INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable

Size: px
Start display at page:

Download "INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable"

Transcription

1 INTEGRATION 1 Integrls of Complex Vlued funtions of REAL vrible If I is n intervl in R (for exmple I = [, b] or I = (, b)) nd h : I C writing h = u + iv where u, v : I C, we n extend ll lulus 1 onepts to h by simply sying H stisfies property P if nd only if u nd v stisfy P. Speifilly: The funtion h is ontinuous on I if nd only if u, v re ontinuous on I. The funtion h is differentible on I if nd only if u, v re differentible on I. In this se we DEFINE for t I. h (t) = u (t) + iv (t) The funtion h is integrble on [, b] if nd only if u, v re integrble on [, b]. In this se we define b = b u(t) dt + i b v(t) dt. The expeted properties hold, even if some re not s obvious s one might think. The fundmentl theorem of lulus still holds, if H = U + iv : [, b] C nd H = h = u + iv, then u = U, v = V nd b = b u(t) dt + i b v(t) dt = U(b) U() + i(v (b) V ()) = (U(b) + iv (b)) (U() + iv ()) = H(b) H(). The formul for hnge of vribles still holds: If ϕ is rel vlued differentible funtion defined on [, b] nd h is defined on the rnge of ϕ, then b h(ϕ(s)) ϕ (s) ds = ϕ(b) ϕ(). A more fmilir version of this formul is the following: In the integrl we n mke the substitution t = ϕ(s), dt = ϕ (s) ds to get d = ϕ 1 (d) ϕ 1 () h(ϕ(s))ϕ (s) ds. d

2 1 INTEGRALS OF COMPLEX VALUED FUNCTIONS OF A REAL VARIABLE2 It should be ler why this is lso vlid if h = u + iv. It should be quite ler lso tht the integrl of sum or differene is the sum or differene of the integrl. It is bit more omplited to verify tht omplex onstnt pulls out of the integrl. Sy α = + ib,, b rel nd h = u + iv s before. Then d α = = = d d d ( + ib)(u(t) + iv(t)) dt = (u(t) bv(t)) dt + i u(t) dt b d ( d = ( + ib) u(t) dt + i d v(t) dt + i d d [(u(t) bv(t)) + i(v(t) + bu(t))] dt (v(t) + bu(t)) dt ( v(t) dt d ) v(t) dt + b = α d d. u(t) dt ) For rel vlued ontinuous funtions h it is true tht b b h(t) dt. (1) The proof is rther immedite; one hs h(t) h(t) h(t) for ll t; integrls of rel vlued funtions preserve the order so tht b h(t) dt b b h(t) dt, whih is preisely (1). Things get hrder in the omplex relm. If h = u + iv, then (1) sttes tht ( 2 ( 2 b b b u(t) dt) + v(t) dt) u(t)2 + v(t) 2 dt. As usul, rewriting omplex vlued funtion in terms of its rel nd imginry prts is the hrdest wy to go. Here is proof (1) if h is omplex vlued. I think it is quite n ingenious nd plesnt proof. We n write, s for every omplex number, where r = b b = re iθ, nd θ [, 2π) is the rgument. Then b b b = r = e iθ = e iθ.

3 2 INTEGRATION OVER PATHS 3 The lst equlity is due to the ft tht onstnt n go in or out of the integrl, s proved bove. We will use now ( ) b b Re = beuse bsolute vlues re rel; we lso use tht Re w w for ll omplex numbers w, tht the rel prt of omplex integrl is the integrl of the rel prt of the integrnd, nd tht integrls of rel vlued funtions preserve the order. We left off bove in b b = e iθ ; tking rel prts of both sides, using the mentioned fts, b b b = Re = Re e iθ = b e iθ h(t) dt = b h(t) dt proving (1); the lst equlity is due to e iθ = 1. b Re ( e iθ h(t) ) dt Finlly, here is useful differentition formul. Suppose h(t) = e αt where α = + ib,, b R. Then h (t) = αe αt ; tht is, just s in the rel se. Here is the proof: d dt eαt = αe αt, (2) d dt eαt = d dt et+ibt = d { e t os bt + ie t sin bt } dt = e t os bt be t sin bt + i sin bt + i ( e t sin bt + be t os bt ) = ( + ib) ( e t os bt + ie t sin bt ) = αe αt. 2 Integrtion over Pths Definition 1 A pieewise smooth pth is ontinuous funtion : [, b] C suh tht there is prtition t,..., t n, = t < t 1 < < t n = b of the losed intervl [, b] suh tht is ontinuously differentible (i.e., differentible with ontinuous derivtive) in eh open intervl (t i 1, t i ) nd (t) hs limit s t pprohes the endpoints of the intervl, from the right t t i 1, from the left t t i. The pth is sid to be losed if () = (b).

