TECHNIQUES OF INTEGRATION

Size: px
Start display at page:

Download "TECHNIQUES OF INTEGRATION"

Transcription

1 7 TECHNIQUES OF INTEGRATION Simpso s Rule estimates itegrals b approimatig graphs with parabolas. Because of the Fudametal Theorem of Calculus, we ca itegrate a fuctio if we kow a atiderivative, that is, a idefiite itegral. We summarize here the most importat itegrals that we have leared so far. d C e d e C d l C a d a l a C si d cos C sec d ta C sec ta d sec C sih d cosh C ta d l sec C cos d si C csc d cot C csc cot d csc C cosh d sih C cot d l si C a d a ta a C sa d si a C I this chapter we develop techiques for usig these basic itegratio formulas to obtai idefiite itegrals of more complicated fuctios. We leared the most importat method of itegratio, the Substitutio Rule, i Sectio 5.5. The other geeral techique, itegratio b parts, is preseted i Sectio 7.. we lear methods that are special to particular classes of fuctios, such as trigoometric fuctios ad ratioal fuctios. Itegratio is ot as straightforward as differetiatio; there are o rules that absolutel guaratee obtaiig a idefiite itegral of a fuctio. Therefore we discuss a strateg for itegratio i Sectio

2 7. INTEGRATION BY PARTS Ever differetiatio rule has a correspodig itegratio rule. For istace, the Substitutio Rule for itegratio correspods to the Chai Rule for differetiatio. The rule that correspods to the Product Rule for differetiatio is called the rule for itegratio b parts. The Product Rule states that if f ad t are differetiable fuctios, the d f t f t tf d I the otatio for idefiite itegrals this equatio becomes f t tf d f t or f t d tf d f t We ca rearrage this equatio as f t d f t tf d Formula is called the formula for itegratio b parts. It is perhaps easier to remember i the followig otatio. Let u f ad v t. the differetials are du f d ad dv t d, so, b the Substitutio Rule, the formula for itegratio b parts becomes udv uv v du EXAPLE Fid si d. SOLUTION USING FORULA Suppose we choose f ad t si. f ad t cos. (For t we ca choose a atiderivative of t.) Thus, usig Formula, we have si d f t tf d cos cos d cos cos d cos si C It s wise to check the aswer b differetiatig it. If we do so, we get si, as epected. 53

3 5 CHAPTER 7 TECHNIQUES OF INTEGRATION SOLUTION USING FORULA Let N It is helpful to use the patter: u dv du v u du d dv si d v cos ad so u d u du si d si d cos cos d cos cos d cos si C NOTE Our aim i usig itegratio b parts is to obtai a simpler itegral tha the oe we started with. Thus i Eample we started with si dad epressed it i terms of the simpler itegral cos d. If we had istead chose u si ad dv d, the du cos dad v, so itegratio b parts gives si d si cos d Although this is true, cos dis a more difficult itegral tha the oe we started with. I geeral, whe decidig o a choice for u ad dv, we usuall tr to choose u f to be a fuctio that becomes simpler whe differetiated (or at least ot more complicated) as log as dv t d ca be readil itegrated to give v. V EXAPLE Evaluate l d. SOLUTION Here we do t have much choice for u ad dv. Let u l dv d du d v Itegratig b parts, we get l d l d N It s customar to write d as d. N Check the aswer b differetiatig it. l d l C Itegratio b parts is effective i this eample because the derivative of the fuctio f l is simpler tha f.

4 SECTION 7. INTEGRATION BY PARTS 55 V EXAPLE 3 Fid t et dt. SOLUTION Notice that t becomes simpler whe differetiated (whereas e t is uchaged whe differetiated or itegrated), so we choose Itegratio b parts gives u t du tdt dv e t dt v e t 3 t e t dt t e t te t dt The itegral that we obtaied, te t dt, is simpler tha the origial itegral but is still ot obvious. Therefore, we use itegratio b parts a secod time, this time with u t ad dv e t dt. du dt, v e t, ad te t dt te t e t dt te t e t C Puttig this i Equatio 3, we get t e t dt t e t te t dt t e t te t e t C t e t te t e t C where C C N A easier method, usig comple umbers, is give i Eercise 5 i Appedi H. V EXAPLE Evaluate e si d. SOLUTION Neither e or si becomes simpler whe differetiated, but we tr choosig u e ad dv si dawa. du e d ad v cos, so itegratio b parts gives e si d e cos e cos d The itegral that we have obtaied, e cos d, is o simpler tha the origial oe, but at least it s o more difficult. Havig had success i the precedig eample itegratig b parts twice, we persevere ad itegrate b parts agai. This time we use u e ad dv cos d. du e d, v si, ad 5 e cos d e si e si d At first glace, it appears as if we have accomplished othig because we have arrived at e si d, which is where we started. However, if we put the epressio for e cos d from Equatio 5 ito Equatio we get e si d e cos e si e si d

