(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P."

Transcription

1 Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time nd the vertil is distne, equl to the instntneous veloity. But the integrl is not so esily interpreted: Why is the re under the urve in ny wy relted to the ntiderivtive? It relly ws stroke of genius by Newton nd Leibnitz to see tht onnetion. In ft, integrtion, in the sense of re or volume, ws invented first, mny enturies erlier: Arhimedes ws involved in trying to ompute the volume of wine sk. The derivtive ws muh hrder onept. In ft, Zeno s prdoxes show how hrd the Greeks found the onept of it in generl. Differentition ould relly only be reognized nd investigted fter Desrtes invented oordinte geometry. As result, few lulus texts, seeking historil pproh to the subjet, over integrtion before differentition. Definition. Let f be bounded funtion on ompt intervl [, b]. (We ll see lter why the underlined words re importnt. For now, just note tht we re not ssuming tht f is even ontinuous. ( A prtition P of [, b] is finite subset of [, b] ontining nd b. If Q is nother prtition nd P Q, then Q is refinement of P. (b Write P = {x 0, x, x 2,..., x n } where = x 0 < x < x 2 < < x n = b. Then U(f, P = L(f, P = n sup{f(x : x i x x i }(x i x i i= n inf{f(x : x i x x i }(x i x i i= re the upper nd lower sums with respet to f nd P. ( The upper nd lower integrls of f on [, b] re U(f = inf{u(f, P } nd L(f = sup{l(f, P }, where the inf nd sup re tken over ll prtitions P of [, b]. (d If U(f = L(f, then f is integrble on [, b], nd their ommon vlue is denoted b f or f(x dx. b Exmple. On the intervl [, b] with prtition P s shown, onsider the funtion f. We hve U(f, P = N(x x 0 + M(x 2 x + N(x 3 x 2 + M(x 4 x 3 + M(x 5 x 4 L(f, P = 0(x x 0 + N(x 2 x + 0(x 3 x 2 + N(x 4 x 3 + 0(x 5 x 4 The blue region shows the re representing the upper sum, nd the pink region the lower sum.

2 With prtition Q tht refines P, the upper sum dereses nd the lower sum inreses: Repling the prtitions with ever finer ones, we see tht the upper sums nd lower sums pproh eh other, with the re of the three tringles s the ommon it: U(f = L(f = b f.

3 We see now why we restrited to bounded funtions: So tht the sup nd inf of f on eh subintervl exists (nd is finite. And similrly we restrited to bounded intervls so tht we do not hve infinitely long subintervls or infinitely mny subintervls, mking the upper nd lower sums muh hrder or impossible to interpret. (We restrited to losed intervls so tht the funtion hs vlues t the endpoints. Lter in lulus ourse integrls of unbounded funtions or over unbounded or unlosed intervls might be llowed, but suh things re lwys lled improper integrls nd interpreted s its of proper integrls, its tht my or my not exist: Exmple. Improper integrls: 0 dx = x ε 0 + x dx = M sin x dx =? M ε M x dx = M The orret version of the lst one is M x dx = sin x dx = [ 2x /2] ε 0 + M [2x /2] M = 2( ε = 2 ε ε 0 + = M 2( M = [ os M x]m M = 0 = 0 M sin x dx = N M M N sin x dx = N [ os M x]m N does not exist The it where the left nd right endpoint pproh nd t the sme rte is the lled the prinipl vlue of the improper integrl. Exmple. A funtion tht is not integrble: The Dirihlet funtion χ Q on [0, ]. Every subintervl in every prtition ontins rtionl numbers, so the supremum of the χ Q -vlues on the subintervl is, so the upper sum for every prtition is, so the upper integrl is. But every subintervl in every prtition lso ontins irrtionl numbers, so the infimum of the χ Q -vlues on the subintervl is 0, so the lower sum for every prtition is 0, so the lower integrl is 0. Lemm. For ny bounded funtion f on the ompt intervl [, b]: ( For ny prtition P of [, b], L(f, P U(f, P. (b For ny prtitions P, Q of [, b], if P Q, then L(f, P L(f, Q nd U(f, P U(f, Q. ( For ny prtition P of [, b], L(f, P U(f Therefore, L(f U(f. Proof. ( On ny subintervl [x i, x i ], inf{f(x : x i x x i } sup{f(x : x i x x i }. (b It is enough to ssume tht Q hs only one more element thn P, sy P = {x 0, x,..., x n } nd Q hs the dditionl division point x between x j nd x j. Then p = inf{f(x : x j x

