Week #13 - Integration by Parts & Numerical Integration Section 7.2
|
|
- Lydia Cooper
- 6 years ago
- Views:
Transcription
1 Week #3 - Inegraion by Pars & Numerical Inegraion Secion 7. From Calculus, Single Variable by Hughes-Halle, Gleason, McCallum e. al. Copyrigh 5 by John Wiley & Sons, Inc. This maerial is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS. sin d 3. Inegrae by pars, selecing: u = dv = sin d so du = d v = cos udv = uv vdu, sin d = cos ( cos ) d = cos + sin + C sin d Inegrae by pars, selecing: u = dv = sin d so du = d v = cos sin d = cos ( cos ) d = cos + cos d This new inegral is simpler (he power of wen from squared o linear), and if we inegrae by pars again, we should ge a solvable inegral: selecing: u = dv = cos d so du = d v = sin [ cos d = sin ] sin d = [ sin + cos ] + C Combining his wih he previous resul, he final inegral evaluaes o sin d = cos + sin + cos + C 9. x 3 ln x dx To inegrae by pars, we mus selec u = ln x, because we don know how o inegrae i:
2 selecing: u = ln x dv = x 3 dx so du = x dx x v = 7. (θ + )sin(θ + ) dθ x 3 ln x dx = x x ln x x x dx = ln x x 3 dx = x ln x 6 x + C Since everyhing seems o be in erms of (θ + ), i migh help o do a subsiuion firs: Le w = (θ +, so dw = dθ, and (θ + )sin(θ + ) dθ = w sin(w)dw This we can evaluae by inegraion by pars, selecing: u = w dv = sin w dw so du = dw v = cos w w sin(w)dw = w cos w ( cos w)dw = w cos w + sinw d = (θ + )cos(θ + ) + sin(θ + ) + C Neiher subsiuion nor inegraion by pars looks o helpful a firs glance. When possible, i can help o separae sums in he numeraor o obain wo simpler inegrals: + 7 d = 5 d 5 }{{} I d } {{ } I The second inegral, I, can be done by subsiuion, or by guess and check: 7 d = 7 5 (5 ) / = (5 ) / + C The firs inegral, I, is more complicaed, bu wih wha we jus found, we should be able o use inegraion by pars: I = d = (5 ) / d 5
3 selecing: u = dv = (5 ) / d so du = d v = (5 ) / I = (5 ) / d = (5 ) / ( (5 ) / ) d = (5 ) / 3 (5 )3/ + C Combining I and I, and combining he consans ino C, we obain 5. xarcan x dx + 7 d = I + I = (5 ) / (5 )3/ (5 ) / +C 5 }{{ 3 }}{{} I I This looks like a classical subsiuion problem, given he x inside he arcan, and is derivaive, x ou fron. Le s sar wih ha subsiuion: Le w = x, so dw = x dx, so xarcan x dx = arcan w dw However, we only know he derivaive of arcan, no is inegral, so we need o use inegraion by pars, selecing: u = arcan w dv = dw so du = +w v = w arcan w dw = (w arcan w w + w dw) This new inegral can also be solved by subsiuion, since he derivaive of he denominaor will involve a w erm. Le z = + w, so dz = wdw: w + w dw = z dz = ln z = ln( + w ) 3
4 Combining all he pars, and hen subsiuing back ino he original x variable, xarcan x dx = (w arcan w w + w dw) = w arcan w ln( + w ) + C = x arcan x ln( + x ) + C ln( + ) d We only know how o differeniae ln, so rying inegraion by pars makes sense, selecing: u = ln( + ) dv = d so du = + d v = 5 ln( + ) d = ln( + ) + d The new inegral can be easily solved wih he subsiuion w = +, so = w and dw = d: w w + d = w dw = w dw w dw = w ln w + C = + ln( + ) + C so ln( + ) d = ln( + ) + d = ln( + ) ( + ln( + ) + C) Wih he limis, 5 = ( + )ln( + ) + C if we le C = C ln( + ) d = ( + )ln( + ) = (6ln 6 6) (ln() ) = 6ln QUIZ PREPARATION PROBLEMS 38. For each of he following inegrals, indicae wheher inegraion by subsiuion or inegraion by pars is more appropriae. Do no evaluae he inegrals.