4 2 INTEGRATION OVER PATHS 4 Writing (t) = x(t) + iy(t), ontinuity mens tht both x(t), y(t) re ontinuous; differentibility tht both x(t), y(t) re differentible in the sense of Clulus I nd (t) = x (t) + iy (t). In priniple the pth is the funtion ; s t rnges from to b it desribes urve in the plne.. This is geometri objet, subset of the plne. We will refer to it s the urve desribed by the pth; osionlly (or frequently) we will be bit reless hoping tht it won t use ny onfusion, nd use the term pth both for the funtion (the prmetriztion) nd the geometri objet. Definition 2 Let : [, b] C be pieewise smooth pth. If f is omplex vlued funtion defined t ll points of the urve desribed by so tht f((t)) mkes sense for ll t [, b] nd f((t)) is ontinuous for t b, we define f(z) dz by f(z) dz = n i=1 ti t i 1 f((t)) (t) dt; where = t < t 1 < < t n = b is prtition suh tht (t) is ontinuously differentible in eh open intervl (t i 1, t i ) nd (t) hs limit s t pprohes the endpoints of the intervl, from the right t t i 1, from the left t t i. I wnt to emphsize: In this ourse, ny time you see the nottion f(z) dz, or vrint thereof (the urve will not lwys be denoted by, nor will the funtion lwys be f, or the vrible of integrtion z) the following onditions must hold, even if not expliitly stted: 1. is pieewise smooth pth. 2. The domin of f ontins the rnge of ; tht is f is defined t ll points desribed by. 3. The omposition of f with is ontinuous. We will dopt nother onvention, to keep the nottion from beoming too umbersome. Suppose, t,..., t n re s bove. We define or write n i=1 ti t i 1 f((t)) (t) dt = so tht we n lwys write f(z) dz = b b f((t)) (t) dt f((t)) (t) dt.

5 2 INTEGRATION OVER PATHS 5 Of ourse, if we tully hve to ompute the integrl by the definition nd is not smooth, just pieewise smooth, we will need to brek up the intervl into portions on whih is smooth. One thing to keep in mind is tht the min use of pths will be to integrte over them. This llows us to mke definition tht, like ll definitions, mens extly wht it sys; Definition 3 Two pths 1 : [, b] C nd 2 : [, d] C will be sid to be equivlent nd we will even buse nottion to the point of writing 1 = 2 if (nd only if): 1. They desribe the sme set of points; tht is, both hve the sme rnge, desribe the sme urve. 2. If we denote this rnge by C, then if f is ny funtion whose domin ontins C nd is ontinuous on C (see below), then f(z) dz = f(z) dz. 1 2 To sy tht f is ontinuous on C mens tht if {z n } is ny sequene of points on C onverging to point w C, then {f(z n )} onverges to f(w). It is lso equivlent to sying tht f((t)) is ontinuous s funtion of t for one nd then lso for every pth desribing C. A few exmples follow. But first omment. Suppose we wnt to integrte funtion f on the irle prmeterized by (t) = e it = os t + i sin t, t 2π. Deomposing everything into rel nd imginry prts is bsilly no help t ll. For exmple, suppose f u + iv is defined on C. Then (plese, hek this!) f(z) dz = = 2π 2π f(os t, sin t)( sin t + i os t) dt ( u(os t, sin t) v(os t, sin t) os t) dt + i 2π (u(os t, sin t) os t v(os t, sin t) sin t) dt The more one deomposes, the worse it gets. The best wy of writing integrls s ordinry Riemnn integrls is to sty in the omplex relm. Using the eqution (2) to ompute (t) we n write f(z) dz = i 2π f(e it )e it dt

