f(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2
|
|
- Carmella Chambers
- 6 years ago
- Views:
Transcription
1 Impope Inegls To his poin we hve only consideed inegls f() wih he is of inegion nd b finie nd he inegnd f() bounded (nd in fc coninuous ecep possibly fo finiely mny jump disconinuiies) An inegl hving eihe n infinie i of inegion o n unbounded inegnd is clled impope. Hee e wo emples + 2 The fis hs n infinie domin of inegion nd he inegnd of he second ends o s ppoches he lef end of he domin of inegion. We ll s wih n emple h illuses he ps h you cn fll ino if you e such inegls sloppily. Then we ll see how o e hem cefully. Emple The compuion = 2 = = 2 is wong. In fc, he nswe is idiculous. The inegnd >, so he inegl hs o be 2 posiive. The flw in he gumen is h he fundmenl heoem of clculus, which sys h if F () = f() hen f() = F(b) F(), is pplicble only when F () eiss nd equls f() fo ll b. In his cse F () = does no eis fo =. The given 2 inegl is impope. We ll see le h he coec nswe is +. Emple The ceful wy o e n inegl like h hs bounded inegnd bu hs + 2 domin of inegion h eends o + is o ppoime he inegl by one wih bounded domin of inegion, like R, nd hen ke he i R. Hee is he + 2 y = f() R foml definiion nd some viions. We ll wok hough some emples in deil sholy. c Joel Feldmn. 25. All ighs eseved. Febuy 5, 25
2 Definiion 2. () If he inegl R f() eiss fo ll R >, hen R f() f() when he i eiss (nd is finie). (b) If he inegl f() eiss fo ll < b, hen when he i eiss (nd is finie). (c) If he inegl R f() f() f() eiss fo ll < R, hen f() c R f() + f() c when boh is eis (nd e finie). Any c cn be used. When he i(s) eis, he inegl is sid o be convegen. Ohewise i is sid o be divegen. The ceful wy o e n inegl like h hs finie domin of inegion bu whose inegnd is unbounded ne one i of inegion is o ppoime he inegl by one wih domin of inegion on which he inegnd is bounded, like, wih >, nd hen ke he i +. Hee is he foml definiion nd some viions. y y = c Joel Feldmn. 25. All ighs eseved. 2 Febuy 5, 25
3 Definiion 3. () If he inegl f() eiss fo ll < < b, hen when he i eiss (nd is finie). (b) If he inegl T f() f() + f() eiss fo ll < T < b, hen when he i eiss (nd is finie). T f() f() T b (c) Le < c < b. If he inegls T f() nd f() eis fo ll < T < c nd c < < b, hen T f() f() + f() T c c+ when boh i eis (nd e finie). When he i(s) eis, he inegl is sid o be convegen. Ohewise i is sid o be divegen. Ifnineglhsmoehnone souceofimpopiey, foemple ninfiniedominof inegion nd n inegnd wih n unbounded inegnd o muliple infinie disconinuiies, hen you spli i up ino sum of inegls wih single souce of impopiey in ech. Fo he inegl, s whole, o convege evey em in h sum hs o convege. Fo emple, he inegl hs domin of inegion h eends o boh + nd nd, ( 2) 2 in ddiion, hs n inegnd which is singul (i.e. becomes infinie) = 2 nd =. So we would wie ( 2) 2 = ( 2) 2 + ( 2) 2 + ( 2) ( 2) ( 2) ( 2) 2 (Hee he bek poin could be eplced by ny numbe sicly less hn ; could be eplced by ny numbe sicly beween nd 2; nd 3 could be eplced by ny numbe sicly bigge hn 2.) On he igh hnd side he fis inegl hs domin of inegion eending o c Joel Feldmn. 25. All ighs eseved. 3 Febuy 5, 25
