6.5 Improper integrals


 Adrian Fields
 8 months ago
 Views:
Transcription
1 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y = f () long the intervl [, ]. By definition, these integrls n only e used to ompute res of ounded regions. In some situtions, however, we re interested in unounded regions these re regions tht etend towrds infinity in t lest one diretion. Yet, mny unounded regions still hve finite re. We strt with si emple of this phenomenon: Prolem 6.6: Wht is the re of the region ordered y the urve y =, the line =, nd the is? Solution for Prolem 6.6: We sketh piture of this region t right. Notie tht this region is unounded: the region etends towrds + s grows lrge. Even though this region is unounded, we n ttempt to determine its re. We ertinly n ompute the re of the portion of the region to the left of = (for ny > ) s the definite integrl y d. As grows lrger, we epet tht the re under the urve on [, ] pprohes the re of the entire region under the urve on [, +). Speifilly, this re is y =! d. The integrl is esy to evlute: d = = Thus, when we tke the it, we get tht the re of the region is. d =!! = =. Note the prdo here: even though the region is unounded, it hs finite re. Prolem 6.6 suggests logil definition: Definition: integrl Let f e ontinuous funtion nd R suh tht (, ) Dom( f ). We define the improper f () d = f () d,! provided the it is defined. If the it is defined nd is not ±, we sy tht the improper integrl onverges. Otherwise, we sy tht the improper integrl diverges. There s n oviously similr definition for improper integrls in the other diretion: 7 Copyrighted Mteril
2 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY Definition: integrl Let f e ontinuous funtion nd R suh tht (, ) Dom( f ). We define the improper f () d = f () d,! provided the it is defined. If the it is defined nd is not ±, we sy tht the improper integrl onverges. Otherwise, we sy tht the improper integrl diverges. Let s generlize Prolem 6.6: Prolem 6.7: Let r e rel numer. Compute r d. Solution for Prolem 6.7: hve: This equls By definition, we ompute the improper integrl y writing it. If r,, then we d =! r!! r. (r ) r. r If r >, then the term pprohes s pprohes. Thus, in this se, the improper integrl onverges r to r. If r <, then the term grows without ound s pprohes. Thus, the integrl diverges. We might lso r write d = if r <. r Our originl integrtion ws not vlid for r =, so we hve to do tht se seprtely:! d =! (log ) =! (log ). As goes towrds infinity, this grows without ound, so the integrl diverges. In summry: r d = 8 >< r if r >, >: diverges if r pple. The net prolem is nother ommon emple of n improper integrl: Prolem 6.8: Compute e d, where is rel numer. Solution for Prolem 6.8: We ompute, for, (we ll investigte = t the end):! e d =! (e ) =! (e ). 8 Copyrighted Mteril
3 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS If is positive, then e =, so the integrl diverges. If is negtive, then e =, so the integrl equls!!. (Note this is positive numer when is negtive, so this nswer mkes sense.) Finlly, if =, then the integrl is R d, whih lerly diverges. Thus, the integrl diverges for nonnegtive eponents, nd onverges for negtive eponents. The result of Prolem 6.8 is typilly written s follows: if r >, then e r d = r. Prolem 6.9: Suppose f nd g re ontinuous funtions on [, ) nd f () pple g() for ll. () Show tht, if R f nd R g oth onverge, then Show tht if oth funtions re positive, nd R () Show tht if oth funtions re positive, nd R Solution for Prolem 6.9: () For ny, we hve (g f )() for ll [, ], thus Therefore, f pple g. g onverges, then R f onverges. f diverges, then R g diverges. (g f )() d. f () d pple g() d, nd sine its preserve nonstrit inequlities, we onlude tht f () d = f () d pple g() d = g() d.!! Define funtion F() = f (t) dt. Note tht F is n inresing funtion (sine f () for ll ), nd tht R eists. Also, sine pple f () pple g() for ll, we hve f () d =! F(), if this it pple F() = f (t) dt pple g(t) dt pple g(t) dt. Thus F is inresing nd hs n upper ound (nmely, R g(t) dt, whih y ssumption onverges), so y the result of Prolem 6.5, the it F() = f (t) dt! eists, so the integrl onverges. 9 Copyrighted Mteril
