Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as


 Britton Walker
 3 years ago
 Views:
Transcription
1 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 below, we might like to know the re below e x from x = to x = : Alterntively, we my need to evlute n integrl over region on which the function hs verticl symptote (in other words, the function is not continuous on the intervl of interest). Below, we might like to know the re below / x from x = to x = ; of course, the function hs verticl symptote t x = : We cll this Type II improper integrl. Thinking of integrls s wy to clculte re, the integrls bove don t seem to mke sense. Don t the curves enclose region of infinite re? Then nswer is, not necessrily. In the first exmple, it my turn out tht the curve gets so close to the xxis so quickly tht, even though we wish to integrte over n infinite region, the enclosed re is ctully finite. Similr behvior cn occur in the second cse. Type I Improper Integrls How do we evlute n integrl such s e x dx? If the top bound of integrtion were number, we could simply find n ntiderivtive nd plug the bounds in. However, it doesn t mke sense to plug in infinity, which is not number to begin with. We solve the problem by using limits: if we wish to evlute e x dx,
2 we begin by finding e x dx, leving b s vrible; then we tke the limit s b : If the limit bove is rel (finite) number, then we sy tht the improper integrl converges; this mens tht the function encloses finite re over n infinite region. Otherwise, we sy tht the integrl diverges, tht is the function encloses infinite re over the region. In summry,. If f(x) is continuous on [, ), then if the limit exists. f(x)dx f(x)dx,. If f(x) is continuous on (, ], then if the limit exists.. If f(x) is continuous on (, ), then for ny rel number. f(x)dx = f(x)dx = lim b b f(x)dx + f(x)dx, f(x)dx, Type II Improper Integrls We cn evlute type II improper integrls using limits, s well. If f(x) hs discontinuity on the intervl over which we wish to integrte, then we define the improper integrls s follows:
3 . If f(x) is continuous on (, b], but discontinuous t x =, then if the limit exists. f(x)dx c + c f(x)dx,. If f(x) is continuous on [, b), but discontinuous t x = b, then if the limit exists. f(x)dx c b c f(x)dx,. If f(x) is continuous on [, b], except t the point x = c with < c < b, then f(x)dx = c f(x)dx + c f(x)dx. Agin, if the limit exist (tht is the limit returns rel number), then we sy tht the improper integrl converges; the improper integrl diverges if the limit does not exist, tht is the function encloses infinite re. Exmples Find dx, if it exists. x By definition, we hve dx x dx x ) ( x ( ) = Since the limit does not exist, the improper integrl diverges. Find e x dx, if it exists.
4 Agin, using the definition, we hve e x dx e x dx e x dx ( e x ) ( e + e ) = e since lim e = (the denomintor pproches infinity, so the frction itself pproches ). So in this cse, the function encloses n re of only e over the infinite region from to. Find v v dv. Agin, since the intervl over which we wish to integrte is infinite, we will need to evlute the improper integrl using limits: v dv v v v dv. We will need prtil frctions to evlute the integrl: since the denomintor fctors s v v = v(v ), the form of the decomposed frction must be Adding the frctions, we see tht A v + B v. A v + B Av A = v v(v ) + Bv v(v ) Av A + Bv = v(v ) Av A + Bv = = v v v v. 4
5 Since Av + Bv A =, we conclude tht A + B = A = (by equting the v terms) (by equting the v terms). Thus A = nd B =. So the frction decomposes s Now we cn evlute the integrl: v dv v v v = v + (v ). v v dv v + (v ) dv ( ln v + ln v ) ( ln + ln + ln ln ) ( ln + ln ) + ln. Since lim ln = nd lim ln =, it is tempting to guess tht the nswer is. However, is n indeterminte form, so we ctully do not know the vlue for the limit. Let s fctor the out of the limit: Now since ln c ln d = ln( c d tht So Since the limit lim lim ( ln + ln ) = lim ( ln + ln ) ), let s try rewriting ln + ln s ln. Then lim ( ln + ln ) = lim (ln ) = ln lim. hs the indeterminte form, we cn use L Hopitl s rule to see LR lim =. ln lim = ln =, 5
6 which mens tht v v dv = lim ( ln + ln ) + ln = + ln = ln. Let p be number so tht p >. For wht vlues of p does dx converge? xp Let s evlute the integrl using limits; we will need two different cses, one when p = nd when when p. Let s begin with the p cse: dx xp ( ( ( ( ( x p dx x p dx b p + x p+ ) ) (x ) p b p p p ) p b p p ) ( p)b p ) p ( p)b p ) + p We need to find lim ( ( p)b p ) = p lim ( b There will be two cses, depending upon the vlue for p:. If p >, then b p s b, so lim ( ) =. ( p)bp p ).. If < p < then p < mens tht b p s b, so p lim ( ) =. bp 6
