SINGLE VARIABLE CALCULUS II
SINGLE VARIABLE CALCULUS II MATH 102
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Calculus 102 Lecture 2 Jeff Dajiang Liu June 7 2005 This is our second lecture I would like to talk about the inverse den39vatz39ve and the Riemann Sum which is the prelude of integration Integration will be our emphasis in this course First of all we would like to talk about the inverse of differentiation which is called antiderivativesiPlease look at the following differential equations dy i I 1 1 f gt lt gt Solving this equation is like the inverse operation of taking derivativesi You are supposed to go from the derivative to the original functionilf we have a solution 91 for l we then call it the antiderivative of Or according to our book we can de ne the antiderivative this way if F then Fzis called the antiderivative of f z But we can see for all the functions of the form Fz C their derivatives are also fx so we have now what the book called the most general Antideriva tives de ned by Cz Fz C where C is also a constant Let us take a look at someother examples 1 IS 312 then we have 13 is the antiderivative of 312 The notation is like f 312 13 C You should always remind yourself to use the most general antideratives when you compute these inde nite integrals 2 65 e from which we know 6 EerC You should never forget about the constantll 3 Well we just give you some derivative formulas you are then able to do integrationi When things are geting compicated you are supposed to remember the certain formulas very carefully in your mind and react quickly Such as given the polynomial 314 812 5 are you able to nd the antiderivatives of this polynomials Hint Use the power ruler Please remind yourself what the power rule is 1remind yourself which is a differential equation Now could you see the power of the formulas Yes learning math also requires some memory D but you are only supposed to know the following formula well other complicated ones are not required you can refer to the book when necessaryll Theorem 1 Some integration formulas k1 k 7 z zdzik1C 2 l cos kzdz E s1n k1 C 3 1 sm kzdz 7 cos k1 C 4 2 1 sec kzdz E tan k1 C 5 2 l csc kzdz 7 cot k1 C 6 l 7dzlnzC 7 z This is the formula you should always remember clearly for these formulas you can not always refer to the book For other complicated ones you can refer to the back or the front of the books Actually you can take those formulas as exercises try some of them Most of them could be got from our elementary formulas But some of those are not easy to get they are not require We have several application of the theorem above Please always pay atten tion that you should never forget about the constant at the end of the integral in the following exercises you should always remind yourself about this 1 nd the integral f 12 sin 31 z 2dz 2 nd the integral l5dz Pay attention here we need to introduce a trick called variable substitution Here is the way We can let u z 1 then we have du dz 1 dl0 so we have du dz that is what we want Then after the substitution we have the following integral fu5du that is exactly the formula 3 nd the integral f Wdz Hint variable subsititution again Sometimes people are given some initial value problem then you can get some speci c value for the constant C let us see the following examples 1 generally the initial value problem is given in the form of diffenrential equations eg solve the initial value problem 2 21 3 yl 2 2 Example the speed of the moving particles satis es the following func tional relations with time7 vt sint 3t7 the initial position 10 37 please nd functional relations between the position and the time Hint 11 Here we would like to summerize about the general procedure you should follow to do integration 1 See if the problem satis es some formulas given above or if they look similar The feeling about whether they are similar is very important 2 Can we do some simple variable substitutions to make them the same If a substitution can change the the problem to the formula up to a constant7 you should make it 3 Integration will largely depend on your experience7 so you should practice as much as you can This is much easier said than done Several more examples about initial value problem and integrations l 2 27 e 2 y0 3 2 mam 2 3f 1 l3dz 4f M iig dz 5Given cos 21 W could you nd the integral fcos2 zdz 1 6Could you explain why and are both antiderivatives of 1 7 could you tell me what this implies about the relations of the two anti derivatives Calculus 102 Lecture 15 Je Dajiang Liu July 12 2005 