Summary of Elementary Calculus


 Martha Riley
 1 years ago
 Views:
Transcription
1 Summry of Elementry Clculus Notes by Wlter Noll (1971) 1 The rel numbers The set of rel numbers is denoted by R. The set R is often visulized geometriclly s numberline nd its elements re often referred to s points. The following specil nottion nd terminology will be used for certin specil subsets of R: The set of nonzero rels is denoted by R := R\{0} = {t R t 0}. The set of positive rels is denoted by P := {t R t 0}. The set of strictly positive rels is denoted by P := P\{0} = {t R t > 0}. Let, b R with b. Then the open intervl ], b, [, the closed intervl [, b], the leftopen intervl ], b ], nd the rightopen intervl [, b [ from to b re defined by ], b [ := {t R < t < b} [, b] := {t R t b} ], b ] := {t R < t b} [, b [ := {t R t < b} The numbers nd b re clled the endpoints of these intervls. If S is ny subset of R we write S := { s s S}. In prticulr, P = {t R t 0} is the set of ll negtive rels nd P = {t R t < 0} the set of ll strictly negtive rels. If S nd T re ny subsets of R, we write 1
2 S + T := {s + t s S, t T }, S T := S + ( T ) = {s t s S, t T }. If R nd T R, we bbrevite + T := {} + T = { + t t T }, T := + ( T ) = { t t T }. The set + P is clled the closed right hlfline nd + P the open right hlfline from R. Similrly, P is clled the closed left hlfline nd P the open left hlfline from. In either cse, we sy tht is the endpoint of the hlfline. Sometimes the nottions [, + [ := + P, ], + [ := + P, ], ] := P, ], [ := P, ], + [ := R re used. The term intervl is pplied not only to intervls from some R to some b R, with b >, s defined before, but lso to hlflines, singleton subsets of R, the whole of R, nd the empty set. One cn prove, then, tht subset I of R is n intervl if nd only if it hs the following property: For ll, b I with < b, [, b] I. The empty set, singletons, nd intervls with endpoints in R re clled bounded intervls. Hlflines nd the whole of R re clled unbounded intervls. We sy tht n intervl is genuine if it is neither empty nor singleton. There re two bsic opertions involving rel numbers: ddition nd multipliction. If, b R, we cn form the sum +b by ddition nd the product b by multipliction. Two specil rel numbers re of centrl importnce: zero 0 nd one 1. Given ny, b R, the eqution?x R, + x = b hs exctly one solution x R, which is denoted by b nd clled the difference between b nd or b minus. The bbrevition := 0 is 2
3 used nd clled minus (insted of zero minus ). Given ny R nd b R the eqution?x R, x = b hs unique solution, which is denoted by b nd clled the quotient of b nd, or b over. One clls 1 the reciprocl of R. If R nd n N, one defines the nth power n in the following wy: 0 := 1, 1 :=, 2 :=, 3 := 2, etc. If R nd m N, one defines m by ( ) m 1 m :=. The set R is totlly ordered (red: less thn ). The corresponding strict order on R is < (red: strictly less thn ). The reverse orders re denoted by ( greter thn ) nd > ( strictly greter thn ). Any nonempty finite subset F or R hs gretest element which is clled the mximum of F nd is denoted by mxf ; it lso hs smllest element, which is clled the minimum of F nd is denoted by minf. The bsolute vlue of P of number R is defined by := mx {, }. 2 Rel Functions Let S be subset of R. A mpping f : S R is clled relvlued function of rel vrible, or simply rel function. The set of ll rel functions on S is denoted by Fun(S) := Mp(S, R) Rel functions cn often be visulized concretely by their grphs. For exmple, the Figure shows the grph of some function f Fun([, b]). 3
4 We cn define two bsic opertions involving rel functions: vluewise ddition nd vluewise multipliction. If f, g Fun(S) we define the vluewise sum f + g Fun(S) by (f + g)(t) := f(t) + g(t) for ll t S, nd we define the vluewise product fg Fun(S) by (fg)(t) := f(t)g(t) for ll t S. We cn lso define vluewise differences of functions by vluewise subtrction: if f, g Fun(S) then f g Fun(S) is defined by (f g)(t) := f(t) g(t) for ll t S. Note tht if f, g Fun(S), then f g is the unique solution of the eqution?x Fun(S), g + x = f. The definition of quotients of functions by vluewise division hs to be hndled with cre. If f, g Fun(S), nd if g(t) 0 for ll t S then f(t) g(t) mkes sense for ll t S nd we cn define f Fun(S) by g ( f g ) (t) := f(t) g(t) for ll t S. In most situtions, however, we hve g(t) = 0 for some vlues of t S but g(t) 0 for other vlues of t S. If this is the cse, we cn still define quotient function f, but this quotient function will hve restricted domin. g To be precise, the set of ll t R for which g(t) 0 is S\g < ({0}) nd we define f g Fun(S\g< ({0})) = Fun(g < (R )) by ( ) f (t) := f(t) g g(t) for ll t S\g< ({0}) = g < (R ). Of prticulr interest is the multipliction of functions by constnts. If f Fun(S) nd R, then f is defined to be the vluewise product of the constnt s R Fun(S) nd the function f. Thus, (f)(t) := (f(t)) for ll t S. We write f for ( 1)f. Then f is the sme s the difference 0 S R f. 4
