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.


 Virgil Shepherd
 3 years ago
 Views:
Transcription
1 Mth 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 be nturlly pplied to function of n vribles. Assume f(x, y) is defined on smooth curve of finite length, prmeterized in terms of rclength s r(s) = x(s), y(s) for t b. For the purposes of introducing this concept, we ssume f(x, y). We will eventully mention wht the difference is if this ssumption is tken wy. We prtition into n smll rcs by forming prtition of [, b] by picking s i s follows: = s < s < < s n = b. Let s i for i =, 2,..., n denote the width of [s i, s i ]. Pick smple points s i from [s i, s i ] for i =, 2,..., n. Then we will pproximte the re of the curtin under the curve x(s), y(s), f(x(s), y(s)) (over the curve the xyplne see the picture t the top of the next pge) by the Riemnn sum n A f(x(s k), y(s k)) s k. k= Geometriclly ech summnd is the re of curvy pnel (to visulize them, tke rectngulr piece of pper nd bend it). Let (n) be the mximum vlue mong the s i for i =, 2,..., n. If the limit of the Riemnn sums s n exists for ny sequence of prtitions chosen such tht (n) s n then this limit is clled the pth integrl of f over (k sclr line integrl), nd it converges to the re of the curtin. See the picture t the top of the next pge. Nottion: When the limit exists, we sy f is integrble on the smooth curve of finite b length nd write f ds = f(r(s)) ds (where r(s) is n rclength prmetriztion of ). IMPORTANT: If we remove the ssumption tht f be nonnegtive, the pth integrl gives the net re (re bove the xyplne minus re below the xyplne). This construction nd these definitions nturlly extend to sclrvlued function f on subset S of R n contining smooth curve of finite length.
2 We will compute the re of the curtin bove, but for now we strt with more simplistic exmple: A thin wire in plne represented by smooth curve with density function ρ(x, y) (units of mss per length) hs Mss M = ρ ds. Suppose thin wire occupies the curve y = x 2 nd hs density ρ(x, y) = y g/cm (ssume x, y crry units of cm). Let s find the mss of this wire! An rclength prmetriztion of the upper hlf of the unit circle is r(s) = cos (s), sin (s) where s π. So the pth integrl is π M = ρ ds = ρ(cos (s), sin (s)) ds = π ρ(cos (s), sin (s)) ds = π sin (s) ds = 2 g.
3 Wht do we do if we hve n rbitrry prmetriztion of smooth curve of finite length? We could try to find n rclength prmetriztion, but often tht cretes n extremely ugly prmetriztion. Insted we will develop formul tht is independent of the prmetriztion! Let smooth curve of finite length nd the vectorvlued function r(t) for t b be n rbitrry differentible prmetriztion (moving long the curve in one direction tht is, not bck nd forth). Recll the rclength function below computes the rclength long from r() to r(t): s(t) = t r (u) du. Differentiting (nd using the Fundmentl Theorem of lculus) gives s (t) = r (t). Then we hve tht ds = s (t)dt = r (t) dt. Then if f is integrble, f ds = b f(r(t)) r (t) dt. So we cn compute the pth integrl with ny prmetriztion! Nturlly we cll r (t) the speed fctor. Let s find the re of the curtin t the top of the lst pge. Let f(x, y) = 3 2 cos (x). Let be given by r(t) = cos (t/3), sin (t/3) for t 2π. So we find f ds: f ds = 2π 3 2 cos (cos (t/3)) sin (t/3), cos (t/3) 3 dt = 2 2π t 3 dt = 2 (2π)2 = π2 3. Note: This mtches the simple solution which is to relize tht this curtin is just bent tringle of bse length 2π 3 nd height π.