4 x j } is less thn or equl to eh of q = inf{f(x : x j x x } nd q 2 = inf{f(x : x x x j } it is equl to t lest one, but no lrger thn either. So the term p(x j x j in L(f, P tht orresponds to [x j, x j ] is t most the sum q (x x j + q 2 (x j x of the two terms in L(f, Q tht orrespond to [x j, x ] nd [x, x j ] beuse x j x j = (x x j + (x j x. Adding up ll the terms for ll the subintervls in eh prtition, we see tht L(f, P L(f, Q. Similrly, U(f, P U(f, Q. ( Assume BWOC tht there is prtition P for whih L(f, P > U(f. Then L(f, P is not lower bound for the set of ll upper sums, so there is prtition Q for whih L(f, P > U(f, Q. But then P Q is refinement of both P nd Q, so by (b nd (, we hve L(f, P L(f, P Q U(f, P Q U(f, Q, ontrdition. Thus, U(f is n upper bound on the set of ll lower sums, it is t lest s lrge s the lest upper bound, nd we hve the Therefore sentene. Here is nother wy to sy integrble, in whih, insted of finding one prtition P tht mkes L(f, P lmost s lrge s possible nd nother, Q, tht mkes U(f, P lmost s smll s possible, we sy tht it is enough to find single prtition tht mkes the upper nd lower sums lose to eh other: Proposition. A bounded funtion f : [, b] R is integrble iff, ε > 0, P prtition of [, b] s.t. U(f, P L(f, P < ε. Proof. ( Assume BWOC tht U(f L(f. Then then U(f L(f = ε > 0. By hypothesis, P s.t. U(f, P L(f, P < ε. But beuse L(f, P L(f nd U(f, P U(f, we hve U(f L(f U(f, P L(f, P < ε = U(f L(f, /\. ( Let ε > 0 be given, nd pik P, Q s.t. U(f, P b f < ε/2 nd b f L(f, Q < ε/2 Then beuse P Q is ommon refinement of P, Q, we hve so L(f, Q L(f, P Q U(f, P Q U(f, P, U(f, P Q L(f, P Q U(f, P L(f, Q ( b = U(f, P f + ( b f L(f, Q < ε. Now we show tht ontinuous funtion (on ompt intervl is integrble. Beuse not ll ontinuous funtions re differentible, we see tht it is hrder for funtion to be differentible thn for it to be integrble. Theorem. If f : [, b] R is ontinuous, then it is integrble. Proof. Let ε > 0 be given. Beuse f is uniformly ontinuous, there is δ > 0 for whih x i x j < δ implies f(x i f(x j < ε/(b. Pik prtition P for whih ll the subintervls [x i, x i ] hve length less thn δ. (For exmple, we might hoose n in N so lrge tht (b /n < δ nd set P = {x i = +i(b /n : i = 0,,..., n}. Then beuse f ttins its inf nd sup on eh [x i, x i ], sy t t i nd u i respetively, we hve t i u i x i x i < δ, so f(t i f(u i < ε/(b, so U(f, P L(f, P = n (f(t i f(u i (x i x i < i= n i= ε b (x i x i = ε b (x n x 0 = ε,