5 (a) (b) (c) (d) (e) (f) (g) xsin x dx x + x 3 dx xe x dx x cos(x 3 ) dx 3x + dx x sin x dx ln x dx (a) This inegral can be evaluaed using inegraion by pars wih u = x, dv = sin x dx. (b) We evaluae his inegral using he subsiuion w = + x 3. (c) We evaluae his inegral using he subsiuion w = x. (d) We evaluae his inegral using he subsiuion w = x 3. (e) We evaluae his inegral using he subsiuion w = 3x +. (f) This inegral can be evaluaed using inegraion by pars wih u = x,dv = sin x dx. (g) This inegral can be evaluaed using inegraion by pars wih u = ln x,dv = dx. 39. Find he exac value of he area under he firs arch of f(x) = xsin(x). You should have some sense of a skech of his graph: i will look like he sine graph, bu he ampliude will be growing wih linearly x because he sin(x) is muliplied by x. Because i is a produc of x and sin(x), he funcion will sill have y = values, crossing he x axis, a he same poins when sin(x) =. The firs arch of he curve is herefore beween x = and x = π, as hese are he firs wo poins when f(x) =. This means he area we re looking for will be represened by he inegral To evaluae his, we ll use inegraion by pars: selecing: u = x dv = sin x dx so du = dx v = cos x π π x sin(x) dx = x( cos x) π π cos(x) dx = x cos(x) + sin(x) π = [( π cos(π) + sin(π)] [( cos() + sin()] = π( ) = π xsin(x) dx. 5
6 7. x n cos ax dx = a xn sin(ax) n a x n sin(ax) dx We inegrae by pars. Le u = x n and dv = cos ax dx, so du = nx n dx and v = a sin ax. Then x n cos axdx = a xn sin ax (nx n )( sin ax) dx a = a xn sin ax n x n sin ax dx. a 5. Esimae f(x)g (x) dx if f(x) = x and g has he values in he following able. x 6 8 g(x) Since f (x) = x, inegraion by pars ells us ha f(x)g (x)dx = f(x)g(x) f (x)g(x)dx = f()g() f()g() xg(x)dx. We can use le and righ Riemann Sums wih x = o approximae xg(x)dx: Lef sum g() x + g() x + g() x + 6 g(6) x + 8 g(8) x = ((.3) + (3.) + (.) + 6(5.5) + 8(5.9)) = 5.6 Righ sum g() x + g() x + 6 g(6) x + 8 g(8) x + g() x = (3.) + (.) + 6(5.5) + 8(5.9) + (6.)) = A good esimae for he inegral is he average of he lef and righ sums, so xg(x)dx Subsiuing values for f and g, we have = f(x)g (x)dx = f()g() f()g() xg(x)dx (6.) (.3) (66.6) =
MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.
MATH 3B: MIDTERM REVIEW JOE HUGHES. Inegraion by Pars. Evaluae 3 e. Soluion: Firs make he subsiuion u =. Then =, hence 3 e = e = ue u Now inegrae by pars o ge ue u = ue u e u + C and subsiue he definiion
More informationMATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence
MATH 433/533, Fourier Analysis Secion 6, Proof of Fourier s Theorem for Poinwise Convergence Firs, some commens abou inegraing periodic funcions. If g is a periodic funcion, g(x + ) g(x) for all real x,
More informationAdvanced Integration Techniques: Integration by Parts We may differentiate the product of two functions by using the product rule:
Avance Inegraion Techniques: Inegraion by Pars We may iffereniae he prouc of wo funcions by using he prouc rule: x f(x)g(x) = f (x)g(x) + f(x)g (x). Unforunaely, fining an anierivaive of a prouc is no
More information23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes
Represening Periodic Funcions by Fourier Series 3. Inroducion In his Secion we show how a periodic funcion can be expressed as a series of sines and cosines. We begin by obaining some sandard inegrals
More informationMA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions
MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firs-order linear ODE is: An inegraing facor is given by
More informationChallenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k
Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,
More information1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.
. Use Simpson s rule wih n 4 o esimae an () +. Soluion: Since we are using 4 seps, 4 Thus we have [ ( ) f() + 4f + f() + 4f 3 [ + 4 4 6 5 + + 4 4 3 + ] 5 [ + 6 6 5 + + 6 3 + ]. 5. Our funcion is f() +.
More informationChapter 2. First Order Scalar Equations
Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.
More informationWeek #6 - Taylor Series, Derivatives and Graphs Section 10.1
Week #6 - Taylor Series, Derivatives and Graphs Section 10.1 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 005 by John Wiley & Sons, Inc. This material is used by
More informationHOMEWORK # 2: MATH 211, SPRING Note: This is the last solution set where I will describe the MATLAB I used to make my pictures.
