2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).


 Ralf Stokes
 11 months ago
 Views:
Transcription
1 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 techniques or comintions thereof: the Power Rule for Antiderivtives recognition usustitution integrtion y prts sustitution with Pythgoren identities trigonometric sustitution prtil frctions nd other methods s specified in clss Approimte integrls with left right midpoint nd trpezoidl Riemnn sums (not new mteril) nd Simpson s Rule Given n error ound formul for ny of the ove pproimtions use it to find necessry numers of suintervls etc.; you need not memorize these formule ut e sure to understnd wht ech prt mens. The formuls re given on pges 557 nd 56. Understnd the two principl types of improper integrls nd how to determine whether ech converges; if n integrl converges now how to evlute it in terms of its (if possile) nd how to determine the error if n pproimtion is mde to the improper integrl with n infinite it. Find the rc length of eplicitlydefined curves (oth y in terms of nd in terms of y) using the formuls for endpoints nd s = + d d or d s = + dy. dy Understnd where these formuls cme from (pges ). Find the surfce re of surfce of revolution: if the surfce is given y function rotted out the is its re is S= πyds where ds is defined y rc length s ove or if the function is rotted out the yis its re is S= π ds. Agin understnd where these formuls cme from (pges 59 59). Prctice Prolems These prolems my e done with clcultor ecept where noted otherwise. The originl test of course required tht you show ll clcultions. The rte t which wter flows out of pipe in gllons per hour is given y differentile function R of time t. The tle elow shows the rtes s mesured every hours for hour period. hr t [ ] ( ) R t gl hr Use midpoint Riemnn sum with suintervls of equl length to pproimte R() t dt. Using correct hr units eplin the mening of your nswer in terms of wter flow. hr Use Simpson s Rule with 8 suintervls of equl length to pproimte R() t dt. c Use the Trpezoid Rule with sudivisions of equl length to pproimte R() t dt. Find the rc length long the grph of y 8 you my use your clcultor to evlute the integrl. hr hr hr = from ( y ) = ( ) to ( ) ( ) hr y = 88. Show the complete setup; Find the re of the surfce of revolution tht is generted y revolving the curve is. Show the complete setup; you my use your clcultor to evlute the integrl. y = e for out the y
2 + sin Let I e defined y I = d. Show using nonclcultor wor tht I converges. + sin Find vlue of for which I d hs n error of less thn.. Justify your nswer using nonclcultor wor. 5 5 Let I e defined y I = ln d. Estimte midpoint sum for I using suintervls of equl length. Determine the numer n of suintervls of equl length needed so tht the error ound for midpoint sum M n given y given intervl.) I K( ) M is less thn or equl to.. (Let K e n upper ound on f ( ) n n on the 6 Let f e the function defined y f ( ) = ln on nd the is. Find the volume of the solid generted y revolving R out the yis or show tht it is infinite. Justify your nswer. d 7 Evlute showing nonclcultor wor. 8 If f ( ) is positive nd f ( ) > on the intervl nd let R e the region etween the grph of y = f ( ) which of the following pproimtions to ( ) must e too lrge? I the right rectngle sum II the midpoint sum III the trpezoidl sum I only II only c I nd II only d III only e II nd III only e d? + + e + e + e c e + e + e + e e e e e e + e + e + e 9 If equl sudivisions of [ ] re used wht is the trpezoidl pproimtion of e e e d ( ) e ( ) f d The epression L + is Riemnn sum pproimtion for d d 5 c d 5 d d 5 5 e 5 d The length of curve from = to = is given y + 9. of the following could e n eqution for this curve? y = + y = 5 + c d y = 6 e y = If the curve contins the point ( ) y = which