5 the seond lst equlity beuse the sum (x x 0 + (x 2 x + + (x n x n telesopes, i.e., ll the terms nel exept the seond nd seond lst. But there re mny disontinuous integrble funtions; our first exmple (the three tringles ws disontinuous t two points but still integrble. We hve seen tht the Dirihlet funtion on [0, ] is not integrble. But we tht the Thome funtion is integrble on this intervl: Exmple. Rell tht the Thome funtion is given by { 0 if x is irrtionl t(x = n if x = m n in lowest terms Beuse every subintervl of every prtition of [0, ] ontins irrtionl numbers, the lower sum of t with respet to every prtition is 0, so the lower integrl of t is 0. Thus, to see tht t is integrble, we need to show tht the upper integrl is 0, i.e., tht, given ε > 0, we n find prtition P with respet to whih U(t, P < ε. We n do tht: There re only finitely mny rtionls x in [0, ] for whih f(x > ε/2. Pik prtition P of [0, ] so tht these finitely mny rtionls re the enters (or, for x = 0 nd x =, the ends of subintervls with totl length ε/2. Then in U(f, P, the terms orresponding to those subintervls dd up to t most times the totl length of these subintervls nd hene less thn ε/2. In the other subintervls the vlue of the funtion is t most ε/2, so those subintervls ontribute to U(f, P totl of less thn (ε/2. It follows tht U(f, P < ε. In ft, it is shown in the text tht bounded funtion on ompt intervl is integrble iff its set of disontinuities hs mesure zero. We won t try to explin this term, beuse the right wy to do tht is to tke totlly different pproh to the ourse, in terms of Lebesgue integrtion. So we will only sy tht ll ountble sets, inluding finite sets, nd the Cntor set hve mesure zero, so tht funtions tht re disontinuous only on these sets re integrble. ( Mesure is onept tht extends the ide of length, so set of length l hs mesure l; but it is hrd to ssign length to set like the rtionls, whih hs mesure 0. Remember tht the Dirihlet funtion is disontinuous everywhere, so its set of disontinuities in [0, ] hs mesure one; but the Thome funtion is disontinuous only on the rtionls, whih hve mesure 0. And sure enough, the Dirihlet funtion is not integrble, while the Thome funtion is. [At this point students re redy to do the eleventh problem set.] Proposition. (Properties of the definite integrl Suppose f, g re integrble on [, b] nd k is onstnt. Then: (0 If < < b, then b f = f + b f. ( b (f + g = b f + b g. (2 b kf = k b f. (3 If f g on [, b], then b f b g. Corollry. (of (3: (3 If m f M on [, b], then m(b b f M(b. (3b f is integrble nd b f b f.

6 The ft tht f is integrble is new informtion the proof is n exerise. To do it, you should verify tht, if m i, M i re the inf nd sup of f on [x i, x i ] nd m i, M i re the sme for f, then M i m i M i m i. To get the seond ssertion in (3b, use f f f on [.b]. Remrks on the proposition: (0 Prt of wht is to be proved here is tht f is integrble on [, b] iff it is integrble on both [, ] nd [, b]. But if we define f = 0 nd b f = b f, then we don t need to dd the If prt of this sttement s long s f is integrble on the intervls determined by, b,. ( The proof is n exerise. (2 Here is proof of the k < 0 se: Let ε > 0 be given, nd pik prtition P of [, b] for whih U(f, P b f < ε/(2 k nd b f L(f, P < ε/(2 k. Then beuse, on eh subintervl [x i, x i ] in P we hve sup{kf(x : x i x x i } = k inf{f(x : x i x x i } nd similrly with sup nd inf reversed, it follows tht U(kf, P L(kf, P = kl(f, P ku(f, P = k (U(f, P L(f, P ( b b k U(f, P f + f L(f, P < ε 2 + ε 2 = ε, so kf is integrble, nd U(kf, P k so k b f = b kf.// b ( f = k L(f, P b ( b f = k f L(f, P < ε 2 < ε, (3 Beuse f g, for ny subintervl [x i, x i ] of ny prtition P of [, b] we hve sup{f(x : x i x x i } sup{g(x : x i x x i } so U(f, P U(g, P, so b f = inf P U(f, P inf P U(g, P = b g. [At this point students re redy to do the twelfth problem set.] Theorem. (Fundmentl Theorem of Clulus ( If f is the derivtive of F on [, b] nd f is integrble, then b f = F (b F (. (2 Let g : [, b] R be integrble, nd define G(t = t g(x dx. Then G : [, b] R is ontinuous. If g is ontinuous t in [, b], then G is differentible t nd G ( = g(. Proof. ( For ny subintervl [x i, x i ] in ny prtition P of [, b], by the Men Vlue Theorem x i [x i, x i ] s.t. F (x i F (x i = f(x i (x i x i, so n i= (F (x i F (x i is between L(f, P nd U(f, P. But the sum is telesoping, to F (b F (, independent of the prtition P, so F (b F ( is between L(f nd U(f, whih re both equl to b f.