6 2 INTEGRATION OVER PATHS 6 nd for onrete funtion f if one n t ompute this integrl s written, one lmost ertinly n t ompute it by breking it up into rel nd imginry prts. Here re the exmples for the definition of equivlene of pths. i. Suppose 1 : [, 2π] C is given by 2 : [, 1] C is given by 3 : [ π, pi] C is given by 1 (t) = e it = os t + i sin t, 2 (t) = e 2πit = os(2πt) + i sin(2πt), 3 (t) = e it = os t + i sin t, nd 4 by 4 (t) = ie it = sin t + i os t. As funtions no two of 1, 2, 3, 4 re the sme; they re different funtions. But in the sense of the definition bove, 1 = 2 = 3 4. The geometri objet desribed by ll four of them is the unit irle C = {z C : z = 1} entered t the origin. Thus the first ondition for equivlene is stisfied for ll of them. It is up to the seond ondition. Using one gin the eqution (2) to ompute i (t) for i = 1, 2, 3, 4, we n write 2π f(z) dz = i f(e it )e it dt, (3) 1 1 f(z) dz = 2πi f(e 2πit )e 2πit dt, (4) 2 π f(z) dz = i f(e it )e it dt, (5) 3 π 2π f(z) dz = f(ie it )e it dt. (6) 4 To see tht pth 1 = 2 (in the sense explined bove!) we n mke the substitution t = 2πs, dt = 2π ds in the first integrl (3); one gets the seond integrl (4). To see tht the first integrl equls the third n be bit more triky. One n use tht the integrnd is periodi of period 2π nd the integrl of periodi funtion over ny intervl of length equl to the period is the sme. Or one n write i 2π f(e it )e it dt = i π f(e it )e it dt + i 2π π f(e it )e it dt

7 2 INTEGRATION OVER PATHS 7 nd in the seond integrl of the right hnd side we mke the substitution t = s + 2π, so s = t 2π, dt = ds. We use tht e 2πi = 1. Then i 2π f(e it )e it dt = i = i π π f(e it )e it dt + i f(e it )e it dt + i 2π 2π π 2π π f(e i(s+2π) )e i(s+2π) dt f(e is )e is dt = i π π f(e it )e it dt. Coming to the fourth integrl, to see it might not be equl to the other three, we n tke onrete f nd evlute. For exmple, tke f(z) = Re z. Then Re (e it ) = os t, Re (ie it ) = sin t so tht 1 f(z) dz = i while 4 f(z) dz = 2π 2π (os t)e it dt = i (sin t)e it dt = 2π 2π Sine iπ iπ, the pths re not equivlent. (os 2 t + i os t sin t) dt = i(π + i) = iπ, (sin t os t i sin 2 t) dt = ( iπ) = iπ. Rule of thumb (mostly works): If you hve two prmetriztions desribing the sme urve nd if the urve is gone through the sme number of times nd in the sme diretion in both ses, then the pths desribing tht urve re equivlent. The most ommon equivlene of pths is by substitution: If : [, b] C is pth, if ϕ : [, d] [, b] is ontinuous, inresing, differentible in the open intervl (, d) nd ϕ() = ϕ(d) = b, then the pth 1 : [, d] C defined by 1 (t) = (ϕ(t)) is equivlent to 1. This is beuse 1(t) = (ϕ(t))ϕ (t) so tht if in d d f(z) dz = f( 1 (t)) 1(t) dt = f((ϕ(t))) (ϕ(t))ϕ (t) dt 1 we mke the hnge of vribles (substitution) s = ϕ(t), ds = ϕ (t) dt, we get d f((ϕ(t))) (ϕ(t))ϕ (t) dt = b f((s)) (s) ds = f(z) dz. One onsequene is: Every pth is equivlent to one prmeterized by the intervl [, 1]. In ft, onsider pth : [.b] C. Define ϕ : [, 1] [, b] by ϕ(s) = + s(b ), whih is nie inresing differentible funtion stisfying ϕ() =, ϕ(1) = b.

8 3 PATH LENGTH 8 3 Pth Length Pths hve lengths. You sw this in Clulus 3. Suppose you hve urve prmeterized by x = x(t), y = y(t), t b. Then the so lled element of r length is ds = x (t) 2 + y (t) 2 dt nd the length of the [pth is defined by Λ() = b ds = b x (t) 2 + y (t) 2 dt. For exmple, the length of the portion of the unit irle ontined in the first qudrnt, whih n be prmeterized by is Λ = = π/2 π/2 ( d x = os t, y = sin t, t π/2, (os t) dt ) 2 ( d + sin 2 t + os 2 t dt = (sin t) dt π/2 ) 2 π/2 dt = ( sin t)2 + (os t) 2 dt π/2 1 dt = dt = π 2. In omplex nottion, we define pths by mp = x + iy : [, b] C; then (t) = x (t) + iy (t) nd (t) = x (t) 2 + y (t) 2 so tht the definition of r length you find in the online text, the one we use here, is extly the sme s the Clulus 3 one, nmely Λ() = b (t) dt = b x (t) 2 + y (t) 2 dt. In omplex nottion, the qurter irle is desribed by (t) = e it, t π/2, then (t) = ie it nd (t) = i e it = 1 so tht one gin Λ = π/2 dt = π 2. One of the most frequent ppernes of r length is in estimting. Suppose : [, b] C is pth of length Λ(). Suppose the urve desribed by (the rnge of ) is C. Suppose f is ontinuous funtion of domin C (or lrger) nd suppose tht f(z) M for ll z C. Then b b f(z) dz = f((t)) (t) dt f((t)) (t) dt M b (t) dt = MΛ(),