4 Everything we hve done here generlizes for function of n vribles: Let f be n integrble sclrvlued function on some subset S of R n contining smooth curve of finite length prmeterized by the differentible vectorvlued function r(t) = x (t), x 2 (t),..., x n (t) (moving long the curve in one direction) for t b. Then f ds = b f(r(t)) r (t) dt = b f(x (t), x 2 (t),..., x n (t)) (x (t)) 2 + (x 2(t)) (x n(t)) 2 dt. Let s clculte the pth integrl for t π 2. z 2 ds = π 2 z 2 ds for the helicl pth given by r(t) = 4 cos (t), 4 sin (t), 3t (3t) 2 π ( ( 4 sin (t)) 2 + (4 cos (t)) 2 + (3) 2 2 π ) 3 dt = 45 t 2 5π 3 dt = 5 = 2 8. Let f be function on subset of R n contining smooth curve of finite length. The verge vlue of f on is given by f ds where l() is the length of. l() Find the verge vlue of f(w, x, y, z) = w 2 + x + 2y + 3z on the line segment joining (,,, 2) to (, 2, 2, ). Let s find prmetriztion of the line segment: r(t) = +t, +t, +t, 2 t for t. The length of this curve is l = = 2. This is lso the speed fctor! f ds = ( t 2 + ( + t) + 2( + t) + 3(2 t) ) (2) dt = 2 So the verge vlue of f on the line segment is ( ) t dt = = 56 3.
5 Now we introduce the concept of line integrl of vector field. First we strt with some bckground. An oriented curve is prmeterized curve for which direction is specified. The positive orienttion is the direction the curve is generted s the prmeter increses; the negtive orienttion is direction s the prmeter decreses. For instnce, the positive orienttion of r(t) = cos (t), sin (t) is counterclockwise. Let F be continuous vector field on some subset of R n tht contin smooth curve of finite length with n rclength prmetriztion r(s). Let T be the unit tngent vector t ech point of (with the positive orienttion given by the prmetriztion). The line integrl of F over curve (with given orienttion) is F T ds. IMPORTANT: F T is the component of F in the direction of T. Hence, geometriclly, the integrl is dding up the components of the vector field tht re tngentil to the curve. As we will see this hs mny pplictions. We immeditely seek wy to evlute line integrl tht does not require n rclength prmetriztion: Let F be vector field on some subset of R n tht contins smooth curve prmeterized by differentible vectorvlued function r(t) for t b such tht r (t) for ll t [, b]. Then T = r r. Then mking use of the fct tht ds = r dt we get b r b F T ds = F r r dt = F r dt. There re other wys to write this: F dr where dr = r dt. f dx + f 2 dx 2 + f n dx n where F = f, f 2,..., f n nd r(t) = x (t), x 2 (t),..., x n (t). Here is the connection: dr = x (t), x 2(t),..., x n(t) dt = dx, dx 2,..., dx n
6 Let s work through some exmples: Let s evlute the line integrl of F = x y, x on the prbol y = x 2 from (, ) to (, ): We cn use the grph prmetriztion r(t) = t, t 2 (which hs the right orienttion). lling the prbol we get F dr = t t 2, t, 2t dt = t + t 2 dt = 5 6. This yields mesure of the strength of the field long the curve in the direction of the curve. Here is picture:
7 Work: If F is vector field in region D of R 3 representing the force (s vector) t points inside D then the work done in moving n object long curve in some direction is the line integrl of F over (using suitbly oriented prmetriztion). An electric chrge t (,, ) produces force field on other point chrges given by k F = x, y, z where k is constnt. Let s clculte the work done in moving (x 2 + y 2 + z 2 ) 3/2 point chrge from (,, ) to (,, ): Using r(t) = t, t, t for t, we get W = k t, t, t,, dt = (3t 2 ) 3/2 k(3t) 3 3t 3 dt = k 3 t dt = k ( ) 2 3 ircultion: Let be closed (ends where it begins) smooth oriented curve in R n prmeterized once round by r(t) for t b (consistent with the orienttion) in region D where continuous vector field F is defined. The circultion of F on is the line integrl F dr. Let s clculte the circultion of F = using counterclockwise orienttion: x2 + y 2 x, y on the circle (x 2)2 + (y 2) 2 = 4 Let s lbel the circle by. First we prmeterize the once counterclockwise with r(t) = cos (t), sin (t) for t 2π. Then we integrte: F dr = = 2 = 2 2π 2π 2π cos (t), sin (t) 2 sin (t), 2 cos (t) dt (2 + 2 cos (t))2 + (2 + 2 sin (t)) 2 cos (t) sin (t) cos (t) + 2 sin (t) dt cos (t) sin (t) cos (t) + 2 sin (t) dt = cos (t) + 2 sin (t) =. 2π The circultion is zero becuse s much of the vector field points long the pth s ginst the pth when trversing. See the picture on the next pge!