5 56 CHAPTER 7 TECHNIQUES OF INTEGRATION N Figure illustrates Eample b showig the graphs of f e si ad F e si cos. As a visual check o our work, otice that f whe F has a maimum or miimum. This ca be regarded as a equatio to be solved for the ukow itegral. Addig e si dto both sides, we obtai e si d e cos e si F Dividig b ad addig the costat of itegratio, we get f e si d e si cos C _3 FIGURE _ 6 If we combie the formula for itegratio b parts with Part of the Fudametal Theorem of Calculus, we ca evaluate defiite itegrals b parts. Evaluatig both sides of Formula betwee a ad b, assumig f ad t are cotiuous, ad usig the Fudametal Theorem, we obtai 6 b a b f t d f t] a b tf d a EXAPLE 5 Calculate ta d. SOLUTION Let u ta du dv d d v So Formula 6 gives N Sice ta for, the itegral i Eample 5 ca be iterpreted as the area of the regio show i Figure. ta d ta ] ta ta d d d =ta! To evaluate this itegral we use the substitutio t (sice u has aother meaig i this eample). dt d, so d dt. Whe, t ; whe, t ; so d dt l t t ] l l l FIGURE Therefore ta d d l

6 SECTION 7. INTEGRATION BY PARTS 57 EXAPLE 6 Prove the reductio formula N Equatio 7 is called a reductio formula because the epoet has bee reduced to ad. 7 si d cos si si d where is a iteger. SOLUTION Let u si du si cos d dv si d v cos so itegratio b parts gives si d cos si si cos d Sice cos si, we have si d cos si si d si d As i Eample, we solve this equatio for the desired itegral b takig the last term o the right side to the left side. Thus we have si d cos si si d or si d cos si si d The reductio formula (7) is useful because b usig it repeatedl we could evetuall epress si di terms of si d(if is odd) or si d d (if is eve). 7. EXERCISES Evaluate the itegral usig itegratio b parts with the idicated choices of u ad dv.. arcta tdt. p 5 l pdp. l d; u l, dv d 3. t sec tdt. s s ds cos d. ; u, dv cos d 5. l d 6. t sih mt dt 3 3 Evaluate the itegral. 7. e si 3 d 8. e cos d 3. cos 5d 5. re r dr 6. t si t dt 7. si d 8. cos m d 9. l d. si d. e d 9.. t cosh tdt. 3. l d. 3 cos d t si 3tdt. e d 9 l s d

7 58 CHAPTER 7 TECHNIQUES OF INTEGRATION 5. d 6. e 7. cos d cos lsi d 3. s3 arcta d l d 3 r 3 s r dr (b) se part (a) to evaluate si 3 dad si 5 d. (c) se part (a) to show that, for odd powers of sie, si d 6. Prove that, for eve powers of sie, l d 3. t e s sit s ds si d irst make a substitutio ad the use itegratio b parts to evaluate the itegral. 33. cos s d 3. t 3 e t dt 35. s s 3 cos d ; 39 Evaluate the ide ite itegral. llustrate, ad check that our aswer is reasoable, b graphig both the fuctio ad its atiderivative (take ). 3. (a) se the reductio formula i Eample to show that l d 38. sil d 39. 3e d. 3 l d. 3 s d. si d (b) se part (a) ad the reductio formula to evaluate si d.. (a) Prove the reductio formula cos d cos si (b) se part (a) to evaluate cos d. (c) se parts (a) ad (b) to evaluate cos d. 5. (a) se the reductio formula i Eample to show that si d si d where is a iteger. si e cos t si t dt si d cos d 7 5 se itegratio b parts to prove the reductio formula se Eercise 7 to d l 3 d. 5. se Eercise to d e d id the area of the regio bouded b the give curves. 53. e.,, 5. l d l l d e d e e d ta d ta ta d sec d ta sec 5 l, ; se a graph to d approimate coordiates of the poits of itersectio of the give curves. d (approi matel) the area of the regio bouded b the curves. 55. si, 56. arcta 3, l 5 sec d 57 6 se the method of clidrical shells to d the volume geerated b rotatig the regio bouded b the give curves about the speci ed ais. 57. cos,, ; about the ais 58. e, e, ; about the ais 59. e,,, ; about 6. e,, ; about the ais

y udv uv y v du 7.1 INTEGRATION BY PARTS

y udv uv y v du 7.1 INTEGRATION BY PARTS 7. INTEGRATION BY PARTS Ever differetitio rule hs correspodig itegrtio rule. For istce, the Substitutio Rule for itegrtio correspods to the Chi Rule for differetitio. The rule tht correspods to the Product

More information

INTEGRATION BY PARTS (TABLE METHOD)

INTEGRATION BY PARTS (TABLE METHOD) INTEGRATION BY PARTS (TABLE METHOD) Suppose you wat to evaluate cos d usig itegratio by parts. Usig the u dv otatio, we get So, u dv d cos du d v si cos d si si d or si si d We see that it is ecessary

More information

x x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0,

x x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0, Math Activity 9( Due with Fial Eam) Usig first ad secod Taylor polyomials with remaider, show that for, 8 Usig a secod Taylor polyomial with remaider, fid the best costat C so that for, C 9 The th Derivative

More information

2 f(x) dx = 1, 0. 2f(x 1) dx d) 1 4t t6 t. t 2 dt i)