7 (2 In ft, we n show tht G is Lipshitz, whih is stronger thn ontinuous: Let M be n upper bound on g in [, b]. Then for ll x, x 2 [, b], we hve 2 x G(x G(x 2 = g g = g M x x 2. Now suppose g is ontinuous t. Then G G(x G( ( = = g x x x x, so we wnt to show, given ε > 0, δ > 0 s.t. 0 < x < δ = g/(x g( < ε. Now there is δ > 0 s.t., if x < δ, then g(x g( < ε, nd for tht δ, if 0 < x < δ, then g ( x g( = x (g(t g(dt x g(t g( dt ε dt = ε. x x (In the middle of this, note tht, if x < 0, then ny x 2 of nonnegtive funtion is lso 0. Here re some exmples of funtions g nd the funtions G defined by their integrls, G(x = g(t dt. Exmple. In these two exmples, = 2, so tht, no mtter wht g is, G( 2 = 2 2 g = 0. We re interested espeilly in the points t whih the definitions of g hnge from one formul to nother. First: { if x < 0 g(x =, if x 0 { G(x = g(t dt = 2 dt = x 2 if x < 0 2 G(0 + 0 dt = 2 + x if x 0 Here, G is differentible exept t the disontinuity x = 0 of g. Seond: { if x g(x =, 2x 3 if x > { x 2 if x G(x = g(t dt = 2 G( + (2t 3dt = 3 + (x2 3x ( 2 = x 2 3x if x > Beuse this g is ontinuous t the point where its definition hnges, the orresponding G is differentible there. Here re the orresponding grphs, with the g s in green nd the G s in red:

8 Though our text doesn t seem to inlude them, I hve seen in other texts exmples of funtions defined by integrls where the upper, nd mybe lso the lower, it(s of integrtion re themselves funtions. In order to find their derivtives, we only need to throw in the Chin Rule: Suppose we re given H(x = f(x g(t dt. We n isolte n intervening funtion: G(u = u g(t dt. Then H(x = G(f(x, so (d/dx(h(x = G (f(x f (x = g(f(xf (x. Exmple. d dx 5x x ( exp( t 2 dt = d 5x 2 +3 exp( t 2 dt dx 0 2x 0 exp( t 2 dt = exp( (5x (0x exp( (2x 2 2 = 0x exp( (5x exp( (2x 2. Here, the intervening funtion is G(u = u 0 exp( t2 dt, so tht G (u = exp( u 2. There is nothing speil bout 0; beuse exp( t 2 is ontinuous everywhere, ny onstnt would hve worked. [At this point students re redy to do the thirteenth problem set.]

T b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.