5 For ny n N, the vluewise nth power f n Fun(S) of f Fun(S) is defined by f n (t) := (f(t)) n for ll t S. For m N the domin of the vluewise mth power f m must be restricted to S\f < ({0}), nd we hve f m := ( ) m 1. f Let f, g Fun(S) nd let h : T S, where S nd T re subsets of R. Then we hve (f + g) h = (f h) + (g h), (fg) h = (f h)(g h), If R, we hve f h (S\g< ({0})) g T \(g h) < ({0}) = f h g h. S R h = T R or simply h = when we do not distinguish, in nottion, between constnts nd their vlues. 3 Elementry Functions Since the identity mpping 1 R of R will be used very often, we use the shorter nottion ι := 1 R. Thus, ι := R R is given by ι(t) := t for ll t R. Using the opertions explined in the previous section, we cn form lrge number of functions from ι nd constnts lone. For exmple, if n N nd 0, 1,..., n R, we cn form wht is clled polynomil function (or simply polynomil) p Fun(R) by p := n ι n + n 1 ι n ι ι
6 The quotient p of two polynomil functions p nd q is clled rtionl q function. The domin of p is q R\q< ({0}), which, in generl, is not the whole of R. Of course, ι is invertible nd it is its own inverse. Most rel functions encountered in elementry mthemtics re not invertible. However, one cn often construct invertible functions from noninvertible ones by suitble djustment of their codomins or domins. For exmple, the squring function ι 2 : R R is not invertible, but the djusted squring function ι 2 P P is. Its inverse, fter extension of the codomin to R, is clled the squre root function nd is denoted by Fun(P), so tht ( ) ι 2 P R P. The vlue of t t P is denoted by t (rther thn the stndrd (t)). We hve ι 2 P =, ι 2 = ι, P where : R R is the bsolutevlue function. If n N is odd, then ι n Fun(R) is invertible. Its inverse is clled the n t h root function nd is denoted by n Fun(R), so tht n := (ι n ). We hve n ι n = ι n n for odd n N. If n N is even, ι n is not invertible, but the djusted function ι n P P is. Its inverse, with codomin extended to R, is clled the nth root function nd is denoted by n Fun(P), so tht n := (ι n ) R. We hve n ι n P =, ι n n = ι P for even n N. The vlue of n t t is written n t (rther thn the stndrd n (t)). Recll tht the composite f g of two functions f nd g mkes sense only if Dom f = Cod g. However, it is often possible to djust the codomin or both domin nd codomin of g in such wy tht f g mkes sense for n pproprite djustment g of g. For exmple, (ι 2 + 1) mkes no sense but ((ι 2 + 1) P ) does. We use then the nottion ι2 + 1 := ((ι 2 + 1) P ). As nother exmple, we see tht ι mkes no sense but ι since ι P P = 1 P, we hve ι P =. P P P does nd 6
7 More generlly, if p is ny polynomil nd if n N we write n p := n p if n is odd, n p := n P p if n is even. p < (P) If n is even, it cn be difficult to determine the domin p < (P) = {t R p(t) 0} of n p. The bsic elementry functions re (i) the constnts, (ii) the identity function ι : R R, (iii) the exponentil function exp : R R, nd (iv) the sine function, sin : R R. Precise definitions of exp nd sin re difficult nd will not be given here. A rel function is clled n elementry function if it cn be obtined from the bsic elementry functions by repeted ppliction of the following opertions: vluewise ddition, subtrction, multipliction, nd division, nd composition nd inversion. When pplying division, composition, nd inversion, one must often suitbly djust the domin or the codomin or both, s we hve seen before. Using poetic license, we often omit the symbols tht indicte these djustments. Also, we do not distinguish, in nottion, between constnt nd its vlue. Exmples of elementry functions re the polynomil nd rtionl functions nd lso the nth root functions mentioned before. The following dditionl elementry functions hve stndrd nmes ttched to them: ( π ) cos := sin 2 ι, tn := sin cos ( ) rcsin := sin [ 1,+1] R [, π s,+ π 2 ] ( ) R rccos := cos, [ 1,+1] [0,π] ( ) R rctn := tn ] π, 2,+ π 2 [ log := (exp P ) sinh := 1 (exp (exp ( ι))), 2 7