Review partial fractions integration by parts Today we are gonna starting a new topic How can you describe certain curves by parameters How can you classify curves more speci cally we are going to study what is called the Conics First of all let us look at several examples Example 1 nd the equation of a circle with certer 34 and ra dius 5 Example 2 nd the equation of the curve whose point is Pxy and the distance to A 11 lAPl is twice its distance to B 374 lBPl Remark Could you tell what curve this is from the equation only De nition 1 A conic curve is a curve formed by intersecting a Cone with a planethe cross sections l Example 3 please nd the locus of the points whose distance to point p 0 is equal to its distance to the line z 7p Example 4 please nd the locus of the points Ppy satisfying lAPl lBPl 2V5 Theorem 1 All the conics can be classi ed into 3 standard type whose standard form are the following 1see my live explanation on the board o Parabola yz kx o Ellipse 2 2 7 77 1 o Hyperbola 2 2 Remark of course7 You can see circle is a special case of Ellipse when a b Now it is the time for us to see a new way to denote these curves We can use what is called polar coordinates to describe these curves 1 can not give a very precise de nition of polar coordinates But I can explain it now The Polar Coordinate of a point is a pair r7197 where r is the distance from the point to the origin and 6 is the angle between line from 07 0 to P and the positive z 7 axis This is well de ned for the points other than the origin Let us try to convert between 2 di erent coordinate system Theorem 2 For the point Ly in usual coordinate system the po lar coordinate is given by r2 2 y27 tand a a Theorem 3 Given a point with polar coordinate r70 the xy 7 coordinate is calculated by z rcosd y rsin0 Example 5 Could you reparametrize the standard conic curve by polar coordinates Calculus 102 Lecture 6 Jeff Dajiang Liu June 15 2005 oday we are going to give a detailed proof of the fundamental theorem of calculus and its applications in the computation of de nite integralsi just give a quick review of what we talked about last time Statement of the theoremi Theorem lzfuudameutal theorem of calculus I if we have a continuous function de ned over a12 we can de ne the area function Fz ftdti Then Fz is the antiderivative of f That is Fz for z E a12i we can also use antiderivatives to evaluate de nite integralsi which is the second part of the fundamental theorem of calculus Theorem 22fuudameutal theorem of calculus U if Cz is my antiderivative of f on a12 then f C12 7 Ca Proof If we can see from inteval union property that F1 limhno ZTh ftdt if we can compute this we will be done By our inequalities we introduced in the previous class and the intermediate value property we know we can nd the point if such that ifgfrh t t Then we have Fl 1an0 i 1quot WW 1an0 m 7 11mm m 7 me Proof Hi refer to our previous notes about using inde nite integrals to evalu ate Riemann lntegralsi actually C12 F12 C f ftdt C Ca Fa C ftdt Ci So C12 7 Ca F12 7 Fa f ftdti Our Fz is the same as de ned in the statement of the theoremi According to our book sometimes fundamental theorem is interpreted as integration and differentiation are inverse processes which is i ftdt Also our second part of the fundamental theorem of calculus can be writ ten as f Gtdt Cz 7 Ca Several applications 1 we have cosz if z 2 0 and 1 7 z ifz 2 0 You are asked to nd the area of the Region bounded above by the graph and below by the Xaxis 2 nd the area of the region bounded up by 13 7 12 7 61below by Xaxis 3 how about 13 7 12 7 Grl 4 another example evaluate the integralff1 13 7 zldz Summerize the way of solving this type of problems 1 you need to nd the zeros of the given functions f 2 nd which area you need to compute Whether the sign of the area is positive or negative 3 according the formulas you learned using fundamental theorem of calcu lus part H nd the de nite integral Sometimes the signs need to be considered Find derivatives of certain functions using fundamental theorem of Calcu luseg hz fox t3 sin tdt By fundamental theorem of calculus the answer is 2 direct But what if you are given hz 01 t3 sin tdt Hint let u 12 then apply the chain rulell We can revisit initial value problem To nd the answer of the initial value problem like the following ya b We can just nd one of the antiderivative by fundamental theorem of calculus using Fz ftdt then let ybFz bfftdt Calculus 102 Lecture 11 JeffDajiang Liu June 267 2005 Today I am still going to talk about integration techniques We