7 Now if p =, then b dx x x dx (ln x) b ) (ln b ln ) (ln b) =, so the integrl diverges. Thus x x dx x p dx { converges to p if p > diverges if p. Find y dy. While it is tempting to integrte immeditely, we must note tht so the integrl is ctully improper. We cn rewrite it s y is not continuous t y =, y dy = y dy + y dy. Let s evlute ech integrl seprtely: since lim =. y dy y (y dy ) ( + ) = 7
8 Similrly, y dy + y +(y ) dy +( ) =. So the finl nswer is Find x dx. y dy = 6. This is n improper integrl becuse the function x is not continuous t x = ; in prticulr, there is verticl symptote t x =. In order to evlute the integrl, we will need to set up dx dx x x x (sin ) (sin sin ). We know tht sin =, so we just need to determine lim sin. We cn use the unit circle to see tht in fct, sin = π : So dx sin ) x (sin = π. 8
9 Testing for Convergence or Divergence We hve noted erlier in the course tht certin integrls cnnot be evluted using elementry methods; for exmple, we cn not hope evlute e x dx becuse we do not know how to find n ntiderivtive of e x. However, we my still be ble to determine whether or not the integrl converges; we will lern two tools in this section tht my help us to do so. Direct Comprison Test If the grph of f(x) lies underneth the grph of g(x), then the re below f(x) must be less thn the re below g(x): In prticulr, if g(x) converges (encloses finite re) then so does f(x). Alterntely, if the grph of f(x) lies bove the grph of g(x), then the re below f(x) must be more thn the re below g(x): So if g(x) diverges (encloses n infinite re), f(x) must s well. These observtions led to the direct comprison test: Theorem... Let f(x) nd g(x) be continuous on [, ). 9
10 . If f(x) g(x) for ll x, nd g(x)dx converges, then f(x)dx does s well.. If g(x) f(x) for ll x, nd g(x)dx diverges, then f(x)dx does s well. Unfortuntely, the theorem does not help us hndle every cse tht could occur. For instnce, if f(x) g(x) for ll x nd or divergence of f(x)dx: g(x)dx diverges, we gin no knowledge bout the convergence Since f(x) g(x), the fct tht g(x) bounds region whose re is infinite does not inform us s to the size of the re under f(x). On the other hnd, if g(x) f(x) for ll x, nd we lern nothing bout the convergence or divergence of g(x)dx converges, then once gin f(x)dx: Since g(x) f(x), the fct tht g(x) bounds region of finite re does not help us determine the size of the re bounded by f(x). To sum up the bove informtion, if you suspect tht n improper integrl of f(x) converges, you verify your clim by finding lrger function g(x) whose integrl converges. Alterntely, if
11 you believe tht n improper integrl of f(x) diverges, you need to find smller function g(x) whose integrl diverges. The hrd prt in using the bove theorems is tht you will need to find functions to use in the comprison; this is not lwys esy or obvious. The following tips my help you to find comprison functions:. Choose comprison function whose convergence or divergence you cn estblish. sin x nd cos x for ny choice of x.. Incresing the size of frction s denomintor while holding the numertor constnt will decrese the size of the frction 4. Decresing the size of frction s denomintor while holding the numertor constnt will increse the size of the frction 5. It is often helpful to simplify some portion of function if tht portion is known to be bounded by constnt (see bove) Exmples Does e x dx converge or diverge? Notice tht we do not know how to find n ntiderivtive of e x, so even if we knew tht the integrl converges, we would not be ble to find its ctul vlue; the best we cn hope for is to be ble to determine convergence or divergence of the improper integrl. We will need to use the direct comprison test to determine convergence, so we will need to be ble to compre e x = with nother function whose convergence or divergence we cn e estblish. It is best to choose x function similr to the originl one for esy comprison. Erlier, we found tht e x dx converges; it is similr to the current function of interest nd so might be helpful. Note tht since e x e x when x, then e x = e x = e x. e x Then since e x dx converges nd e x e x, e x dx must converge s well.