8 cosh := 1 (exp +(exp ( ι))), 2 tnh := sinh cosh. For every α R, the α t h power function ι α : P R, with vlues ι α (t) = t α, is defined by ι α := exp (α log). In the cse when α is n integer, it coincides with restriction to P of the vluewise power function of ι ccording to the definition t the end of Sect. 2. If α is rtionl, so tht α = n m with n Z nd m N, we hve ι n m = m ι n P = m P ιn P P. 4 Limits nd continuity Let f be rel function nd let R be number, which my or my not belong to the domin of f. We ssume, however, tht Dom f includes genuine intervl which hs s n endpoint. Then ] δ, +δ[ Dom f\{} is nonempty for ll δ P. Definition: We sy tht f hs the limit λ R t R nd we write lim f = λ or lim t f(t) = λ if for every ε P there exists δ P such tht i.e., f(t) λ < ε for ll t ] δ, + δ[ (Dom f\{}). f(t) λ < ε whenever 0 < t < δ nd t Dom f. This definition gives precise mening to the intuitive ide tht f(t) pproches λ s t pproches. Roughly speking, the definition sys tht f(t) comes rbitrrily close to λ when t moves sufficiently close to. 8
9 The desired closeness of f(t) to λ is described by ε nd the required closeness of t to is described by δ. One cn prove tht f cnnot hve more thn one limit t. It is for this reson tht the nottion lim f = limf(t) t is legitimte. Exmples: 1. The nturl domin of the function ι2 1 is R\{1}, becuse the denomintor vnishes t 1. The domin R\{1} includes, for exmple, the ι 1 intervls 1 + P nd 1 P, which hve 1 s n endpoint. Hence it mkes sense to sk whether ι2 1 hs limit t 1, even though 1 does ι 1 not belong to the domin. Since ( ) ι 2 1 (t) = t2 1 ι 1 t 1 = (t + 1)(t 1) t 1 = t + 1 for ll t R\{1} (but not for t = 1, it is cler tht for every ε P we hve ( ) ι 2 1 (t) 2 ι 1 = t = t 1 < ε whenever t 1 < δ := ε nd t 1. We conclude tht ι 2 1 lim 1 ι 1 = lim t 2 1 t 1 t 1 = The nturl domin of the function 1 is ι R, becuse the denomintor vnishes t 0. As in 1, it mkes sense to sk whether 1 hs limit t ι 0. If 1 hd limit λ t 0, we could find, for every choice of ε ι P, δ P such tht 1 t λ < ε whenever t 0 = t < δ nd t 0. If we choose ε := 1, we then hve 1 t λ < 1 whenever t < δ nd t 0. 9
10 This is impossible when t is smller thn both δ nd 1 λ +27. Hence 1 ι hs no limit t 0. (Note tht 27 is whimsicl choice; ny other rel number greter then 1 would lso do.) Let f : I R be rel function whose domin I is genuine intervl, nd let t I. It is mengful to sk whether or not f hs limit t t. We sy tht f is continuous t t if the following two conditions re stisfied: (i) f hs limit t t. (ii) The limit of f t t is equl to the vlue of f t t. The two conditions together mount to the sttement lim t f = f(t). A function f Fun(I) is sid to be continuous if it is continuous t every t I. If the domin of f is genuine intervl, the grph of f is often connected curve which cn be drwn with pencil without lifting it. If this is the cse, then f is continuous. However, there re weird continuous functions whose grph cnnot esily be represented by pencil drwing. Let f be rel function whose domin is the union of collection of genuine intervls. (For exmple, tn hs this property.) We then sy tht f is continuous if the restriction f I is continuous for every intervl I included in the domin. The nottion is often used. C(I) := {f Fun(I) f is continuous} All elementry functions re continuous, but most continuous functions re not elementry. 10
11 5 Differentibility, derivtives Let f : I R be rel function whose domin I is genuine intervl nd let t I be number. Consider the new function (f (t + ι)) f(t) : ( t + I)\{0} R. ι Here, we hve omitted the symbols tht indicte necessry djustments of domin or codomin. Also, we hve denoted constnts by their vlues. Hd we not done so, this function would hve been written f ( t ( t+i) R + ι ( t+i) ) I f(t)( t+i) R ι ( t+i) Clerly, it mkes sense to sk whether this new function hs limit t 0. Definition: We sy tht f : I R is differentible t t I if the limit f (t + ι) f(t) t f := lim 0 ι exists. If this is the cse, then t f is clled the derivtive of f t t. We sy tht f is differentible if it is differentible t every t I. If this is the cse, the derivtive f : I R of f is defined by Remrk: df, df(t) d, dt dt dy dx f (t) := t f for ll t I. There re vrious other populr nottions for derivtives: f(t), f dt t, or f t for t f nd f, f or Df for f. The nottion when y = f(x), which is very populr, doesn t relly mke ny sense. If f is differentible t t I then f is lso continuous t t. The converse if flse: A function cn be continuous t point without being differentible t tht point. For exmple, the bsolutevlue function (t t ) : R R is continuous but not differentible t 0. One cn even construct function f :]0, 1[ R which is continuous but not differentible t every t ]0, 1[. (The construction is very difficult.) Let f be rel function whose domin is the union of collection of genuine intervls. We then sy tht f is differentible if f I is differentible 11