have already coverd Integration By Parts and variable substitu twmtoday we are going to study how to combine them when you see speci c problems So today7s class will be example intensive You are expected to learn from these examples more experience Very often7 you might see examples involving variable substitution of trigonometric functions This is a very special part of variable substitution7 you will see a lot of examples from the following The rst important property of trigonometric function is 5239an 005 1 1 This is sometimes used to extract things from Please try the following examples of dx of dx of leds of 1 dx 17m2 When you see some nonlinear things inside the square root you should consider using this type of information to extract things Sometimes7 there are also some applications for equation 17 and you also need the following sin x cos x 2 cos x 7 sin z 3 please look at the following examples 0 fsinsxcoszxdz o fcos5 sdz o fsin2 sdx o fcosz sdz YOU MAY FIND YOURSELF LACK OF HELP for the last 2 prob lems It may also seem to you that cos5 x is easier to deal with than cos2 x So I can remind you the helpful formula you SHOULD know sin 2x 2sinxcos 4 cos 2x cos2 x 7 sin2 z 5 We can look at the following examples 1 f dx 239 f sint cos3 t 3 fsin4dz Calculus 102 Lecture 14 Jeff Dajiang Liu July 77 2005 The nal is approaching7 and I am going to give you some exercise at every class7 and let you review what you have learnt in the past7 and be prepared for the materials in the nal Today I am going to start studying a new type of function7 which is called linear differential equations This is the hardest functions we are going to study in our course You should understand the integrating factor7 and know how to apply them in speci c problems De nition 1 A linear rst order di erential equation is one that could be written as y 7 P z z 1 ch lt gt21 Q gt lt gt Solving this is tricky You should pay attention to my steps now Here comes our rst example Example 1 d agiy zzy 2x3 1 dx You probably have no idea about how to do this at the rst glance Anyway you can rearrange the function to the form in Equation1 by dividing both side by 3 Only after you rearranging the terrn7 can you apply the integrating factor 6fPmdm You then multiply both side of the equation by You can verify7 you will have the following equality efpmdmy QW 6fPmdm if you integrate both side7 you will get yefPmdz C You answer will be we ff WW Qltzgtef Wen 0 lt2 Please see the following examples Example 2 d i By 2m 3w Example 3 CL 7 267 dx y i 8 Example 4 dig 7 2x emz dx y 7 Example 5 dy 7 dx y 7 You can also solve some initial value problems for this kind of linear differential equation Please work on the following examples Example6 dy 773 3 0 1 zdx y mm Example7 dy M 2w 10 72 Of course for different types of differential equations7 you could also consider initial value problem Example 8 7y 1 7 Cosx 7r 2 m y 71 Calculus 102 Lecture 19 Jeff Dajiang Liu July 21 2005 Let us see a great person Ramanujan7 who is superior in using series You can see the following formula 1 4n 110326390n 7 1 7r 9801 W0 nl4 3964 l It is almost impossible to imagine a thing like that A person with formal mathematics education will probably never be able to invent a creature like that At least I think And it really converges to the answer very fast So I type it out here Hope you enjoy it We can describe an in nite sequence like an ordered in nite list 017 027 037 7 am or we can have some more concise description for an in nite sequence like anff1 or simply can You have a lot of ways to de ne an in nite sequence7 like the following 1 explicitly enumerate 1234757 6 2 use functions where f z 3 use recursive method fn fwl 1 Well7 I just gave you an trivial sequence7 but do not think real sequences are trivial Let us see harder examples F I 39 1 F L a j fn which may be de ned by f117 f217 and fn1fnfn71 fOTnSQ Example 2 Look at this recursiuely de ned sequence 11 67 an1 V6an Please write out the sequence explicitly The basic thing we consider about in nite sequence is its limit We have the following de nition De nition 1 The limit of the sequence an is L denoted by lirn an L Hoe if and only iffor any 6 gt 0 there eccists an N such that for every ngtN laniLllte Example 3 Please nd the limit of the following sequence by any means you know 1 Slnn an7 n 2 q 1 an nsini n 3 i n33n22n1 L7 3z35z26z7 4 n10 an7 671 De nition 2 in nite series of the form Zana1a2an n1 De nition 3 For an in nite series we can de ne the partial sums You can see the partial sums Sn is an in nite