T b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions. Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene

More information

The Riemann-Stieltjes Integral

The Riemann-Stieltjes Integral Chpter 6 The Riemnn-Stieltjes Integrl 6.1. Definition nd Eistene of the Integrl Definition 6.1. Let, b R nd < b. ( A prtition P of intervl [, b] is finite set of points P = { 0, 1,..., n } suh tht = 0

More information

Chapter 6. Riemann Integral

Chapter 6. Riemann Integral Introduction to Riemnn integrl Chpter 6. Riemnn Integrl Won-Kwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl

More information

Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1)

Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1) Green s Theorem Mth 3B isussion Session Week 8 Notes Februry 8 nd Mrh, 7 Very shortly fter you lerned how to integrte single-vrible funtions, you lerned the Fundmentl Theorem of lulus the wy most integrtion

More information

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES

MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES MA10207B: ANALYSIS SECOND SEMESTER OUTLINE NOTES CHARLIE COLLIER UNIVERSITY OF BATH These notes hve been typeset by Chrlie Collier nd re bsed on the leture notes by Adrin Hill nd Thoms Cottrell. These

More information

More Properties of the Riemann Integral

More Properties of the Riemann Integral More Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologil Sienes nd Deprtment of Mthemtil Sienes Clemson University Februry 15, 2018 Outline More Riemnn Integrl Properties The Fundmentl

More information

For a continuous function f : [a; b]! R we wish to define the Riemann integral

For a continuous function f : [a; b]! R we wish to define the Riemann integral Supplementry Notes for MM509 Topology II 2. The Riemnn Integrl Andrew Swnn For continuous function f : [; b]! R we wish to define the Riemnn integrl R b f (x) dx nd estblish some of its properties. This

More information

6.5 Improper integrals

6.5 Improper integrals Eerpt from "Clulus" 3 AoPS In. www.rtofprolemsolving.om 6.5. IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y =

More information

Math 554 Integration

Math 554 Integration Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

More information

MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.

MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals. MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded

More information

MATH 409 Advanced Calculus I Lecture 22: Improper Riemann integrals.

MATH 409 Advanced Calculus I Lecture 22: Improper Riemann integrals. MATH 409 Advned Clulus I Leture 22: Improper Riemnn integrls. Improper Riemnn integrl If funtion f : [,b] R is integrble on [,b], then the funtion F(x) = x f(t)dt is well defined nd ontinuous on [,b].

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

7.2 The Definition of the Riemann Integral. Outline

7.2 The Definition of the Riemann Integral. Outline 7.2 The Definition of the Riemnn Integrl Tom Lewis Fll Semester 2014 Upper nd lower sums Some importnt theorems Upper nd lower integrls The integrl Two importnt theorems on integrbility Outline Upper nd

More information

MATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral.

MATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral. MATH 409 Advnced Clculus I Lecture 18: Drboux sums. The Riemnn integrl. Prtitions of n intervl Definition. A prtition of closed bounded intervl [, b] is finite subset P [,b] tht includes the endpoints

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

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, 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. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

Math Calculus with Analytic Geometry II

Math 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 x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

More information

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

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

38 Riemann sums and existence of the definite integral.

38 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 x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

More information

Calculus Cheat Sheet. Integrals Definitions. where F( x ) is an anti-derivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx

Calculus Cheat Sheet. Integrals Definitions. where F( x ) is an anti-derivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx Clulus Chet Sheet Integrls Definitions Definite Integrl: Suppose f ( ) is ontinuous Anti-Derivtive : An nti-derivtive of f ( ) on [, ]. Divide [, ] into n suintervls of is funtion, F( ), suh tht F = f.

More information

MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL

MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA-302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition

More information

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

The 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 information

Line Integrals and Entire Functions

Line Integrals and Entire Functions Line Integrls nd Entire Funtions Defining n Integrl for omplex Vlued Funtions In the following setions, our min gol is to show tht every entire funtion n be represented s n everywhere onvergent power series

More information

6.1 Definition of the Riemann Integral

6.1 Definition of the Riemann Integral 6 The Riemnn Integrl 6. Deinition o the Riemnn Integrl Deinition 6.. Given n intervl [, b] with < b, prtition P o [, b] is inite set o points {x, x,..., x n } [, b], lled grid points, suh tht x =, x n

More information

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q. Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the

More information

Math 360: A primitive integral and elementary functions

Math 360: A primitive integral and elementary functions Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