9 4 ADDING AND SUBTRACTING PATHS 9 briefly, f(z) dz MΛ(). (7) 4 Adding nd Subtrting Pths If we hve two pths 1, 2 nd 1 ends where 2 begins, then we n form the pth onsisting of 1 followed by 2. I ll give forml definition, but then get down to the prtil one. So suppose 1 : [, b] C, 2 : [, d] C nd 1 (b) = 2 (). Then is ny pth equivlent to the following one: { 1 (t), if t < b, ( )(t) = 2 (t + b), if b t b + d. The min property of this new pth 1 + 2, defined on the intervl [, b+d ] is tht if f is ny ontinuous funtion defined on the union of the urve desribed by 1 with the urve desribed by 2, then f(z) dz = 1+ 2 f(z) dz + 1 f(z) dz. 2 (8) This formul is esy to verify by simple substitution. Suppose now tht we hve three urves 1, 2, 3 defined respetively on [, b], [, d], nd [p, q]. Suppose 1 (b) = 2 () nd 2 (d) = 3 (p). We n then form ( ) + 3 nd 1 + ( ). It is esy to see tht ( ) + 3 = 1 + ( ). However, depending on the textbook, you my find definitions of the sum of two urves tht turns out not to be ssoitive in this strit sense. For exmple, it is populr to reprmeterize ll urves so the prmeter intervl is lwys the intervl [, 1] nd then the stndrd definition of is tht we go through 1 in the intervl [, 1/2] nd through 2 in the intervl [1/2, 1]. Addition eses to be ssoitive. But tht is not importnt, espeilly not to us. The min property of the ddition is given by (8). We n use ddition to define some simple urves. For exmple if we denote (s I probbly will from now on) the line segment from z C to z 1 C by L z,z 1 ; tht is, L z,z 1 is the pth L z,z 1 (t) = z + t(z 1 z ), t 1; we n define the tringle T z,z 1,z 2 with verties z, z 1, z 2 by T z,z 1,z 2 = L z,z 1 + L z1,z 2 + L z2,z. Exmple: Compute Re z dz, where T = T (, 1, 1 + i) is the tringle of T verties, 1, 1 + i (gone through in tht order). To void too mny subindies, let L 1 = L,1, L 2 = L 1,1+i, L 3 = L 1+i,.

10 4 ADDING AND SUBTRACTING PATHS 1 We hve L 1 (t) = t, Re L 1 (t) = t, L 1(t) = 1, hene Re z dz = L 1 1 t dt = 1 2. Next L 2 (t) = 1 + ti, Re L 2 (t) = 1, L 2(t) = i, so tht L 2 Re z dz = i 1 dt = 1 2 i. Finlly L 3 (t) = 1 + i t(1 + i), Re L 2 (t) = 1 t, L 2(t) = (1 + i), L 3 Re z dz = (1 + i) 1 (1 t) dt = 1 (1 + i). 2 Thus Re z dz = T Re z dz + L 1 Re z dz + L 2 Re z dz =. L 3 A ouple of shortuts to onsider re: 1. If the pth is horizontl line segment, the integrl n be immeditely written s Riemnn integrl. A horizontl line segment will be segment from point z = + i to z 1 = b + i. Then L z,z 1 f(z) dz = b f(x + i) dx In ft, if we prmeterize the intervl by the stndrd prmetriztion; tht is, by L z,z 1 (t) = z +t(z 1 z ) = +i+t(b ) = +t(b )+i, L (t) = b, t 1, we hve t first n integrl from to 1; nmely L z,z 1 f(z) dz = (b ) 1 f( + t(b ) + i) dt.

11 4 ADDING AND SUBTRACTING PATHS 11 The substitution x = + t(b ), dx = (b ) dt shows this integrl is equl to b f(x + i) dx. In prtiulr, if the pth is the segment from to b on the rel line, then f dz = b f(x) dx. 2. Similr onsidertion hold for vertil line segments. If is the line segment from + i to + id, then d f(z) dz = f( + iy) dy. Exmple Compute z 2 dz where is the pth pitured below. We n write = SC + L where SC is the semiirle from 2 to 2, gone through positively, nd L is the line segment from 2 to 2. We ompute t one 2 z 2 dz = x 2 dx = L 2 We n prmeterize the semiirle, for exmple, by SC(t) = 2e it, t π; SC (t) = 2ie it, so π π z 2 dz = 2e it 2 (2ie it ) dt = 8i e it dt = 16. SC It follows tht z 2 dz = = If is pth we will denote by the pth gone through in the opposite diretion. So, for exmple, if desribes urve strting t z nd ending t z 1, then desribes the sme urve, but strting t z 1 ending t z. If : [, b] C, possible prmetriztion for is (t) = (b t + ), t b.