4 he second inegl hs n inegnd h becomes unbounded s, he hid inegl hs n inegnd h becomes unbounded s +, he fouh inegl hs n inegnd h becomes unbounded s 2, he fifh inegl hs n inegnd h becomes unbounded s 2+, nd he ls inegl hs domin of inegion eending o +. We e now edy fo some (impon) emples. Emple 4 ( p ) Fi ny p >. The domin of he inegl eends o + nd he inegnd is p p coninuous nd bounded on he whole domin. So he definiion of his inegl is R p p When p, one nideivive of p is p+, nd when p = nd >, one nideivive p+ of p is ln. So p p ] R p if p lnr if p = As R, lnr ends o nd R p ends o when p > (i.e. he eponen is posiive) nd ends o when p < (i.e. he eponen is negive). So { divegen if p = p if p > p Emple 4 Emple 5 ( p ) Agin fi ny p >. The domin of inegion of he inegl is finie, bu he inegnd becomes unbounded s ppoches he lef end,, of he domin of inegion. p p So he definiion of his inegl is p + p When p, one nideivive of p is p+, nd when p = nd >, one nideivive p+ of p is ln. So ] p if p p p + ln if p = c Joel Feldmn. 25. All ighs eseved. 4 Febuy 5, 25
5 As +, ln ends o nd p ends o when p > (i.e. he eponen is posiive) nd ends o when p < (i.e. he eponen is negive). So { = if p < p p divegen if p Emple 5 Emple 6 ( p ) Ye gin fi p >. This ime he domin of inegion of he inegl eends o p +, nd in ddiion he inegnd becomes unbounded s ppoches he lef end,, p of he domin of inegion. So we hve o spli i up. p = p + We sw, in Emple 5, h he fis inegl diveged wheneve p, nd we lso sw, in Emple 4, h he second inegl diveged wheneve p. So he inegl diveges p fo ll vlues of p. Emple 6 p Emple 7 ( ) This is pey suble emple. I looks fom he skech below h he signed e o he y y = lef of he y is should ecly cncel he e o he igh of he y is mking he vlue of he inegl ecly zeo. Bu boh of he inegls ] ln = + + T T + T ln ] T + ln ln T = T c Joel Feldmn. 25. All ighs eseved. 5 Febuy 5, 25
6 divege so diveges. Don mke he miske of hinking h =. I is undefined nd fo good eson. Fo emple, we hve jus seen h he e o he igh of he y is is + = + nd h he e o he lef of he y is is (subsiue 7 fo T bove) + 7 = If =, he following i should be. 7 ] + + ln ] + +ln 7 ln ] + +ln(7) ln7 + = ln7 This ppes o give = ln7. Of couse he numbe 7 ws picked ndom. You cn mke be ny numbe ll, by mking suible eplcemen fo 7. Emple 7 Emple 8 (Emple evisied) The ceful compuion of he inegl of Emple is 2 T T T = ] T 2 + ] + Emple 8 Since Emple 9 ( ) + 2 R ] R + cn cn The inegl conveges nd kes he vlue π. + 2 ] cnr = π 2 cn = π 2 Emple 9 c Joel Feldmn. 25. All ighs eseved. 6 Febuy 5, 25
7 Emple Fo wh vlues of p does e convege? (ln) p Soluion. Fo e, he denomino (ln) p is neve zeo. So he inegnd is bounded on he enie domin of inegion nd his inegl is impope only becuse he domin of inegion eends o + nd we poceed s usul. We hve e R (ln) p e (ln) p lnr du u p = p (lnr) p wih u = ln, du = ln(lnr) if p = { divegen if p if p > p ] if p jus s in Emple 4, bu wih R eplced by lnr. Emple c Joel Feldmn. 25. All ighs eseved. 7 Febuy 5, 25
An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples.
Improper Inegrls To his poin we hve only considered inegrls f(x) wih he is of inegrion nd b finie nd he inegrnd f(x) bounded (nd in fc coninuous excep possibly for finiely mny jump disconinuiies) An inegrl
More informationINTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).