4 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY () This is just the ontrpositive sttement to prt, so there is nothing dditionl to prove. WARNING!! j We n only use the omprison tests in prts nd () of Prolem 6.9 if oth funtions re positive. As trivil emple, if f () = nd g() =, then for ny R, R g =, so R g onverges, ut R f diverges. Thus fr in this setion, we hve looked t improper integrls tht ompute res of regions tht re unounded in the diretion. There is nother type of improper integrl tht ours when the region tht we re emining is unounded in the ydiretion, s in the following emple: Prolem 6.: Compute p d. Solution for Prolem 6.: Skething the grph will immeditely show the issue. We hve! + p =. So the re under y = p is potentilly infinite (nd in ft the funtion is not even defined t ). We n do essentilly the sme thing we did for improper integrls with it of integrtion of ±. We define y y = p p d =! + p d. Note the + sine we only re out the intervl (,], we only re out wht hppens to the right of. This integrl is now esy to ompute: p d = p = p. As! +, this pprohes. Hene p d =. One gin, seemingly infinite re turns out to e finite. We n generlize the definition from Prolem 6.: Definition: integrl Suppose f is funtion, ontinuous on (, ], suh tht f () = ±. We define the improper! + f () d = f () d,! + provided this it is defined. If the it is defined, we sy tht this improper integrl onverges, nd if it is undefined, we sy tht the improper integrl diverges. Of ourse, we n do the sme thing if the funtion hs it of ± t the end of [, ). (We will omit writing out the forml definition.) Copyrighted Mteril
5 Eerpt from "Clulus" 3 AoPS In IMPROPER INTEGRALS Sidenote: Note tht the ove definition is onsistent with our usul (nonimproper) integrls. In prtiulr, if R f is defined, then y the Fundmentl Theorem of Clulus, the funtion g() = is di erentile, hene ontinuous, nd thus f (t) dt f (t) dt = g() = g() = f (t) dt.! +! + We know tht for regulr (not improper) integrls, we n rek them prt t ny point into two seprte integrls. Speifilly, if (, ), then f = f + f. This is lso how we evlute integrls tht re improper t oth ends, s in the following emple: Prolem 6.: Compute d for ll r > (or determine when it diverges). r Solution for Prolem 6.: The orret thing to do with n integrl tht is improper t oth ends is to split it somewhere in the middle. For emple, we n write r d = r d + r d. (We didn t hve to pik = s the point t whih to split them, ut it seems onvenient sine r is niely ehved t =.) We lredy know y Prolem 6.7 tht R d onverges if nd only if r >. The other integrl is r! d = d = = r!+ r!+ (r ) r r.!+ r If r >, then the frtion gets ritrrily lrge, so the it is infinite. Thus R d diverges for r >. r Hene our originl doulyimproper integrl is never onvergent: the integrl on (, ] diverges for r >, nd the integrl on [, ) diverges for r pple. Importnt: If (, ) Dom( f ) nd f (t) dt is improper t oth ends of (, ), then for ny (, ). f (t) dt = f (t) dt + f (t) dt,! +! As noted in the solution to Prolem 6., it doesn t mtter t whih point we rek up the doulyimproper integrl. Copyrighted Mteril
6 Eerpt from "Clulus" 3 AoPS In. CHAPTER 6. INFINITY Conept: We n rek n integrl prt s f = f + f t ny (, ) tht we hoose. Thus, hoose to e s onvenient s possile. We will leve it s n eerise to prove this. Also, it is not orret to try to tke shortut nd del with oth ends of douleimproper integrl t one. In prtiulr: WARNING!! j f () d is not the sme s f () d.! The orret wy to evlute n integrl over ll of R is to hoose R, nd then ompute f () d = f () d + f () d = f () d + f () d.!!+ We will leve it s n eerise to eplore this further. We lso hve to e it utious when deling with funtions with domins tht re not ll of R. Integrls of suh funtions might e improper ut not immeditely pper so. For emple: 3 Prolem 6.: Compute d. Solution for Prolem 6.: If you weren t pying lose ttention, you might do this: Bogus Solution: 3 d = 3 = 3 + = 6. We n t do this, euse the funtion is not defined t! To e little more preise, the funtion does not hve n ntiderivtive on the intervl [, 3], euse it is not defined t =, so we nnot pply the Fundmentl Theorem of Clulus. In order to evlute the integrl, we need to rek it up into sum of two improper integrls t the point t whih the funtion is undefined: 3 d = 3 d + d. As we sw in Prolem 6., oth of these diverge. Thus, the originl integrl itself diverges. d More generlly, when omputing something like, it might e tempting to sy is n odd funtion, so the integrl from to will nel out the integrl from to, nd thus the overll integrl is. This is lso the result tht nive lultion will give: Bogus Solution: d = log = log() log() =. Copyrighted Mteril