sequence if the limit CCElSt we say the in nite series converges We will see several special examples Example 4 The geometh series is de ned by a0 a7 an Tanil 2 0 37 For geometric series we have the following theorem Theorem 1 The geometric series converges and the sum is a 5ar 17 In rare cases could you nd out the exact answers to an in nite sequence most of the time you can only judge whether it converges or not You can not nd out the explicit answer for a given in nite sequencell We have several theorems to follow Theorem 2 The nth Term Test for Divergence if either 113an t 0 or the limit does not CCElSt the in nite series with general term an diverges People like series better than other types of functions so they seek ways to express every function using series but it is not possible for every function to have a series expression But we still have the important theorem here Theorem 3 The nth degree Taylor polynomial for fd is de ned by r fWa k z7 aK RM k 0 431an we Calculus 102 Lecture 12 Jeff Dajiang Liu June 307 2005 today we will move on to introduce some new integration tech niques Very often7 we will need to integrate a very special type of function which is called rational functions Here is the de nition De nition 1 A rational function is the function of the form Pz QW where Pz and Q are both polynomials Rz The way we will apply is called method of partial fraction De nition 2 Method of Partial Fraction The method ofpar tial fraction is the algebraic technique that decomposes R into a sum of terms 39 PW QW where p is a polynomial and is afraction that can be inte grated without di culties Rz p F1z F2z Fkz7 1 We know from basic algebra that the only inreducible polynomials with coef cient in R are quadratic and linear polynomials Believe it or now7 Using this principle7 we can show every rational function can be written as equation 17 with being either A W 2 or BpC ax bx c 3 Fractions of these forms are called partial fractions Equation1 is called partial fraction decomposition We need to learn to deal with the rst type of partial fractions now We need a de ntion rst De nition 3 A fractiori is called proper if arid only if the de gree of Pz is strictly less than that of The harder part is to deal with proper fractions7 but now we need to deal with fraction that is not proper The technique is called long division Please see the following examples Example 1 Please nd f 215195 23 z2d Example 2 please ridf m 1 z So I need you to remember now7 the very rst step to integrate a rational function is to apply long division to reduce the improper fraction to proper fraction Then we need to factorize the denominator As we just said7 the only inreducible polynomial is quadratic and linear polynomial7 so you should express the bottom as the product of linear and quadratic lRREDUClBLE polynomials The rst rule you should apply is Rule 1 The part of the partial fractiori decompsitiori that corre sporids to the liriear factor ax b with multiplicity n is A1 A2 An axbaxb2quot39axb 4 of course the capital Ai is coristarits Example 3 please nd fmdz Example 4 please nd f 3 z23z2d39 Here you will have to FACTORIZE the denominator yourself Let us see another more dif cult problem Example 5 please nd f The O 4m273m74 13z272z d idea is you can list the factors according to rule 17 and then you need to nd the common denominator of these partial fractions which is of course the original denomina torbelieve or not7 then write the partial fractions using this common denomina tor 7 arrange the terms and match the coef cients with that of the original numerator7 you will get a list of simultaneous equa tions solve them to get the coef cient Ala integrate it Example 6 please nd x21 d x3423 z Example 7 please nd x3 dx x33z23z1 Example 8 please nd x374x71 d z 713 z Example 9 please nd 3 7 4x 7 1 da 3 x2 7 2x Rule 2 The part of the partial fraction decompsition that corre sponds to the quadratic factor of bx c with multiplicity n is B1 Cl Bg 02 Bni On 5 az2bc ax2bzc2 ax2bxc of course the capital B1 and Cl is constants Here sometimes you will see very complicated examples if you were given some very large multiplicities for the quadratic factor I was tortured that way when I rst learnt this But you can be assured it will not happen to you What you need to pay attention here is that the numerator is a linear polynomial instead of a constant Do not confuse yourself with the linear factor case Here is several examples for you to do Example 10 5z373x2271 4 2 Example 11 a d 96 12 1 Example 12 z 4 d a 3 4x As the book suggests7 