INTEGRALS JOHN QUIGG Eercise. Le f : [, b] R be bounded, nd le P nd Q be priions of [, b]. Prove h if P Q hen U(P ) U(Q) nd L(P ) L(Q). Soluion: Le P = {,..., n }. Since Q is obined from P by dding finiely
More information4.8 Improper Integrals
4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls
More informationReinforcement learning
CS 75 Mchine Lening Lecue b einfocemen lening Milos Huskech milos@cs.pi.edu 539 Senno Sque einfocemen lening We wn o len conol policy: : X A We see emples of bu oupus e no given Insed of we ge feedbck
More informationREAL ANALYSIS I HOMEWORK 3. Chapter 1
REAL ANALYSIS I HOMEWORK 3 CİHAN BAHRAN The quesions re from Sein nd Shkrchi s e. Chper 1 18. Prove he following sserion: Every mesurble funcion is he limi.e. of sequence of coninuous funcions. We firs
More informatione t dt e t dt = lim e t dt T (1 e T ) = 1
Improper Inegrls There re wo ypes of improper inegrls - hose wih infinie limis of inegrion, nd hose wih inegrnds h pproch some poin wihin he limis of inegrion. Firs we will consider inegrls wih infinie
More informationMathematics 805 Final Examination Answers
. 5 poins Se he Weiersrss M-es. Mhemics 85 Finl Eminion Answers Answer: Suppose h A R, nd f n : A R. Suppose furher h f n M n for ll A, nd h Mn converges. Then f n converges uniformly on A.. 5 poins Se
More informationAddition & Subtraction of Polynomials
Addiion & Sucion of Polynomil Addiion of Polynomil: Adding wo o moe olynomil i imly me of dding like em. The following ocedue hould e ued o dd olynomil 1. Remove enhee if hee e enhee. Add imil em. Wie
More information() t. () t r () t or v. ( t) () () ( ) = ( ) or ( ) () () () t or dv () () Section 10.4 Motion in Space: Velocity and Acceleration
Secion 1.4 Moion in Spce: Velociy nd Acceleion We e going o dive lile deepe ino somehing we ve ledy inoduced, nmely () nd (). Discuss wih you neighbo he elionships beween posiion, velociy nd cceleion you
More informationHomework 5 for BST 631: Statistical Theory I Solutions, 09/21/2006
Homewok 5 fo BST 63: Sisicl Theoy I Soluions, 9//6 Due Time: 5:PM Thusy, on 9/8/6. Polem ( oins). Book olem.8. Soluion: E = x f ( x) = ( x) f ( x) + ( x ) f ( x) = xf ( x) + xf ( x) + f ( x) f ( x) Accoing
More informationContraction Mapping Principle Approach to Differential Equations
epl Journl of Science echnology 0 (009) 49-53 Conrcion pping Principle pproch o Differenil Equions Bishnu P. Dhungn Deprmen of hemics, hendr Rn Cmpus ribhuvn Universiy, Khmu epl bsrc Using n eension of
More informationf t f a f x dx By Lin McMullin f x dx= f b f a. 2
Accumulion: Thoughs On () By Lin McMullin f f f d = + The gols of he AP* Clculus progrm include he semen, Sudens should undersnd he definie inegrl s he ne ccumulion of chnge. 1 The Topicl Ouline includes
More informationSeptember 20 Homework Solutions
College of Engineering nd Compuer Science Mechnicl Engineering Deprmen Mechnicl Engineering A Seminr in Engineering Anlysis Fll 7 Number 66 Insrucor: Lrry Creo Sepember Homework Soluions Find he specrum
More informationCircuits 24/08/2010. Question. Question. Practice Questions QV CV. Review Formula s RC R R R V IR ... Charging P IV I R ... E Pt.
4/08/00 eview Fomul s icuis cice s BL B A B I I I I E...... s n n hging Q Q 0 e... n... Q Q n 0 e Q I I0e Dischging Q U Q A wie mde of bss nd nohe wie mde of silve hve he sme lengh, bu he dimee of he bss
More informationENGR 1990 Engineering Mathematics The Integral of a Function as a Function
ENGR 1990 Engineering Mhemics The Inegrl of Funcion s Funcion Previously, we lerned how o esime he inegrl of funcion f( ) over some inervl y dding he res of finie se of rpezoids h represen he re under
More informationMath 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:
Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial
More information5.1-The Initial-Value Problems For Ordinary Differential Equations
5.-The Iniil-Vlue Problems For Ordinry Differenil Equions Consider solving iniil-vlue problems for ordinry differenil equions: (*) y f, y, b, y. If we know he generl soluion y of he ordinry differenil
More informationMotion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.