7 Eerpt from "Clulus" 3 AoPS In. REVIEW PROBLEMS But this is not orret! The only wy leglly to evlute this integrl is to rek it up into its improper prts. Neither prt onverges, so the integrl diverges. Eerises 6.5. Compute the following improper integrls: () 6.5. () 3 d ( ) Compute + d. Compute Compute + d. e d. (log ) d d = d + d. () e d (d) 4 d Show tht it doesn t mtter t whih point we rek up doulyimproper integrl. Speifilly, show tht, for ny, d (, ), if R f nd R f onverge, then R d f nd R f lso onverge, nd d d f + f = f + f. d Hints: 3, ? () Show tht if f () d onverges, then f () d = f () d. Hints: 65! Show tht the onverse of prt () is not true; tht is, it is possile tht f () d diverges. Hints: 8, 5! f () d onverges ut tht Review Prolems 6.3 Compute the following: ()! log os! 3! () e 4 (Soure: Rie) 6.4 sin t tn t Suppose nd re nonzero rel numers. Find nd. Hints: 3 t! sin t t! tn t 6.5 Compute () e d 3 d () d Hints: Copyrighted Mteril
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 informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationMath 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1)
Green s Theorem Mth 3B isussion Session Week 8 Notes Februry 8 nd Mrh, 7 Very shortly fter you lerned how to integrte singlevrible funtions, you lerned the Fundmentl Theorem of lulus the wy most integrtion
More informationNumbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point
GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply
More informationSection 4.4. Green s Theorem
The Clulus of Funtions of Severl Vriles Setion 4.4 Green s Theorem Green s theorem is n exmple from fmily of theorems whih onnet line integrls (nd their higherdimensionl nlogues) with the definite integrls
More informationProject 6: Minigoals Towards Simplifying and Rewriting Expressions
MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy
More information(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.
Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More informationMore Properties of the Riemann Integral
More Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologil Sienes nd Deprtment of Mthemtil Sienes Clemson University Februry 15, 2018 Outline More Riemnn Integrl Properties The Fundmentl
More informationThe RiemannStieltjes Integral
Chpter 6 The RiemnnStieltjes Integrl 6.1. Definition nd Eistene of the Integrl Definition 6.1. Let, b R nd < b. ( A prtition P of intervl [, b] is finite set of points P = { 0, 1,..., n } suh tht = 0
More information6.1 Definition of the Riemann Integral
6 The Riemnn Integrl 6. Deinition o the Riemnn Integrl Deinition 6.. Given n intervl [, b] with < b, prtition P o [, b] is inite set o points {x, x,..., x n } [, b], lled grid points, suh tht x =, x n
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More information10. AREAS BETWEEN CURVES
. AREAS BETWEEN CURVES.. Ares etween curves So res ove the xxis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in
More informationSection 7.1 Area of a Region Between Two Curves
Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationSolutions to Assignment 1
MTHE 237 Fll 2015 Solutions to Assignment 1 Problem 1 Find the order of the differentil eqution: t d3 y dt 3 +t2 y = os(t. Is the differentil eqution liner? Is the eqution homogeneous? b Repet the bove
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationMath 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 informationf (x)dx = f(b) f(a). a b f (x)dx is the limit of sums
Green s Theorem If f is funtion of one vrible x with derivtive f x) or df dx to the Fundmentl Theorem of lulus, nd [, b] is given intervl then, ording This is not trivil result, onsidering tht b b f x)dx
More informationNONDETERMINISTIC FSA
Tw o types of nondeterminism: NONDETERMINISTIC FS () Multiple strtsttes; strtsttes S Q. The lnguge L(M) ={x:x tkes M from some strtstte to some finlstte nd ll of x is proessed}. The string x = is
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationChapter Gauss Quadrature Rule of Integration
Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to
More informationMTH 505: Number Theory Spring 2017
MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of $ nd $ s two denomintions of coins nd $c
More informationQUADRATIC EQUATION. Contents
QUADRATIC EQUATION Contents Topi Pge No. Theory 004 Exerise  0509 Exerise  093 Exerise  3 45 Exerise  4 6 Answer Key 78 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,
More informationSolutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite!