3 Answers 58.6 gl 57. gl c 55 gl see solutions = ln.59 5 n = 6 π 6 V = 7 ln5 8 d 9 e Solutions Firstly ny integrl of rte gives totl mount. Therefore this integrl gives n pproimtion to the totl mount of wter tht flowed out of the pipe from t = hr to t = hr. To find the pproimtion to the integrl note tht ech suintervl hs length 6 hr ecuse the intervl s totl length is hr nd we re dividing it into hr prts; then = 6 hr. Therefore our suintervls if they re to e of equl length re t ( hr6 hr ) t ( 6 hr hr ) t ( hr8 hr ) nd t ( 8 hr hr ). Then ecuse we re using the midpoint rule our smple points re t = hr t = 9 hr t = 5 hr nd t = hr. So the suintervl midpoint pproimtion to the R t dt 6 hr. gl hr +. gl hr +. gl hr +. gl hr = 58.6 gl. hr integrl is () ( )( ) hr hr This time our suintervls hve length hr ecuse = hr. The smple points re ll the given vlues of t. 8 A formul for Simpson s Rule is given on the top of pge 56; using this the pproimtion to the integrl is hr R() t dt ( hr )( R( ) + R( ) + R( 6 ) + R( 9 ) + R( ) + R( 5 ) + R( 8 ) + R( ) + R( )) hr = hr = 57. gl. ( )( ) c We re c to suintervls of length 6 hr s in prt. However the smple points re now the endpoints of the suintervls with ll ut the first nd lst ones counted twice ech nd the totl divided y two (verging the two ses of ech trpezoid). Thus the trpezoidl pproimtion to the integrl in question is given y hr R t dt 6 R + R 6 + R + R 8 + R = ( ) () ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) hr = 55 gl. dy First solve the eqution for y eplicitly in terms of : y =. Therefore d the rc length formul from to is s = + d d we hve 6 s8 = + d 9 unplesnt to evlute y hnd ut ny CAS will gldly tell you tht it s equl to 5 nd you cn lso 7 pproimte it with This is est done in terms of where = ln y nd. y = The its will e e e = [ e ] = y the Power Rule. Since 8. This is thoroughly. Then the re is given e y S= π ln y + dy. y The ntiderivtive of this is not n elementry function ut cn e pproimted; CAS gives n pproimtion to the integrl s 7.55.
4 + sin for [ ) nd f ( ) d cn e shown to converge. The numertor of the originl integrnd + sin rnges 5 + sin in vlue from to 5 since sin rnges from to ; therefore lwys nd it is firly convenient to show tht d converges: it is equl to d. This integrl simplifies to sin = + = which is finite. Therefore d converges. + sin + sin + sin We cn divide d into two prts: d d + in which the second term represents the error of the pproimtion. Thus we need to find such tht.. + sin d However n + sin ntiderivtive of cnnot e epressed in terms of elementry functions so we insted consider 5 5 d. which is vlid ecuse the ltter is lwys greter mening the error in pproimting d will lwys e greter thn the error in pproimting the integrl in which we re ctully interested. Since d = = + = we re fced with the rther simple ts of finding such 5 tht.. Now ll we need to do is divide oth sides y 5 giving. nd then te the reciprocl of the squre root of oth sides. Therefore = 5. We show this using the Comprison Test; tht is we must find function f such tht f ( ) 5 Ech suintervl hs length ; the suintervls re ( ) nd ( 5 ) so the smple points re = nd =. 