5.7 Improper Integrals

Size: px
Start display at page:

Download "5.7 Improper Integrals"

Transcription

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.

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

The Regulated and Riemann Integrals

The Regulated and Riemann Integrals Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

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

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

More information

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

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

More information

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

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral. Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:

More information

Math 113 Exam 2 Practice

Math 113 Exam 2 Practice Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

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

More information

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

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

MAA 4212 Improper Integrals

MAA 4212 Improper Integrals Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

More information

5.2 Volumes: Disks and Washers

5.2 Volumes: Disks and Washers 4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of cross-section or slice. In this section, we restrict

More information

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

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

More information

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

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

More information

Chapter 0. What is the Lebesgue integral about?

Chapter 0. What is the Lebesgue integral about? Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

More information

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

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given

More information

Chapter 6 Notes, Larson/Hostetler 3e

Chapter 6 Notes, Larson/Hostetler 3e Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

More information

Definite integral. Mathematics FRDIS MENDELU

Definite integral. Mathematics FRDIS MENDELU Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

More information

Review of Calculus, cont d

Review of Calculus, cont d Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

More information

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30 Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

More information

Lecture 1. Functional series. Pointwise and uniform convergence.

Lecture 1. Functional series. Pointwise and uniform convergence. 1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

More information

1 The Riemann Integral

1 The Riemann Integral The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

More information

Improper Integrals, and Differential Equations

Improper Integrals, and Differential Equations Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

More information

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

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

More information

and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

and that at t = 0 the object is at position 5. Find the position of the object at t = 2. 7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

More information

Math 8 Winter 2015 Applications of Integration

Math 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 information

Homework 11. Andrew Ma November 30, sin x (1+x) (1+x)

Homework 11. Andrew Ma November 30, sin x (1+x) (1+x) Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin

More information

Math 131. Numerical Integration Larson Section 4.6

Math 131. Numerical Integration Larson Section 4.6 Mth. Numericl Integrtion Lrson Section. This section looks t couple of methods for pproimting definite integrls numericlly. The gol is to get good pproimtion of the definite integrl in problems where n

More information

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

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

More information

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

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

More information

Math& 152 Section Integration by Parts

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

More information

INTRODUCTION TO INTEGRATION

INTRODUCTION TO INTEGRATION INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide

More information

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004 Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when

More information

1 Part II: Numerical Integration

1 Part II: Numerical Integration Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble

More information

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

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

More information

Math 113 Exam 2 Practice

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

More information

MATH 144: Business Calculus Final Review

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

More information

Chapter 6 Techniques of Integration

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

More information

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

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

More information

Overview of Calculus I

Overview of Calculus I Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

More information

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

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

More information

NUMERICAL INTEGRATION

NUMERICAL INTEGRATION NUMERICAL INTEGRATION How do we evlute I = f (x) dx By the fundmentl theorem of clculus, if F (x) is n ntiderivtive of f (x), then I = f (x) dx = F (x) b = F (b) F () However, in prctice most integrls

More information

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1 The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

More information

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

More information

5.5 The Substitution Rule

5.5 The Substitution Rule 5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n nti-derivtive is not esily recognizble, then we re in

More information

Riemann Sums and Riemann Integrals

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

More information

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

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

More information

7.2 The Definite Integral

7.2 The Definite Integral 7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

More information

Riemann Sums and Riemann Integrals

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

More information

4.4 Areas, Integrals and Antiderivatives

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

More information

Numerical Analysis: Trapezoidal and Simpson s Rule

Numerical Analysis: Trapezoidal and Simpson s Rule nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =

More information

Improper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Improper Integrals. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics Improper Integrls MATH 2, Clculus II J. Robert Buchnn Deprtment of Mthemtics Spring 28 Definite Integrls Theorem (Fundmentl Theorem of Clculus (Prt I)) If f is continuous on [, b] then b f (x) dx = [F(x)]

More information

Section 6: Area, Volume, and Average Value

Section 6: Area, Volume, and Average Value Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

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

More information

Section 4: Integration ECO4112F 2011

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

More information

AP Calculus Multiple Choice: BC Edition Solutions

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

More information

Summer MTH142 College Calculus 2. Section J. Lecture Notes. Yin Su University at Buffalo

Summer MTH142 College Calculus 2. Section J. Lecture Notes. Yin Su University at Buffalo Summer 6 MTH4 College Clculus Section J Lecture Notes Yin Su University t Bufflo yinsu@bufflo.edu Contents Bsic techniques of integrtion 3. Antiderivtive nd indefinite integrls..............................................

More information

!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.

!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

More information

HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016

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

More information

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

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

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 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)

More information

Section 6.1 Definite Integral

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

More information

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

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

More information

Week 10: Line Integrals

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.

More information

Suppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = -2.

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

More information

APPROXIMATE INTEGRATION

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

More information

practice How would you find: e x + e x e 2x e x 1 dx 1 e today: improper integrals

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

More information

Riemann is the Mann! (But Lebesgue may besgue to differ.)

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 >

More information

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).

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

More information

Lecture 12: Numerical Quadrature

Lecture 12: Numerical Quadrature 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

More information

MA 124 January 18, Derivatives are. Integrals are.

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,

More information

Example Sheet 6. Infinite and Improper Integrals

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

More information

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

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

More information

7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series

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

More information

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?

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

More information

Bernoulli Numbers Jeff Morton

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

More information

The practical version

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

More information

Sections 5.2: The Definite Integral

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)

More information

x = b a N. (13-1) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is

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

More information

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

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

More information

4.5 THE FUNDAMENTAL THEOREM OF CALCULUS

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

More information

Week 10: Riemann integral and its properties

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

More information

The Henstock-Kurzweil integral

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

More information

6.2 CONCEPTS FOR ADVANCED MATHEMATICS, C2 (4752) AS

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.

More information

Chapter 9. Arc Length and Surface Area

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

More information

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom

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

More information

1 Probability Density Functions

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

More information

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1

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

More information

Main topics for the First Midterm

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

More information

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.

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

More information

Stuff You Need to Know From Calculus

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

More information

Duality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.

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

More information

APPLICATIONS OF THE DEFINITE INTEGRAL

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

More information

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

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

More information

Main topics for the Second Midterm

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

More information

Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015

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

More information

50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS

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

More information

c n φ n (x), 0 < x < L, (1) n=1

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

More information

Section 6.1 INTRO to LAPLACE TRANSFORMS

Section 6.1 INTRO to LAPLACE TRANSFORMS 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

More information

Lecture 14: Quadrature

Lecture 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 rel-vlues nd smooth The pproximtion of n integrl by numericl

More information

Not for reproduction

Not for reproduction AREA OF A SURFACE OF REVOLUTION cut h FIGURE FIGURE πr r r l h FIGURE A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundry of solid of revolution of the type

More information

Definite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +

Definite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 + Definite Integrls --5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the left-hnd

More information