you might consider using the following for mula7 the reason I put them here is to help you do problem faster But I will assume you can nd it yourself when you see them in the problem dz 1nx2 k2 O 1 1 mdx Earctang C Calculus 102 Lecture 4 Jeff Dajiang Liu June 127 2005 We are trying to evaluate Riemann Integral by using inde nite integration formulas instead of using Riemann Sum1 You will nd your life much easier soon1 Following the book7 we introduce an area function here1 141 ftdt1 By looking at the graph7 you can see the change of the area is approxiamately just the height times A11 Written out in equation7 A14 m So taking the derivative of 141 by which is just From this7 and the definition of Antiderivatives7 we can see 14 141 is the antiderivative of We know if C1 is another antiderivatives of 141 C1 C1 Al though the antiderivative itself has many different forms differs by a constant there diffenrence is well defined We can see 14a ftdt 07 the area of a line is naturally considered to zero And 141 fl7ftdt7 so we have f ftdt 141 7 14a C12 7 Ca It is independent of which repre sentative of antiderivative you choose1You can simply let your constant in the indefinite integral be zero7 and evaluate the definite integral in the above way We have the following theorem1 Theorem 1 Evaluation of Integrals C1 is an arbitrary antiderivative of a continuous function on the in terval an then f C12 7 Ca We will use the following notations in our evaluation of Riemann Integrals b b mm 7 Gem 7 602 7 0a lt1 We have several examples here 11 0313d1 21 OW cos 1d1 and How about fog cos 1d1 3i evaluate ff xz 3dr Continued from last time we still start from evaluating some de nite inte grals for our class You should be able to reproduce them after class Repro ducing the solution is a good way for you to know if you really understand the ideas involved in the computation First of all remind yourself what the Riemann integral is what the geomet ric meaning is Secondly for a integral like f fzdz what should you do to them to evaluate Take a concrete example could you evaluate the following integrals ll OW xSz 641 COS5IdI 2 How are about a function de ned like l31lz 6 711 From the above 2 cases we want to give out some general theorem in eval uating integralsi Theorem 2 trivial case what if you integrate a constant like f cdz H i the linear propertyi f Bgzdz Affzdz B fgzdz to the inteval Union Property if we have a lt c lt b then f mm ff mm 9 4 Comparison Property 1 if S 91 for all z 6 ab then f S f gzdzi 5 Comparison Property H ifm S S M then ma 7 a S f S Mb 7 a We would like to emphasize the evaluation of integrals in our course We need to address more examples here 1 ff sinz cos zdz 2i ff sin2 1 cos zdz 3i OW cos6z Calculus 102 Lecture 17 Je Dajiang Liu July 13 2005 First of all7 let me give the de nition of the parametric curves De nition 1 A parametric curve C in the plane is a pair offunc tions 96 N y 900 That gives cc and y coordinate the parameter t where ft and gt are continuous functions of the real number t in some interval I We have already seen examples about parametric curves eg z cost7 y sin0 could you remind yourself what curve this is For the same curve you can di erent parameterizations7 they may be similar They may also be totally di erent Example 1 z cost97 y sin0 is a circle But how about z cos26 y sin26 the answer is yeah this is also a circle But they look similar anyway Believe it or not 1 i t2 2t z 1 t2 y 1t2 is also another parametrization of a circle How can you verify that Actually7 my research deals with parametric curves in algebraic geometry7 which still has a lot of unsolved problems with parametric curves We are trying to nd what is the naturalparametrization of an implicit equation og 0 You should pay attention z cosat97 y sinb6 when a 31 b is of crazy shape7 you can nd the picture in gure 944 in our textbook We have some silly ways to nd parameterized curves like this The curve you get is useless however If you have y x7 you can trivially let x t then you have the following parametrization 96 t y f t 7 you can verify that with our de nition You can also eliminate the parameters from the parameterized curve Example 2 xt71 y2t274t1 092 In our calculus course7 derivative is the most important de nition7 you may want to nd the tangent line of the parameterized curve by means of differentiating the parameterized curve Here is the way Theorem 1 The slope of the tangent line of the curve is given by dy dx dt dt I will give a pretty straight forward proof based on the famous Chain Rule You