5 Thus the pproimtion is ln d ( ln + ln ) = ln + ln( ) = ln + ln + ln = 6 ln The integrnd s second derivtive is f ( ) = =. Therefore we re interested in finding n given or. n.5 n ; integer vlue of n so n = 6. therefore ( ) ( ) nd n upper ound for its solute vlue on the intervl [ ].5n or ( ) 5 is 5.. The right side simplifies: n n % n giving 5.8. Nturlly we wnt n 6 This must e done with cylindricl shells. Ech one hs dv = πrhdr = π( ln ) d so = ( ) V π ln d. For simplicity s se we will momentrily ignore the fctor of π. To find n ntiderivtive for this use integrtion y prts with u= ln nd dv = d. This gives du = nd v=. Therefore d d uv vdu = ln + = ln + d = ln + + C. Evluting this t is esy it gives ut t we hve n symptote. Therefore we must concern ourselves with ( ln + ). The second term clerly its to zero nd the first term my e nlyzed with L Hôpitl s Rule. We first rewrite it s ln. Now we hve n indeterminte form of type. Differentiting the numertor nd denomintor
5 Clerly this is lso zero. Therefore evluting the integrl hs given us = which we must multiply y π to give π. gives = ( ). 7 The integrnd s denomintor fctors to ( )( ) eecuted s follows: we epect something of the form the denomintors tht A( ) ( ) + nd integrtion y prtil frctions of A B + = + ( + )( ) + B + =. This is rerrnged to give ( A B) A B d is ( + )( ) so we now from clering + + = which llows A+ B= us to set up the system. This solves to ( AB ) = ( ). Thus the originl integrnd is equivlent to +. Now we cn integrte ech term giving ln( + ) + ln( ) which cn e rewrit A+ B= + ten s ln =. Now we evlute ln + s follows (ignoring the ; we ll put tht c in t the + = end): ln ln ln ln = = ln5 + ln. Since the function f ( ) = ln is continuous for very lrge vlues of we cn pull out the nturl logrithm function from the it: ln5 + ln = ln5 + ln = ln5. Now we need to return the + to its rightful plce; the nswer is ln5. > on the intervl in question mens tht the function is concve up. Therefore secnt line such s tht used in the trpezoid sum will lwys e ove the function itself ming the trpezoid pproimtion too lrge. Therefore III must e one of the correct nswers; options nd c re incorrect. Without nowing nything more out the shpe of the function it is impossile to conclude whether tngent line to the midpoint of suintervl will lie entirely ove or elow the function or neither so we cnnot me conclusion out the ccurcy of the midpoint sum. Therefore d is the correct nswer. 8 The fct tht f ( ) 9 The suintervls re [ ] [ ] nd [ ] ; ech hs length. With the trpezoidl pproimtion we count ech suintervl s endpoint twice ecept for the first nd lst one s outer points nd divide y on the outside to indicte verging ech trpezoid s ses. Thus the integrl s trpezoidl pproimtion is given y e ( ) e e e ( = ) = choice e. ( e e e e ) ( e e e e ) 5 This cn e rewritten s. The division y 5 indictes tht there re 5 suintervls nd the fct tht 5 = 5 the coefficient s numertor is indictes tht the difference of the its is. Rememer tht the summnd tes ( ) the form f + n which tells us tht the lower it of the integrl is. Since we now tht the difference of the its is the upper it is nd since n = 5 the function is f ( ) =. This mens tht the cor rect response is. The rc length formul for y on [ ] is tht y C = + for some C; since ( ) ( ) s = + d. d Here dy dy = 9 so =. This mens d d y = 6 is on the curve c is not possile nswer nd only is correct.
Math 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 informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
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 informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
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 informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
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 informationx ) dx dx x sec x over the interval (, ).