12 for every intervl I included in the domin. If this is the cse, the derivtive f : Dom f R cn be defined in such wy tht (f I ) = f I for ll intervls I included in the domin. Given S R, we sy tht f : I S is differentible if f R is differentible nd we write f := (f R ). Not ll elementry functions re differentible. For exmple, 3 : R R nd : P R re not differentible t 0. However, these nd ll other elementry functions become differentible fter few exceptionl points re deleted from their domins. Let I be genuine intervl or union of collection of such. We write Diff (I) := {f Fun(I) f is differentible}. We now list, without proof, the bsic fcts of differentil clculus. Here, I nd J denote genuine intervls or unions of collections of such. Theorem 5.1. (linerity rule) Given f, g Diff(I) nd c R, we hve f + g Diff(I), cf Diff(I) nd (f + g) = f + g, (cf) = c(f ). Theorem 5.2. (product rule) IGiven f, g Diff(I) we hve fg Diff(I) nd (fg) = f(g ) + (f )g. Theorem 5.3. (chin rule) Given f Diff(I) with Rngf J nd g Diff(J), we hve g f J Diff(I) nd (g f J) = ( g f J) f. Theorem 5.4. (inversion rule) Assume tht f : I J is invertible nd differentible, nd tht its derivtive f : I R is continuous. Let Then f J I := I\f < ({0} = {t I f (t) 0}, J := f > (I ). is differentible nd (f J ) = 1 f ( f I J ). 12
13 Note tht there re two sttements in ech of these theorems. The first sserts tht certin function is differentible, the second gives formul tht tells how to obtin the derivtive. It is evident, from the definition, tht ll constnts re differentible nd hve the sme derivtive, nmely the constnt zero. The following converse if much hrder to prove: Theorem 5.5. Let genuine intervl I nd f Diff(I) be given. If f = 0, then f is constnt. We sy tht function F Diff(I) is n ntiderivtive of given function f Fun(I) if F = f. If I is genuine intervl nd f hs such n ntiderivtive F, then {F + c c is constnt} is the set of ll ntiderivtives of f. This is consequence of Theorems 1 nd 5. 6 Derivtives nd ntiderivtives of elementry functions It follows immeditely from the definition tht the identity function ι is differentible nd its derivtive is the constnt 1: (ι ) = 1. (1) One cn prove, on the bsis of the definition, tht 1 : ι R R is differentible nd tht its derivtive is given by ( ) 1 = 1 ι ι. (2) 2 The bsic elementry functions sin nd exp re differentible nd their derivtives re given by ( π ) sin = cos := sin 2 ι (3) exp = exp. (4) 13
14 Using only the formuls (1)  (4), one cn determine, in routine mnner, the derivtives of ll elementry functions with the help of Theorems 14 of the previous section, without ever going bck to the definition of derivtive. The following fcts re helpful: If the given function f : I R is differentible, then 1 : I\f < ({0}) R cn be represented in the form f 1 = 1 f R. Hence, by the chin rule (Theorem 3), 1 f ι I\f < f ({0}) nd ( ) 1 = f f f. 2 is differentible Let f, g Diff(I) be given. If we pply the product rule (Theorem 2) to the cse when g is replced by 1 nd I by g I\g< ({0}) we obtin the quotient rule: f : g I\g< ({0}) R is differentible nd ( f g ) = f g fg g 2. Exmple: We wish to determine rctn. Since cos := sin ( π ι) we cn 2 use the chin rule to obtin ( ( π )) ( π ) ( π ) cos = sin 2 ι 2 ι = cos 2 ι. Since ( π ι) ( π ι) = ι, we hve 2 2 ( π ) ( ( π )) ( π ) cos 2 ι = sin 2 ι 2 ι = sin nd hence The quotient rule gives cos = sin. tn = ( ) sin = sin cos sin cos = cos2 + sin 2 = 1 + tn 2. cos cos 2 cos 2 Using the inversion rule, we finlly get 14
15 ] rctn 1 π 2 = 1 + (tn ] π 2,+ π 2 [ ) rctn,+ π 2 [ = ι. 2 For the record, we list the derivtives of the elementry functions mentioned t the end of Sect. 3: cos = sin, tn = 1 + tn 2 = 1 cos 2, rcsin = 1 1 ι 2, rccos = 1 1 ι 2, rctn = 1 1+ι 2, log = 1 ι, sin h = cos h, cos h = sin h, tn h = 1 tn h 2 = 1 cos h 2 (ι α ) = αι α 1 for ll α R. The derivtives of elementry functions re gin elementry functions. But not every elementry function hs ntiderivtives tht re elementry functions. Even if it hs, it is, in generl, very much hrder to find n ntiderivtive thn it is to compute the derivtive. The following tricks work sometimes: (i) Given function f, find functions h nd g nd number α R\{1} such tht f = hg nd such tht n ntiderivtive K of k := h g + αf cn be determined. Then F := 1 (hg K) 1 α 15
16 is n ntiderivtive of f, s is esily proved using the product rule. Exmple: If f := cos 2, we my tke h := cos, g := sin, nd α := 1. Then k = cos sin cos 2 = sin 2 cos 2 = 1, which hs the ntiderivtive K := ι. Hence is n ntiderivtive of cos 2. F := 1 (cos sin +ι) 2 (ii) Given function f, find functions h nd g such tht f = (h g)g nd such tht n ntiderivtive H of h cn be determined. Then F := H g is n ntiderivtive of f, s is evident from the chin rule. Exmple: If f := 1, we my tke g := exp, so tht 1+exp f = 1 ( ) ( ) 1 exp ι exp = 1 + ι exp exp = (1 + ι)ι exp exp Thus we hve f = (h g)g with h := 1 = 1 1, which hs the ntiderivtive H := log log (ι + 1). Hence F := H g = ι log (exp +1) is (1+ι)ι ι ι+1 n ntiderivtive of f. 7 Integrls Let S be subset of R. We sy tht f Fun(S) is positive nd write f 0 if f(t) 0 for ll t S. Let I be n intervl nd let, b I, with b. Given positive continuous function f Fun(I), we hve n intuitive notion of the re 16