12 5 THE FUNDAMENTAL THEOREM OF CALCULUS DECIDES TO SHOW UP12 One then hs, s is esily verified by hnge of vribles, f(z) dz = f(z) dz. Exmple Let be the retngle of verties, 3, 3 + i, i, gone through one in the order indited by the verties. Compute We hve Using the shortuts mentioned bove 1 (z 2 z 2 ) dz = xy dz. 4i = L,3 + L 3,3+i L i,3+i L,i. L,3 xy dz = xy dz = L 3,3+i xy dz = L i,3+i xy dz = L,i x dx =, 3y dy = 3 2, x dx = 1 2, y dx =. Thus xy dz = = 1. 5 The Fundmentl Theorem of Clulus Deides to Show Up The following result is simple onsequene of the Fundmentl Theorem of Clulus.

13 5 THE FUNDAMENTAL THEOREM OF CALCULUS DECIDES TO SHOW UP13 Theorem 1 Assume f is ontinuous in domin U nd let desribe urve ompletely ontined in U, strting t point z, ending t point z 1. Suppose there is n nlyti funtion F in U suh tht F (z) = f(z) for ll z U. Then f(z) dz = F (z 1 ) F (z ). In prtiulr, if is losed urve (s z 1 = z ), then f(z) dz =. Fun Ft: Mny Physiists nd Engineers (nd probbly lso some mthemtiins) like to write f(z) dz insted of f(z) dz if is losed pth. If you like this nottion, plese feel free to mke it your own. The theorem is esily justified. Assume F is nlyti in U, : [, b] C nd (t) U for ll t [, b] I will ssume tht is smooth; if it is only pieewise smooth some minor djustments re required. Then F is funtion defined in the intervl [, b] nd the sme wy one proves the hin rule in Clulus 1 (ssuming one proves it) or the hin rule for ompositions of nlyti funtions, one sees tht d dt (F )(t) = d dt (F ((t)) = F ((t)) (t) = f((t) (t), where F is the omplex derivtive of F nd (t) the Clulus 1 derivtive of ; tht is, if (t) = x(t)+iy(t), then (t) = x (t)+iy (t), where x, y re Clulus 1 derivtives. We thus hve f(z) dz = b f((t)) (t) dt = b d dt (F )(t) dt = F ((b)) F (()) F (z 1) F (z ). Exmple Let be the r of the prbol of eqution x = y 2 from (4, 2) = 4+2i to (, ) =. Compute z 2 dz. Here is doing it by prmeterizing the urve. Notiing the diretion in whih goes we write = 1, where 1 (t) = t + i t, t 4, 1(t) = 1 + i 2 t.

14 5 THE FUNDAMENTAL THEOREM OF CALCULUS DECIDES TO SHOW UP14 Then 4 z 2 dz = z 2 dt = (t + i ( t) i ) 1 2 dt t 4 ( ( 5 = t 2 2t + i 2 t3/2 1 )) 2 t1/2 dt = 1 3 t2 + t 2 (t 5/2 13 ) 4 t3/2 = 1 ( i). 3 This wsn t so hrd. But suppose the urve hd been muh more omplited one? Beuse z 2 is the derivtive of n nlyti funtion, it does not mtter how we went from 4 + 2i to. Tking F (z) = 1 3 z3 so tht F (z) = z 2 we hve z 2 dz = (4 + 2i)3 = 1 3 ( i 48 8i) = 1 ( i) 3 Exmple Compute z =R z n dz, for n =, ±1, ±2,.... Nottionl onvention It is stndrd to denote by f(z) dz z z =R the integrl over the positively oriented irle of rdius R entered t z ; the pth tht n be prmeterized by (t) = z + Re it, t 2π. Conerning the exmple, we sw in lss tht { z n, if n n integer, n 1, dz = 2πi, ifn = 1. z =R Using our just gined knowledge, we n get the sme result without hving to ompute ny integrl. The pth of integrtion is losed urve. It is ontined in the open set {z C : z }. In this set, if n 1, z n is the derivtive of n nlyti funtion, nmely z n = d ( ) 1 dz n + 1 zn+1. Thus z =R z n dz =

