Math 2260 Written HW #8 Solutions

Size: px
Start display at page:

Download "Math 2260 Written HW #8 Solutions"

Transcription

1 Mth 60 Written HW #8 Soltions. Sppose nd b re two fied positive nmbers. Find the re enclosed by the ellipse + y b =. [Hint: yo might wnt to find the re of the hlf of the ellipse bove the -is nd then doble it.] Answer: Here is pictre of the ellipse, which hs -intercepts t = ± nd y-intercepts t y = ±b. I cn find the re of the top hlf of the ellipse nd then doble it. To do so, I ll solve for y s fnction of. Certinly, y b =, so nd hence ( ) y = b ( ) y = ± b = ±b. The top hlf of the ellipse corresponds to choosing the positive sqre root in the bove. For lter convenience, notice tht = ( ), nd hence the top hlf of the ellipse is given by y = b.

2 Since the ellipse etends from = to =, the re nder the bove crve (which is the re of the top hlf of the ellipse) is eql to b d. Therefore, the re of the whole ellipse is eql to b d. To evlte this integrl, mke the trig sbstittion = sin θ. Then d = cos θ dθ. Also, since sin θ = /, we see tht sin θ = / = when =, mening tht θ =. Likewise, when =, sin θ = / =, so θ = π/. Ptting this ll together, then, the bove integrl is eql to b π/ sin θ cos θ dθ = b π/ cos θ cos θ dθ Therefore the re of the ellipse is πb. = b = b = b π/ π/ π/ cos θ dθ + cos(θ) dθ ( + cos(θ)) dθ [ = b θ + sin(θ) ] π/ [( π ) ( π )] = b = πb.. Evlte the integrl e 4t + e t e t e t dt. + Answer: First, mke the -sbstittion = e t. Then d = e t dt nd so the bove integrl is eql to ( e 3t + e t ) e t dt 3 + e t = + d. + Dividing nmertor by denomintor yields )

3 so or integrl is eql to ( + ) d = + ( + + ) d = + d+ + d + d. The first integrl is esy. For the second, let v = +, then dv = d nd so + d = dv v = ln v + C = ln( + ) + C. For the third integrl, let = tn θ. Then d = sec θ dθ nd so + d = sec θ tn θ + sec θ dθ = sec θ dθ = θ + C = rctn() + C. Therefore, the bove -integrl is eql to + ln( + ) rctn + C, where I ve combined ll the constnts into the single constnt C. Now, since = e t, this mens tht the originl integrl is eql to e t + ln(et + ) rctn(e t ) + C. 3. Sociologists sometimes se the phrse socil diffsion to describe the wy informtion spreds throgh popltion. The informtion cold be nything: scientific brekthrogh, news of ntrl disster, litercy, etc. In sfficiently lrge popltion, the nmber of people who hve the informtion is treted s differentible fnction of time t, nd the rte of diffsion, d/dt, is ssmed to be proportionl to the nmber of people who hve the informtion times the nmber of people who do not. This leds to the eqtion d dt = k(n ), where N is the nmber of people in the popltion. The constnt k determines how fst the informtion spreds: yo wold epect k to be reltively lrge for, sy, celebrity gossip, which is ssimilted lmost instntly, wheres it wold be qite smll for informtion tht only dissemintes very slowly (like how to compte integrls!). Sppose t is in dys, k = /50, nd two people strt rmor t time t = 0 in popltion of N = 000 people. () Find s fnction of t. [Hint: Logrithm identities re yor friends!] Answer: First, we seprte vribles nd integrte: d (000 ) = dt. ( ) 50 3

4 The right hnd side will be esy, so let s focs on the left hnd side. Using the method of prtil frctions, we gess tht Clering denomintors yields or, eqivlently, which yields the system of eqtions The second eqtion implies tht A = Therefore, ( /000 d (000 ) = (000 ) = A + B 000. = A(000 ) + B, = (B A) + 000A, 0 = B A = 000A. 000, while the first implies tht B = A = /000 ) d = ln ln(000 ) 000 (we don t hve to worry bot bsolte vles, becse we re only interested in vles of between 0 nd 000). Sbstitting this into eqtion ( ) nd integrting the right hnd side gives s Mltiply both sides by 000 to get 000 ln ln(000 ) = t + C. ln ln(000 ) = 4t + 4C. Now here s where the logrithm identity ln ln b = ln ( ) b comes in hndy. We cn se this identity to re-write the left hnd side: ( ) ln = 4t + 4C. 000 Since we wnt to solve for, eponentite both sides to get 000 = e4t+4c = e 4C e 4t = Ae 4t, where I ve let A = e 4C. This is still little nplesnt, bt it gets nicer if I tke the reciprocl of both sides: 000 = Ae 4t. 4