need to know how to do that in speci c problems Example 3 nd the tangent line of the curve through speci ed points in the following examples z2t21 y3t32 t1 7T zcos3t ysin3t tZ 7r z tsmt y ztcost7 t 5 Calculus 102 Lecture 7 Je Dajiang Liu June 16 2005 After nishing fundamental theorem of calculus We have already nished all the de nitions and theories of our course Basically our course could be named theory and quot quot of quotm 39 39 and39 t quot 77 Now all of our material in the rest of the semester will be applications of integration plus some necessary techniques The de nition of integration came from the computation of areas of the regions the graph of certain functions You can of course expect the rst application be the calculation of areas Today we will start talking about the computation of the area of the region got from the crosssecitons You will learn the material from various kinds of examples The rst application is nd the area of the region between 2 curves Here is the de nition De nition 1 and 91 are 2 continuous functions with 2 91 for 1 6 ab Then the area of the region bounded by the curves 9 and y 91 and by the vertical lines 1 a 1 b is 7 91d1 Example 1 nd the area of the region bounded by the lines 9 1 and 1 2 and by the curve 9 Example 2 nd the area of the region bounded by y 1 and y 3 7 12 The solution of the above 2 examples are exactly following the de ntion of The Area Between Two Curves Let us summerize the way of dealing with this kind problems 1 nd the bound for the integationi Namely sometimes you will need to nd the intersection of 2 curves by using simultaneous equations 2 nd which function is greater PS this is important in order to avoid negative answers 3 integrate by A 7 91d1 Sometimes7 we need to use union interval property of de nite integrals to nd the area of certain graphi As we have shown last time7 if you have a function which has different de nitions over different intervals7 you will need to integra tion separatelyi Here is a reVisit of the problem we worked last time Example 3 if we have a function like the following cosz I S 0 171 120 Example 4 Find the area of the region bounded by the line y z and the parabola y2 8 7 z The above is a tough example You have to either consider the function sepa rately in 2 intervals or look at the graph the other around Here follows our second de nition De nition 2 The area between 2 curves Letf and g be 2 continuousfunc tions of y with 2 gy for y 6 ad Then the area bounded by the horizontal lines y c y d and by the curves 1 z gy is A 1m 7 may Example 5 Redo the previous example using the new way introduced in our previous de ntion Calculus 102 Lecture 16 Je Dajiang Liu July 13 2005 Review Linear First Order Equations7 and several integration by parts Today we are going to study several types of curves in polar coordi nates7 and learn how to compute the area by using polar coordinates First of all7 we are going to study the most basic curves in polar coordinatesiC39irele Here is an example Example 1 Change the following polar coordinate equation to dy coordinate 1 r300s0 r 4 sin 6 Example 2 please nd what the general curve is O r 2acost9 r 2asin6 We now start talking about how you can compute the area using the polar coordinate Theorem 1 The area of the region bounded by the lines 6 046 and the curve r f0 is given by 12 AiDtirdd 1 Please pay attention to what I will explain on the board7 about how you can interpret this formula Please do the following example Example 3 nd the area of the region bounded by the curve of equa tion r 3 2 cos 6 where 7 0 S 6 S 27139 Example 4 nd the area of the region bounded by r 3 cos 6 There are very weird polar curves7 you can consult the book on page 642 We will not waste time on this It seems useless in your future study Before we start parametric curves7 I would like to state another theorem Theorem 2 The area between 2 curves r f6 and r g6and 2 lines 6 oz and 6 B is given by integral A giro e i9lt6gti2d6 Example 5 nd the area between r 4 and r 2 cos 6 Calculus 102 Lecture 3 Jeff Dajiang Liu June 9 2005 Here we are going to talk about Riemann Sums and De nite Integralsi The idea of de nite integrals comes from the calculations of areas of the region below of the graph of certain functions Let us take a look at the following function 12 which is a curve called the parabola 1 We just need to look at the graph within the interval 01 for our com putational convenience we can divide the interval into n subintervals of equal length These intervals are denoted by 10 11 11 12 12 13 i i i 171 