HOMEWORK # 2: MATH 2, SPRING 25 TJ HITCHMAN Noe: This is he las soluion se where I will describe he MATLAB I used o make my picures.. Exercises from he ex.. Chaper 2.. Problem 6. We are o show ha y() =
More information( ) 2. Review Exercise 2. cos θ 2 3 = = 2 tan. cos. 2 x = = x a. Since π π, = 2. sin = = 2+ = = cotx. 2 sin θ 2+
Review Eercise sin 5 cos sin an cos 5 5 an 5 9 co 0 a sinθ 6 + 4 6 + sin θ 4 6+ + 6 + 4 cos θ sin θ + 4 4 sin θ + an θ cos θ ( ) + + + + Since π π, < θ < anθ should be negaive. anθ ( + ) Pearson Educaion
More informationu(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x
. 1 Mah 211 Homework #3 February 2, 2001 2.4.3. y + (2/x)y = (cos x)/x 2 Answer: Compare y + (2/x) y = (cos x)/x 2 wih y = a(x)x + f(x)and noe ha a(x) = 2/x. Consequenly, an inegraing facor is found wih
More informationTHE SINE INTEGRAL. x dt t
THE SINE INTEGRAL As one learns in elemenary calculus, he limi of sin(/ as vanishes is uniy. Furhermore he funcion is even and has an infinie number of zeros locaed a ±n for n1,,3 Is plo looks like his-
More information(π 3)k. f(t) = 1 π 3 sin(t)
Mah 6 Fall 6 Dr. Lil Yen Tes Show all our work Name: Score: /6 No Calculaor permied in his par. Read he quesions carefull. Show all our work and clearl indicae our final answer. Use proper noaion. Problem
More information( ) = 0.43 kj = 430 J. Solutions 9 1. Solutions to Miscellaneous Exercise 9 1. Let W = work done then 0.
Soluions 9 Soluions o Miscellaneous Exercise 9. Le W work done hen.9 W PdV Using Simpson's rule (9.) we have. W { 96 + [ 58 + 6 + 77 + 5 ] + [ + 99 + 6 ]+ }. kj. Using Simpson's rule (9.) again: W.5.6
More informationSolutions to Assignment 1
MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we
More informationMath 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:
Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial
More informationSection 5: Chain Rule
Chaper The Derivaive Applie Calculus 11 Secion 5: Chain Rule There is one more ype of complicae funcion ha we will wan o know how o iffereniae: composiion. The Chain Rule will le us fin he erivaive of
More informationdy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page
Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources,
More informationDifferential Equations
Mah 21 (Fall 29) Differenial Equaions Soluion #3 1. Find he paricular soluion of he following differenial equaion by variaion of parameer (a) y + y = csc (b) 2 y + y y = ln, > Soluion: (a) The corresponding
More informationIntegration by Parts
Calculus 2 Lia Vas Integration by Parts Using integration by parts one transforms an integral of a product of two functions into a simpler integral. Divide the initial function into two parts called u
More informationSMT 2014 Calculus Test Solutions February 15, 2014 = 3 5 = 15.
SMT Calculus Tes Soluions February 5,. Le f() = and le g() =. Compue f ()g (). Answer: 5 Soluion: We noe ha f () = and g () = 6. Then f ()g () =. Plugging in = we ge f ()g () = 6 = 3 5 = 5.. There is a
More informationMath 116 Second Midterm March 21, 2016
Mah 6 Second Miderm March, 06 UMID: EXAM SOLUTIONS Iniials: Insrucor: Secion:. Do no open his exam unil you are old o do so.. Do no wrie your name anywhere on his exam. 3. This exam has pages including
More informationAP CALCULUS AB 2003 SCORING GUIDELINES (Form B)
SCORING GUIDELINES (Form B) Quesion A blood vessel is 6 millimeers (mm) long Disance wih circular cross secions of varying diameer. x (mm) 6 8 4 6 Diameer The able above gives he measuremens of he B(x)
More informationMethods of Integration
Methods of Integration Professor D. Olles January 8, 04 Substitution The derivative of a composition of functions can be found using the chain rule form d dx [f (g(x))] f (g(x)) g (x) Rewriting the derivative
More information3.6 Derivatives as Rates of Change
3.6 Derivaives as Raes of Change Problem 1 John is walking along a sraigh pah. His posiion a he ime >0 is given by s = f(). He sars a =0from his house (f(0) = 0) and he graph of f is given below. (a) Describe
More informationTHE 2-BODY PROBLEM. FIGURE 1. A pair of ellipses sharing a common focus. (c,b) c+a ROBERT J. VANDERBEI
THE 2-BODY PROBLEM ROBERT J. VANDERBEI ABSTRACT. In his shor noe, we show ha a pair of ellipses wih a common focus is a soluion o he 2-body problem. INTRODUCTION. Solving he 2-body problem from scrach
More informationMath 106: Review for Final Exam, Part II. (x x 0 ) 2 = !