Curve on 6 For , () Evlute the integrl, n (b) check your nswer by ifferentiting. ( ). ( ). ( ).. 6. sin cos 7. sec csccot 8. sec (sec tn ) 9. sin csc. Evlute the integrl sin by multiplying the numertor
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 information5.1 Estimating with Finite Sums Calculus
5.1 ESTIMATING WITH FINITE SUMS Emple: Suppose from the nd to 4 th hour of our rod trip, ou trvel with the cruise control set to ectl 70 miles per hour for tht two hour stretch. How fr hve ou trveled during
More informationThe Trapezoidal Rule
SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion Approimte
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 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 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 information10. AREAS BETWEEN CURVES
. AREAS BETWEEN CURVES.. Ares etween curves So res ove the xxis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationSection 7.1 Area of a Region Between Two Curves
Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region
More informationUnit #10 De+inite Integration & The Fundamental Theorem Of Calculus
Unit # De+inite Integrtion & The Fundmentl Theorem Of Clculus. Find the re of the shded region ove nd explin the mening of your nswer. (squres re y units) ) The grph to the right is f(x) = x + 8x )Use
More informationKOÇ UNIVERSITY MATH 106 FINAL EXAM JANUARY 6, 2013
KOÇ UNIVERSITY MATH 6 FINAL EXAM JANUARY 6, 23 Durtion of Exm: 2 minutes INSTRUCTIONS: No clcultors nd no cell phones my be used on the test. No questions, nd tlking llowed. You must lwys explin your nswers
More informationCHAPTER : INTEGRATION Content pge Concept Mp 4. Integrtion of Algeric Functions 4 Eercise A 5 4. The Eqution of Curve from Functions of Grdients. 6 Ee
ADDITIONAL MATHEMATICS FORM 5 MODULE 4 INTEGRATION CHAPTER : INTEGRATION Content pge Concept Mp 4. Integrtion of Algeric Functions 4 Eercise A 5 4. The Eqution of Curve from Functions of Grdients. 6 Eercise
More informationMTH 122 Fall 2008 Essex County College Division of Mathematics Handout Version 10 1 October 14, 2008
MTH 22 Fll 28 Essex County College Division of Mthemtics Hndout Version October 4, 28 Arc Length Everyone should be fmilir with the distnce formul tht ws introduced in elementry lgebr. It is bsic formul
More informationAP Calculus BC Review Applications of Integration (Chapter 6) noting that one common instance of a force is weight
AP Clculus BC Review Applictions of Integrtion (Chpter Things to Know n Be Able to Do Fin the re between two curves by integrting with respect to x or y Fin volumes by pproximtions with cross sections:
More informationLINEAR ALGEBRA APPLIED
5.5 Applictions of Inner Product Spces 5.5 Applictions of Inner Product Spces 7 Find the cross product of two vectors in R. Find the liner or qudrtic lest squres pproimtion of function. Find the nthorder
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 informationImproper Integrals with Infinite Limits of Integration
6_88.qd // : PM Pge 578 578 CHAPTER 8 Integrtion Techniques, L Hôpitl s Rule, nd Improper Integrls Section 8.8 f() = d The unounded region hs n re of. Figure 8.7 Improper Integrls Evlute n improper integrl
More informationMATH 573 FINAL EXAM. May 30, 2007
MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems.
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 informationFarey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University
U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions
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 informationu( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 218, pp 4448): Determine the equation of the following graph.
nlyzing Dmped Oscilltions Prolem (Medor, exmple 218, pp 4448): Determine the eqution of the following grph. The eqution is ssumed to e of the following form f ( t) = K 1 u( t) + K 2 e!"t sin (#t + $
More informationFORM FIVE ADDITIONAL MATHEMATIC NOTE. ar 3 = (1) ar 5 = = (2) (2) (1) a = T 8 = 81
FORM FIVE ADDITIONAL MATHEMATIC NOTE CHAPTER : PROGRESSION Arithmetic Progression T n = + (n ) d S n = n [ + (n )d] = n [ + Tn ] S = T = T = S S Emple : The th term of n A.P. is 86 nd the sum of the first
More information[ ( ) ( )] Section 6.1 Area of Regions between two Curves. Goals: 1. To find the area between two curves
Gols: 1. To find the re etween two curves Section 6.1 Are of Regions etween two Curves I. Are of Region Between Two Curves A. Grphicl Represention = _ B. Integrl Represention [ ( ) ( )] f x g x dx = C.
