Notes on Calculus. Dinakar Ramakrishnan Caltech Pasadena, CA Fall 2001

Transcription

1 Notes on Clculus by Dinkr Rmkrishnn Cltech Psden, CA Fll 21 1

2 Contents Logicl Bckground 2.1 Sets Functions Crdinlity EquivlenceReltions Rel nd Complex Numbers DesiredProperties Nturl Numbers, Well Ordering, nd Induction Integers RtionlNumbers OrderedFields RelNumbers AbsoluteVlue ComplexNumbers Sequences nd Series Convergenceofsequences Cuchy scriterion ConstructionofRelNumbersrevisited Infiniteseries TestsforConvergence Alterntingseries Bsics of Integrtion Open, closed nd compct sets in R Integrls of bounded functions Integrbility of monotone functions Computtion of x s dx Exmple of non-integrble, bounded function Propertiesofintegrls The integrl of x m revisited,ndpolynomils Continuous functions, Integrbility LimitsndContinuity Sometheoremsoncontinuousfunctions Integrbility of continuous functions Trigonometricfunctions Functionswithdiscontinuities

3 5 Improper Integrls, Ares, Polr Coordintes, Volumes ImproperIntegrls Ares Polrcoordintes Volumes Theintegrltestforinfiniteseries Differentition, Properties, Tngents, Extrem Derivtives Rulesofdifferentition,consequences Proofsoftherules Tngents Extremofdifferentiblefunctions Themenvluetheorem The Fundmentl Theorems of Clculus, Methods of Integrtion The fundmentl theorems Theindefiniteintegrl Integrtionbysubstitution Integrtionbyprts

4 7 The Fundmentl Theorems of Clculus, Methods of Integrtion So fr we hve seprtely lernt the bsics of integrtion nd differentition. But they re not unrelted. In fct, they re inverse opertions. This is wht we will try toexplore in the first section, vi the two fundmentl theorems of Clculus. After tht we will discuss the two min methods one uses for integrting somewht complicted functions, nmely integrtion by substitution nd integrtion by prts. 7.1 The fundmentl theorems Suppose f is n integrble function on closed intervl [, b]. Then we cn consider the signed re function A on[, b] (reltive to f) defined by the definite integrl of f from to x, i.e., (7.1.) A(x) = x f(t)dt. The reson for the signed re terminology is tht f is not ssumed to be, sopriori A(x) could be negtive. It is extremely interesting to know how A(x) vries with x. Wht conditions does one need toput on f tomke sure tht A is continuous, or even differentible? The continuity prt of the question is esy to nswer. Lemm Let f,a be s bove. Then A is continuous function on [, b]. Proof. Let c be ny point in [, b]. Then f is continuous t c iff we hve lim A(c + h) =A(c). h Of course, i tking the limit, we consider ll smll enough h for which c + h lies in [, b], nd then let h gotozero. By the dditivity of the integrl, we hve (using (7.1.1)), A(c + h) A(c) = f(t)dt, where I(c, h) denotes the closed intervl between c nd c + h. Clerly, I(c, h) is[c, c + h], resp. [c + h, c], if h is positive, resp. negtive. When h goes to zero, I(c, h) shrinks tothe point {c}, ndso lim A(c + h) A(c) =, h which is wht we needed toshow. I(c,h) 89

5 The question of differentibility of A is more subtle, nd the complete nswer is given by the following importnt result: Theorem (The first fundmentl theorem of Clculus) Let f be n integrble function on [, b], ndleta be the function defined by (7.1.). Pick ny point c in (, b), nd suppose tht f is continuous t c. ThenA is differentible there nd moreover, Some would write this symboliclly s A (c) =f(c). (7.1.3) d dx x f(t)dt = f(x). In plin words, this sys tht differentiting the integrl gives bck the originl s long s the originl function is continuous t the point in question. Proof. To kno w if A(x) is differentible t c, we need toevlute the limit A(c + h) A(c) (7.1.4) L = lim. h h By the dditivity of the definite integrl, we hve (7.1.5) A(c + h) A(c) = f(x)dx, where I(c, h) is s in the proof of Lemm Denote by M(c, h), resp. m(c, h), the supremum, resp.infemum, ofthevluesoff over I(c, h). Then the following bounds evidently hold: (7.1.6) hm(c, h) f(x)dx hm(c, h). I(c,h) I(c,h) Combining (7.1.4), (7.1.5) nd (7.1.6), we get for ll smll h, (7.1.7) lim h m(c, h) L lim h M(c, h). But by hypothesis, f is continuous t c. Thenbothm(c, h) ndm(c, h) will tend to f(c) s h goes to, which proves the Theorem in view of (7.1.7). Let f benyfunctiononnopenintervli. Suppose there is differentible function φ on I such tht φ (x) =f(x) forllx in I. Then we will cll φ primitive of f on I. Note tht the primitive is not unique. Indeed, for ny constnt α, the function φ + α will hve the sme derivtive s φ. Intuitively, one feels immeditely tht the notion of primitive 9