Solutions for HW9 Exerise 28. () Drw C 6, W 6 K 6, n K 5,3. C 6 : W 6 : K 6 : K 5,3 : () Whih of the following re iprtite? Justify your nswer. Biprtite: put the re verties in V 1 n the lk in V 2. Biprtite:
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More informationHow do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?
XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=
More informationLine Integrals and Entire Functions
Line Integrls nd Entire Funtions Defining n Integrl for omplex Vlued Funtions In the following setions, our min gol is to show tht every entire funtion n be represented s n everywhere onvergent power series
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationLesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem.
27 Lesson 2: The Pythgoren Theorem nd Similr Tringles A Brief Review of the Pythgoren Theorem. Rell tht n ngle whih mesures 90º is lled right ngle. If one of the ngles of tringle is right ngle, then we
More informationUsing 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 informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06073 HIKARI Ltd, www.mhikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More informationu(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.
Lecture 4 Complex Integrtion MATHGA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex
More informationQUADRATIC EQUATION EXERCISE  01 CHECK YOUR GRASP
QUADRATIC EQUATION EXERCISE  0 CHECK YOUR GRASP. Sine sum of oeffiients 0. Hint : It's one root is nd other root is 8 nd 5 5. tn other root 9. q 4p 0 q p q p, q 4 p,,, 4 Hene 7 vlues of (p, q) 7 equtions
More informationThis enables us to also express rational numbers other than natural numbers, for example:
Overview Study Mteril Business Mthemtis 0506 Alger The Rel Numers The si numers re,,3,4, these numers re nturl numers nd lso lled positive integers. The positive integers, together with the negtive integers
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationThe Dirichlet Problem in a Two Dimensional Rectangle. Section 13.5
The Dirichlet Prolem in Two Dimensionl Rectngle Section 13.5 1 Dirichlet Prolem in Rectngle In these notes we will pply the method of seprtion of vriles to otin solutions to elliptic prolems in rectngle
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More information2 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 informationAlgorithm Design and Analysis
Algorithm Design nd Anlysis LECTURE 8 Mx. lteness ont d Optiml Ching Adm Smith 9/12/2008 A. Smith; sed on slides y E. Demine, C. Leiserson, S. Rskhodnikov, K. Wyne Sheduling to Minimizing Lteness Minimizing
More informationfor all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx
Applitions of Integrtion Are of Region Between Two Curves Ojetive: Fin the re of region etween two urves using integrtion. Fin the re of region etween interseting urves using integrtion. Desrie integrtion
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More information1 Nondeterministic Finite Automata
1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationImproper Integrals with Infinite Limits of Integration
6_88.qd // : PM Pge 578 578 CHAPTER 8 Integrtion Techniques, L Hôpitl s Rule, nd Improper Integrls Section 8.8 f() = d The unounded region hs n re of. Figure 8.7 Improper Integrls Evlute n improper integrl
More informationState Minimization for DFAs
Stte Minimiztion for DFAs Red K & S 2.7 Do Homework 10. Consider: Stte Minimiztion 4 5 Is this miniml mchine? Step (1): Get rid of unrechle sttes. Stte Minimiztion 6, Stte is unrechle. Step (2): Get rid
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More information( ) as a fraction. Determine location of the highest
AB/ Clulus Exm Review Sheet Solutions A Prelulus Type prolems A1 A A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f( x) Set funtion equl to Ftor or use qudrti eqution if qudrti Grph to
More informationSTEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.
STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t
More informationa) Read over steps (1) (4) below and sketch the path of the cycle on a P V plot on the graph below. Label all appropriate points.