Mah 6: Review for Final Exam, Par II. Use a second-degree Taylor polynomial o esimae 8. We choose f(x) x and x 7 because 7 is he perfec cube closes o 8. f(x) x / f(7) f (x) x / f (7) x / 7 / 7 f (x) 9
More information15. Vector Valued Functions
1. Vecor Valued Funcions Up o his poin, we have presened vecors wih consan componens, for example, 1, and,,4. However, we can allow he componens of a vecor o be funcions of a common variable. For example,
More information!!"#"$%&#'()!"#&'(*%)+,&',-)./0)1-*23)
"#"$%&#'()"#&'(*%)+,&',-)./)1-*) #$%&'()*+,&',-.%,/)*+,-&1*#$)()5*6$+$%*,7&*-'-&1*(,-&*6&,7.$%$+*&%'(*8$&',-,%'-&1*(,-&*6&,79*(&,%: ;..,*&1$&$.$%&'()*1$$.,'&',-9*(&,%)?%*,('&5
More informationWeek #16 - Differential Equations (Euler s Method) Section 11.3
Week #16 - Differential Equations (Euler s Method) Section 11.3 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used
More informationIntegration by parts (product rule backwards)
Integration by parts (product rule backwards) The product rule states Integrating both sides gives f(x)g(x) = d dx f(x)g(x) = f(x)g (x) + f (x)g(x). f(x)g (x)dx + Letting f(x) = u, g(x) = v, and rearranging,
More information23.5. Half-Range Series. Introduction. Prerequisites. Learning Outcomes
Half-Range Series 2.5 Inroducion In his Secion we address he following problem: Can we find a Fourier series expansion of a funcion defined over a finie inerval? Of course we recognise ha such a funcion
More informationCircuit Variables. AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters per second to miles per second: 1 ft 12 in
Circui Variables 1 Assessmen Problems AP 1.1 Use a produc of raios o conver wo-hirds he speed of ligh from meers per second o miles per second: ( ) 2 3 1 8 m 3 1 s 1 cm 1 m 1 in 2.54 cm 1 f 12 in 1 mile
More informationPredator - Prey Model Trajectories and the nonlinear conservation law
Predaor - Prey Model Trajecories and he nonlinear conservaion law James K. Peerson Deparmen of Biological Sciences and Deparmen of Mahemaical Sciences Clemson Universiy Ocober 28, 213 Ouline Drawing Trajecories
More informationSOLUTIONS TO ECE 3084
SOLUTIONS TO ECE 384 PROBLEM 2.. For each sysem below, specify wheher or no i is: (i) memoryless; (ii) causal; (iii) inverible; (iv) linear; (v) ime invarian; Explain your reasoning. If he propery is no
More informationChapter 7: Solving Trig Equations
Haberman MTH Secion I: The Trigonomeric Funcions Chaper 7: Solving Trig Equaions Le s sar by solving a couple of equaions ha involve he sine funcion EXAMPLE a: Solve he equaion sin( ) The inverse funcions
More informationdt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.
Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies
More informationa. Show that these lines intersect by finding the point of intersection. b. Find an equation for the plane containing these lines.
Mah A Final Eam Problems for onsideraion. Show all work for credi. Be sure o show wha you know. Given poins A(,,, B(,,, (,, 4 and (,,, find he volume of he parallelepiped wih adjacen edges AB, A, and A.
More informationMath 115 Final Exam December 14, 2017
On my honor, as a suden, I have neiher given nor received unauhorized aid on his academic work. Your Iniials Only: Iniials: Do no wrie in his area Mah 5 Final Exam December, 07 Your U-M ID # (no uniqname):
More informationMath 1b. Calculus, Series, and Differential Equations. Final Exam Solutions
Mah b. Calculus, Series, and Differenial Equaions. Final Exam Soluions Spring 6. (9 poins) Evaluae he following inegrals. 5x + 7 (a) (x + )(x + ) dx. (b) (c) x arcan x dx x(ln x) dx Soluion. (a) Using
More informationon the interval (x + 1) 0! x < ", where x represents feet from the first fence post. How many square feet of fence had to be painted?