2 f(x) dx = 1, 0. 2f(x 1) dx d) 1 4t t6 t. t 2 dt i) Math PracTest Be sure to review Lab (ad all labs) There are lots of good questios o it a) State the Mea Value Theorem ad draw a graph that illustrates b) Name a importat theorem where the Mea Value Theorem

More information

MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of

MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of MATHEMATICS 6 The differetial equatio represetig the family of curves where c is a positive parameter, is of Order Order Degree (d) Degree (a,c) Give curve is y c ( c) Differetiate wrt, y c c y Hece differetial

More information

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said

More information

Honors Calculus Homework 13 Solutions, due 12/8/5

Honors Calculus Homework 13 Solutions, due 12/8/5 Hoors Calculus Homework Solutios, due /8/5 Questio Let a regio R i the plae be bouded by the curves y = 5 ad = 5y y. Sketch the regio R. The two curves meet where both equatios hold at oce, so where: y

More information

MATH 10550, EXAM 3 SOLUTIONS

MATH 10550, EXAM 3 SOLUTIONS MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece,

More information

September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1

September 2012 C1 Note. C1 Notes (Edexcel) Copyright   - For AS, A2 notes and IGCSE / GCSE worksheets 1 September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright

More information

PRACTICE FINAL/STUDY GUIDE SOLUTIONS

PRACTICE FINAL/STUDY GUIDE SOLUTIONS Last edited December 9, 03 at 4:33pm) Feel free to sed me ay feedback, icludig commets, typos, ad mathematical errors Problem Give the precise meaig of the followig statemets i) a f) L ii) a + f) L iii)

More information

MATH Exam 1 Solutions February 24, 2016

MATH Exam 1 Solutions February 24, 2016 MATH 7.57 Exam Solutios February, 6. Evaluate (A) l(6) (B) l(7) (C) l(8) (D) l(9) (E) l() 6x x 3 + dx. Solutio: D We perform a substitutio. Let u = x 3 +, so du = 3x dx. Therefore, 6x u() x 3 + dx = [

More information

1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute

1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute Math, Calculus II Fial Eam Solutios. 5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute 4 d. The check your aswer usig the Evaluatio Theorem. ) ) Solutio: I this itegral,

More information

Taylor Polynomials and Taylor Series

Taylor Polynomials and Taylor Series Taylor Polyomials ad Taylor Series Cotets Taylor Polyomials... Eample.... Eample.... 4 Eample.3... 5 Eercises... 6 Eercise Solutios... 8 Taylor s Iequality... Eample.... Eample.... Eercises... 3 Eercise

More information

MATH 1A FINAL (7:00 PM VERSION) SOLUTION. (Last edited December 25, 2013 at 9:14pm.)

MATH 1A FINAL (7:00 PM VERSION) SOLUTION. (Last edited December 25, 2013 at 9:14pm.) MATH A FINAL (7: PM VERSION) SOLUTION (Last edited December 5, 3 at 9:4pm.) Problem. (i) Give the precise defiitio of the defiite itegral usig Riema sums. (ii) Write a epressio for the defiite itegral

More information

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 113, Calculus II Winter 2007 Final Exam Solutions Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this

More information

Calculus with Analytic Geometry 2

Calculus with Analytic Geometry 2 Calculus with Aalytic Geometry Fial Eam Study Guide ad Sample Problems Solutios The date for the fial eam is December, 7, 4-6:3p.m. BU Note. The fial eam will cosist of eercises, ad some theoretical questios,

More information

f t dt. Write the third-degree Taylor polynomial for G

f t dt. Write the third-degree Taylor polynomial for G AP Calculus BC Homework - Chapter 8B Taylor, Maclauri, ad Power Series # Taylor & Maclauri Polyomials Critical Thikig Joural: (CTJ: 5 pts.) Discuss the followig questios i a paragraph: What does it mea

More information

Chapter 4. Fourier Series

Chapter 4. Fourier Series Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,

More information

Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist.

Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist. Topic 5 [44 marks] 1a (i) Fid the rage of values of for which eists 1 Write dow the value of i terms of 1, whe it does eist Fid the solutio to the differetial equatio 1b give that y = 1 whe = π (cos si

More information

Example 2. Find the upper bound for the remainder for the approximation from Example 1.

Example 2. Find the upper bound for the remainder for the approximation from Example 1. Lesso 8- Error Approimatios 0 Alteratig Series Remaider: For a coverget alteratig series whe approimatig the sum of a series by usig oly the first terms, the error will be less tha or equal to the absolute

More information

MAT136H1F - Calculus I (B) Long Quiz 1. T0101 (M3) Time: 20 minutes. The quiz consists of four questions. Each question is worth 2 points. Good Luck!