5 The left hnd side is eql to 000, so I hve tht 000 = + Ae4t + = Ae4t Ae 4t (where I fond common denomintor nd dded the terms on the right hnd side). Tking the reciprocl gin yields nd so 000 = Ae4t + Ae 4t, Ae4t = Ae 4t. Since the rmor strted with two people, we know tht (0) =. We cn se this to solve for the constnt A: Dividing both sides by 000, we hve Ae4(0) (0) = Ae 4(0) = 000 A + A 500 = A + A. To solve for A, tke the reciprocl of both sides: so nd hence Finlly, then, this mens tht 500 = + A A = A +, A = 499 A = 499. (t) = 000 Here s grph s fnction of t: 499 e4t e4t + = e4t e 4t. 5

6 (b) When will hlf the popltion hve herd the rmor? (This is when the rmor will be spreding the fstest.) Answer: The qestion is: t wht time t 0 will we hve tht (t 0 ) = 500? Well, evlte the fnction t t = t 0 : Diving both sides by 000 yields Tke the reciprocl of both sides to get so we hve nd so (t 0 ) = = e 4t. 0 = e 4t = e4t 0 Tke the ntrl logrithm of both sides: = 499 = t 0 = ln(499), = 499 +, nd hence ln 499 t 0 = Therefore, it tke bot one nd hlf dys for hlf the popltion to hve herd the rmor. 6

We are looking for ways to compute the integral of a function f(x), f(x)dx.

We are looking for ways to compute the integral of a function f(x), f(x)dx. INTEGRATION TECHNIQUES Introdction We re looking for wys to compte the integrl of fnction f(x), f(x)dx. To pt it simply, wht we need to do is find fnction F (x) sch tht F (x) = f(x). Then if the integrl

More information

spring from 1 cm to 2 cm is given by

spring from 1 cm to 2 cm is given by Problem [8 pts] Tre or Flse. Give brief explntion or exmple to jstify yor nswer. ) [ pts] Given solid generted by revolving region bot the line x, if we re sing the shell method to compte its volme, then

More information

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

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.

. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =. Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos( - 1 2 ) = rcsin( 1 2 ) = rcsin( - 1 2 ) = Cn you do similr problems? Review of Bsic Concepts

More information

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

Polynomial 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 information

. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =

. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) = Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos( - 1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin( - 1 ) = -π 2 6 2 6 Cn you do similr problems?

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivatives/Indefinite Integrals of Basic Functions Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second

More information

ntegration (p3) Integration by Inspection When differentiating using function of a function or the chain rule: If y = f(u), where in turn u = f(x)

ntegration (p3) Integration by Inspection When differentiating using function of a function or the chain rule: If y = f(u), where in turn u = f(x) ntegrtion (p) Integrtion by Inspection When differentiting using function of function or the chin rule: If y f(u), where in turn u f( y y So, to differentite u where u +, we write ( + ) nd get ( + ) (.

More information

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

More information

Chapter 8: Methods of Integration

Chapter 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 information

HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016

HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016 HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M

More information

Problem set 5: Solutions Math 207B, Winter r(x)u(x)v(x) dx.

Problem set 5: Solutions Math 207B, Winter r(x)u(x)v(x) dx. Problem set 5: Soltions Mth 7B, Winter 6. Sppose tht p : [, b] R is continosly differentible fnction sch tht p >, nd q, r : [, b] R re continos fnctions sch tht r >, q. Define weighted inner prodct on

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper 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 information

Calculus AB. For a function f(x), the derivative would be f '(

Calculus AB. For a function f(x), the derivative would be f '( lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:

More information

5.7 Improper Integrals

5.7 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 information

Math Calculus with Analytic Geometry II

Math Calculus with Analytic Geometry II orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

More information

Solution Set 2. y z. + j. u + j

Solution Set 2. y z. + j. u + j Soltion Set 2. Review of Div, Grd nd Crl. Prove:. () ( A) =, where A is ny three dimensionl vector field. i j k ( Az A = y z = i A A y A z y A ) ( y A + j z z A ) ( z Ay + k A ) y ( A) = ( Az y A ) y +