Calculus II MAT 46 Improper Inegrals A mahemaician asked a fence painer o complee he unique ask of paining one side of a fence whose face could be described by he funcion y f (x on he inerval (x + x
More informationAP Calculus BC Chapter 10 Part 1 AP Exam Problems
AP Calculus BC Chaper Par AP Eam Problems All problems are NO CALCULATOR unless oherwise indicaed Parameric Curves and Derivaives In he y plane, he graph of he parameric equaions = 5 + and y= for, is a
More informationMath 333 Problem Set #2 Solution 14 February 2003
Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial
More information2.7. Some common engineering functions. Introduction. Prerequisites. Learning Outcomes
Some common engineering funcions 2.7 Inroducion This secion provides a caalogue of some common funcions ofen used in Science and Engineering. These include polynomials, raional funcions, he modulus funcion
More informationMEI STRUCTURED MATHEMATICS 4758
OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Cerificae of Educaion Advanced General Cerificae of Educaion MEI STRUCTURED MATHEMATICS 4758 Differenial Equaions Thursday 5 JUNE 006 Afernoon
More informationReview - Quiz # 1. 1 g(y) dy = f(x) dx. y x. = u, so that y = xu and dy. dx (Sometimes you may want to use the substitution x y
Review - Quiz # 1 (1) Solving Special Tpes of Firs Order Equaions I. Separable Equaions (SE). d = f() g() Mehod of Soluion : 1 g() d = f() (The soluions ma be given implicil b he above formula. Remember,
More informationUCLA: Math 3B Problem set 3 (solutions) Fall, 2018
UCLA: Mah 3B Problem se 3 (soluions) Fall, 28 This problem se concenraes on pracice wih aniderivaives. You will ge los of pracice finding simple aniderivaives as well as finding aniderivaives graphically
More informationt + t sin t t cos t sin t. t cos t sin t dt t 2 = exp 2 log t log(t cos t sin t) = Multiplying by this factor and then integrating, we conclude that
ODEs, Homework #4 Soluions. Check ha y ( = is a soluion of he second-order ODE ( cos sin y + y sin y sin = 0 and hen use his fac o find all soluions of he ODE. When y =, we have y = and also y = 0, so
More information1 st order ODE Initial Condition
Mah-33 Chapers 1-1 s Order ODE Sepember 1, 17 1 1 s order ODE Iniial Condiion f, sandard form LINEAR NON-LINEAR,, p g differenial form M x dx N x d differenial form is equivalen o a pair of differenial
More information6.003 Homework #8 Solutions
6.003 Homework #8 Soluions Problems. Fourier Series Deermine he Fourier series coefficiens a k for x () shown below. x ()= x ( + 0) 0 a 0 = 0 a k = e /0 sin(/0) for k 0 a k = π x()e k d = 0 0 π e 0 k d
More informationAP Calculus BC - Parametric equations and vectors Chapter 9- AP Exam Problems solutions
AP Calculus BC - Parameric equaions and vecors Chaper 9- AP Exam Problems soluions. A 5 and 5. B A, 4 + 8. C A, 4 + 4 8 ; he poin a is (,). y + ( x ) x + 4 4. e + e D A, slope.5 6 e e e 5. A d hus d d
More informationAP CALCULUS AB 2004 SCORING GUIDELINES (Form B)
4 SCORING GUIDELINES (Form B) Quesion A es plane flies in a sraigh line wih (min) 5 1 15 5 5 4 posiive velociy v (), in miles per v ()(mpm) 7. 9. 9.5 7. 4.5.4.4 4. 7. minue a ime minues, where v is a differeniable
More informationContinuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0.