MAT136H1F - Calculus I (B) Long Quiz 1. T0101 (M3) Time: 20 minutes. The quiz consists of four questions. Each question is worth 2 points. Good Luck! MAT36HF - Calculus I (B) Log Quiz. T (M3) Time: 2 miutes Last Name: Studet ID: First Name: Please mark your tutorial sectio: T (M3) T2 (R4) T3 (T4) T5 (T5) T52 (R5) The quiz cosists of four questios. Each

More information

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew

Z ß cos x + si x R du We start with the substitutio u = si(x), so du = cos(x). The itegral becomes but +u we should chage the limits to go with the ew Problem ( poits) Evaluate the itegrals Z p x 9 x We ca draw a right triagle labeled this way x p x 9 From this we ca read off x = sec, so = sec ta, ad p x 9 = R ta. Puttig those pieces ito the itegralrwe

More information

CHAPTER 4 Integration

CHAPTER 4 Integration CHAPTER Itegratio Sectio. Atierivatives a Iefiite Itegratio......... 77 Sectio. Area............................. 8 Sectio. Riema Sums a Defiite Itegrals........... 88 Sectio. The Fuametal Theorem of Calculus..........

More information

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b) Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig

More information

EDEXCEL NATIONAL CERTIFICATE UNIT 4 MATHEMATICS FOR TECHNICIANS OUTCOME 4 - CALCULUS

EDEXCEL NATIONAL CERTIFICATE UNIT 4 MATHEMATICS FOR TECHNICIANS OUTCOME 4 - CALCULUS EDEXCEL NATIONAL CERTIFICATE UNIT 4 MATHEMATICS FOR TECHNICIANS OUTCOME 4 - CALCULUS TUTORIAL 1 - DIFFERENTIATION Use the elemetary rules of calculus arithmetic to solve problems that ivolve differetiatio

More information

1 Cabin. Professor: What is. Student: ln Cabin oh Log Cabin! Professor: No. Log Cabin + C = A Houseboat!

1 Cabin. Professor: What is. Student: ln Cabin oh Log Cabin! Professor: No. Log Cabin + C = A Houseboat! MATH 4 Sprig 0 Exam # Tuesday March st Sectios: Sectios 6.-6.6; 6.8; 7.-7.4 Name: Score: = 00 Istructios:. You will have a total of hour ad 50 miutes to complete this exam.. A No-Graphig Calculator may

More information

Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations

Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations Chapter The Solutio of Numerical Algebraic ad Trascedetal Equatios Itroductio I this chapter we shall discuss some umerical methods for solvig algebraic ad trascedetal equatios. The equatio f( is said

More information

Seunghee Ye Ma 8: Week 5 Oct 28

Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value

More information

( a) ( ) 1 ( ) 2 ( ) ( ) 3 3 ( ) =!

( a) ( ) 1 ( ) 2 ( ) ( ) 3 3 ( ) =! .8,.9: Taylor ad Maclauri Series.8. Although we were able to fid power series represetatios for a limited group of fuctios i the previous sectio, it is ot immediately obvious whether ay give fuctio has

More information

Calculus 2 - D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.)

Calculus 2 - D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.) Calculus - D Yue Fial Eam Review (Versio //7 Please report ay possible typos) NOTE: The review otes are oly o topics ot covered o previous eams See previous review sheets for summary of previous topics

More information

Complete Solutions to Supplementary Exercises on Infinite Series

Complete Solutions to Supplementary Exercises on Infinite Series Coplete Solutios to Suppleetary Eercises o Ifiite Series. (a) We eed to fid the su ito partial fractios gives By the cover up rule we have Therefore Let S S A / ad A B B. Covertig the suad / the by usig

More information

MA1200 Exercise for Chapter 7 Techniques of Differentiation Solutions. First Principle 1. a) To simplify the calculation, note. Then. lim h.

MA1200 Exercise for Chapter 7 Techniques of Differentiation Solutions. First Principle 1. a) To simplify the calculation, note. Then. lim h. MA00 Eercise for Chapter 7 Techiques of Differetiatio Solutios First Priciple a) To simplify the calculatio, ote The b) ( h) lim h 0 h lim h 0 h[ ( h) ( h) h ( h) ] f '( ) Product/Quotiet/Chai Rules a)

More information

AP Calculus BC Review Applications of Derivatives (Chapter 4) and f,

AP Calculus BC Review Applications of Derivatives (Chapter 4) and f, AP alculus B Review Applicatios of Derivatives (hapter ) Thigs to Kow ad Be Able to Do Defiitios of the followig i terms of derivatives, ad how to fid them: critical poit, global miima/maima, local (relative)

More information

CALCULUS BASIC SUMMER REVIEW

CALCULUS BASIC SUMMER REVIEW CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=

More information

Taylor Series (BC Only)

Taylor Series (BC Only) Studet Study Sessio Taylor Series (BC Oly) Taylor series provide a way to fid a polyomial look-alike to a o-polyomial fuctio. This is doe by a specific formula show below (which should be memorized): Taylor

More information

CHAPTER 10 INFINITE SEQUENCES AND SERIES

CHAPTER 10 INFINITE SEQUENCES AND SERIES CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece

More information

Calculus II exam 1 6/18/07 All problems are worth 10 points unless otherwise noted. Show all analytic work.