More informationChapter Gauss Quadrature Rule of Integration
Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to
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 informationLecture 6: Singular Integrals, Open Quadrature rules, and Gauss Quadrature
Lecture notes on Vritionl nd Approximte Methods in Applied Mthemtics  A Peirce UBC Lecture 6: Singulr Integrls, Open Qudrture rules, nd Guss Qudrture (Compiled 6 August 7) In this lecture we discuss the
More informationRAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IITJEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties
M.Sc. (Mths), B.Ed, M.Phil (Mths) MATHEMATICS Mob. : 947084408 9546359990 M.Sc. (Mths), B.Ed, M.Phil (Mths) RAM RAJYA MORE, SIWAN XI th, XII th, TARGET IITJEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII
More informationFrom the Numerical. to the Theoretical in. Calculus
From the Numericl to the Theoreticl in Clculus Teching Contemporry Mthemtics NCSSM Ferury 67, 003 Doug Kuhlmnn Phillips Acdemy Andover, MA 01810 dkuhlmnn@ndover.edu How nd Why Numericl Integrtion Should
More information1 Error Analysis of Simple Rules for Numerical Integration
cs41: introduction to numericl nlysis 11/16/10 Lecture 19: Numericl Integrtion II Instructor: Professor Amos Ron Scries: Mrk Cowlishw, Nthnel Fillmore 1 Error Anlysis of Simple Rules for Numericl Integrtion
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 informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
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 informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationAdd and Subtract Rational Expressions. You multiplied and divided rational expressions. You will add and subtract rational expressions.
TEKS 8. A..A, A.0.F Add nd Subtrct Rtionl Epressions Before Now You multiplied nd divided rtionl epressions. You will dd nd subtrct rtionl epressions. Why? So you cn determine monthly cr lon pyments, s
More informationTest , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes
Test 2 8., 8.2, 8.4 (density only), 8.5 (work only), 9., 9.2 nd 9.3 relted test mteril nd mteril from prior clsses Locl to Globl Perspectives Anlyze smll pieces to understnd the big picture. Exmples: numericl
More informationMETHODS OF APPROXIMATING THE RIEMANN INTEGRALS AND APPLICATIONS
Journl of Young Scientist Volume III 5 ISSN 448; ISSN CDROM 449; ISSN Online 445; ISSNL 44 8 METHODS OF APPROXIMATING THE RIEMANN INTEGRALS AND APPLICATIONS An ALEXANDRU Scientific Coordintor: Assist
More informationHomework Assignment 3 Solution Set
Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution.
More informationPrecalculus Spring 2017
Preclculus Spring 2017 Exm 3 Summry (Section 4.1 through 5.2, nd 9.4) Section P.5 Find domins of lgebric expressions Simplify rtionl expressions Add, subtrct, multiply, & divide rtionl expressions Simplify
More information( ) as a fraction. Determine location of the highest
AB/ Clulus Exm Review Sheet Solutions A Prelulus Type prolems A1 A A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f( x) Set funtion equl to Ftor or use qudrti eqution if qudrti Grph to
More informationThe problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.
ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion
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 informationHomework Solution  Set 5 Due: Friday 10/03/08
CE 96 Introduction to the Theory of Computtion ll 2008 Homework olution  et 5 Due: ridy 10/0/08 1. Textook, Pge 86, Exercise 1.21. () 1 2 Add new strt stte nd finl stte. Mke originl finl stte nonfinl.
More informationNot for reproduction
AREA OF A SURFACE OF REVOLUTION cut h FIGURE FIGURE πr r r l h FIGURE A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundry of solid of revolution of the type
More informationCHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6. VOLUMES USING CROSSSECTIONS. A() ;, ; (digonl) ˆ Èˆ È V A() d d c d 6 (dimeter) c d c d c ˆ 6. A() ;, ; V A() d d. A() (edge) È Š È Š È ;, ; V A() d d 8
More informationx = a To determine the volume of the solid, we use a definite integral to sum the volumes of the slices as we let!x " 0 :
Clculus II MAT 146 Integrtion Applictions: Volumes of 3D Solids Our gol is to determine volumes of vrious shpes. Some of the shpes re the result of rotting curve out n xis nd other shpes re simply given