17 mesure of the region between the grph of f, the horizontl xis R {0}, nd the verticl lines {} R nd {b} R. (The region is shded in Figure 1.) The integrl f should be defined in such wy tht it gives this remesure. In prticulr, f should be zero. If f Fun(I) is continuous but not necessrily positive, we consider the functions f + nd f defined by f + (t) := mx {f(t), 0}, f (t) := mx { f(t), 0} for ll t I. It is obvious tht f + nd f re positive nd tht f = f + f. Also, it is esy to prove tht f + nd f re continuous. The grphs of f + nd f re indicted in Figure 2 by the dotted nd the dshed curves, respectively). The integrl of f from I to b I, with b, is defined in such wy tht f := f + where the two terms on the right cn be interpreted s remesures. Note tht b is defined by f, f = 0 for ll I. If, b I nd > b, the integrl of f from to f := This eqution then holds for ll, b I. The considertion bove reduces the definition of integrls to tht of remesures. However, rigorous definition for remesure is not ny esier thn rigorous definition for integrls. All we hve gined is some intuitive guide. A rigorous definition of integrls will not be given here. b f. 17
18 On the surfce, derivtives nd remesures seem to be unrelted. It is somewht surprising, therefore, tht there is connection: F (t) := t f for ll t I is n ntiderivtive of f, i.e., F = f. Of course, proof of this theorem cn be given only fter integrls hve been rigorously defined. It follows from Theorem 7.1 tht ll continuous functions hve ntiderivtives;(but noncontinuous functions sometimes hve ntiderivtives, too.) Theorem 7.1 reduces, for continuous functions, the problem of finding ntiderivtives to the problem of finding integrls. The following corollry of Theorem 7.1 does the reverse: it reduces the problem of finding integrls to the problem of finding ntiderivtives. Theorem 7.1. Let I be n intervl, let f : I R be continuos nd let F be n ntiderivtive of f. Then for every, b I, we hve Proof. By Theorem 7.1, t f = F (b) F (). t f is n ntiderivtive of f. Hence, by Theorem 5.5, it cn differ from the given ntiderivtive F only by constnt, sy c, so tht t f F (t) = c for ll t I. Evluting this t t := nd then t t := b nd tking the difference, we obtin the desired result. The following rules re useful for deling with integrls. The first two re vlid when I is n intervl nd, b I, f, g C(I) nd c R. The third requires, in ddition, tht f, g Diff(I) nd f, g C(I). 18
19 (f + g) = f + g, (cf) = c f, f g = (fg)(b) (fg)() (integrtion by prts). The fourth rule is vlid when I is n intervl,, b I, g Diff(I) with g C(I), nd f C(J), where J is n intervl such tht Rng g J : g(b), re given. g(b) g() f = (f g)g (substitution rule). These rules re immedite consequences of Theorem 7.2 nd Theorems 13 of Section 5. fg 19
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 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 informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
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 information7.2 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 informationBernoulli 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 informationARITHMETIC 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 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 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 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 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 informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More 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 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 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 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 informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
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 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 informationStuff 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 informationMath& 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 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 informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