Calculus II exam 1 6/18/07 All problems are worth 10 points unless otherwise noted. Show all analytic work. 9.-0 Calculus II exam 6/8/07 All problems are worth 0 poits uless otherwise oted. Show all aalytic work.. (5 poits) Prove that the area eclosed i the circle. f( x) = x +, 0 x. Use the approximate the area

More information

y = f x x 1. If f x = e 2x tan -1 x, then f 1 = e 2 2 e 2 p C e 2 D e 2 p+1 4

y = f x x 1. If f x = e 2x tan -1 x, then f 1 = e 2 2 e 2 p C e 2 D e 2 p+1 4 . If f = e ta -, the f = e e p e e p e p+ 4 f = e ta -, so f = e ta - + e, so + f = e p + e = e p + e or f = e p + 4. The slope of the lie taget to the curve - + = at the poit, - is - 5 Differetiate -

More information

2 ) 5. (a) (1)(3) + (1)(2) = 5 (b) {area of shaded region in Fig. 24b} < 5

2 ) 5. (a) (1)(3) + (1)(2) = 5 (b) {area of shaded region in Fig. 24b} < 5 Odd Aswers: Chapter Four Cotemporary Calculus PROBLEM ANSWERS Chapter Four Sectio 4.. (a) ()() + (8)(4) = 5 (b) ()() ()(8) = 76. bh + b(h h) = bh + bh bh = b ( h + H ) 5. (a) ()() + ()() = 5 (b) {area

More information

MATH 2411 Spring 2011 Practice Exam #1 Tuesday, March 1 st Sections: Sections ; 6.8; Instructions:

MATH 2411 Spring 2011 Practice Exam #1 Tuesday, March 1 st Sections: Sections ; 6.8; Instructions: MATH 411 Sprig 011 Practice Exam #1 Tuesday, March 1 st Sectios: Sectios 6.1-6.6; 6.8; 7.1-7.4 Name: Score: = 100 Istructios: 1. You will have a total of 1 hour ad 50 miutes to complete this exam.. A No-Graphig

More information

Representing Functions as Power Series. 3 n ...

Representing Functions as Power Series. 3 n ... Math Fall 7 Lab Represetig Fuctios as Power Series I. Itrouctio I sectio.8 we leare the series c c c c c... () is calle a power series. It is a uctio o whose omai is the set o all or which it coverges.

More information

MATH CALCULUS II Objectives and Notes for Test 4

MATH CALCULUS II Objectives and Notes for Test 4 MATH 44 - CALCULUS II Objectives ad Notes for Test 4 To do well o this test, ou should be able to work the followig tpes of problems. Fid a power series represetatio for a fuctio ad determie the radius

More information

18.01 Calculus Jason Starr Fall 2005

18.01 Calculus Jason Starr Fall 2005 Lecture 18. October 5, 005 Homework. Problem Set 5 Part I: (c). Practice Problems. Course Reader: 3G 1, 3G, 3G 4, 3G 5. 1. Approximatig Riema itegrals. Ofte, there is o simpler expressio for the atiderivative

More information

n n 2 + 4i = lim 2 n lim 1 + 4x 2 dx = 1 2 tan ( 2i 2 x x dx = 1 2 tan 1 2 = 2 n, x i = a + i x = 2i

n n 2 + 4i = lim 2 n lim 1 + 4x 2 dx = 1 2 tan ( 2i 2 x x dx = 1 2 tan 1 2 = 2 n, x i = a + i x = 2i . ( poits) Fid the limits. (a) (6 poits) lim ( + + + 3 (6 poits) lim h h h 6 微甲 - 班期末考解答和評分標準 +h + + + t3 dt. + 3 +... + 5 ) = lim + i= + i. Solutio: (a) lim i= + i = lim i= + ( i ) = lim x i= + x i =

More information

In algebra one spends much time finding common denominators and thus simplifying rational expressions. For example:

In algebra one spends much time finding common denominators and thus simplifying rational expressions. For example: 74 The Method of Partial Fractios I algebra oe speds much time fidig commo deomiators ad thus simplifyig ratioal epressios For eample: + + + 6 5 + = + = = + + + + + ( )( ) 5 It may the seem odd to be watig

More information

Topic 1 2: Sequences and Series. A sequence is an ordered list of numbers, e.g. 1, 2, 4, 8, 16, or

Topic 1 2: Sequences and Series. A sequence is an ordered list of numbers, e.g. 1, 2, 4, 8, 16, or Topic : Sequeces ad Series A sequece is a ordered list of umbers, e.g.,,, 8, 6, or,,,.... A series is a sum of the terms of a sequece, e.g. + + + 8 + 6 + or... Sigma Notatio b The otatio f ( k) is shorthad

More information

2018 MAΘ National Convention Mu Individual Solutions ( ) ( ) + + +

2018 MAΘ National Convention Mu Individual Solutions ( ) ( ) + + + 8 MΘ Natioal ovetio Mu Idividual Solutios b a f + f + + f + + + ) (... ) ( l ( ) l ( ) l ( 7) l ( ) ) l ( 8) ) a( ) cos + si ( ) a' ( ) cos ( ) a" ( ) si ( ) ) For a fuctio to be differetiable at c, it

More information

Fundamental Concepts: Surfaces and Curves

Fundamental Concepts: Surfaces and Curves UNDAMENTAL CONCEPTS: SURACES AND CURVES CHAPTER udametal Cocepts: Surfaces ad Curves. INTRODUCTION This chapter describes two geometrical objects, vi., surfaces ad curves because the pla a ver importat

More information

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =

More information

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet

More information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This

More information

CHAPTER 4 Integration

CHAPTER 4 Integration CHAPTER Itegratio Sectio. Atierivatives a Iefiite Itegratio......... Sectio. Area............................. Sectio. Riema Sums a Defiite Itegrals........... Sectio. The Fuametal Theorem of Calculus..........