Prole 3: Crnot Cyle of n Idel Gs In this prole, the strting pressure P nd volue of n idel gs in stte, re given he rtio R = / > of the volues of the sttes nd is given Finlly onstnt γ = 5/3 is given You
More informationRiemann Integrals and the Fundamental Theorem of Calculus
Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums
More informationNon Right Angled Triangles
Non Right ngled Tringles Non Right ngled Tringles urriulum Redy www.mthletis.om Non Right ngled Tringles NON RIGHT NGLED TRINGLES sin i, os i nd tn i re lso useful in nonright ngled tringles. This unit
More informationMT Integral equations
MT58  Integrl equtions Introduction Integrl equtions occur in vriety of pplictions, often eing otined from differentil eqution. The reson for doing this is tht it my mke solution of the prolem esier or,
More informationON THE INEQUALITY OF THE DIFFERENCE OF TWO INTEGRAL MEANS AND APPLICATIONS FOR PDFs
ON THE INEQUALITY OF THE DIFFERENCE OF TWO INTEGRAL MEANS AND APPLICATIONS FOR PDFs A.I. KECHRINIOTIS AND N.D. ASSIMAKIS Deprtment of Eletronis Tehnologil Edutionl Institute of Lmi, Greee EMil: {kehrin,
More informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
More information1 From NFA to regular expression
Note 1: How to convert DFA/NFA to regulr expression Version: 1.0 S/EE 374, Fll 2017 Septemer 11, 2017 In this note, we show tht ny DFA cn e converted into regulr expression. Our construction would work
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More informationPhysics 9 Fall 2011 Homework 2  Solutions Friday September 2, 2011
Physics 9 Fll 0 Homework  s Fridy September, 0 Mke sure your nme is on your homework, nd plese box your finl nswer. Becuse we will be giving prtil credit, be sure to ttempt ll the problems, even if you
More informationPythagoras theorem and surds
HPTER Mesurement nd Geometry Pythgors theorem nd surds In IEEM Mthemtis Yer 8, you lernt out the remrkle reltionship etween the lengths of the sides of rightngled tringle. This result is known s Pythgors
More information(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35
7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know
More informationMath RE  Calculus II Area Page 1 of 12
Mth RE  Clculus II re Pge of re nd the Riemnn Sum Let f) be continuous function nd = f) f) > on closed intervl,b] s shown on the grph. The Riemnn Sum theor shows tht the re of R the region R hs re=
More informationx = a To determine the volume of the solid, we use a definite integral to sum the volumes of the slices as we let!x " 0 :
Clculus II MAT 146 Integrtion Applictions: Volumes of 3D Solids Our gol is to determine volumes of vrious shpes. Some of the shpes re the result of rotting curve out n xis nd other shpes re simply given
More informationu( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 218, pp 4448): Determine the equation of the following graph.
nlyzing Dmped Oscilltions Prolem (Medor, exmple 218, pp 4448): Determine the eqution of the following grph. The eqution is ssumed to e of the following form f ( t) = K 1 u( t) + K 2 e!"t sin (#t + $
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More information4402 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam
4402 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP
More informationIf u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du
Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find ntiderivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible
More informationdx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω.
Chpter 8 Stility theory We discuss properties of solutions of first order two dimensionl system, nd stility theory for specil clss of liner systems. We denote the independent vrile y t in plce of x, nd
More informationHomework Assignment 3 Solution Set
Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution.
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationFarey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University
U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions
More informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
More informationProving the Pythagorean Theorem
Proving the Pythgoren Theorem W. Bline Dowler June 30, 2010 Astrt Most people re fmilir with the formul 2 + 2 = 2. However, in most ses, this ws presented in lssroom s n solute with no ttempt t proof or
More information11.1 Exponential Functions
. Eponentil Functions In this chpter we wnt to look t specific type of function tht hs mny very useful pplictions, the eponentil function. Definition: Eponentil Function An eponentil function is function
More informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
More informationCIT 596 Theory of Computation 1. Graphs and Digraphs
CIT 596 Theory of Computtion 1 A grph G = (V (G), E(G)) onsists of two finite sets: V (G), the vertex set of the grph, often enote y just V, whih is nonempty set of elements lle verties, n E(G), the ege
More informationLINEAR ALGEBRA APPLIED
5.5 Applictions of Inner Product Spces 5.5 Applictions of Inner Product Spces 7 Find the cross product of two vectors in R. Find the liner or qudrtic lest squres pproimtion of function. Find the nthorder
More informationGeometry of the Circle  Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272.