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 information7 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 informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationMath 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 informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/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 informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationSYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus
SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is
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 informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
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 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 informationIntegral points on the rational curve
Integrl points on the rtionl curve y x bx c x ;, b, c integers. Konstntine Zeltor Mthemtics University of Wisconsin  Mrinette 750 W. Byshore Street Mrinette, WI 5443453 Also: Konstntine Zeltor P.O. Box
More informationImproper 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 informationAP 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 informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationThe 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 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 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 informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
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 informationNUMERICAL 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 informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
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 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 informationx = b a N. (131) 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 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 informationP 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 informationChapter 1: Fundamentals
Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,
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 informationSection 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 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 informationHow 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 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 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 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 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 informationSection 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 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 information7. Indefinite Integrals
7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find
More informationUNIFORM 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 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 informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
More 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 informationNumerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
More informationSTEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.
STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t
More 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 informationMath Solutions to homework 1
Mth 75  Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors PreChpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
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 informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More informationThe usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1
Mth50 Introduction to Differentil Equtions Brief Review of Complex Numbers Complex Numbers No rel number stisfies the eqution x =, since the squre of ny rel number hs to be nonnegtive. By introducing
More informationand 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 informationc 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 information4402 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam
4402 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationFurther 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 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 informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition 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 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 informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More informationRAM RAJYA MORE, SIWAN. XI th, XII th, TARGET IITJEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII (PQRS) INDEFINITE INTERATION & Their Properties
M.Sc. (Mths), B.Ed, M.Phil (Mths) MATHEMATICS Mob. : 947084408 9546359990 M.Sc. (Mths), B.Ed, M.Phil (Mths) RAM RAJYA MORE, SIWAN XI th, XII th, TARGET IITJEE (MAIN + ADVANCE) & COMPATETIVE EXAM FOR XII
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 informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More information