More information

The type of limit that is used to find TANGENTS and VELOCITIES gives rise to the central idea in DIFFERENTIAL CALCULUS, the DERIVATIVE.

The type of limit that is used to find TANGENTS and VELOCITIES gives rise to the central idea in DIFFERENTIAL CALCULUS, the DERIVATIVE. NOTES : LIMITS AND DERIVATIVES Name: Date: Period: Iitial: LESSON.1 THE TANGENT AND VELOCITY PROBLEMS Pre-Calculus Mathematics Limit Process Calculus The type of it that is used to fid TANGENTS ad VELOCITIES

More information

Student s Printed Name:

Student s Printed Name: Studet s Prited Name: Istructor: XID: C Sectio: No questios will be aswered durig this eam. If you cosider a questio to be ambiguous, state your assumptios i the margi ad do the best you ca to provide

More information

Additional Notes on Power Series

Additional Notes on Power Series Additioal Notes o Power Series Mauela Girotti MATH 37-0 Advaced Calculus of oe variable Cotets Quick recall 2 Abel s Theorem 2 3 Differetiatio ad Itegratio of Power series 4 Quick recall We recall here

More information

MATH spring 2008 lecture 3 Answers to selected problems. 0 sin14 xdx = x dx. ; (iv) x +

MATH spring 2008 lecture 3 Answers to selected problems. 0 sin14 xdx = x dx. ; (iv) x + MATH - sprig 008 lecture Aswers to selected problems INTEGRALS. f =? For atiderivatives i geeral see the itegrals website at http://itegrals.wolfram.com. (5-vi (0 i ( ( i ( π ; (v π a. This is example

More information

HKDSE Exam Questions Distribution

HKDSE Exam Questions Distribution HKDSE Eam Questios Distributio Sample Paper Practice Paper DSE 0 Topics A B A B A B. Biomial Theorem. Mathematical Iductio 0 3 3 3. More about Trigoometric Fuctios, 0, 3 0 3. Limits 6. Differetiatio 7

More information

( ) ( ) ( ) ( ) ( + ) ( )

( ) ( ) ( ) ( ) ( + ) ( ) LSM Nov. 00 Cotet List Mathematics (AH). Algebra... kow ad use the otatio!, C r ad r.. kow the results = r r + + = r r r..3 kow Pascal's triagle. Pascal's triagle should be eteded up to = 7...4 kow ad

More information

1 Approximating Integrals using Taylor Polynomials

1 Approximating Integrals using Taylor Polynomials Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios................................................

More information

Calculus I Practice Test Problems for Chapter 5 Page 1 of 9

Calculus I Practice Test Problems for Chapter 5 Page 1 of 9 Calculus I Practice Test Problems for Chapter 5 Page of 9 This is a set of practice test problems for Chapter 5. This is i o way a iclusive set of problems there ca be other types of problems o the actual

More information

MATH2007* Partial Answers to Review Exercises Fall 2004

MATH2007* Partial Answers to Review Exercises Fall 2004 MATH27* Partial Aswers to Review Eercises Fall 24 Evaluate each of the followig itegrals:. Let u cos. The du si ad Hece si ( cos 2 )(si ) (u 2 ) du. si u 2 cos 7 u 7 du Please fiish this. 2. We use itegratio

More information

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense, 3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [

More information

LIMIT. f(a h). f(a + h). Lim x a. h 0. x 1. x 0. x 0. x 1. x 1. x 2. Lim f(x) 0 and. x 0

LIMIT. f(a h). f(a + h). Lim x a. h 0. x 1. x 0. x 0. x 1. x 1. x 2. Lim f(x) 0 and. x 0 J-Mathematics LIMIT. INTRODUCTION : The cocept of it of a fuctio is oe of the fudametal ideas that distiguishes calculus from algebra ad trigoometr. We use its to describe the wa a fuctio f varies. Some

More information

COMPUTING SUMS AND THE AVERAGE VALUE OF THE DIVISOR FUNCTION (x 1) + x = n = n.