Geometry of the irle  hords nd ngles Geometry of the irle hord nd ngles urriulum Redy MMG: 272 www.mthletis.om hords nd ngles HRS N NGLES The irle is si shpe nd so it n e found lmost nywhere. This setion
More informationC1M14. Integrals as Area Accumulators
CM Integrls s Are Accumultors Most tetbooks do good job of developing the integrl nd this is not the plce to provide tht development. We will show how Mple presents Riemnn Sums nd the ccompnying digrms
More informationUniversitaireWiskundeCompetitie. Problem 2005/4A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that
Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More informationBoolean Algebra. Boolean Algebra
Boolen Alger Boolen Alger A Boolen lger is set B of vlues together with:  two inry opertions, commonly denoted y + nd,  unry opertion, usully denoted y ˉ or ~ or,  two elements usully clled zero nd
More informationSummary of Elementary Calculus
Summry of Elementry Clculus Notes by Wlter Noll (1971) 1 The rel numbers The set of rel numbers is denoted by R. The set R is often visulized geometriclly s numberline nd its elements re often referred
More informationLine and Surface Integrals: An Intuitive Understanding
Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of
More informationHomework Solution  Set 5 Due: Friday 10/03/08
CE 96 Introduction to the Theory of Computtion ll 2008 Homework olution  et 5 Due: ridy 10/0/08 1. Textook, Pge 86, Exercise 1.21. () 1 2 Add new strt stte nd finl stte. Mke originl finl stte nonfinl.
More informationBIFURCATIONS IN ONEDIMENSIONAL DISCRETE SYSTEMS
BIFRCATIONS IN ONEDIMENSIONAL DISCRETE SYSTEMS FRANCESCA AICARDI In this lesson we will study the simplest dynmicl systems. We will see, however, tht even in this cse the scenrio of different possible
More informationChapter 4 StateSpace Planning
Leture slides for Automted Plnning: Theory nd Prtie Chpter 4 StteSpe Plnning Dn S. Nu CMSC 722, AI Plnning University of Mrylnd, Spring 2008 1 Motivtion Nerly ll plnning proedures re serh proedures Different
More informationSurface Integrals of Vector Fields
Mth 32B iscussion ession Week 7 Notes Februry 21 nd 23, 2017 In lst week s notes we introduced surfce integrls, integrting sclrvlued functions over prmetrized surfces. As with our previous integrls, we
More informationECO 317 Economics of Uncertainty Fall Term 2007 Notes for lectures 4. Stochastic Dominance
Generl structure ECO 37 Economics of Uncertinty Fll Term 007 Notes for lectures 4. Stochstic Dominnce Here we suppose tht the consequences re welth mounts denoted by W, which cn tke on ny vlue between
More informationIntro to Nuclear and Particle Physics (5110)
Intro to Nucler nd Prticle Physics (5110) Feb, 009 The Nucler Mss Spectrum The Liquid Drop Model //009 1 E(MeV) n n(n1)/ E/[ n(n1)/] (MeV/pir) 1 C 16 O 0 Ne 4 Mg 7.7 14.44 19.17 8.48 4 5 6 6 10 15.4.41
More informationILLUSTRATING THE EXTENSION OF A SPECIAL PROPERTY OF CUBIC POLYNOMIALS TO NTH DEGREE POLYNOMIALS
ILLUSTRATING THE EXTENSION OF A SPECIAL PROPERTY OF CUBIC POLYNOMIALS TO NTH DEGREE POLYNOMIALS Dvid Miller West Virgini University P.O. BOX 6310 30 Armstrong Hll Morgntown, WV 6506 millerd@mth.wvu.edu
More informationMATH 573 FINAL EXAM. May 30, 2007
MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems.
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More informationQuadratic reciprocity
Qudrtic recirocity Frncisc Bozgn Los Angeles Mth Circle Octoer 8, 01 1 Qudrtic Recirocity nd Legendre Symol In the eginning of this lecture, we recll some sic knowledge out modulr rithmetic: Definition
More informationContinuity. Recall the following properties of limits. Theorem. Suppose that lim. f(x) =L and lim. lim. [f(x)g(x)] = LM, lim
Recll the following properties of limits. Theorem. Suppose tht lim f() =L nd lim g() =M. Then lim [f() ± g()] = L + M, lim [f()g()] = LM, if M = 0, lim f() g() = L M. Furthermore, if f() g() for ll, then
More informationLinear Systems with Constant Coefficients
Liner Systems with Constnt Coefficients 4305 Here is system of n differentil equtions in n unknowns: x x + + n x n, x x + + n x n, x n n x + + nn x n This is constnt coefficient liner homogeneous system
More informationNaming the sides of a rightangled triangle
6.2 Wht is trigonometry? The word trigonometry is derived from the Greek words trigonon (tringle) nd metron (mesurement). Thus, it literlly mens to mesure tringle. Trigonometry dels with the reltionship
More information