More 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

Lecture 3. Limits of Functions and Continuity

Lecture 3. Limits of Functions and Continuity Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live

More information

Solutions to Assignment 1

Solutions to Assignment 1 MTHE 237 Fll 2015 Solutions to Assignment 1 Problem 1 Find the order of the differentil eqution: t d3 y dt 3 +t2 y = os(t. Is the differentil eqution liner? Is the eqution homogeneous? b Repet the bove

More information

f (x)dx = f(b) f(a). a b f (x)dx is the limit of sums

f (x)dx = f(b) f(a). a b f (x)dx is the limit of sums Green s Theorem If f is funtion of one vrible x with derivtive f x) or df dx to the Fundmentl Theorem of lulus, nd [, b] is given intervl then, ording This is not trivil result, onsidering tht b b f x)dx

More information

Chapter 4. Lebesgue Integration

Chapter 4. Lebesgue Integration 4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition.

More information

Project 6: Minigoals Towards Simplifying and Rewriting Expressions

Project 6: Minigoals Towards Simplifying and Rewriting Expressions MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy

More information

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

Interpreting Integrals and the Fundamental Theorem

Interpreting Integrals and the Fundamental Theorem Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

More information

QUADRATIC EQUATION. Contents

QUADRATIC EQUATION. Contents QUADRATIC EQUATION Contents Topi Pge No. Theory 0-04 Exerise - 05-09 Exerise - 09-3 Exerise - 3 4-5 Exerise - 4 6 Answer Key 7-8 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,

More information

Math 324 Course Notes: Brief description

Math 324 Course Notes: Brief description Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd

More information

Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.

Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer. Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points

More information

Introduction to Olympiad Inequalities

Introduction to Olympiad Inequalities Introdution to Olympid Inequlities Edutionl Studies Progrm HSSP Msshusetts Institute of Tehnology Snj Simonovikj Spring 207 Contents Wrm up nd Am-Gm inequlity 2. Elementry inequlities......................

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

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

10. AREAS BETWEEN CURVES

10. AREAS BETWEEN CURVES . AREAS BETWEEN CURVES.. Ares etween curves So res ove the x-xis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in

More information

Riemann Integrals and the Fundamental Theorem of Calculus

Riemann 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 information

5: The Definite Integral

5: The Definite Integral 5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce

More information

Mathematical Analysis: Supplementary notes I

Mathematical Analysis: Supplementary notes I Mthemticl Anlysis: Supplementry notes I 0 FIELDS The rel numbers, R, form field This mens tht we hve set, here R, nd two binry opertions ddition, + : R R R, nd multipliction, : R R R, for which the xioms

More information

Big idea in Calculus: approximation

Big 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 information

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1 MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further

More information

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx Applitions of Integrtion Are of Region Between Two Curves Ojetive: Fin the re of region etween two urves using integrtion. Fin the re of region etween interseting urves using integrtion. Desrie integrtion

More information

Section 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

Section 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 information

Integrals along Curves.

Integrals 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 information

Line Integrals. Partitioning the Curve. Estimating the Mass

Line Integrals. Partitioning the Curve. Estimating the Mass Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to

More information

Section 4.4. Green s Theorem

Section 4.4. Green s Theorem The Clulus of Funtions of Severl Vriles Setion 4.4 Green s Theorem Green s theorem is n exmple from fmily of theorems whih onnet line integrls (nd their higher-dimensionl nlogues) with the definite integrls

More information

Lecture 1 - Introduction and Basic Facts about PDEs

Lecture 1 - Introduction and Basic Facts about PDEs * 18.15 - Introdution to PDEs, Fll 004 Prof. Gigliol Stffilni Leture 1 - Introdution nd Bsi Fts bout PDEs The Content of the Course Definition of Prtil Differentil Eqution (PDE) Liner PDEs VVVVVVVVVVVVVVVVVVVV

More information

Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction

Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCK-KURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When rel-vlued