COMPUTING SUMS AND THE AVERAGE VALUE OF THE DIVISOR FUNCTION (x 1) + x = n = n. COMPUTING SUMS AND THE AVERAGE VALUE OF THE DIVISOR FUNCTION Abstract. We itroduce a method for computig sums of the form f( where f( is ice. We apply this method to study the average value of d(, where

More information

Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5

Ma 4121: Introduction to Lebesgue Integration Solutions to Homework Assignment 5 Ma 42: Itroductio to Lebesgue Itegratio Solutios to Homework Assigmet 5 Prof. Wickerhauser Due Thursday, April th, 23 Please retur your solutios to the istructor by the ed of class o the due date. You

More information

When dealing with series, n is always a positive integer. Remember at every, sine has a value of zero, which means

When dealing with series, n is always a positive integer. Remember at every, sine has a value of zero, which means Fourier Series Some Prelimiar Ideas: Odd/Eve Fuctios: Sie is odd, which meas si ( ) si Cosie is eve, which meas cos ( ) cos Secial values of siie a cosie at Whe dealig with series, is alwas a ositive iteger.

More information

Polynomial Functions and Their Graphs

Polynomial Functions and Their Graphs Polyomial Fuctios ad Their Graphs I this sectio we begi the study of fuctios defied by polyomial expressios. Polyomial ad ratioal fuctios are the most commo fuctios used to model data, ad are used extesively

More information

Castiel, Supernatural, Season 6, Episode 18

Castiel, Supernatural, Season 6, Episode 18 13 Differetial Equatios the aswer to your questio ca best be epressed as a series of partial differetial equatios... Castiel, Superatural, Seaso 6, Episode 18 A differetial equatio is a mathematical equatio

More information

Subject: Differential Equations & Mathematical Modeling-III

Subject: Differential Equations & Mathematical Modeling-III Power Series Solutios of Differetial Equatios about Sigular poits Subject: Differetial Equatios & Mathematical Modelig-III Lesso: Power series solutios of differetial equatios about Sigular poits Lesso

More information

FINALTERM EXAMINATION Fall 9 Calculus & Aalytical Geometry-I Questio No: ( Mars: ) - Please choose oe Let f ( x) is a fuctio such that as x approaches a real umber a, either from left or right-had-side,

More information

For use only in Badminton School November 2011 C2 Note. C2 Notes (Edexcel)

For use only in Badminton School November 2011 C2 Note. C2 Notes (Edexcel) For use oly i Badmito School November 0 C Note C Notes (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets For use oly i Badmito School November 0 C Note Copyright www.pgmaths.co.uk

More information

Diploma Programme. Mathematics HL guide. First examinations 2014

Diploma Programme. Mathematics HL guide. First examinations 2014 Diploma Programme First eamiatios 014 33 Topic 6 Core: Calculus The aim of this topic is to itroduce studets to the basic cocepts ad techiques of differetial ad itegral calculus ad their applicatio. 6.1

More information

Strategy for Testing Series

Strategy for Testing Series Strategy for Testig Series We ow have several ways of testig a series for covergece or divergece; the problem is to decide which test to use o which series. I this respect testig series is similar to itegratig

More information

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is Calculus BC Fial Review Name: Revised 7 EXAM Date: Tuesday, May 9 Remiders:. Put ew batteries i your calculator. Make sure your calculator is i RADIAN mode.. Get a good ight s sleep. Eat breakfast. Brig:

More information

Subject: Differential Equations & Mathematical Modeling -III. Lesson: Power series solutions of Differential Equations. about ordinary points

Subject: Differential Equations & Mathematical Modeling -III. Lesson: Power series solutions of Differential Equations. about ordinary points Power series solutio of Differetial equatios about ordiary poits Subject: Differetial Equatios & Mathematical Modelig -III Lesso: Power series solutios of Differetial Equatios about ordiary poits Lesso

More information

CS 270 Algorithms. Oliver Kullmann. Growth of Functions. Divide-and- Conquer Min-Max- Problem. Tutorial. Reading from CLRS for week 2

CS 270 Algorithms. Oliver Kullmann. Growth of Functions. Divide-and- Conquer Min-Max- Problem. Tutorial. Reading from CLRS for week 2 Geeral remarks Week 2 1 Divide ad First we cosider a importat tool for the aalysis of algorithms: Big-Oh. The we itroduce a importat algorithmic paradigm:. We coclude by presetig ad aalysig two examples.

More information

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet

More information

WELCOME. Welcome to the Course. to MATH 104: Calculus I

WELCOME. Welcome to the Course. to MATH 104: Calculus I WELCOME to MATH : Calculus I Welcome to the Course. Pe Math Calculus I. Topics: quick review of high school calculus, methods ad applicatios of itegratio, ifiite series ad applicatios, some fuctios of

More information

Exponential and Trigonometric Functions Lesson #1

Exponential and Trigonometric Functions Lesson #1 Epoetial ad Trigoometric Fuctios Lesso # Itroductio To Epoetial Fuctios Cosider a populatio of 00 mice which is growig uder plague coditios. If the mouse populatio doubles each week we ca costruct a table

More information

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as

More information

Riemann Sums y = f (x)

Riemann Sums y = f (x) Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, o-egative fuctio o the closed iterval [a, b] Fid

More information

MEI Conference 2009 Stretching students: A2 Core

MEI Conference 2009 Stretching students: A2 Core MEI Coferece 009 Stretchig studets: A Core Preseter: Berard Murph berard.murph@mei.org.uk Workshop G How ca ou prove that these si right-agled triagles fit together eactl to make a 3-4-5 triagle? What