8 Think of it like the vector field is the wind nd you wlk round. In this cse, s much is t your bck s t your fce... Let F = x 2, 2xy + x, z nd let be the circle x 2 + y 2 = oriented counterclockwise in the plne z =. Let s find the circultion of F on : Let s use the prmetriztion r(t) = cos (t), sin (t),. Then F dr = = = = 2π 2π 2π 2π cos 2 (t), 2 cos (t) sin (t) + cos (t), sin (t), cos (t), dt sin (t) cos 2 (t) + 2 cos 2 (t) sin (t) + cos 2 (t) dt cos 2 (t) sin (t) + cos 2 (t) dt cos 2 (t) sin (t) + ( + cos (2t)) dt 2 (t + 2 ) 2π sin (2t) = 3 cos3 (t) + 2 = π. Note: ircultion my be computed on closed curve mde up of piecewise smooth curves by just computing the line integrl on ech curve nd dding up the results.
9 Flux: The line integrl of vector field F dds up the components of F long smooth curve of finite length. Wht if we were to do the sme thing, but with the norml components? Tht is clled the flux of F cross. We will only focus on how this is developed for vector field in plne: To come up with the flux integrl in plne, we first hve to determine wy to find the unit norml vector to smooth curve of finite length. Let r(t) = x(t), y(t) for t b be differentible prmetriztion of (with some fixed orienttion) such tht r for ll t b. Recll: T = r r is the unit tngent vector to in the direction of the prmetriztion. Second we embed the xyplne into R 3, where the vector k =,, is orthogonl to the xyplne. Then unit vector orthogonl to in the xyplne would be orthogonl to both T nd k. So we set N = T k (the cross product of two orthogonl unit vectors gives unit vector orthogonl to both inputs) nd drop its third component, which is, so tht it ultimtely lives in R 2. So the integrl for the flux of F cross is F N ds where s is rclength. IMPORTANT: F N is the norml component of F to the curve. In the cse where is smooth closed curve using counterclockwise orienttion, N points outwrd to, nd thus, wherever F points outwrd cross there is positive contribution to the flux integrl, while wherever F point inwrd cross there is negtive contribution to the flux integrl. Our first gol is to find nicer formul for flux: Embedded in R 3, T = (x (t)) 2 +(y (t)) 2 x (t), y (t),. Then N = T k = (x (t)) 2 +(y (t)) 2 y (t), x (t),. After dropping the third component, N = (x (t)) 2 + (y (t)) 2 y (t), x (t). Then with F = f, g y (t), x (t) F N ds = F (x (t)) 2 + (y (t)) 2 dt = (x (t)) 2 + (y (t)) 2 fdy gdx.
10 Let s find nd simplify the flux integrl of F = x, y on the circle x2 + y2 (x 2) 2 + (y 2) 2 = 4 using counterclockwise orienttion: Let s lbel the circle by. First we prmeterize the once counterclockwise with r(t) = cos (t), sin (t) for t 2π. Then we integrte: F N ds = = 2 2π 2π 4.8. (2 + 2 cos (t))(2 cos (t)) (2 + 2 sin (t))( 2 sin (t)) (2 + 2 cos (t))2 + (2 + 2 sin (t)) 2 + cos (t) + sin (t) cos (t) + 2 sin (t) dt Note: The integrl ws pproximted numericlly with Mthemtic. Anywy, the positive result mens tht more of the vector field flows out of the region thn into the region (just tke look t the picture couple pges bck). dt Let s clculte the upwrd flux of the rottionl vector field F = y, x cross the curve given by y = x for x 4. While we could use the grph prmetriztion, it turns out to be nicer to use r(t) = t 2, t for t 2. Then r, 2t, 2t (t) = 2t,, which mkes for norml N = ± = ± r (t) 4t2 +. For n upwrd norml we choose the minus sign. Then F N ds = 2 t, t 2, 2t dt = 2 t + 2t 3 dt =. See the picture below!
