# 5.7 Improper Integrals

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1 458 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 moon: R+H R Gm 2 d = Gm R+H R = Gm R Gm R + H Notice tht s the height H grows very lrge, the second term in this nswer becomes very smll nd the totl work pproches Gm R. We cn write: Gm lim H R Gm = Gm R + H R Here we re tking limit of n epression tht rose s the vlue of definite integrl, so we cn lso write: Gm Gm R H R Gm R+H Gm R + H H R 2 d We could write this lst integrl, t lest informlly, s: R Gm 2 We cll this new type of integrl n improper integrl becuse the intervl of integrtion is infinite, violting n ssumption we mde when d originlly developing the definite integrl b f () d using Riemnn sums tht the length of the intervl of integrtion,, b, ws finite. Emple. Represent the re of the infinite region between f () = 2 nd the -is for (see mrgin) s n improper integrl. Solution. We cn represent the re of region (which hs infinite length) s: 2 d We don t yet know whether this re is finite or infinite. Prctice. Represent the volume swept out when the infinite region between f () = nd the -is for 4 is revolved bout the -is (see mrgin) using n improper integrl. Generl Strtegy for Improper Integrls In the lifting--pylod emple bove, we defined our first improper integrl s the limit of proper integrl over finite intervl s the length of the intervl becme lrger nd lrger.

2 5.7 improper integrls 459 Our generl pproch to evlute improper integrls over infinitely long intervls s well s nother type of improper integrl introduced lter in this section will mimic this strtegy: Shrink the intervl of integrtion so you hve (proper) definite integrl you cn evlute, then let the intervl grow to pproch the desired intervl of integrtion. The vlue of the improper integrl will be the limiting vlue of the (proper) definite integrls s the intervls grow to the intervl you wnt, provided tht this limit eists. Infinitely Long Intervls of Integrtion To evlute n improper integrl on n infinitely long intervl: replce the infinitely long intervl with finite intervl evlute the integrl on the finite intervl let the finite intervl grow longer nd longer, pproching the originl infinitely long intervl Emple 2. Evlute d (see mrgin). 2 Solution. The intervl, ) is infinitely long, but we cn evlute the integrl on finite intervls such s, 2,,,, nd, more generlly,, where is some mssive positive number: 2 2 d = 2 2 d = 2 d = nd, more generlly, = = = 2 2 = 2 = d 2 = = 9 = = d = so: The vlue of the improper integrl is. d 2 = We sy tht the improper integrl d in the Emple 2 is 2 convergent nd tht it converges to. Furthermore, from Emple, we know tht this improper integrl represents the re of n infinitely long region. We now hve n emple which you my find highly counterintuitive of region with infinite length but finite re. Not ll improper integrls converge, however.

3 46 pplictions of definite integrls Emple. Evlute ech improper integrl. (See mrgin for grphicl interprettions of these integrls s res of unbounded regions.) () + 2 d (b) d (c) cos() d Solution. () Replcing the upper limit of the improper integrl with mssive positive number : d d 2 = π 2 rctn() rctn() so the improper integrl is convergent nd converges to π 2. (b) Replcing the upper limit of the improper integrl with mssive positive number : d d ln() ln() = Becuse this limit diverges, we sy the improper integrl is divergent or tht it diverges. (c) Once gin replcing with in the upper limit of the integrl: lim cos() d sin() sin() As grows without bound, the vlues of sin() oscillte between nd, never pproching single vlue, so the limit does not eist; we sy tht this improper integrl diverges. Prctice 2. Evlute: () d (b) sin() d Definition: For ny integrble function f () defined for ll nd ny integrble function g() defined for ll b: b f () d g() d = lim N b N f () d g() d If the limit in question eists nd is finite, we sy tht the corresponding improper integrl converges or is convergent nd define the vlue of the improper integrl to be the vlue of the limit. If the limit in question does not eist, we sy tht the corresponding improper integrl diverges or is divergent.