15 5 THE FUNDAMENTAL THEOREM OF CALCULUS DECIDES TO SHOW UP15 for n =, 1, 2, 2, 3, 3,.... The se n = 1 needs bit more work beuse there is no nlyti funtion in {z } whose derivtive is 1/z. So wht we n do is to notie tht dz z =R z = lim dz ɛ + ɛ z where gm ɛ is the pth prmeterized by ɛ (t) = Re it, π + ɛ t π ɛ Here s piture of the pth. The pth ɛ is now inside n open set in whih Logz is nlyti nd 1 z = d dz Logz. Thus dz z ɛ = Log(Re i(π ɛ) ) Log(Re i( π+ɛ) ) = (ln R + i(π ɛ)) (ln R + i( π + ɛ)) = 2πi 2ɛ. Letting ɛ + we get z =R dz z = 2πi.

INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable

INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable INTEGRATION NOTE: These notes re supposed to supplement Chpter 4 of the online textbook. 1 Integrls of Complex Vlued funtions of REAL vrible If I is n intervl in R (for exmple I = [, b] or I = (, b)) nd

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

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

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e Green s Theorem. Let be the boundry of the unit squre, y, oriented ounterlokwise, nd let F be the vetor field F, y e y +, 2 y. Find F d r. Solution. Let s write P, y e y + nd Q, y 2 y, so tht F P, Q. Let

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

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

AP Calculus AB Unit 4 Assessment

AP Calculus AB Unit 4 Assessment Clss: Dte: 0-04 AP Clulus AB Unit 4 Assessment Multiple Choie Identify the hoie tht best ompletes the sttement or nswers the question. A lultor my NOT be used on this prt of the exm. (6 minutes). The slope

More information

Part 4. Integration (with Proofs)

Part 4. Integration (with Proofs) Prt 4. Integrtion (with Proofs) 4.1 Definition Definition A prtition P of [, b] is finite set of points {x 0, x 1,..., x n } with = x 0 < x 1

More information

(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.

(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. 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

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

Notes on length and conformal metrics

Notes on length and conformal metrics Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued

More 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

p(t) dt + i 1 re it ireit dt =

p(t) dt + i 1 re it ireit dt = Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)

More 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

Section 3.6. Definite Integrals

Section 3.6. Definite Integrals The Clulus of Funtions of Severl Vribles Setion.6 efinite Integrls We will first define the definite integrl for funtion f : R R nd lter indite how the definition my be extended to funtions of three or

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

MAT 403 NOTES 4. f + f =

MAT 403 NOTES 4. f + f = MAT 403 NOTES 4 1. Fundmentl Theorem o Clulus We will proo more generl version o the FTC thn the textook. But just like the textook, we strt with the ollowing proposition. Let R[, ] e the set o Riemnn

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

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1. 1 [(y ) 2 + yy + y 2 ] dx,

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1. 1 [(y ) 2 + yy + y 2 ] dx, MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions 1 Exmples 5 Qu.1 Show tht the extreml funtion of the funtionl I[y] = 1 0 [(y ) + yy + y ] dx, where y(0) = 0 nd y(1) = 1, is y(x)

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

RIEMANN INTEGRATION. Throughout our discussion of Riemann integration. B = B [a; b] = B ([a; b] ; R)

RIEMANN INTEGRATION. Throughout our discussion of Riemann integration. B = B [a; b] = B ([a; b] ; R) RIEMANN INTEGRATION Throughout our disussion of Riemnn integrtion B = B [; b] = B ([; b] ; R) is the set of ll bounded rel-vlued funtons on lose, bounded, nondegenerte intervl [; b] : 1. DEF. A nite set

More information

II. Integration and Cauchy s Theorem

II. Integration and Cauchy s Theorem MTH6111 Complex Anlysis 2009-10 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.

More information

Type 2: Improper Integrals with Infinite Discontinuities

Type 2: Improper Integrals with Infinite Discontinuities mth imroer integrls: tye 6 Tye : Imroer Integrls with Infinite Disontinuities A seond wy tht funtion n fil to be integrble in the ordinry sense is tht it my hve n infinite disontinuity (vertil symtote)

More information

INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012

INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012 Lecture 6: Line Integrls INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Anlysis Autumn 2012 August 8, 2012 Lecture 6: Line Integrls Lecture 6: Line Integrls Lecture 6: Line Integrls Integrls of complex

More information

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

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

Section 17.2 Line Integrals

Section 17.2 Line Integrals Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We