Time-Domain Sysem Analysis Coninuous Time. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 1. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 2 Le a sysem be described by a 2 y ( ) + a 1
More informationCalculus II Practice Test Problems for Chapter 7 Page 1 of 6
Calculus II Practice Test Problems for Chapter 7 Page of 6 This is a set of practice test problems for Chapter 7. This is in no way an inclusive set of problems there can be other types of problems on
More informationFinal Spring 2007
.615 Final Spring 7 Overview The purpose of he final exam is o calculae he MHD β limi in a high-bea oroidal okamak agains he dangerous n = 1 exernal ballooning-kink mode. Effecively, his corresponds o
More informationMorning Time: 1 hour 30 minutes Additional materials (enclosed):
ADVANCED GCE 78/0 MATHEMATICS (MEI) Differenial Equaions THURSDAY JANUARY 008 Morning Time: hour 30 minues Addiional maerials (enclosed): None Addiional maerials (required): Answer Bookle (8 pages) Graph
More information72 Calculus and Structures
72 Calculus and Srucures CHAPTER 5 DISTANCE AND ACCUMULATED CHANGE Calculus and Srucures 73 Copyrigh Chaper 5 DISTANCE AND ACCUMULATED CHANGE 5. DISTANCE a. Consan velociy Le s ake anoher look a Mary s
More information4.6 One Dimensional Kinematics and Integration
4.6 One Dimensional Kinemaics and Inegraion When he acceleraion a( of an objec is a non-consan funcion of ime, we would like o deermine he ime dependence of he posiion funcion x( and he x -componen of
More informationThe equation to any straight line can be expressed in the form:
Sring Graphs Par 1 Answers 1 TI-Nspire Invesigaion Suden min Aims Deermine a series of equaions of sraigh lines o form a paern similar o ha formed by he cables on he Jerusalem Chords Bridge. Deermine he
More informationY 0.4Y 0.45Y Y to a proper ARMA specification.
HG Jan 04 ECON 50 Exercises II - 0 Feb 04 (wih answers Exercise. Read secion 8 in lecure noes 3 (LN3 on he common facor problem in ARMA-processes. Consider he following process Y 0.4Y 0.45Y 0.5 ( where
More informationIntegration by Parts. MAT 126, Week 2, Thursday class. Xuntao Hu
MAT 126, Week 2, Thursday class Xuntao Hu Recall that the substitution rule is a combination of the FTC and the chain rule. We can also combine the FTC and the product rule: d dx [f (x)g(x)] = f (x)g (x)
More informationEECE 301 Signals & Systems Prof. Mark Fowler
EECE 3 Signals & Sysems Prof. Mark Fowler Noe Se #2 Wha are Coninuous-Time Signals??? Reading Assignmen: Secion. of Kamen and Heck /22 Course Flow Diagram The arrows here show concepual flow beween ideas.
More information8. Basic RL and RC Circuits
8. Basic L and C Circuis This chaper deals wih he soluions of he responses of L and C circuis The analysis of C and L circuis leads o a linear differenial equaion This chaper covers he following opics
More information1998 Calculus AB Scoring Guidelines
AB{ / BC{ 1999. The rae a which waer ows ou of a pipe, in gallons per hour, is given by a diereniable funcion R of ime. The able above shows he rae as measured every hours for a {hour period. (a) Use a
More informationy = (y 1)*(y 3) t
MATH 66 SPR REVIEW DEFINITION OF SOLUTION A funcion = () is a soluion of he differenial equaion d=d = f(; ) on he inerval ff < < fi if (d=d)() =f(; ()) for each so ha ff
More informationAP CALCULUS AB 2017 SCORING GUIDELINES
AP CALCULUS AB 17 SCORING GUIDELINES 16 SCORING GUIDELINES Quesion For, a paricle moves along he x-axis. The velociy of he paricle a ime is given by v ( ) = 1 + sin. The paricle is a posiion x = a ime.