More information

Line and Surface Integrals: An Intuitive Understanding

Line 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 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

We know that if f is a continuous nonnegative function on the interval [a, b], then b

We 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 information

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE 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 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

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

Mapping the delta function and other Radon measures

Mapping the delta function and other Radon measures Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support

More information

Continuous Random Variables

Continuous Random Variables STAT/MATH 395 A - PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is rel-vlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht

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

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1 3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

More information

FINALTERM EXAMINATION 2011 Calculus &. Analytical Geometry-I

FINALTERM EXAMINATION 2011 Calculus &. Analytical Geometry-I FINALTERM EXAMINATION 011 Clculus &. Anlyticl Geometry-I Question No: 1 { Mrks: 1 ) - Plese choose one If f is twice differentible function t sttionry point x 0 x 0 nd f ''(x 0 ) > 0 then f hs reltive...

More information

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0. STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t

More information

Convex Sets and Functions

Convex Sets and Functions B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line

More information

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

More information

ON THE INEQUALITY OF THE DIFFERENCE OF TWO INTEGRAL MEANS AND APPLICATIONS FOR PDFs

ON THE INEQUALITY OF THE DIFFERENCE OF TWO INTEGRAL MEANS AND APPLICATIONS FOR PDFs ON THE INEQUALITY OF THE DIFFERENCE OF TWO INTEGRAL MEANS AND APPLICATIONS FOR PDFs A.I. KECHRINIOTIS AND N.D. ASSIMAKIS Deprtment of Eletronis Tehnologil Edutionl Institute of Lmi, Greee EMil: {kehrin,

More information

MTH 122 Fall 2008 Essex County College Division of Mathematics Handout Version 10 1 October 14, 2008

MTH 122 Fall 2008 Essex County College Division of Mathematics Handout Version 10 1 October 14, 2008 MTH 22 Fll 28 Essex County College Division of Mthemtics Hndout Version October 4, 28 Arc Length Everyone should be fmilir with the distnce formul tht ws introduced in elementry lgebr. It is bsic formul

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

Calculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties

Calculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ fuehr@mth.rwth-chen.de Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009 Overview 1 Motivtion:

More information

Math 107H Topics for the first exam. csc 2 x dx = cot x + C csc x cotx dx = csc x + C tan x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C

Math 107H Topics for the first exam. csc 2 x dx = cot x + C csc x cotx dx = csc x + C tan x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C Integrtion Mth 07H Topics for the first exm Bsic list: x n dx = xn+ + C (provided n ) n + sin(kx) dx = cos(kx) + C k sec x dx = tnx + C sec x tnx dx = sec x + C /x dx = ln x + C cos(kx) dx = sin(kx) +

More information

Arrow s Impossibility Theorem

Arrow s Impossibility Theorem Rep Voting Prdoxes Properties Arrow s Theorem Arrow s Impossiility Theorem Leture 12 Arrow s Impossiility Theorem Leture 12, Slide 1 Rep Voting Prdoxes Properties Arrow s Theorem Leture Overview 1 Rep

More information

The Wave Equation I. MA 436 Kurt Bryan

The Wave Equation I. MA 436 Kurt Bryan 1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string

More information

Chapter Gauss Quadrature Rule of Integration

Chapter Gauss Quadrature Rule of Integration Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to

More information

NOTES AND PROBLEMS: INTEGRATION THEORY

NOTES AND PROBLEMS: INTEGRATION THEORY NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFS-I to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents

More information

Properties of the Riemann Stieltjes Integral

Properties of the Riemann Stieltjes Integral Properties of the Riemnn Stieltjes Integrl Theorem (Linerity Properties) Let < c < d < b nd A,B IR nd f,g,α,β : [,b] IR. () If f,g R(α) on [,b], then Af +Bg R(α) on [,b] nd [ ] b Af +Bg dα A +B (b) If

More information

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by. NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

More information

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find nti-derivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible

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

Test 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher).