More information

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the

More information

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into

More information

Complex variables lecture 5: Complex integration

Complex variables lecture 5: Complex integration omplex vribles lecture 5: omplex integrtion Hyo-Sung Ahn School of Mechtronics Gwngju Institute of Science nd Technology (GIST) 1 Oryong-dong, Buk-gu, Gwngju, Kore Advnced Engineering Mthemtics omplex

More information

The Regulated and Riemann Integrals

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

More information

Riemann Sums and Riemann Integrals

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

More information

Riemann Sums and Riemann Integrals

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

More information

MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 5 SOLUTIONS. cos t cos at dt + i

MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 5 SOLUTIONS. cos t cos at dt + i MATH 85: COMPLEX ANALYSIS FALL 9/ PROBLEM SET 5 SOLUTIONS. Let R nd z C. () Evlute the following integrls Solution. Since e it cos t nd For the first integrl, we hve e it cos t cos t cos t + i t + i. sin

More information

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

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

More information

Line Integrals. Chapter Definition

Line Integrals. Chapter Definition hpter 2 Line Integrls 2.1 Definition When we re integrting function of one vrible, we integrte long n intervl on one of the xes. We now generlize this ide by integrting long ny curve in the xy-plne. It

More information

MATH Final Review

MATH Final Review MATH 1591 - Finl Review November 20, 2005 1 Evlution of Limits 1. the ε δ definition of limit. 2. properties of limits. 3. how to use the diret substitution to find limit. 4. how to use the dividing out

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

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

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

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

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

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

More information

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

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

More information

arxiv: v1 [math.ca] 21 Aug 2018

arxiv: v1 [math.ca] 21 Aug 2018 rxiv:1808.07159v1 [mth.ca] 1 Aug 018 Clulus on Dul Rel Numbers Keqin Liu Deprtment of Mthemtis The University of British Columbi Vnouver, BC Cnd, V6T 1Z Augest, 018 Abstrt We present the bsi theory of

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

We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.

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

More information

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

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1 Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution

More information

] dx (3) = [15x] 2 0

] dx (3) = [15x] 2 0 Leture 6. Double Integrls nd Volume on etngle Welome to Cl IV!!!! These notes re designed to be redble nd desribe the w I will eplin the mteril in lss. Hopefull the re thorough, but it s good ide to hve

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

Final Exam Review. [Top Bottom]dx =

Final Exam Review. [Top Bottom]dx = Finl Exm Review Are Between Curves See 7.1 exmples 1, 2, 4, 5 nd exerises 1-33 (odd) The re of the region bounded by the urves y = f(x), y = g(x), nd the lines x = nd x = b, where f nd g re ontinuous nd

More information

CHAPTER V INTEGRATION, AVERAGE BEHAVIOR A = πr 2.

CHAPTER V INTEGRATION, AVERAGE BEHAVIOR A = πr 2. CHAPTER V INTEGRATION, AVERAGE BEHAVIOR A πr 2. In this hpter we will derive the formul A πr 2 for the re of irle of rdius r. As mtter of ft, we will first hve to settle on extly wht is the definition

More information

Tutorial Worksheet. 1. Find all solutions to the linear system by following the given steps. x + 2y + 3z = 2 2x + 3y + z = 4.

Tutorial Worksheet. 1. Find all solutions to the linear system by following the given steps. x + 2y + 3z = 2 2x + 3y + z = 4. Mth 5 Tutoril Week 1 - Jnury 1 1 Nme Setion Tutoril Worksheet 1. Find ll solutions to the liner system by following the given steps x + y + z = x + y + z = 4. y + z = Step 1. Write down the rgumented mtrix

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

The usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1

The 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 non-negtive. By introducing

More information

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus

SYDE 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 information

Lecture Summaries for Multivariable Integral Calculus M52B

Lecture Summaries for Multivariable Integral Calculus M52B These leture summries my lso be viewed online by liking the L ion t the top right of ny leture sreen. Leture Summries for Multivrible Integrl Clulus M52B Chpter nd setion numbers refer to the 6th edition.

More information

MAA 4212 Improper Integrals

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

More information

The Double Integral. The Riemann sum of a function f (x; y) over this partition of [a; b] [c; d] is. f (r j ; t k ) x j y k

The Double Integral. The Riemann sum of a function f (x; y) over this partition of [a; b] [c; d] is. f (r j ; t k ) x j y k The Double Integrl De nition of the Integrl Iterted integrls re used primrily s tool for omputing double integrls, where double integrl is n integrl of f (; y) over region : In this setion, we de ne double

More information

ODE: Existence and Uniqueness of a Solution

ODE: 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 information

Chapter 0. What is the Lebesgue integral about?

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

More information

Complex integration. L3: Cauchy s Theory.