More information

MAT 271 Project: Partial Fractions for certain rational functions

MAT 271 Project: Partial Fractions for certain rational functions MAT 7 Project: Partial Fractios for certai ratioal fuctios Prerequisite kowledge: partial fractios from MAT 7, a very good commad of factorig ad complex umbers from Precalculus. To complete this project,

More information

Chapter 8. Euler s Gamma function

Chapter 8. Euler s Gamma function Chapter 8 Euler s Gamma fuctio The Gamma fuctio plays a importat role i the fuctioal equatio for ζ(s that we will derive i the ext chapter. I the preset chapter we have collected some properties of the

More information

Section 13.3 Area and the Definite Integral

Section 13.3 Area and the Definite Integral Sectio 3.3 Area ad the Defiite Itegral We ca easily fid areas of certai geometric figures usig well-kow formulas: However, it is t easy to fid the area of a regio with curved sides: METHOD: To evaluate

More information

Math 12 Final Exam, May 11, 2011 ANSWER KEY. 2sinh(2x) = lim. 1 x. lim e. x ln. = e. (x+1)(1) x(1) (x+1) 2. (2secθ) 5 2sec2 θ dθ.

Math 12 Final Exam, May 11, 2011 ANSWER KEY. 2sinh(2x) = lim. 1 x. lim e. x ln. = e. (x+1)(1) x(1) (x+1) 2. (2secθ) 5 2sec2 θ dθ. Math Fial Exam, May, ANSWER KEY. [5 Poits] Evaluate each of the followig its. Please justify your aswers. Be clear if the it equals a value, + or, or Does Not Exist. coshx) a) L H x x+l x) sihx) x x L

More information

Chapter 10: Power Series

Chapter 10: Power Series Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because

More information

Math 21B-B - Homework Set 2

Math 21B-B - Homework Set 2 Math B-B - Homework Set Sectio 5.:. a) lim P k= c k c k ) x k, where P is a partitio of [, 5. x x ) dx b) lim P k= 4 ck x k, where P is a partitio of [,. 4 x dx c) lim P k= ta c k ) x k, where P is a partitio

More information

Properties and Tests of Zeros of Polynomial Functions

Properties and Tests of Zeros of Polynomial Functions Properties ad Tests of Zeros of Polyomial Fuctios The Remaider ad Factor Theorems: Sythetic divisio ca be used to fid the values of polyomials i a sometimes easier way tha substitutio. This is show by

More information

6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.

6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer. 6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio

More information

f x x c x c x c... x c...

f x x c x c x c... x c... CALCULUS BC WORKSHEET ON POWER SERIES. Derive the Taylor series formula by fillig i the blaks below. 4 5 Let f a a c a c a c a4 c a5 c a c What happes to this series if we let = c? f c so a Now differetiate

More information

Exponents. Learning Objectives. Pre-Activity

Exponents. Learning Objectives. Pre-Activity Sectio. Pre-Activity Preparatio Epoets A Chai Letter Chai letters are geerated every day. If you sed a chai letter to three frieds ad they each sed it o to three frieds, who each sed it o to three frieds,

More information

LESSON 2: SIMPLIFYING RADICALS

LESSON 2: SIMPLIFYING RADICALS High School: Workig with Epressios LESSON : SIMPLIFYING RADICALS N.RN.. C N.RN.. B 5 5 C t t t t t E a b a a b N.RN.. 4 6 N.RN. 4. N.RN. 5. N.RN. 6. 7 8 N.RN. 7. A 7 N.RN. 8. 6 80 448 4 5 6 48 00 6 6 6

More information

where c is a scaling constant, 0, 0,. r c sinh cos csinh cos cos, csinh cos sin, ccosh sin U csinh sin sin, csinh sin cos,0

where c is a scaling constant, 0, 0,. r c sinh cos csinh cos cos, csinh cos sin, ccosh sin U csinh sin sin, csinh sin cos,0 MATH 38:Partial Differetial Equatios Solutios to The Secod Midterm DEGENERATE ELLIPSOID COORDINATES. Problem PROOF: Give csih si cos, y csihsi si, z ccoshcos, where c is a scalig costat,,,. r We compute

More information

U8L1: Sec Equations of Lines in R 2

U8L1: Sec Equations of Lines in R 2 MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (-D) Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie

More information

Calculus II. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed

Calculus II. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at. The olie versio of this documet is available at http://tutorial.math.lamar.edu.

More information

CHAPTER I: Vector Spaces

CHAPTER I: Vector Spaces CHAPTER I: Vector Spaces Sectio 1: Itroductio ad Examples This first chapter is largely a review of topics you probably saw i your liear algebra course. So why cover it? (1) Not everyoe remembers everythig

More information

2. Fourier Series, Fourier Integrals and Fourier Transforms

2. Fourier Series, Fourier Integrals and Fourier Transforms Mathematics IV -. Fourier Series, Fourier Itegrals ad Fourier Trasforms The Fourier series are used for the aalysis of the periodic pheomea, which ofte appear i physics ad egieerig. The Fourier itegrals

More information