Test 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher). Test 3 Review Jiwen He Test 3 Test 3: Dec. 4-6 in CASA Mteril - Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec. 14-17 in CASA You Might Be Interested

More information

Solutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite!

Solutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite! Solutions for HW9 Exerise 28. () Drw C 6, W 6 K 6, n K 5,3. C 6 : W 6 : K 6 : K 5,3 : () Whih of the following re iprtite? Justify your nswer. Biprtite: put the re verties in V 1 n the lk in V 2. Biprtite:

More information

Best Approximation. Chapter The General Case

Best Approximation. Chapter The General Case Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

New Expansion and Infinite Series

New Expansion and Infinite Series Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06-073 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University

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

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

x = 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 information

Necessary and Sufficient Conditions for Differentiating Under the Integral Sign

Necessary and Sufficient Conditions for Differentiating Under the Integral Sign Necessry nd Sufficient Conditions for Differentiting Under the Integrl Sign Erik Tlvil 1. INTRODUCTION. When we hve n integrl tht depends on prmeter, sy F(x f (x, y dy, it is often importnt to know when

More information

(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35

(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35 7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know

More information

Riemann Stieltjes Integration - Definition and Existence of Integral

Riemann Stieltjes Integration - Definition and Existence of Integral - Definition nd Existence of Integrl Dr. Adity Kushik Directorte of Distnce Eduction Kurukshetr University, Kurukshetr Hryn 136119 Indi. Prtition Riemnn Stieltjes Sums Refinement Definition Given closed

More information

u(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.

u(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C. Lecture 4 Complex Integrtion MATH-GA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex

More information

where the box contains a finite number of gates from the given collection. Examples of gates that are commonly used are the following: a b

where the box contains a finite number of gates from the given collection. Examples of gates that are commonly used are the following: a b CS 294-2 9/11/04 Quntum Ciruit Model, Solovy-Kitev Theorem, BQP Fll 2004 Leture 4 1 Quntum Ciruit Model 1.1 Clssil Ciruits - Universl Gte Sets A lssil iruit implements multi-output oolen funtion f : {0,1}

More information

The problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.

The problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests. ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion

More information

ILLUSTRATING THE EXTENSION OF A SPECIAL PROPERTY OF CUBIC POLYNOMIALS TO NTH DEGREE POLYNOMIALS

ILLUSTRATING THE EXTENSION OF A SPECIAL PROPERTY OF CUBIC POLYNOMIALS TO NTH DEGREE POLYNOMIALS ILLUSTRATING THE EXTENSION OF A SPECIAL PROPERTY OF CUBIC POLYNOMIALS TO NTH DEGREE POLYNOMIALS Dvid Miller West Virgini University P.O. BOX 6310 30 Armstrong Hll Morgntown, WV 6506 millerd@mth.wvu.edu

More information

CS667 Lecture 6: Monte Carlo Integration 02/10/05

CS667 Lecture 6: Monte Carlo Integration 02/10/05 CS667 Lecture 6: Monte Crlo Integrtion 02/10/05 Venkt Krishnrj Lecturer: Steve Mrschner 1 Ide The min ide of Monte Crlo Integrtion is tht we cn estimte the vlue of n integrl by looking t lrge number of

More information

Summary of Elementary Calculus

Summary of Elementary Calculus Summry of Elementry Clculus Notes by Wlter Noll (1971) 1 The rel numbers The set of rel numbers is denoted by R. The set R is often visulized geometriclly s number-line nd its elements re often referred

More information

Math 3B Final Review

Math 3B Final Review Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:45-10:45m SH 1607 Mth Lb Hours: TR 1-2pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems

More information

Functions of bounded variation

Functions of bounded variation Division for Mthemtics Mrtin Lind Functions of bounded vrition Mthemtics C-level thesis Dte: 2006-01-30 Supervisor: Viktor Kold Exminer: Thoms Mrtinsson Krlstds universitet 651 88 Krlstd Tfn 054-700 10

More information