4 5.7 improper integrls 46 Functions Undefined t n Endpoint of the Intervl of Integrtion Consider the grph of on the intervl (, (see mrgin) nd compre this region to the grph from Emple 2. It ppers we cn generte the new region by reflecting the old region cross y = nd dding rectngle (of re ) t the bottom, so we might resonbly ssume tht the integrl d is finite number. This integrl is over finite intervl,,, but we hve new problem: the integrnd is undefined t =, one of the endpoints of the intervl of integrtion. This violtes nother ssumption we mde when developing the definition of definite integrl s limit of Riemnn sums. If the function you wnt to integrte is unbounded t one of the endpoints of n intervl of finite length, s in this sitution, you cn shrink the intervl of integrtion so tht the function is bounded t both endpoints of the new, smller intervl, then evlute the integrl over the smller intervl, nd finlly let the smller intervl grow to pproch the originl intervl. Emple 4. Evlute d. Solution. The function is not defined t =, the lower endpoint of integrtion, but the function is bounded on intervls such s.6,,.9, nd, more generlly, on the intervl c, for ny c > :.6.9 nd, in generl: d = 2 d = 2 c.6 = = 2.2 =.8.9 = = 2.6 =.4 d = 2 = 2 2 c = 2 2 c c so, tking the limit s c decreses towrd : lim d 2 2 c = 2 c + c c + which is wht you should hve epected bsed on the grph. A region of re plus rectngle of re should hve n re of + = 2. Definition: For ny function f () defined nd continuous on (, b nd ny function g() defined nd continuous on, b): b b b f () d c + g() d c b c c f () d g() d

5 462 pplictions of definite integrls See Problems for prctice with integrls of this type. If the limit eists, we sy the integrl converges nd define the vlue of the integrl to be the vlue of the limit. If the limit does not eist, we sy tht the integrl diverges. Prctice. Show tht () d = 6 nd (b) d diverges. If n integrnd is unbounded t one or more points inside the intervl of integrtion, you cn split the originl improper integrl into two or more improper integrls over subintervls where the integrnd is unbounded t only one endpoint of ech subintervl. Testing for Convergence: The P-Test nd the Comprison Test Sometimes we cre only whether or not n improper integrl converges. We now consider two methods for testing the convergence of n improper integrl. Neither method gives you the ctul vlue of the integrl, but ech enbles you to determine whether or not certin improper integrls converge. The Comprison Test for Integrls enbles you to determine the convergence (or divergence) of certin integrls by compring them with other (esier) integrls. The P-Test involves specil cses often used with the Comprison Test for Integrls. P-Test for integrls: For ny >, the improper integrl converges if p > nd diverges if p. = p d Proof. It is esiest to consider three cses rther thn two: p =, p > nd p <. If p = then: p d = d d ln ( ) ln() ln() so the improper integrl diverges. For the other two cses, p =, so: lim p p+ p d p + p p p If p >, then p < so lim p p = nd: p+ p d p + p+ p + = p p which is finite number, so the improper integrl converges. If p <, then p > p so lim p = nd: p d so the improper integrl diverges. p+ p + p+ = p +

6 5.7 improper integrls 46 Emple 5. Determine the convergence or divergence of ech integrl. () 5 2 d (b) d (c) 8 d Solution. () The integrl mtches the form required by the P-Test with p = 2 >, so the improper integrl converges. The P-Test does not tell us the vlue of the integrl. (b) The integrl mtches the form required by the P-Test with p = 2 <, so the improper integrl diverges. (c) This is not n improper integrl, so the P-Test does not pply, but: 8 8 d = 8 d = 2 2 = = = 9 2 so the vlue of the integrl is 4.5. The following Comprison Test enbles us to determine the convergence or divergence of n improper integrl of positive function by compring this function with functions whose improper integrls we lredy know converge or diverge. Comprison Test for Integrls of Positive Functions: Suppose f () nd g() re defined nd integrble for ll with f () g(). Then: g() d converges f () d converges. f () d diverges g() d diverges. The proof involves strightforwrd ppliction of the definition of n improper integrl nd vrious fcts bout limits, but the grph in the mrgin provides geometriclly intuitive wy of understnding why these results must hold. If g() d converges, then the re under the grph of g() is finite, so the (smller) re under the grph of f () must lso be finite, nd f () d must converge s well. If f () d diverges, then the re under the grph of f () is infinite, so the (bigger) re under the grph of g() must lso be infinite, nd g() d must lso diverge. Just s importnt s understnding wht this Comprison Test does tell us is relizing wht the Comprison Test does not tell us. If g() d diverges, or if f () d converges, the Comprison Test tells us bsolutely nothing bout the convergence or divergence of the other integrl. Geometriclly, if g() d diverges, then the re under the grph of g() is infinite, but the (smller) re under the