More information

Spring 2017 Exam 1 MARK BOX HAND IN PART PIN: 17

Spring 2017 Exam 1 MARK BOX HAND IN PART PIN: 17 Spring 07 Exm problem MARK BOX points HAND IN PART 0 5-55=x5 0 NAME: Solutions 3 0 0 PIN: 7 % 00 INSTRUCTIONS This exm comes in two prts. () HAND IN PART. Hnd in only this prt. () STATEMENT OF MULTIPLE

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

a 2 +x 2 x a 2 -x 2 Figure 1: Triangles for Trigonometric Substitution

a 2 +x 2 x a 2 -x 2 Figure 1: Triangles for Trigonometric Substitution I.B Trigonometric Substitution Uon comletion of the net two sections, we will be ble to integrte nlyticlly ll rtionl functions of (t lest in theory). We do so by converting binomils nd trinomils of the

More information

Math 113 Exam 2 Practice

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 information

AP Calculus Multiple Choice: BC Edition Solutions

AP Calculus Multiple Choice: BC Edition Solutions AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this

More information

Chapter 0. What is the Lebesgue integral about?

Chapter 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 information

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.

Improper 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 information

( x )( x) dx. Year 12 Extension 2 Term Question 1 (15 Marks) (a) Sketch the curve (x + 1)(y 2) = 1 2

( x )( x) dx. Year 12 Extension 2 Term Question 1 (15 Marks) (a) Sketch the curve (x + 1)(y 2) = 1 2 Yer Etension Term 7 Question (5 Mrks) Mrks () Sketch the curve ( + )(y ) (b) Write the function in prt () in the form y f(). Hence, or otherwise, sketch the curve (i) y f( ) (ii) y f () (c) Evlute (i)

More information

f(a+h) f(a) x a h 0. This is the rate at which

f(a+h) f(a) x a h 0. This is the rate at which M408S Concept Inventory smple nswers These questions re open-ended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnk-out-n-nswer problems! (There re plenty of those in the

More information

Math 31S. Rumbos Fall Solutions to Assignment #16

Math 31S. Rumbos Fall Solutions to Assignment #16 Mth 31S. Rumbos Fll 2016 1 Solutions to Assignment #16 1. Logistic Growth 1. Suppose tht the growth of certin niml popultion is governed by the differentil eqution 1000 dn N dt = 100 N, (1) where N(t)

More information

Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...

Chapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx... Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting

More information

n 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

n 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 information

Chapter 6 Techniques of Integration

Chapter 6 Techniques of Integration MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

More information

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

More information

y b y y sx 2 y 2 z CHANGE OF VARIABLES IN MULTIPLE INTEGRALS

y b y y sx 2 y 2 z CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ECION.8 CHANGE OF VAIABLE IN MULIPLE INEGAL 73 CA tive -is psses throgh the point where the prime meridin (the meridin throgh Greenwich, Englnd) intersects the eqtor. hen the ltitde of P is nd the longitde

More information

We divide the interval [a, b] into subintervals of equal length x = b a n

We divide the interval [a, b] into subintervals of equal length x = b a n Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:

More information

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40

Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40 Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since

More information

Antiderivatives Introduction

Antiderivatives Introduction Antierivtives 0. Introuction So fr much of the term hs been sent fining erivtives or rtes of chnge. But in some circumstnces we lrey know the rte of chnge n we wish to etermine the originl function. For

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

Trignometric Substitution

Trignometric Substitution Trignometric Substitution Trigonometric substitution refers simply to substitutions of the form x sinu or x tnu or x secu It is generlly used in conjunction with the trignometric identities to sin θ+cos

More information

5.5 The Substitution Rule

5.5 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 nti-derivtive is not esily recognizble, then we re in

More information

The Wave Equation I. MA 436 Kurt Bryan

The 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 information

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

2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ). AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following

More information

Math 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions

Math 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,

More information

Physics I Math Assessment with Answers

Physics I Math Assessment with Answers Physics I Mth Assessment with Answers The prpose of the following 10 qestions is to ssess some mth skills tht yo will need in Physics I These qestions will help yo identify some mth res tht yo my wnt to

More information

Lecture 1: Introduction to integration theory and bounded variation

Lecture 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

The 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 First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

ES.182A Topic 32 Notes Jeremy Orloff

ES.182A Topic 32 Notes Jeremy Orloff ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In

More information

Fall 2017 Exam 1 MARK BOX HAND IN PART PIN: 17

Fall 2017 Exam 1 MARK BOX HAND IN PART PIN: 17 Fll 7 Exm problem MARK BOX points HAND IN PART 3-5=x5 NAME: Solutions PIN: 7 % INSTRUCTIONS This exm comes in two prts. () HAND IN PART. Hnd in only this prt. () STATEMENT OF MULTIPLE CHOICE PROBLEMS.