Moion Accelerion Pr : Consn Accelerion Accelerion Accelerion Accelerion is he re of chnge of velociy. = v - vo = Δv Δ ccelerion = = v - vo chnge of velociy elpsed ime Accelerion is vecor, lhough in one-dimensionl
More information0 for t < 0 1 for t > 0
8.0 Sep nd del funcions Auhor: Jeremy Orloff The uni Sep Funcion We define he uni sep funcion by u() = 0 for < 0 for > 0 I is clled he uni sep funcion becuse i kes uni sep = 0. I is someimes clled he Heviside
More informationMATH 351 Solutions: TEST 3-B 23 April 2018 (revised)
MATH Soluions: TEST -B April 8 (revised) Par I [ ps each] Each of he following asserions is false. Give an eplici couner-eample o illusrae his.. If H: (, ) R is coninuous, hen H is unbounded. Le H() =
More informationHow to Prove the Riemann Hypothesis Author: Fayez Fok Al Adeh.
How o Prove he Riemnn Hohesis Auhor: Fez Fok Al Adeh. Presiden of he Srin Cosmologicl Socie P.O.Bo,387,Dmscus,Sri Tels:963--77679,735 Emil:hf@scs-ne.org Commens: 3 ges Subj-Clss: Funcionl nlsis, comle
More informationThe solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.
[~ o o :- o o ill] i 1. Mrices, Vecors, nd Guss-Jordn Eliminion 1 x y = = - z= The soluion is ofen represened s vecor: n his exmple, he process of eliminion works very smoohly. We cn elimine ll enries
More information( ) ( ) ( ) ( ) ( ) ( y )
8. Lengh of Plne Curve The mos fmous heorem in ll of mhemics is he Pyhgoren Theorem. I s formulion s he disnce formul is used o find he lenghs of line segmens in he coordine plne. In his secion you ll
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum Squred-Error Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > for ll smples y i solve sysem of liner inequliies MSE procedure y i = i for ll smples
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum Squred-Error Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > 0 for ll smples y i solve sysem of liner inequliies MSE procedure y i i for ll smples
More informationWeek 8. Topic 2 Properties of Logarithms
Week 8 Topic 2 Popeties of Logithms 1 Week 8 Topic 2 Popeties of Logithms Intoduction Since the esult of ithm is n eponent, we hve mny popeties of ithms tht e elted to the popeties of eponents. They e
More informationHow to prove the Riemann Hypothesis
Scholrs Journl of Phsics, Mhemics nd Sisics Sch. J. Phs. Mh. S. 5; (B:5-6 Scholrs Acdemic nd Scienific Publishers (SAS Publishers (An Inernionl Publisher for Acdemic nd Scienific Resources *Corresonding
More informationPHYSICS 102. Intro PHYSICS-ELECTROMAGNETISM
PHYS 0 Suen Nme: Suen Numbe: FAUTY OF SIENE Viul Miem EXAMINATION PHYSIS 0 Ino PHYSIS-EETROMAGNETISM Emines: D. Yoichi Miyh INSTRUTIONS: Aemp ll 4 quesions. All quesions hve equl weighs 0 poins ech. Answes
More informationMTH 146 Class 11 Notes
8.- Are of Surfce of Revoluion MTH 6 Clss Noes Suppose we wish o revolve curve C round n is nd find he surfce re of he resuling solid. Suppose f( ) is nonnegive funcion wih coninuous firs derivive on he
More informationPhysics 2A HW #3 Solutions
Chper 3 Focus on Conceps: 3, 4, 6, 9 Problems: 9, 9, 3, 41, 66, 7, 75, 77 Phsics A HW #3 Soluions Focus On Conceps 3-3 (c) The ccelerion due o grvi is he sme for boh blls, despie he fc h he hve differen
More informationMATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.