More informationProblem Set 7-7. dv V ln V = kt + C. 20. Assume that df/dt still equals = F RF. df dr = =
20. Assume ha df/d sill equals = F + 0.02RF. df dr df/ d F+ 0. 02RF = = 2 dr/ d R 0. 04RF 0. 01R 10 df 11. 2 R= 70 and F = 1 = = 0. 362K dr 31 21. 0 F (70, 30) (70, 1) R 100 Noe ha he slope a (70, 1) is
More informationPROBLEMS FOR MATH 162 If a problem is starred, all subproblems are due. If only subproblems are starred, only those are due. SLOPES OF TANGENT LINES
PROBLEMS FOR MATH 6 If a problem is sarred, all subproblems are due. If onl subproblems are sarred, onl hose are due. 00. Shor answer quesions. SLOPES OF TANGENT LINES (a) A ball is hrown ino he air. Is
More informationTime: 1 hour 30 minutes
Paper Reference(s) 6666/0 Edexcel GCE Core Mahemaics C4 Silver Level S4 Time: hour 30 minues Maerials required for examinaion papers Mahemaical Formulae (Green) Iems included wih quesion Nil Candidaes
More informationMath 105 Second Midterm March 16, 2017
Mah 105 Second Miderm March 16, 2017 UMID: Insrucor: Iniials: Secion: 1. Do no open his exam unil you are old o do so. 2. Do no wrie your name anywhere on his exam. 3. This exam has 9 pages including his
More informationGraphs of Antiderivatives, Substitution Integrals
Unit #10 : Graphs of Antiderivatives, Substitution Integrals Goals: Relationship between the graph of f(x) and its anti-derivative F (x) The guess-and-check method for anti-differentiation. The substitution
More informationln 2 1 ln y x c y C x
Lecure 14 Appendi B: Some sample problems from Boas Here are some soluions o he sample problems assigned for Chaper 8 8: 6 Soluion: We wan o find he soluion o he following firs order equaion using separaion
More informationChapter 8 The Complete Response of RL and RC Circuits
Chaper 8 The Complee Response of RL and RC Circuis Seoul Naional Universiy Deparmen of Elecrical and Compuer Engineering Wha is Firs Order Circuis? Circuis ha conain only one inducor or only one capacior
More informationTwo Coupled Oscillators / Normal Modes
Lecure 3 Phys 3750 Two Coupled Oscillaors / Normal Modes Overview and Moivaion: Today we ake a small, bu significan, sep owards wave moion. We will no ye observe waves, bu his sep is imporan in is own
More informationThe Relation between the Integral and the Derivative Graphs. Unit #10 : Graphs of Antiderivatives, Substitution Integrals
Graphs of Antiderivatives - Unit #0 : Graphs of Antiderivatives, Substitution Integrals Goals: Relationship between the graph of f(x) and its anti-derivative F (x) The guess-and-check method for anti-differentiation.
More informationAP CALCULUS AB 2003 SCORING GUIDELINES (Form B)
SCORING GUIDELINES (Form B) Quesion A ank conains 15 gallons of heaing oil a ime =. During he ime inerval 1 hours, heaing oil is pumped ino he ank a he rae 1 H ( ) = + ( 1 + ln( + 1) ) gallons per hour.
More information4.8 Improper Integrals
4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls
More informationSection 5.5 More Integration Formula (The Substitution Method) 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I
Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures College of Science MATHS : Calculus I (University of Bahrain) Integrals / 7 The Substitution Method Idea: To replace a relatively
More information3, so θ = arccos
Mahemaics 210 Professor Alan H Sein Monday, Ocober 1, 2007 SOLUTIONS This problem se is worh 50 poins 1 Find he angle beween he vecors (2, 7, 3) and (5, 2, 4) Soluion: Le θ be he angle (2, 7, 3) (5, 2,
More informationBe able to sketch a function defined parametrically. (by hand and by calculator)
Pre Calculus Uni : Parameric and Polar Equaions (7) Te References: Pre Calculus wih Limis; Larson, Hoseler, Edwards. B he end of he uni, ou should be able o complee he problems below. The eacher ma provide
More informationSolutions from Chapter 9.1 and 9.2
Soluions from Chaper 9 and 92 Secion 9 Problem # This basically boils down o an exercise in he chain rule from calculus We are looking for soluions of he form: u( x) = f( k x c) where k x R 3 and k is
More informationAssignment 6 Solution. Please do not copy and paste my answer. You will get similar questions but with different numbers!
Assignment 6 Solution Please do not copy and paste my answer. You will get similar questions but with different numbers! This question tests you the following points: Integration by Parts: Let u = x, dv
More information6.003 Homework #9 Solutions
6.00 Homework #9 Soluions Problems. Fourier varieies a. Deermine he Fourier series coefficiens of he following signal, which is periodic in 0. x () 0 0 a 0 5 a k sin πk 5 sin πk 5 πk for k 0 a k 0 πk j
More information-e x ( 0!x+1! ) -e x 0!x 2 +1!x+2! e t dt, the following expressions hold. t
4 Higher and Super Calculus of Logarihmic Inegral ec. 4. Higher Inegral of Eponenial Inegral Eponenial Inegral is defined as follows. Ei( ) - e d (.0) Inegraing boh sides of (.0) wih respec o repeaedly
More informationMA Study Guide #1
MA 66 Su Guide #1 (1) Special Tpes of Firs Order Equaions I. Firs Order Linear Equaion (FOL): + p() = g() Soluion : = 1 µ() [ ] µ()g() + C, where µ() = e p() II. Separable Equaion (SEP): dx = h(x) g()
More informationCh.1. Group Work Units. Continuum Mechanics Course (MMC) - ETSECCPB - UPC
Ch.. Group Work Unis Coninuum Mechanics Course (MMC) - ETSECCPB - UPC Uni 2 Jusify wheher he following saemens are rue or false: a) Two sreamlines, corresponding o a same insan of ime, can never inersec
More informationTHE WAVE EQUATION. part hand-in for week 9 b. Any dilation v(x, t) = u(λx, λt) of u(x, t) is also a solution (where λ is constant).
