/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 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 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 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 informationMath 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
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 informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 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
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 information0.1 Chapters 1: Limits and continuity
1 REVIEW SHEET FOR CALCULUS 140 Some of the topics hve smple problems from previous finls indicted next to the hedings. 0.1 Chpters 1: Limits nd continuity Theorem 0.1.1 Sndwich Theorem(F 96 # 20, F 97
More informationReversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b
Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite
More informationf(a+h) f(a) x a h 0. This is the rate at which
M408S Concept Inventory smple nswers These questions re openended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnkoutnnswer problems! (There re plenty of those in the
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 informationFinal Review, Math 1860 Thomas Calculus Early Transcendentals, 12 ed
Finl Review, Mth 860 Thoms Clculus Erly Trnscendentls, 2 ed 6. Applictions of Integrtion: 5.6 (Review Section 5.6) Are between curves y = f(x) nd y = g(x), x b is f(x) g(x) dx nd similrly for x = f(y)
More informationa < 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
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 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 informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through
More informationThe Product Rule state that if f and g are differentiable functions, then
Chpter 6 Techniques of Integrtion 6. Integrtion by Prts Every differentition rule hs corresponding integrtion rule. For instnce, the Substitution Rule for integrtion corresponds to the Chin Rule for differentition.
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 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 informationTest 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher).
Test 3 Review Jiwen He Test 3 Test 3: Dec. 46 in CASA Mteril  Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec. 1417 in CASA You Might Be Interested
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 informationChapter 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 informationES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1  Thurs 28th Sept 17 Review of trigonometry and basic calculus
ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1  Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems
More informationTechniques of Integration
Chpter 8 Techniques of Integrtion 8. Integrtion by Prts Some Exmples of Integrtion Exmple 8... Use π/4 +cos4x. cos θ = +cosθ. Exmple 8... Find secx. The ide is to multiply secx+tnx both the numertor nd
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 informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.
Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos(  1 2 ) = rcsin( 1 2 ) = rcsin(  1 2 ) = Cn you do similr problems? Review of Bsic Concepts
More informationMath 118: Honours Calculus II Winter, 2005 List of Theorems. L(P, f) U(Q, f). f exists for each ǫ > 0 there exists a partition P of [a, b] such that
Mth 118: Honours Clculus II Winter, 2005 List of Theorems Lemm 5.1 (Prtition Refinement): If P nd Q re prtitions of [, b] such tht Q P, then L(P, f) L(Q, f) U(Q, f) U(P, f). Lemm 5.2 (Upper Sums Bound
More informationFinal Exam  Review MATH Spring 2017
Finl Exm  Review MATH 5  Spring 7 Chpter, 3, nd Sections 5.5.5, 5.7 Finl Exm: Tuesdy 5/9, :37:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.
More information1 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.
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
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 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 informationcos 3 (x) sin(x) dx 3y + 4 dy Math 1206 Calculus Sec. 5.6: Substitution and Area Between Curves
Mth 126 Clculus Sec. 5.6: Substitution nd Are Between Curves I. USubstitution for Definite Integrls A. Th m 6Substitution in Definite Integrls: If g (x) is continuous on [,b] nd f is continuous on the
More informationMath 190 Chapter 5 Lecture Notes. Professor Miguel Ornelas
Mth 19 Chpter 5 Lecture Notes Professor Miguel Ornels 1 M. Ornels Mth 19 Lecture Notes Section 5.1 Section 5.1 Ares nd Distnce Definition The re A of the region S tht lies under the grph of the continuous
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 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 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 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 informationMath 113 Exam 2 Practice
Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This
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 informationIntroduction and Review
Chpter 6A Notes Pge of Introuction n Review Derivtives y = f(x) y x = f (x) Evlute erivtive t x = : y = x x= f f(+h) f() () = lim h h Geometric Interprettion: see figure slope of the line tngent to f t
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationB Veitch. Calculus I Study Guide
Clculus I Stuy Guie This stuy guie is in no wy exhustive. As stte in clss, ny type of question from clss, quizzes, exms, n homeworks re fir gme. There s no informtion here bout the wor problems. 1. Some
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 informationFINALTERM EXAMINATION 9 (Session  ) Clculus & Anlyticl GeometryI Question No: ( Mrs: )  Plese choose one f ( x) x According to PowerRule of differentition, if d [ x n ] n x n n x n n x + ( n ) x n+
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 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 informationSpring 2017 Exam 1 MARK BOX HAND IN PART PIN: 17
Spring 07 Exm problem MARK BOX points HAND IN PART 0 555=x5 0 NAME: Solutions 3 0 0 PIN: 7 % 00 INSTRUCTIONS This exm comes in two prts. () HAND IN PART. Hnd in only this prt. () STATEMENT OF MULTIPLE
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
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 informationn=0 ( 1)n /(n + 1) converges, but not n=100 1/n2, is at most 1/100.
Mth 07H Topics since the second exm Note: The finl exm will cover everything from the first two topics sheets, s well. Absolute convergence nd lternting series A series n converges bsolutely if n converges.