Complex integration. L3: Cauchy s Theory. MM Vercelli. L3: Cuchy s Theory. Contents: Complex integrtion. The Cuchy s integrls theorems. Singulrities. The residue theorem. Evlution of definite integrls. Appendix: Fundmentl theorem of lgebr. Discussions

More information

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space. Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)

More 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

Overview of Calculus I

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

More information

7.2 The Definite Integral

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

More information

Applications of Definite Integral

Applications of Definite Integral Chpter 5 Applitions of Definite Integrl 5.1 Are Between Two Curves In this setion we use integrls to find res of regions tht lie between the grphs of two funtions. Consider the region tht lies between

More information

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

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

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

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

More information

Lecture 1: Introduction to integration theory and bounded variation

Lecture 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 information

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

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

More information

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

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

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

Applications of Definite Integral

Applications of Definite Integral Chpter 5 Applitions of Definite Integrl 5.1 Are Between Two Curves In this setion we use integrls to find res of regions tht lie between the grphs of two funtions. Consider the region tht lies between

More information

University of Sioux Falls. MAT204/205 Calculus I/II

University of Sioux Falls. MAT204/205 Calculus I/II University of Sioux Flls MAT204/205 Clulus I/II Conepts ddressed: Clulus Textook: Thoms Clulus, 11 th ed., Weir, Hss, Giordno 1. Use stndrd differentition nd integrtion tehniques. Differentition tehniques

More information

Formula for Trapezoid estimate using Left and Right estimates: Trap( n) If the graph of f is decreasing on [a, b], then f ( x ) dx

Formula for Trapezoid estimate using Left and Right estimates: Trap( n) If the graph of f is decreasing on [a, b], then f ( x ) dx Fill in the Blnks for the Big Topis in Chpter 5: The Definite Integrl Estimting n integrl using Riemnn sum:. The Left rule uses the left endpoint of eh suintervl.. The Right rule uses the right endpoint

More information

Integrals Depending on a Parameter

Integrals Depending on a Parameter Universidd Crlos III de Mdrid Clulus II Mrin Delgdo Téllez de Ceped Unit 3 Integrls Depending on Prmeter Definition 3.1. Let f : [,b] [,d] R, if for eh fixed t [,d] the funtion f(x,t) is integrble over

More information

MATH , Calculus 2, Fall 2018

MATH , Calculus 2, Fall 2018 MATH 36-2, 36-3 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 information

10 Vector Integral Calculus

10 Vector Integral Calculus Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve

More information

Math Advanced Calculus II

Math Advanced Calculus II Mth 452 - Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused

More information

1. On some properties of definite integrals. We prove

1. On some properties of definite integrals. We prove This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.

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

INTRODUCTION TO INTEGRATION

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

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS 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 information

Improper Integrals, and Differential Equations

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

More information

11 An introduction to Riemann Integration

11 An introduction to Riemann Integration 11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in

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

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

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

More information

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

W. 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 information

Electromagnetism Notes, NYU Spring 2018

Electromagnetism Notes, NYU Spring 2018 Eletromgnetism Notes, NYU Spring 208 April 2, 208 Ation formultion of EM. Free field desription Let us first onsider the free EM field, i.e. in the bsene of ny hrges or urrents. To tret this s mehnil system

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

We identify two complex numbers z and w if and only if Rez = Rew and Imz = Imw. We also write

We identify two complex numbers z and w if and only if Rez = Rew and Imz = Imw. We also write Chpter 1 The Complex Plne 1.1. The Complex Numbers A complex number is n expression of the form z = x + iy = x + yi, where x, y re rel numbers nd i is symbol stisfying i 2 = ii = i i = 1. Here, x is clled

More information

AP Calculus Multiple Choice: BC Edition Solutions

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

More information

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE

MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE MATH 311: COMPLEX ANALYSIS INTEGRATION LECTURE Contents 1. Introduction 1 2. A fr-reching little integrl 4 3. Invrince of the complex integrl 5 4. The bsic complex integrl estimte 6 5. Comptibility 8 6.

More information

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

Physics 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 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

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

For a, b, c, d positive if a b and. ac bd. Reciprocal relations for a and b positive. If a > b then a ab > b. then

For a, b, c, d positive if a b and. ac bd. Reciprocal relations for a and b positive. If a > b then a ab > b. then Slrs-7.2-ADV-.7 Improper Definite Integrls 27.. D.dox Pge of Improper Definite Integrls Before we strt the min topi we present relevnt lger nd it review. See Appendix J for more lger review. Inequlities:

More information

Definite integral. Mathematics FRDIS MENDELU

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

More information