More informationSTRAND B: NUMBER THEORY
Mthemtics SKE, Strnd B UNIT B Indices nd Fctors: Tet STRAND B: NUMBER THEORY B Indices nd Fctors Tet Contents Section B. Squres, Cubes, Squre Roots nd Cube Roots B. Inde Nottion B. Fctors B. Prime Fctors,
More informationAnonymous Math 361: Homework 5. x i = 1 (1 u i )
Anonymous Mth 36: Homewor 5 Rudin. Let I be the set of ll u (u,..., u ) R with u i for ll i; let Q be the set of ll x (x,..., x ) R with x i, x i. (I is the unit cube; Q is the stndrd simplex in R ). Define
More informationDistance And Velocity
Unit #8  The Integrl Some problems nd solutions selected or dpted from HughesHllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl
More informationMTH 505: Number Theory Spring 2017
MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of $ nd $ s two denomintions of coins nd $c
More informationf[a] x + f[a + x] x + f[a +2 x] x + + f[b x] x =
Chpter 3 Symbolic Integrtion This chpter contins the bsic tricks of the symbolic integrtion trde. The gol of this chpter is not to mke you slow inncurte integrtion softwre, but rther to help you understnd
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 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 information3.1 Review of Sine, Cosine and Tangent for Right Angles
Foundtions of Mth 11 Section 3.1 Review of Sine, osine nd Tngent for Right Tringles 125 3.1 Review of Sine, osine nd Tngent for Right ngles The word trigonometry is derived from the Greek words trigon,
More informationFINALTERM EXAMINATION 2011 Calculus &. Analytical GeometryI
FINALTERM EXAMINATION 011 Clculus &. Anlyticl GeometryI Question No: 1 { Mrks: 1 )  Plese choose one If f is twice differentible function t sttionry point x 0 x 0 nd f ''(x 0 ) > 0 then f hs reltive...
More informationUSA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year
1/1/21. Fill in the circles in the picture t right with the digits 18, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.
More informationNumerical integration
2 Numericl integrtion This is pge i Printer: Opque this 2. Introduction Numericl integrtion is problem tht is prt of mny problems in the economics nd econometrics literture. The orgniztion of this chpter
More informationFinal Exam Study Guide
Finl Exm Study Guide Includes. Integrls & Antiderivtive Rules 2. Definite Integrls (Integrls with bounds) 3. Are Between Two Curves  Region Bounded by Two Curves 4. Consumer nd Producer Surplus. USubstitution.
More informationIntegration by Substitution. Pattern Recognition
SECTION Integrtion b Substitution 9 Section Integrtion b Substitution Use pttern recognition to find n indefinite integrl Use chnge of vribles to find n indefinite integrl Use the Generl Power Rule for
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 informationThermal Diffusivity. Paul Hughes. Department of Physics and Astronomy The University of Manchester Manchester M13 9PL. Second Year Laboratory Report
Therml iffusivity Pul Hughes eprtment of Physics nd Astronomy The University of nchester nchester 3 9PL Second Yer Lbortory Report Nov 4 Abstrct We investigted the therml diffusivity of cylindricl block
More informationMATHS NOTES. SUBJECT: Maths LEVEL: Higher TEACHER: Aidan Roantree. The Institute of Education Topics Covered: Powers and Logs
MATHS NOTES The Institute of Eduction 06 SUBJECT: Mths LEVEL: Higher TEACHER: Aidn Rontree Topics Covered: Powers nd Logs About Aidn: Aidn is our senior Mths techer t the Institute, where he hs been teching
More informationSection 14.3 Arc Length and Curvature
Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in
More informationMT Integral equations
MT58  Integrl equtions Introduction Integrl equtions occur in vriety of pplictions, often eing otined from differentil eqution. The reson for doing this is tht it my mke solution of the prolem esier or,
More informationChapter 1  Functions and Variables
Business Clculus 1 Chpter 1  Functions nd Vribles This Acdemic Review is brought to you free of chrge by preptests4u.com. Any sle or trde of this review is strictly prohibited. Business Clculus 1 Ch 1:
More informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
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 informationNumerical Integration