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 informationSection 17.2 Line Integrals
Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We
More informationMath 32B Discussion Session Session 7 Notes August 28, 2018
Mth 32B iscussion ession ession 7 Notes August 28, 28 In tody s discussion we ll tlk bout surfce integrls both of sclr functions nd of vector fields nd we ll try to relte these to the mny other integrls
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 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 informationRiemann 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 informationRiemann 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 information10 Vector Integral Calculus
Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve
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 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 informationDefinite 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 informationDefinite 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 informationProperties 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 informationJim Lambers MAT 280 Spring Semester Lecture 17 Notes. These notes correspond to Section 13.2 in Stewart and Section 7.2 in Marsden and Tromba.
Jim Lmbers MAT 28 Spring Semester 29 Lecture 7 Notes These notes correspond to Section 3.2 in Stewrt nd Section 7.2 in Mrsden nd Tromb. Line Integrls Recll from singlevrible clclus tht if constnt force
More informationMath Advanced Calculus II
Mth 452  Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More information1 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 informationImproper 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 informationGoals: 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 informationSections 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 informationINDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012
Lecture 6: Line Integrls INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Anlysis Autumn 2012 August 8, 2012 Lecture 6: Line Integrls Lecture 6: Line Integrls Lecture 6: Line Integrls Integrls of complex
More information( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that
Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we
More informationSection 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 informationSection 14.3 Arc Length and Curvature
Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More informationNote 16. Stokes theorem Differential Geometry, 2005
Note 16. Stokes theorem ifferentil Geometry, 2005 Stokes theorem is the centrl result in the theory of integrtion on mnifolds. It gives the reltion between exterior differentition (see Note 14) nd integrtion
More informationn 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 informationdf dt f () b f () a dt
Vector lculus 16.7 tokes Theorem Nme: toke's Theorem is higher dimensionl nlogue to Green's Theorem nd the Fundmentl Theorem of clculus. Why, you sk? Well, let us revisit these theorems. Fundmentl Theorem
More informationSpace Curves. Recall the parametric equations of a curve in xyplane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xyplne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)
More informationDefinite 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 lefthnd
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 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 informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationLecture 3: Curves in Calculus. Table of contents
Mth 348 Fll 7 Lecture 3: Curves in Clculus Disclimer. As we hve textook, this lecture note is for guidnce nd supplement only. It should not e relied on when prepring for exms. In this lecture we set up
More informationChapter 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 informationMath 426: Probability Final Exam Practice
Mth 46: Probbility Finl Exm Prctice. Computtionl problems 4. Let T k (n) denote the number of prtitions of the set {,..., n} into k nonempty subsets, where k n. Argue tht T k (n) kt k (n ) + T k (n ) by
More informationThe 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 informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationA 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 informationINTRODUCTION 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 informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationFINAL REVIEW. 1. Vector Fields, Work, and Flux Suggested Problems:
FINAL EVIEW 1. Vector Fields, Work, nd Flux uggested Problems: { 14.1 7, 13, 16 14.2 17, 25, 27, 29, 36, 45 We dene vector eld F (x, y) to be vector vlued function tht mps ech point in the plne to two
More informationWeek 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 informationBig idea in Calculus: approximation
Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:
More 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 informationRiemann 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 informationWe 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 informationMATH 13 FINAL STUDY GUIDE, WINTER 2012
MATH 13 FINAL TUY GUI, WINTR 2012 This is ment to be quick reference guide for the topics you might wnt to know for the finl. It probbly isn t comprehensive, but should cover most of wht we studied in
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationIntegrals along Curves.
Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationNotes on length and conformal metrics
Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued
More informationDIRECT CURRENT CIRCUITS
DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through
More informationf(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 informationThe Fundamental Theorem of Calculus
The Fundmentl Theorem of Clculus MATH 151 Clculus for Mngement J. Robert Buchnn Deprtment of Mthemtics Fll 2018 Objectives Define nd evlute definite integrls using the concept of re. Evlute definite integrls
More informationUNIFORM 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 informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
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 informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the righthnd side limit equls to the lefthnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationAdvanced 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 informationSUMMER KNOWHOW STUDY AND LEARNING CENTRE
SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18
More informationAPPLICATIONS 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 crosssection in plne pssing through
More informationMath 116 Calculus II
Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationReview on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.
Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5.  5.3) Remrks on the course. Slide Review: Sec. 5.5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description
More informationUnit #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 informationOverview 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 informationOne dimensional integrals in several variables
Chpter 9 One dimensionl integrls in severl vribles 9.1 Differentition under the integrl Note: less thn 1 lecture Let f (x,y be function of two vribles nd define g(y : b f (x,y dx Suppose tht f is differentible
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More information4.4 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 informationMath 6A Notes. Written by Victoria Kala SH 6432u Office Hours: R 12:30 1:30pm Last updated 6/1/2016
Prmetric Equtions Mth 6A Notes Written by Victori Kl vtkl@mth.ucsb.edu H 6432u Office Hours: R 12:30 1:30pm Lst updted 6/1/2016 If x = f(t), y = g(t), we sy tht x nd y re prmetric equtions of the prmeter
More informationMATH Summary of Chapter 13
MATH 21259 ummry of hpter 13 1. Vector Fields re vector functions of two or three vribles. Typiclly, vector field is denoted by F(x, y, z) = P (x, y, z)i+q(x, y, z)j+r(x, y, z)k where P, Q, R re clled
More informationContents. 4.1 Line Integrals. Calculus III (part 4): Vector Calculus (by Evan Dummit, 2018, v. 3.00) 4 Vector Calculus
lculus III prt 4): Vector lculus by Evn Dummit, 8, v. 3.) ontents 4 Vector lculus 4. Line Integrls................................................. 4. Surfces nd Surfce Integrls........................................
More informationConservation Law. Chapter Goal. 5.2 Theory
Chpter 5 Conservtion Lw 5.1 Gol Our long term gol is to understnd how mny mthemticl models re derived. We study how certin quntity chnges with time in given region (sptil domin). We first derive the very
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More informationl 2 p2 n 4n 2, the total surface area of the
Week 6 Lectures Sections 7.5, 7.6 Section 7.5: Surfce re of Revolution Surfce re of Cone: Let C be circle of rdius r. Let P n be n nsided regulr polygon of perimeter p n with vertices on C. Form cone
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More information1 Line Integrals in Plane.
MA213 thye Brief Notes on hpter 16. 1 Line Integrls in Plne. 1.1 Introduction. 1.1.1 urves. A piece of smooth curve is ssumed to be given by vector vlued position function P (t) (or r(t) ) s the prmeter
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More 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 informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationThe 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 informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More 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 informationLecture 13  Linking E, ϕ, and ρ
Lecture 13  Linking E, ϕ, nd ρ A Puzzle... InnerSurfce Chrge Density A positive point chrge q is locted offcenter inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More informationdt. However, we might also be curious about dy
Section 0. The Clculus of Prmetric Curves Even though curve defined prmetricly my not be function, we cn still consider concepts such s rtes of chnge. However, the concepts will need specil tretment. For
More informationMATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The halfangle formula cos 2 θ = 1 2
MATH 53 WORKSHEET MORE INTEGRATION IN POLAR COORDINATES ) Find the volume of the solid lying bove the xyplne, below the prboloid x + y nd inside the cylinder x ) + y. ) We found lst time the set of points
More informationII. Integration and Cauchy s Theorem
MTH6111 Complex Anlysis 200910 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.
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 informationSection Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled?
Section 5.  Ares nd Distnces Exmple : Suppose cr trvels t constnt 5 miles per hour for 2 hours. Wht is the totl distnce trveled? Exmple 2: Suppose cr trvels 75 miles per hour for the first hour, 7 miles
More informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationNumerical 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