THE WAVE EQUATION 43. (S) Le u(x, ) be a soluion of he wave equaion u u xx = 0. Show ha Q43(a) (c) is a. Any ranslaion v(x, ) = u(x + x 0, + 0 ) of u(x, ) is also a soluion (where x 0, 0 are consans).
More informationd 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3
and d = c b - b c c d = c b - b c c This process is coninued unil he nh row has been compleed. The complee array of coefficiens is riangular. Noe ha in developing he array an enire row may be divided or
More informationAP CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES. Question 1. 1 : estimate = = 120 liters/hr
AP CALCULUS AB/CALCULUS BC 16 SCORING GUIDELINES Quesion 1 (hours) R ( ) (liers / hour) 1 3 6 8 134 119 95 74 7 Waer is pumped ino a ank a a rae modeled by W( ) = e liers per hour for 8, where is measured
More informationk B 2 Radiofrequency pulses and hardware
1 Exra MR Problems DC Medical Imaging course April, 214 he problems below are harder, more ime-consuming, and inended for hose wih a more mahemaical background. hey are enirely opional, bu hopefully will
More informationSection 7.4 Modeling Changing Amplitude and Midline
488 Chaper 7 Secion 7.4 Modeling Changing Ampliude and Midline While sinusoidal funcions can model a variey of behaviors, i is ofen necessary o combine sinusoidal funcions wih linear and exponenial curves
More informationMathcad Lecture #8 In-class Worksheet Curve Fitting and Interpolation
Mahcad Lecure #8 In-class Workshee Curve Fiing and Inerpolaion A he end of his lecure, you will be able o: explain he difference beween curve fiing and inerpolaion decide wheher curve fiing or inerpolaion
More informationSolutionbank Edexcel AS and A Level Modular Mathematics
Page of 4 Soluionbank Edexcel AS and A Level Modular Mahemaics Exercise A, Quesion Quesion: Skech he graphs of (a) y = e x + (b) y = 4e x (c) y = e x 3 (d) y = 4 e x (e) y = 6 + 0e x (f) y = 00e x + 0
More informationThe average rate of change between two points on a function is d t
SM Dae: Secion: Objecive: The average rae of change beween wo poins on a funcion is d. For example, if he funcion ( ) represens he disance in miles ha a car has raveled afer hours, hen finding he slope
More informationKEY. Math 334 Midterm I Fall 2008 sections 001 and 003 Instructor: Scott Glasgow
1 KEY Mah 4 Miderm I Fall 8 secions 1 and Insrucor: Sco Glasgow Please do NOT wrie on his eam. No credi will be given for such work. Raher wrie in a blue book, or on our own paper, preferabl engineering
More informationWeek 1 Lecture 2 Problems 2, 5. What if something oscillates with no obvious spring? What is ω? (problem set problem)
Week 1 Lecure Problems, 5 Wha if somehing oscillaes wih no obvious spring? Wha is ω? (problem se problem) Sar wih Try and ge o SHM form E. Full beer can in lake, oscillaing F = m & = ge rearrange: F =
More informationMath 4600: Homework 11 Solutions
Mah 46: Homework Soluions Gregory Handy [.] One of he well-known phenomenological (capuring he phenomena, bu no necessarily he mechanisms) models of cancer is represened by Gomperz equaion dn d = bn ln(n/k)
More informationy = x 3 and y = 2x 2 x. 2x 2 x = x 3 x 3 2x 2 + x = 0 x(x 2 2x + 1) = 0 x(x 1) 2 = 0 x = 0 and x = (x 3 (2x 2 x)) dx
Millersville University Name Answer Key Mathematics Department MATH 2, Calculus II, Final Examination May 4, 2, 8:AM-:AM Please answer the following questions. Your answers will be evaluated on their correctness,
More informationINDEX. Transient analysis 1 Initial Conditions 1
INDEX Secion Page Transien analysis 1 Iniial Condiions 1 Please inform me of your opinion of he relaive emphasis of he review maerial by simply making commens on his page and sending i o me a: Frank Mera
More information