Chpter 1 Numericl Integrtion Numericl differentition methods compute pproximtions to the derivtive of function from known vlues of the function. Numericl integrtion uses the sme informtion to compute numericl
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More informationMATH 222 Second Semester Calculus. Fall 2015
MATH Second Semester Clculus Fll 5 Typeset:August, 5 Mth nd Semester Clculus Lecture notes version. (Fll 5) This is self contined set of lecture notes for Mth. The notes were written by Sigurd Angenent,
More informationWorksheet A EXPONENTIALS AND LOGARITHMS PMT. 1 Express each of the following in the form log a b = c. a 10 3 = 1000 b 3 4 = 81 c 256 = 2 8 d 7 0 = 1
C Worksheet A Epress ech of the following in the form log = c. 0 = 000 4 = 8 c 56 = 8 d 7 0 = e = f 5 = g 7 9 = 9 h 6 = 6 Epress ech of the following using inde nottion. log 5 5 = log 6 = 4 c 5 = log 0
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationLinear Systems with Constant Coefficients
Liner Systems with Constnt Coefficients 4305 Here is system of n differentil equtions in n unknowns: x x + + n x n, x x + + n x n, x n n x + + nn x n This is constnt coefficient liner homogeneous system
More information2. VECTORS AND MATRICES IN 3 DIMENSIONS
2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2dimensionl Vectors x A point in 3dimensionl spce cn e represented y column vector of the form y z zxis yxis z x y xxis Most of the
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 information18.01 Single Variable Calculus Fall 2006
MIT OpenCourseWre http://ocw.mit.edu 8. Single Vrible Clculus Fll 6 For informtion bout citing these mterils or our Terms of Use, visit: http://ocw.mit.edu/terms. Lecture 8 8. Fll 6 Lecture 8: Definite
More informationLecture 2 : Propositions DRAFT
CS/Mth 240: Introduction to Discrete Mthemtics 1/20/2010 Lecture 2 : Propositions Instructor: Dieter vn Melkeeek Scrie: Dlior Zelený DRAFT Lst time we nlyzed vrious mze solving lgorithms in order to illustrte
More informationMath 017. Materials With Exercises
Mth 07 Mterils With Eercises Jul 0 TABLE OF CONTENTS Lesson Vriles nd lgeric epressions; Evlution of lgeric epressions... Lesson Algeric epressions nd their evlutions; Order of opertions....... Lesson
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationPreSession Review. Part 1: Basic Algebra; Linear Functions and Graphs
PreSession Review Prt 1: Bsic Algebr; Liner Functions nd Grphs A. Generl Review nd Introduction to Algebr Hierrchy of Arithmetic Opertions Opertions in ny expression re performed in the following order:
More informationSTEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.
STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t
More information31.2. Numerical Integration. Introduction. Prerequisites. Learning Outcomes
Numericl Integrtion 3. Introduction In this Section we will present some methods tht cn be used to pproximte integrls. Attention will be pid to how we ensure tht such pproximtions cn be gurnteed to be
More informationContinuity. Recall the following properties of limits. Theorem. Suppose that lim. f(x) =L and lim. lim. [f(x)g(x)] = LM, lim
Recll the following properties of limits. Theorem. Suppose tht lim f() =L nd lim g() =M. Then lim [f() ± g()] = L + M, lim [f()g()] = LM, if M = 0, lim f() g() = L M. Furthermore, if f() g() for ll, then
More informationThe Dirichlet Problem in a Two Dimensional Rectangle. Section 13.5
The Dirichlet Prolem in Two Dimensionl Rectngle Section 13.5 1 Dirichlet Prolem in Rectngle In these notes we will pply the method of seprtion of vriles to otin solutions to elliptic prolems in rectngle
More informationLine and Surface Integrals: An Intuitive Understanding
Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of
More informationChapter 4 Contravariance, Covariance, and Spacetime Diagrams
Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More informationThe Fundamental Theorem of Algebra
The Fundmentl Theorem of Alger Jeremy J. Fries In prtil fulfillment of the requirements for the Mster of Arts in Teching with Speciliztion in the Teching of Middle Level Mthemtics in the Deprtment of Mthemtics.
More informationSome Methods in the Calculus of Variations
CHAPTER 6 Some Methods in the Clculus of Vritions 6. If we use the vried function ( α, ) α sin( ) + () Then d α cos ( ) () d Thus, the totl length of the pth is d S + d d α cos ( ) + α cos ( ) d Setting
More information