MATH 3B: MIDTERM REVIEW JOE HUGHES. Inegraion by Pars. Evaluae 3 e. Soluion: Firs make he subsiuion u =. Then =, hence 3 e = e = ue u Now inegrae by pars o ge ue u = ue u e u + C and subsiue he definiion
More informationMath 4318 : Real Analysis II Mid-Term Exam 1 14 February 2013
Mth 4318 : Rel Anlysis II Mid-Tem Exm 1 14 Febuy 2013 Nme: Definitions: Tue/Flse: Poofs: 1. 2. 3. 4. 5. 6. Totl: Definitions nd Sttements of Theoems 1. (2 points) Fo function f(x) defined on (, b) nd fo
More informationClass Summary. be functions and f( D) , we define the composition of f with g, denoted g f by
Clss Summy.5 Eponentil Functions.6 Invese Functions nd Logithms A function f is ule tht ssigns to ech element D ectly one element, clled f( ), in. Fo emple : function not function Given functions f, g:
More informationScience Advertisement Intergovernmental Panel on Climate Change: The Physical Science Basis 2/3/2007 Physics 253
Science Adeisemen Inegoenmenl Pnel on Clime Chnge: The Phsicl Science Bsis hp://www.ipcc.ch/spmfeb7.pdf /3/7 Phsics 53 hp://www.fonews.com/pojecs/pdf/spmfeb7.pdf /3/7 Phsics 53 3 Sus: Uni, Chpe 3 Vecos
More information10 Statistical Distributions Solutions
Communictions Engineeing MSc - Peliminy Reding 1 Sttisticl Distiutions Solutions 1) Pove tht the vince of unifom distiution with minimum vlue nd mximum vlue ( is ) 1. The vince is the men of the sques
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 informationProperties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x)
Properies of Logrihms Solving Eponenil nd Logrihmic Equions Properies of Logrihms Produc Rule ( ) log mn = log m + log n ( ) log = log + log Properies of Logrihms Quoien Rule log m = logm logn n log7 =
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 left-hnd
More informationPreviously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system
436-459 Advnced contol nd utomtion Extensions to bckstepping contolle designs Tcking Obseves (nonline dmping) Peviously Lst lectue we looked t designing nonline contolles using the bckstepping technique
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 informationSections 3.1 and 3.4 Exponential Functions (Growth and Decay)
Secions 3.1 and 3.4 Eponenial Funcions (Gowh and Decay) Chape 3. Secions 1 and 4 Page 1 of 5 Wha Would You Rahe Have... $1million, o double you money evey day fo 31 days saing wih 1cen? Day Cens Day Cens
More informationME 141. Engineering Mechanics
ME 141 Engineeing Mechnics Lecue 13: Kinemics of igid bodies hmd Shhedi Shkil Lecue, ep. of Mechnicl Engg, UET E-mil: sshkil@me.bue.c.bd, shkil6791@gmil.com Websie: eche.bue.c.bd/sshkil Couesy: Veco Mechnics
More informationChapter Direct Method of Interpolation
Chper 5. Direc Mehod of Inerpolion Afer reding his chper, you should be ble o:. pply he direc mehod of inerpolion,. sole problems using he direc mehod of inerpolion, nd. use he direc mehod inerpolns o
More informationMAT 266 Calculus for Engineers II Notes on Chapter 6 Professor: John Quigg Semester: spring 2017
MAT 66 Clculus for Engineers II Noes on Chper 6 Professor: John Quigg Semeser: spring 7 Secion 6.: Inegrion by prs The Produc Rule is d d f()g() = f()g () + f ()g() Tking indefinie inegrls gives [f()g
More informationInvert and multiply. Fractions express a ratio of two quantities. For example, the fraction
Appendi E: Mnipuling Fions Te ules fo mnipuling fions involve lgei epessions e el e sme s e ules fo mnipuling fions involve numes Te fundmenl ules fo omining nd mnipuling fions e lised elow Te uses of
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 informationLecture 10. Solution of Nonlinear Equations - II
Fied point Poblems Lectue Solution o Nonline Equtions - II Given unction g : R R, vlue such tht gis clled ied point o the unction g, since is unchnged when g is pplied to it. Whees with nonline eqution
More informationChapter 7. Kleene s Theorem. 7.1 Kleene s Theorem. The following theorem is the most important and fundamental result in the theory of FA s:
Chpte 7 Kleene s Theoem 7.1 Kleene s Theoem The following theoem is the most impotnt nd fundmentl esult in the theoy of FA s: Theoem 6 Any lnguge tht cn e defined y eithe egul expession, o finite utomt,
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 information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationCh.4 Motion in 2D. Ch.4 Motion in 2D
Moion in plne, such s in he sceen, is clled 2-dimensionl (2D) moion. 1. Posiion, displcemen nd eloci ecos If he picle s posiion is ( 1, 1 ) 1, nd ( 2, 2 ) 2, he posiions ecos e 1 = 1 1 2 = 2 2 Aege eloci
More informationAverage & instantaneous velocity and acceleration Motion with constant acceleration
Physics 7: Lecure Reminders Discussion nd Lb secions sr meeing ne week Fill ou Pink dd/drop form if you need o swich o differen secion h is FULL. Do i TODAY. Homework Ch. : 5, 7,, 3,, nd 6 Ch.: 6,, 3 Submission
More informationDuality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.
Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we
More informationLECTURE 5. is defined by the position vectors r, 1. and. The displacement vector (from P 1 to P 2 ) is defined through r and 1.
LECTURE 5 ] DESCRIPTION OF PARTICLE MOTION IN SPACE -The displcemen, veloci nd cceleion in -D moion evel hei veco nue (diecion) houh he cuion h one mus p o hei sin. Thei full veco menin ppes when he picle
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationLecture 3. Electrostatics
Lecue lecsics In his lecue yu will len: Thee wys slve pblems in elecsics: ) Applicin f he Supepsiin Pinciple (SP) b) Applicin f Guss Lw in Inegl Fm (GLIF) c) Applicin f Guss Lw in Diffeenil Fm (GLDF) C
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 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 informationFaraday s Law. To be able to find. motional emf transformer and motional emf. Motional emf
Objecie F s w Tnsfome Moionl To be ble o fin nsfome. moionl nsfome n moionl. 331 1 331 Mwell s quion: ic Fiel D: Guss lw :KV : Guss lw H: Ampee s w Poin Fom Inegl Fom D D Q sufce loop H sufce H I enclose
More informationThe shortest path between two truths in the real domain passes through the complex domain. J. Hadamard
Complex Analysis R.G. Halbud R.Halbud@ucl.ac.uk Depamen of Mahemaics Univesiy College London 202 The shoes pah beween wo uhs in he eal domain passes hough he complex domain. J. Hadamad Chape The fis fundamenal
More informationAnswers to test yourself questions
Answes to test youself questions opic Descibing fields Gm Gm Gm Gm he net field t is: g ( d / ) ( 4d / ) d d Gm Gm Gm Gm Gm Gm b he net potentil t is: V d / 4d / d 4d d d V e 4 7 9 49 J kg 7 7 Gm d b E
More informationD zone schemes
Ch. 5. Enegy Bnds in Cysls 5.. -D zone schemes Fee elecons E k m h Fee elecons in cysl sinα P + cosα cosk α cos α cos k cos( k + π n α k + πn mv ob P 0 h cos α cos k n α k + π m h k E Enegy is peiodic
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 informationJim Lambers MAT 169 Fall Semester Lecture 4 Notes
Jim Lmbers MAT 169 Fll Semester 2009-10 Lecture 4 Notes These notes correspond to Section 8.2 in the text. Series Wht is Series? An infinte series, usully referred to simply s series, is n sum of ll of
More informationLecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.
Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one
More informationECE Microwave Engineering. Fall Prof. David R. Jackson Dept. of ECE. Notes 10. Waveguides Part 7: Transverse Equivalent Network (TEN)
EE 537-635 Microwve Engineering Fll 7 Prof. Dvid R. Jcson Dep. of EE Noes Wveguides Pr 7: Trnsverse Equivlen Newor (N) Wveguide Trnsmission Line Model Our gol is o come up wih rnsmission line model for
More informationgraph of unit step function t
.5 Piecewie coninuou forcing funcion...e.g. urning he forcing on nd off. The following Lplce rnform meril i ueful in yem where we urn forcing funcion on nd off, nd when we hve righ hnd ide "forcing funcion"
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 informationA 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m
PHYS : Soluions o Chper 3 Home Work. SSM REASONING The displcemen is ecor drwn from he iniil posiion o he finl posiion. The mgniude of he displcemen is he shores disnce beween he posiions. Noe h i is onl
More informationSections 2.2 & 2.3 Limit of a Function and Limit Laws
Mah 80 www.imeodare.com Secions. &. Limi of a Funcion and Limi Laws In secion. we saw how is arise when we wan o find he angen o a curve or he velociy of an objec. Now we urn our aenion o is in general
More informationReinforcement learning
Lecue 3 Reinfocemen leaning Milos Hauskech milos@cs.pi.edu 539 Senno Squae Reinfocemen leaning We wan o lean he conol policy: : X A We see examples of x (bu oupus a ae no given) Insead of a we ge a feedback
More informationSection P.1 Notes Page 1 Section P.1 Precalculus and Trigonometry Review
Secion P Noe Pge Secion P Preclculu nd Trigonomer Review ALGEBRA AND PRECALCULUS Eponen Lw: Emple: 8 Emple: Emple: Emple: b b Emple: 9 EXAMPLE: Simplif: nd wrie wi poiive eponen Fir I will flip e frcion