More information

CALCULUS WITHOUT LIMITS

CALCULUS WITHOUT LIMITS CALCULUS WITHOUT LIMITS The current stndrd for the clculus curriculum is, in my opinion, filure in mny spects. We try to present it with the modern stndrd of mthemticl rigor nd comprehensiveness but of

More information

Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b

Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite

More information

VERIFYING AND DISCOVERING BBP-TYPE FORMULAS. Submitted by. Melissa Larson. Applied and Computational Mathematics

VERIFYING AND DISCOVERING BBP-TYPE FORMULAS. Submitted by. Melissa Larson. Applied and Computational Mathematics VERIFYING AND DISCOVERING BBP-TYPE FORMULAS Sbmitted by Meliss Lrson Applied nd Compttionl Mthemtics In prtil flfillment of the reqirements For the Degree of Mster of Science University of Minnesot Dlth

More information

4.4 Areas, Integrals and Antiderivatives

4.4 Areas, Integrals and Antiderivatives . res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order

More information

1 Techniques of Integration

1 Techniques of Integration November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.

More information

P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)

P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0) 1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this

More information

PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com C Integration - By sbstittion PhysicsAndMathsTtor.com. Using the sbstittion cos +, or otherwise, show that e cos + sin d e(e ) (Total marks). (a) Using the sbstittion cos, or otherwise, find the eact vale

More information

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems

More information

R. I. Badran Solid State Physics

R. I. Badran Solid State Physics I Bdrn Solid Stte Physics Crystl vibrtions nd the clssicl theory: The ssmption will be mde to consider tht the men eqilibrim position of ech ion is t Brvis lttice site The ions oscillte bot this men position

More information

Loudoun Valley High School Calculus Summertime Fun Packet

Loudoun Valley High School Calculus Summertime Fun Packet Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember!

More information

Calculus II: Integrations and Series

Calculus II: Integrations and Series Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]

More information

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

More information

Math 211/213 Calculus III-IV. Directions. Kenneth Massey. September 17, 2018

Math 211/213 Calculus III-IV. Directions. Kenneth Massey. September 17, 2018 Mth 211/213 Clculus -V Kenneth Mssey Crson-Newmn University September 17, 2018 C-N Mth 211 - Mssey, 1 / 1 Directions You re t the origin nd giving directions to the point (4, 3). 1. n Mnhttn: go est 4

More information

Math 360: A primitive integral and elementary functions

Math 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 information

If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x)

If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x) Mth 50 The method of prtil frction decomposition (PFD is used to integrte some rtionl functions of the form p(x, where p/q is in lowest terms nd deg(num < deg(denom. q(x If deg(num deg(denom, then we should

More information

FOURIER ANALYSIS AND THE UNCERTAINTY PRINCIPLE

FOURIER ANALYSIS AND THE UNCERTAINTY PRINCIPLE FOUIE ANALYSIS AND THE UNCETAINTY PINCIPLE WENFEI DU Abstrct. Forier nlysis is not only sefl tool in mthemtics, bt hs pplictions in other fields s well. Specificlly, it cn be sed to nlyze signls nd solve

More information

MATH 144: Business Calculus Final Review

MATH 144: Business Calculus Final Review MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

ES.181A Topic 8 Notes Jeremy Orloff

ES.181A Topic 8 Notes Jeremy Orloff ES.8A Topic 8 Notes Jeremy Orloff 8 Integrtion: u-substitution, trig-substitution 8. Integrtion techniques Only prctice will mke perfect. These techniques re importnt, but not the intellectul hert of the

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

More information

Topics Covered AP Calculus AB

Topics Covered AP Calculus AB Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.

More information

Unit 5. Integration techniques

Unit 5. Integration techniques 18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A-1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd

More information

Calculus AB Bible. (2nd most important book in the world) (Written and compiled by Doug Graham)

Calculus AB Bible. (2nd most important book in the world) (Written and compiled by Doug Graham) PG. Clculus AB Bile (nd most importnt ook in the world) (Written nd compiled y Doug Grhm) Topic Limits Continuity 6 Derivtive y Definition 7 8 Derivtive Formuls Relted Rtes Properties of Derivtives Applictions

More information

Using integration tables

Using integration tables Using integrtion tbles Integrtion tbles re inclue in most mth tetbooks, n vilble on the Internet. Using them is nother wy to evlute integrls. Sometimes the use is strightforwr; sometimes it tkes severl