6 should be tied up with the notion of n integrl. The following very importnt nd oft-used result mkes this expected reltionship precise. Theorem (The second fundmentl theorem of Clculus) Suppose f,φ re functions on [, b], withf integrble on [, b] nd φ primitive of f on (, b), withφ defined nd continuous t the endpoints, b. Then φ(b) φ() = One cn rewrite this, perhps more expressively, s φ(b) φ() = f(x)dx. d dx φdx. Proof. Choose ny prtition P : = t <t 1 <...<t n = b, nd set, for ech j {1, 2,...,n}, M j =sup(f([t j 1,t j ])) nd m j =inf(f([t j 1,t j ])). By definition, (7.1.9) n (t j t j 1 )m j j=1 f(x)dx n (t j t j 1 )m j. On the other hnd, the Men Vlue Theorem gives us, for ech j, numberc j in [t j 1,t j ] such tht (7.1.1) φ (c j )= φ(t j) φ(t j 1 ). t j t j 1 Since φ is by hypothesis the primitive of f on(, b), f(c j )=φ (c j )forechj. Moreover, (7.1.11) m j f(c j ) M j, nd n (7.1.12) φ(t j ) φ(t j 1 )=φ(b) φ(). j=1 Combining (7.1.1), (7.1.11) nd (7.1.12), we obtin n (7.1.13) (t j t j 1 )m j φ(b) φ() j=1 j=1 n (t j t j 1 )m j. Since (7.1.9) nd (7.1.13) hold for every prtition P, nd since f is integrble on [, b], the ssertion of the Theorem follows. j=1 91

7 7.2 The indefinite integrl Suppose φ is primitive of function f onnopenintervli. Itisnotunusulto following Leibniz, (7.2.1) f(x)dx = φ(x). set, This is clled n indefinite integrl becuse there re nolimits nd φ is non-unique. So one cn think of such n indefinite integrl s function of x which is unique only up to ddition of n rbitrry constnt. One hs, in other words, n equlity for ll sclrs C f(x)dx = f(x)dx + C. It could be bit unsettling to work with such n indefinite, nebulous function t first, but one lerns soon enough tht it is useful concept to be wre of. In mny Clculus texts one finds formuls like cos xdx =sinx + C nd 1 dx =logx + C. x All they men is tht sin x nd log x re the primitives of cos x nd 1, i.e., x d sin x =cosx dx nd d dx log x = 1 x. Of course the sitution is completely different in the cse of definite integrls. 7.3 Integrtion by substitution There re host of techniques which re useful in evluting vrious definite integrls. We will single out two of them in this chpter nd nlyze them. The first one is the method of substitution, which one should lwys try first before trying others. Theorem Let [, b] be closed intervl nd g function differentible on n open intervl contining [, b], withg continuous on [, b]. Alsoletf be continuous function on g([, b]). Then we hve the identity f(g(x))g (x)dx = g(b) g() f(u)du. 92

8 Proof. Let φ denote primitive of f, which exists becuse the continuity ssumption on f mkes it integrble on [, b]. Then we hve, by the second fundmentl theorem of Clculus, (7.3.2) g(b) f(u)du = φ(g(b)) φ(g()) = (φ g)(b) (φ g)(). g() On the other hnd, by the chin rule pplied to the composite function φ g, wehve (φ g) (x) =(φ g)(x) g (x) =(f g)(x) g (x). Consequently, (7.3.3) f(g(x))g (x)dx = (φ g) (x)dx. Applying the second fundmentl theorem of Clculus gin, the right hnd side of (7.3.3) is the sme s (7.3.4) (φ g)(b) (φ g)(). The Theorem now follows by combining (7.3.2), (7.3.3) nd (7.3.4). Before giving some exmples let us note tht powers of sin x nd cos x, swellspolynomils, re differentible on R with continuous derivtives. In fct we cn differentite them ny number of times; one sys they re infinitely differentible. The sme holds for rtios of such functions or their combintions, s long s the denomintor is non-zero in the intervl of interest. Exmples 7.3.5: (1) Let I = sin 3 x cos xdx. Thnks tothe remrk bove on the infinite differentibility of the functions in the integrnd, we re llowed to pply Theorem here, with g(x) = sin x nd f(u) =u 3. Then, since g (x) =cosx (s proved erlier, g() = nd g(π/2) = 1, we obtin I = 1 u 3 du =