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 informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationWe will see what is meant by standard form very shortly
THEOREM: For fesible liner progrm in its stndrd form, the optimum vlue of the objective over its nonempty fesible region is () either unbounded or (b) is chievble t lest t one extreme point of the fesible
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 informationSolutions to Problems from Chapter 2
Soluions o Problems rom Chper Problem. The signls u() :5sgn(), u () :5sgn(), nd u h () :5sgn() re ploed respecively in Figures.,b,c. Noe h u h () :5sgn() :5; 8 including, bu u () :5sgn() is undeined..5
More informationProperties of the Riemann Stieltjes Integral
Properties of the Riemnn Stieltjes Integrl Theorem (Linerity Properties) Let < c < d < b nd A,B IR nd f,g,α,β : [,b] IR. () If f,g R(α) on [,b], then Af +Bg R(α) on [,b] nd [ ] b Af +Bg dα A +B (b) If
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 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 informationProblem Set 4: Solutions Math 201A: Fall 2016
Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be one-to-one, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true
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 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 informationA LOG IS AN EXPONENT.
Ojeives: n nlze nd inerpre he ehvior of rihmi funions, inluding end ehvior nd smpoes. n solve rihmi equions nlill nd grphill. n grph rihmi funions. n deermine he domin nd rnge of rihmi funions. n deermine
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 informationthe king's singers And So It Goes the colour of song Words and Vusic by By Joel LEONARD Arranged by Bob Chilcott
085850 SATB div cppell US $25 So Goes Wods nd Vusic by By Joel Anged by Bob Chilco he king's singes L he colou of song A H EXCLUSVELY DSTRBUTED BY LEONARD (Fom The King's Singes 25h Annivesy Jubilee) So
More informationANSWERS TO EVEN NUMBERED EXERCISES IN CHAPTER 2
ANSWERS TO EVEN NUMBERED EXERCISES IN CHAPTER Seion Eerise -: Coninuiy of he uiliy funion Le λ ( ) be he monooni uiliy funion defined in he proof of eisene of uiliy funion If his funion is oninuous y hen
More informationTopic 1 Notes Jeremy Orloff
Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble
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 x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationP441 Analytical Mechanics - I. Coupled Oscillators. c Alex R. Dzierba
Lecure 3 Mondy - Deceber 5, 005 Wrien or ls upded: Deceber 3, 005 P44 Anlyicl Mechnics - I oupled Oscillors c Alex R. Dzierb oupled oscillors - rix echnique In Figure we show n exple of wo coupled oscillors,
More informationAns: In the rectangular loop with the assigned direction for i2: di L dt , (1) where (2) a) At t = 0, i1(t) = I1U(t) is applied and (1) becomes
omewok # P7-3 ecngul loop of widh w nd heigh h is siued ne ve long wie cing cuen i s in Fig 7- ssume i o e ecngul pulse s shown in Fig 7- Find he induced cuen i in he ecngul loop whose self-inducnce is
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 informationTechnical Vibration - text 2 - forced vibration, rotational vibration
Technicl Viion - e - foced viion, oionl viion 4. oced viion, viion unde he consn eenl foce The viion unde he eenl foce. eenl The quesion is if he eenl foce e is consn o vying. If vying, wh is he foce funcion.
More informationSome Inequalities variations on a common theme Lecture I, UL 2007
Some Inequliies vriions on common heme Lecure I, UL 2007 Finbrr Hollnd, Deprmen of Mhemics, Universiy College Cork, fhollnd@uccie; July 2, 2007 Three Problems Problem Assume i, b i, c i, i =, 2, 3 re rel
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 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 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 informationChallenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k
Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More information