More information

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

More information

r = cos θ + 1. dt ) dt. (1)

r = cos θ + 1. dt ) dt. (1) MTHE 7 Proble Set 5 Solutions (A Crdioid). Let C be the closed curve in R whose polr coordintes (r, θ) stisfy () Sketch the curve C. r = cos θ +. (b) Find pretriztion t (r(t), θ(t)), t [, b], of C in polr

More information

5.3 The Fundamental Theorem of Calculus

5.3 The Fundamental Theorem of Calculus CHAPTER 5. THE DEFINITE INTEGRAL 35 5.3 The Funmentl Theorem of Clculus Emple. Let f(t) t +. () Fin the re of the region below f(t), bove the t-is, n between t n t. (You my wnt to look up the re formul

More information

Integration Techniques

Integration Techniques Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u

More information

7.2 The Definite Integral

7.2 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 information

Appendix 3, Rises and runs, slopes and sums: tools from calculus

Appendix 3, Rises and runs, slopes and sums: tools from calculus Appendi 3, Rises nd runs, slopes nd sums: tools from clculus Sometimes we will wnt to eplore how quntity chnges s condition is vried. Clculus ws invented to do just this. We certinly do not need the full

More information

AP Calculus AB Summer Packet

AP Calculus AB Summer Packet AP Clculus AB Summer Pcket Nme: Welcome to AP Clculus AB! Congrtultions! You hve mde it to one of the most dvnced mth course in high school! It s quite n ccomplishment nd you should e proud of yourself

More information

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick

More information

p(t) dt + i 1 re it ireit dt =

p(t) dt + i 1 re it ireit dt = Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)

More information

Topic 1 Notes Jeremy Orloff

Topic 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 information

Math 3B Final Review

Math 3B Final Review Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:45-10:45m SH 1607 Mth Lb Hours: TR 1-2pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties 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 information

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is

More information

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

More information

Anonymous Math 361: Homework 5. x i = 1 (1 u i )

Anonymous 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 information

Lecture 1. Functional series. Pointwise and uniform convergence.

Lecture 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 information

Section 7.1 Integration by Substitution

Section 7.1 Integration by Substitution Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find

More information

MATH , Calculus 2, Fall 2018

MATH , Calculus 2, Fall 2018 MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly

More information

Bernoulli Numbers Jeff Morton

Bernoulli Numbers Jeff Morton Bernoulli Numbers Jeff Morton. We re interested in the opertor e t k d k t k, which is to sy k tk. Applying this to some function f E to get e t f d k k tk d k f f + d k k tk dk f, we note tht since f

More information

Practice final exam solutions

Practice final exam solutions University of Pennsylvni Deprtment of Mthemtics Mth 26 Honors Clculus II Spring Semester 29 Prof. Grssi, T.A. Asher Auel Prctice finl exm solutions 1. Let F : 2 2 be defined by F (x, y (x + y, x y. If

More information

Big idea in Calculus: approximation

Big idea in Calculus: approximation Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:

More information

Hung problem # 3 April 10, 2011 () [4 pts.] The electric field points rdilly inwrd [1 pt.]. Since the chrge distribution is cylindriclly symmetric, we pick cylinder of rdius r for our Gussin surfce S.

More information

Electricity and Magnetism Electric Dipole Continuous Distribution of Charge

Electricity and Magnetism Electric Dipole Continuous Distribution of Charge Electricit nd Mgnetism Electric Dipole Continos Distribtion of Chrge Ln heridn De Anz College Jn 16, 2018 Lst time electric field lines electric field from point chrge net electric field from mn chrges

More information

Identify graphs of linear inequalities on a number line.

Identify graphs of linear inequalities on a number line. COMPETENCY 1.0 KNOWLEDGE OF ALGEBRA SKILL 1.1 Identify grphs of liner inequlities on number line. - When grphing first-degree eqution, solve for the vrible. The grph of this solution will be single point

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.-3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This

More information

Interpreting Integrals and the Fundamental Theorem

Interpreting Integrals and the Fundamental Theorem Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

More information

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8 Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite

More information

Kinematics equations, some numbers

Kinematics equations, some numbers Kinemtics equtions, some numbers Kinemtics equtions: x = x 0 + v 0 t + 1 2 t2, v = v 0 + t. They describe motion with constnt ccelertion. Brking exmple, = 1m/s. Initil: x 0 = 10m, v 0 = 10m/s. x(t=1s)

More information

Stuff You Need to Know From Calculus

Stuff You Need to Know From Calculus Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you

More information

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic

More information