9 (2) Put I = π/4 cos 2 xdx. Recll tht Since cos 2 x +sin 2 x =1,weget cos 2 x sin 2 x =cos2x. cos 2 x = 1+cos2x. 2 Using this nd the esy integrl π/4 dx = π/4, we get I = π 8 + J, with J = 1 2 cos 2xdx. Put g(x) =2x nd f(u) =cosu, which re both infinitely differentible on ll of R, nd use Theorem to conclude tht, since g (x) =2,g() = nd g(π/4) = π/2, This implies tht J = 1 4 π/4 cos udu = sin(π/4) sin ) 4 I = π = π = 1 4 2, (3) Evlute I = 1 1 x2 dx. Here we use the substitution theorem in the reverse direction. The bsic ide is tht 1 x 2 would simplify if x were sin t orcost. Put g(t) = sin t nd f(u) = u. Then g is differentible everywhere with g (t) =cost being continuous on [, 1]. We chose the intervl [, 1] becuse g() = nd g(π/2) = 1, giving us the limits of integrtion of I. Also, f is continuous on g([,π/2]) = [, 1]. (At the end point, the continuity of f mens it is right continuous there. This is good, becuse f is not defined to the left of.) So we hve stisfied ll the hypotheses of Theorem nd we my pply it to get I = f(g(t))g (t)dt = 94 1 sin 2 t cos tdt.

10 But 1 sin 2 t = cos 2 t = cos t, which is just cos t, becuse the cosine function is non-negtive in the intervl [,π/2]. Hence I = π/2 cos 2 tdt. We just evluted the integrl of cos 2 t in the previous exmple, lbeit with different limits. In ny cse, proceeding s in tht exmple, we get I = π 4 + sin(π/2) sin 4 = π Integrtion by prts Some consider this the most importnt theorem of Clculus. Its use is pervsive. Theorem Let [, b] be closed intervl nd let f,g be differentible functions in n open intervl round [, b] such tht f,g continuous on [, b]. Then we hve f(x)g (x)dx = f(x)g(x) b f (x)g(x)dx. Here f(x)g(x) b denotes f(b)g(b) f()g(). Proof. By the product rule, (7.4.2) (fg) (x) =f(x)g (x)+f (x)g (x) for ll x where f nd g re both differentible. Subtrcting f (x)g(x) from both sides nd integrting over [, b] we get (7.4.3) f(x)g (x)dx = (fg) (x)dx f (x)g(x)dx. But by the second fundmentl theorem of Clculus, (7.4.4) (fg) (x)dx =(fg)(b) (fg)(). The ssertion now follows by combining (7.4.3) nd (7.4.4). 95

11 Exmple 7.4.5: Evlute, for ny integer n, I n = cos n xdx. First note tht I = 1 dx = π 2 nd I 1 = cos xdx =sin(π/2) sin = 1, becuse sin(π/2) = 1 nd sin =. We hve lredy solved the n = 2 cse in the previous section using substitution, but we will not use it here. Integrtion by prts is more powerful! Sowe my suppose tht n>1. Put f(x) =cos n 1 x nd g(x) = sin x. Then, s noted erlier, f nd g re infinitely differentible on ll of R, with f (x) =(n 1) cos n 2 x ( sin x) nd g (x) =cosx, where the first formul comes from the chin rule. Now we my pply Theorem nd obtin I n = ( cos n 1 (x)sinx ) π/2 (n 1) cos n 2 x( sin x) sin xdx. Note tht n 1 sn>1ndcosθ sin θ is if θ is or π/2. Therefore the first term on the right is zero, nd we get Since sin 2 x =1 cos 2 x,weget I n =(n 1) cos n 2 x sin 2 xdx. I n =(n 1)I n 2 +(n 1)I n, which trnsltes intothe net recursive reltion I n = n 1 n I n 2. 96

12 In prticulr, nd I 2 = 1 2 I = π 4,I 4 = 3 4 I 2 = 3π 16,... I 3 = 2 3 I 1 = 2 3,I 5 = 4 5 I 3 = It will be left s nice exercise for the reder to find exct formule for I 2n nd I 2n 1. 97