7 464 pplictions of definite integrls grph of f () could be either finite or infinite, so we cn t conclude nything bout the convergence or divergence of f () d. Likewise, if f () d converges, then the re under the grph of f () is finite, but the (bigger) re under the grph of g() could be either finite or infinite, so we cn t conclude nything bout the convergence or divergence of g() d. Emple 6. Determine whether ech of these integrls is convergent or divergent by compring it with n pproprite integrl tht you lredy know converges or diverges. () d (b) Solution. () We know tht 5 > nd so: + sin() 9 2 d (c) d > < + 5 < < < 7 The numertor of the originl integrnd is constnt nd the dominnt term in the denomintor of tht integrnd is, so it should mke sense to compre the originl integrnd with. The numertor of the originl integrnd fluctutes between 2 nd 4 while the dominnt (nd only) term in its denomintor is 2, so it should mke sense to compre the originl integrnd with 2. We lso know, by the P-Test with p = >, tht d con- verges, so 7 d = 7 d lso converges. By the Com- 7 d must converge s + 5 prison Test, the smller integrl well. (b) We know tht sin(), so: 2 + sin() 4 < + sin() = 4 2 By the P-Test with p = 2 >, 2 d converges, so 4 2 d = 4 d lso converges. By the Comprison Test, the smller 2 + sin() integrl 2 d must converge s well. (c) We know tht u is n incresing function, so: 5 < 5 < 5 > By the P-Test with p = 2 <, integrl 6 6 d 5 d must lso diverge. diverges, so the bigger

8 5.7 improper integrls Problems In Problems 26, evlute ech improper integrl, or show why it diverges d e π 2 π d 4. 5 ln() d 2 d e ( 2) d. ( + 2) 2 d 2. d 4. 4 d d 8. 2 sin() d 2. tn() d 22. tn() d d 26. π 5 ln() 2 d 2 e d + 2 d ( 2) 2 d + 2 d ( + 2) d d 2 d 2 8 d sin() d 2 d d 2 d e π 7 4 d d + d 7 + ln() d + cos() 2 d d d d 2 d 2 d d d 45. Emple (b) showed tht d grew rbitrrily lrge s grew rbitrrily lrge, so no finite mount of pint would cover the region bounded by the -is nd the grph of f () = for > : Show tht the volume of the solid obtined when the region grphed bove is revolved bout the -is: In Problems 27 44, determine whether ech improper integrl converges or diverges, but do not evlute the integrl d d d d d 4 7 d is finite, so the -dimensionl trumpet-shped region cn be filled with finite mount of pint. Does this present contrdiction?

9 466 pplictions of definite integrls 46. Determine whether or not the volume of the solid obtined by revolving the region between the - is nd the grph of f () = sin() for for (see below) bout the -is is finite. 5. Use the figure below left to help determine which A A is lrger: d or k. k= 47. Compute the volume of the solid obtined when the region in the first qudrnt between the positive -is nd the grph of f () = 2 + (see below) is revolved bout the -is. 52. Use the figure bove right to help determine A A which is lrger: d or k. k=2 5. Use the figure below left to help determine which is lrger: A A 2 d or k= k Compute the volume of the solid obtined when the region in the first qudrnt between the positive -is nd the grph of f () = e is revolved bout the -is. 49. Compute the volume of the solid obtined when the region in the first qudrnt between the positive -is nd the grph of f () = 2 + (see below) is revolved bout the y-is. 5. Compute the volume of the solid obtined when the region in the first qudrnt between the positive -is nd the grph of f () = e is revolved bout the y-is. 54. Use the figure bove right to help determine which is lrger: A 2 d or A k=2 k 2. The Lplce trnsform of function f (t) is defined using n improper integrl involving prmeter s: F(s) = e st f (t) dt Lplce trnsforms re often used to solve differentil equtions. 55. Compute the Lplce trnsform of the constnt function f (t) =. 56. Compute the Lplce trnsform of f (t) = e 4t. 57. Define function g(t) by: { if t < 2 g(t) = if t 2 Compute the Lplce trnsform of g(t). 58. Define function h(t) by: { if t < h(t) = if t Compute the Lplce trnsform of h(t).

10 5.7 improper integrls 467 b 59. Devise Q-Test to determine whether q d converges or diverges for ny number b >. 6. Use the result of the previous problem to test the e π convergence of d nd d. 5.7 Prctice Answers ( ) 2. π d = π d 2. () d d = 2 2 (b) Replcing with in the upper limit of the integrl: sin() d sin() d cos() + cos() cos() This limit does not eist (the vlues of cos() oscillte between nd 2 nd never pproch ny fied number) so the improper integrl diverges.. () The integrl is improper t its upper limit, where =, so: d = c lim ( ) 2 d c 2 c c c 2 c = + 2 = 6 (b) The integrl is improper t its lower limit, where =, so: d d ln ( ) c + c ln() ln(c) = c + c c + so the integrl diverges.