In all with length Actually we will see we can divide the interval arbitrarily according to our own preferencei If the maximal length of these subinterval is arbitrarily small the limit of the Riemann Sum will turn out to be the same As we can from the graph on the blackboard the inscribed rectangle has height fzi1 and the circumscribed rectangle has height Adding these inscribed triangle together we have An ELI fzi1Azi And adding these circumscribled triangle we have A ELI fziAz we can denote the actuall area of the region by A then we have the following inequality An S A S Ani Of course this is because the function is increasing Question if we have de creasing functions what will happen to these inequality If n happens to be large then the difference between AAnA is going to be small then we can actually compute the actual area A by taking limit of An A limnnm21fzi1Az limnn00 The two limit will turn out to be equal if the function is integrable We can compute the limit n 3 nn12n1 5 i nowi We will need the following formula Ei1 2 After this we can introduce the formal de nition of Riemann Sum nowi DEFINITION Riemann Sum let be a function de ned on the interval abi if P is a partition of ab and S is a selection for P then the Riemann Sum for fdetermined by P and S is R ELI where lies in II1 1sorry I will have to use handdrawn picture so far I will try to learn how to add pictures to my ETEX le as soon as possible From the de nition you can see actually it has nothing really new it is just a summation process And also geometrically it is the area of the region under the graph of certain functions There is another example of the Riemann Sums Please nd out An and A for 31 de ned over 01 Here please remember the selection of is just the left end or the right end of the subinterval Now we come to the de nition of De nite lntegrali This is the center of our course Most of our subsequent topics will be around the computation and application of de nite integrals De ntion The De nite lntegral The de nite integral of the function f from a to b is the number 1 lim Z zgmz lPHO 11 If the limit exists we say f is integrable on abi Of course you need to remind yourself of the e 7 6 language For each number 6 there will exist a 6 which will depend on e such that lli 21le lt 6 i1 for each partition P of the interval a b for which lpl lt 6 The usual notation we will take is f There is an important theorem about which function is integrable Namely it is concerning which function can make the above limit existi Theorem 1 Existence of lntegrali If the function is continuous on the interval a b then f is integrable on a 12 Of course a lot of functions we see at present are continous ones Like the sin and cos function the parabola and so on But there is also a lot of func tions that are not continuous like a fuction de ned like this 3 when I lt l 4 when I 2 1 But you should pay attention here this function is integrable which means the area under the graph is computablei This is of course truer Right But there are some weired functions mathematicians created which is not continuous and also not integrable Like the following function xz 1 when I E Q xz 0 when I g It is not integrable Because for every subinterval you use in the Riemann Sum you will have 2 choices for your selection One is choose 1 such that 1 or you can choose such that 0 So you will have 2 limits of the Riemann Sum over interval a by One is b7 a The other is 0 So The limit does not exist It is to say the function is not integrable Generally people will consider 3 types of functions continuous diffrentiable and integrable We now now diffentiable function Q continuous continuous Q integrable Actually there are even more weired functions which are continuous everywhere but not differentiable everywhere Also there are functions which are integrable but not differentiable Do you knowHint It appeared previously in this notes Later people tried to improve Riemann lntegration which is the limit of Rie mann Sums to what is called the Lebesgue lntegrationi Then the Lebesgue integal of xz over ab is always 0 A lot of problem unsolved in Riemann lntegration can be solved with Lebesgue lntegrationi We will not talk about this in our class The text book contains some examples about how to evaluate Riemann ln tegrals by taking the limit of Riemann Sumsl This is not necessary Actually it will be trivial after we learn how to evaluate Riemann lntegral by Inde nite lntegral from our last class So we will skip themi
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