More informationMath 142: Final Exam Formulas to Know
Mth 4: Finl Exm Formuls to Know This ocument tells you every formul/strtegy tht you shoul know in orer to o well on your finl. Stuy it well! The helpful rules/formuls from the vrious review sheets my be
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 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 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 informationStuff 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
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More information( ) Same as above but m = f x = f x  symmetric to yaxis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.
AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find
More informationf(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
More informationDisclaimer: 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
More informationFunctions of Several Variables
Functions of Severl Vribles Sketching Level Curves Sections Prtil Derivtives on every open set on which f nd the prtils, 2 f y = 2 f y re continuous. Norml Vector x, y, 2 f y, 2 f y n = ± (x 0,y 0) (x
More informationTABLE OF CONTENTS 3 CHAPTER 1
TABLE OF CONTENTS 3 CHAPTER 1 Set Lnguge & Nottion 3 CHAPTER 2 Functions 3 CHAPTER 3 Qudrtic Functions 4 CHAPTER 4 Indices & Surds 4 CHAPTER 5 Fctors of Polynomils 4 CHAPTER 6 Simultneous Equtions 4 CHAPTER
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
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 informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two
More information( ) as a fraction. Determine location of the highest
AB Clculus Exm Review Sheet  Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if
More information( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
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 informationntegration (p3) Integration by Inspection When differentiating using function of a function or the chain rule: If y = f(u), where in turn u = f(x)
ntegrtion (p) Integrtion by Inspection When differentiting using function of function or the chin rule: If y f(u), where in turn u f( y y So, to differentite u where u +, we write ( + ) nd get ( + ) (.
More informationChapters 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
More informationF is n ntiderivtive èor èindeæniteè integrlè off if F 0 èxè =fèxè. Nottion: F èxè = ; it mens F 0 èxè=fèxè ëthe integrl of f of x dee x" Bsic list: xn
Mth 70 Topics for third exm Chpter 3: Applictions of Derivtives x7: Liner pproximtion nd diæerentils Ide: The tngent line to grph of function mkes good pproximtion to the function, ner the point of tngency.
More informationAB Calculus Review Sheet
AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is
More informationStudent Handbook for MATH 3300
Student Hndbook for MATH 3300 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 0.5 0 0.5 0.5 0 0.5 If people do not believe tht mthemtics is simple, it is only becuse they do not relize how complicted life is. John Louis
More informationBig idea in Calculus: approximation
Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:
More informationImproper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Improper Integrls MATH 2, Clculus II J. Robert Buchnn Deprtment of Mthemtics Spring 28 Definite Integrls Theorem (Fundmentl Theorem of Clculus (Prt I)) If f is continuous on [, b] then b f (x) dx = [F(x)]
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 informationNotes on Calculus II Integral Calculus. Miguel A. Lerma
Notes on Clculus II Integrl Clculus Miguel A. Lerm November 22, 22 Contents Introduction 5 Chpter. Integrls 6.. Ares nd Distnces. The Definite Integrl 6.2. The Evlution Theorem.3. The Fundmentl Theorem
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 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 information1. Find the derivative of the following functions. a) f(x) = 2 + 3x b) f(x) = (5 2x) 8 c) f(x) = e2x
I. Dierentition. ) Rules. *product rule, quotient rule, chin rule MATH 34B FINAL REVIEW. Find the derivtive of the following functions. ) f(x) = 2 + 3x x 3 b) f(x) = (5 2x) 8 c) f(x) = e2x 4x 7 +x+2 d)
More informationl 2 p2 n 4n 2, the total surface area of the
Week 6 Lectures Sections 7.5, 7.6 Section 7.5: Surfce re of Revolution Surfce re of Cone: Let C be circle of rdius r. Let P n be n nsided regulr polygon of perimeter p n with vertices on C. Form cone
More informationThe 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
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 informationIf deg(num) deg(denom), then we should use longdivision of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x)
Mth 50 The method of prtil frction decomposition (PFD is used to integrte some rtionl functions of the form p(x, where p/q is in lowest terms nd deg(num < deg(denom. q(x If deg(num deg(denom, then we should
More informationf a L Most reasonable functions are continuous, as seen in the following theorem:
Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f
More informationThe Fundamental Theorem of Calculus
The Fundmentl Theorem of Clculus Professor Richrd Blecksmith richrd@mth.niu.edu Dept. of Mthemticl Sciences Northern Illinois University http://mth.niu.edu/ richrd/mth229. The Definite Integrl We define
More informationA. Limits  L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c
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 informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationTrignometric Substitution
Trignometric Substitution Trigonometric substitution refers simply to substitutions of the form x sinu or x tnu or x secu It is generlly used in conjunction with the trignometric identities to sin θ+cos
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 informationMAT187H1F 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
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 informationSummer Work Packet for MPH Math Classes
Summer Work Pcket for MPH Mth Clsses Students going into Preclculus AC Sept. 018 Nme: This pcket is designed to help students sty current with their mth skills. Ech mth clss expects certin level of number
More informationMath Bootcamp 2012 Calculus Refresher
Mth Bootcmp 0 Clculus Refresher Exponents For ny rel number x, the powers of x re : x 0 =, x = x, x = x x, etc. Powers re lso clled exponents. Remrk: 0 0 is indeterminte. Frctionl exponents re lso clled
More informationx 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx
. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute
More information