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### 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

### Section 4: Integration ECO4112F 2011 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

### 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

### !0 f(x)dx + lim!0 f(x)dx. The latter is sometimes also referred to as improper integrals of the. k=1 k p converges for p>1 and diverges otherwise. Chpter 7 Improper integrls 7. Introduction The gol of this chpter is to meningfully extend our theory of integrls to improper integrls. There re two types of so-clled improper integrls: the first involves

### 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

### 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

### Further integration. x n nx n 1 sinh x cosh x log x 1/x cosh x sinh x e x e x tan x sec 2 x sin x cos x tan 1 x 1/(1 + x 2 ) cos x sin x Further integrtion Stndrd derivtives nd integrls The following cn be thought of s list of derivtives or eqully (red bckwrds) s list of integrls. Mke sure you know them! There ren t very mny. f(x) f (x)

### Section 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

### . 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?

### Week 10: Line Integrals Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.

### Suppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = -2. Mth 43 Section 6. Section 6.: 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

### APPROXIMATE INTEGRATION APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

### practice How would you find: e x + e x e 2x e x 1 dx 1 e today: improper integrals prctice How would you find: dx e x + e x e 2x e x 1 dx e 2x 1 e x dx 1. Let u=e^x. Then dx=du/u. Ans = rctn ( e^x ) + C 2. Let u=e^x. Becomes u du / (u-1), divide to get u/(u-1)=1+1/(u-1) Ans = e^x + ln

### Riemann is the Mann! (But Lebesgue may besgue to differ.) Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >

### different 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 Lecture 12: Numericl Qudrture J.K. Ryn@tudelft.nl WI3097TU Delft Institute of Applied Mthemtics Delft University of Technology 5 December 2012 () Numericl Qudrture 5 December 2012 1 / 46 Outline 1 Review

### MA 124 January 18, Derivatives are. Integrals are. MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,

### Example Sheet 6. Infinite and Improper Integrals Sivkumr Exmple Sheet 6 Infinite nd Improper Integrls MATH 5H Mteril presented here is extrcted from Stewrt s text s well s from R. G. Brtle s The elements of rel nlysis. Infinite Integrls: These integrls

### . 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

### 7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series 7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts

### How can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function? Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those

### 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

### The practical version Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht

### Sections 5.2: The Definite Integral Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)

### x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According

### THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS. THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

### 4.5 THE FUNDAMENTAL THEOREM OF CALCULUS 4.5 The Funmentl Theorem of Clculus Contemporry Clculus 4.5 THE FUNDAMENTAL THEOREM OF CALCULUS This section contins the most importnt n most use theorem of clculus, THE Funmentl Theorem of Clculus. Discovere

### Week 10: Riemann integral and its properties Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the

### The Henstock-Kurzweil integral fculteit Wiskunde en Ntuurwetenschppen The Henstock-Kurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft

### 6.2 CONCEPTS FOR ADVANCED MATHEMATICS, C2 (4752) AS 6. CONCEPTS FOR ADVANCED MATHEMATICS, C (475) AS Objectives To introduce students to number of topics which re fundmentl to the dvnced study of mthemtics. Assessment Emintion (7 mrks) 1 hour 30 minutes.

### Chapter 9. Arc Length and Surface Area Chpter 9. Arc Length nd Surfce Are In which We ppl integrtion to stud the lengths of curves nd the re of surfces. 9. Arc Length (Tet 547 553) P n P 2 P P 2 n b P i ( i, f( i )) P i ( i, f( i )) distnce

### Continuous 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

### 1 Probability Density Functions Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our

### Section 4.8. D v(t j 1 ) t. (4.8.1) j=1 Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions

### Main topics for the First Midterm Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

### We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b. Mth 255 - Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn

### Duality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below. Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we

### 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

### APPLICATIONS OF THE DEFINITE INTEGRAL APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its cross-section in plne pssing through

### 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

### Main topics for the Second Midterm Min topics for the Second Midterm The Midterm will cover Sections 5.4-5.9, Sections 6.1-6.3, nd Sections 7.1-7.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the

### Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015 Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n

### 50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS 68 CHAPTE MULTIPLE INTEGALS 46. e da, 49. Evlute tn 3 4 da, where,. [Hint: Eploit the fct tht is the disk with center the origin nd rdius is smmetric with respect to both es.] 5. Use smmetr to evlute 3

### c n φ n (x), 0 < x < L, (1) n=1 SECTION : Fourier Series. MATH4. In section 4, we will study method clled Seprtion of Vribles for finding exct solutions to certin clss of prtil differentil equtions (PDEs. To do this, it will be necessry Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform 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 rel-vlues nd smooth The pproximtion of n integrl by numericl