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
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 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 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 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 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 informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
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 informationLecture 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
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 informationCalculus 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]
More informationpractice How would you find: e x + e x e 2x e x 1 dx 1 e today: improper integrals
prctice How would you find: dx e x + e x e 2x e x 1 dx e 2x 1 e x dx 1. Let u=e^x. Then dx=du/u. Ans = rctn ( e^x ) + C 2. Let u=e^x. Becomes u du / (u1), divide to get u/(u1)=1+1/(u1) Ans = e^x + ln
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& 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 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 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 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 informationMORE 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
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 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 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 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 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 informationThe graphs of Rational Functions
Lecture 4 5A: The its of Rtionl Functions s x nd s x + The grphs of Rtionl Functions The grphs of rtionl functions hve severl differences compred to power functions. One of the differences is the behvior
More informationChapter 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
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 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 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 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 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 Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 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
More information( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that
Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we
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 informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More informationapproaches 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.
More informationMath 231E, Lecture 33. Parametric Calculus
Mth 31E, Lecture 33. Prmetric Clculus 1 Derivtives 1.1 First derivtive Now, let us sy tht we wnt the slope t point on prmetric curve. Recll the chin rule: which exists s long s /. = / / Exmple 1.1. Reconsider
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 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 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 information!0 f(x)dx + lim!0 f(x)dx. The latter is sometimes also referred to as improper integrals of the. k=1 k p converges for p>1 and diverges otherwise.
Chpter 7 Improper integrls 7. Introduction The gol of this chpter is to meningfully extend our theory of integrls to improper integrls. There re two types of soclled improper integrls: the first involves
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 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 informationDefinite 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
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 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 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 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 informationUNIFORM 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,
More information5.2 Exponent Properties Involving Quotients
5. Eponent Properties Involving Quotients Lerning Objectives Use the quotient of powers property. Use the power of quotient property. Simplify epressions involving quotient properties of eponents. Use
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
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 informationDefinite 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
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 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 informationc n φ n (x), 0 < x < L, (1) n=1
SECTION : Fourier Series. MATH4. In section 4, we will study method clled Seprtion of Vribles for finding exct solutions to certin clss of prtil differentil equtions (PDEs. To do this, it will be necessry
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
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 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 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 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 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 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 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 informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
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 informationAPPM 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
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 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 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 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 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 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 information(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35
7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know
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 information13.4. Integration by Parts. Introduction. Prerequisites. Learning Outcomes
Integrtion by Prts 13.4 Introduction Integrtion by Prts is technique for integrting products of functions. In this Section you will lern to recognise when it is pproprite to use the technique nd hve 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 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 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 information1 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
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 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 information5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship
5.4, 6.1, 6.2 Hnout As we ve iscusse, the integrl is in some wy the opposite of tking erivtive. The exct reltionship is given by the Funmentl Theorem of Clculus: The Funmentl Theorem of Clculus: If f is
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 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 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 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 informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
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 informationMath 116 Calculus II
Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................
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 informationAdvanced 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
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 2 MATH00040 SEMESTER /2018
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH4 SEMESTER 7/8 DR. ANTHONY BROWN 4.. Introduction to Integrtion. 4. Integrl Clculus As ws the cse with the chpter on differentil clculus,
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 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 informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationRiemann 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 >
More informationAPPROXIMATE INTEGRATION
APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose ntiderivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be
More informationFurther integration. x n nx n 1 sinh x cosh x log x 1/x cosh x sinh x e x e x tan x sec 2 x sin x cos x tan 1 x 1/(1 + x 2 ) cos x sin x
Further integrtion Stndrd derivtives nd integrls The following cn be thought of s list of derivtives or eqully (red bckwrds) s list of integrls. Mke sure you know them! There ren t very mny. f(x) f (x)
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
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 information