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by: Otilia Murray I
Otilia Murray I
GPA 3.88

Matthias Koeppe

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Matthias Koeppe
Class Notes
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This 22 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 021B at University of California - Davis taught by Matthias Koeppe in Fall. Since its upload, it has received 33 views. For similar materials see /class/187383/mat-021b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15
Calculus 21 B Lecture 53 Matthias K ppe mkoeppe21b math ucdavis edu April 6 2009 Organizational details Exams 0 All midterms and the final exam will be written inclass exams o Midterm 1 Friday May 1 in class 0 Midterm 2 Wednesday May 20 in class a Final Saturday June 6 at 1 pm a No calculators or notes are allowed The course website now includes an anonymous feedback form 0 Please let me know what you like and don t like and what should be improved 0 Thanks for the feedback already received rgence of Lower and Upper Sums Reminder o Reminder We proved for the function fX 1 7X2 overthe interval 0 1 the convergence of lower and upper sums when we use a partition of the interval into subintervals of equal length Note we had to work hard here use summation formulas for squares etc o It is mathematically not convincing to base this definition on a partition of the interval into subintervals of equal length 0 Instead we will be using arbitrary partitions subdivisions of the interval into subintervals and their associated Riemann sums 0 Then it is not sufficient to just have many subintervals For example in the partition 1 11 1 1 11 2 1 711 if we let n a 00 we cannot expect the lower and upper sums converge to the same value because the first subinterval always stays long Notati n for Riemann sums a When subdividing the interval ab into subintervals we choose n71 points X1Xn1 with aX0ltX1ltX2ltltXn1ltXnb We oallthe set P X07X17X27n7Xn71zxn a partition of 3 b a The norm ofthe partition P denoted by H PM is the largest length of a subinterval X0X1 X1X2 Xn1xn 0 So a small norm indicates a uniformly fine subdivision Towards the Riemann integral v The Riemann sum of a 39 quotcmquot partition P will be denoted by or In on Sp i fCkXk qu k1 n Z fCkAXk k1 Note that it depends also on the points ck k 1n FIGURE 59 The recmngles appmxlmme the legion beween the graph unite function y x and iluwaxis Note that if flakes positive and negative values the Riemann sum approximates the signed area between the graph and the Xaxis this is mathematically more natural and more useful If we are really interested in the total unsigned area between the graph and the Xaxis we need to break the function into pieces where the sign is constant 5 A remark about conv gence e We are talking about convergence of Riemann sums here but this is not just convergence of sequences of numbers or the limit of a function of one variable 0 So we cannot just use our notation lim S 777H0 P 0 Let s start from scratch and use an 87 5ster definition of convergence The Riemann integral DEFINITION The Definite Integral as a Limit of Riemann Sums Lct f39 70 a function dcfincd on it closed interval la b We say that u number is the definite integral off over 11 III and that I is the limit of thc Riemann sums Elf lflcil ANA il39thc following condition is sniisl icu iivcn any number 6 gt 0 thi is a corresponding number 5 lquot 0 such that forever pzmition P 39r of 1 II with lPll lti 53nd anychoicc of q in X i xiv we have lz flql AN 7 I 3911 6 11 In this situation a we use the notation lim fck AXkI llF llHOkgi a we say that the Riemann sums converge to the definite integral I a we call the function fRiemann integrable over a b The de nite Integral Notation H a number mm these properues exms we use We notanon tor n Th Muir i the integrand Upper lnml of unegmnon v is the variable a imegmmn When you nd um value m 39 x b Inmgml gn f dx a h 1quot mm H H I mmmed the ntegml We say the mtegrat trom a to b 0t flt dee x Integrable and nonintegrable functions Theorem Continuous functions are integrable If a function f is continuous on an interval a b then it is integrable over a b More classes of functions are integrable 0 increasing functions a discontinuous functions with only finitely many jumps Example of a function that is not Riemannintegrable over 01 due to Dirichlet 1 iins rational O iins irrational Show that upper and lower sums converge to different values More sophisticated notions of integration can integrate this function and give the answer 0 because there are way more irrational numbers than rational numbers This is the domain of measure theory a beautiful area of modern mathematics 539 Properties of definite integrals IAELE 53 Rules satis ed by de nite integrals a b 1 Orderuflmegmliun x dx if x 1x newan h a 2 Zero mdlhlmervnl x dx 0 Msm Dcmmnn a h n 3 Cammm Multiple kxlx k fxdlt m mmm a u I r xdxxdx u a n a b n 4 SumnndDi mnce x igrdx mom i gmdx 44 n n c E 5 Audizimy x dx um 10 a l u s Muvainlnezuulilys 1r has maximum value max 1 and minimum value min f on 1 b than 139 minfb in s xst maxfb 7a h b 7 Dumirmmm x 2 gx on 12 1 mm 2 gxdx u u h x 2 0 on 2 7 gt x dx 2 O rswmlusuv 5340 Properties of definite integrals a Zen w Interval fxdx 0 The area over a pain is n V y n u b b Comm Muln39ple hitxMx fbfumx Shawnfazk z from mm FIGURE 511 i minivibia5 max Smaxfbil y 3 10 m Areas add y fx f Domination X f 3 0H 11 172 0mg gum 534i Calculus 21 B Lecture 51 Matthias K ppe mkoeppe21b math ucdavis edu April 1 2009 O ganizational details 0 0 Slides are posted on the course website Lecture on this Friday will be given by my colleague Peter Malkin No office hours on this Friday afternoon I can make up for this next week it necessary Please enroll in the course on MyMathLab homework will be assigned starting this evening Homework continuous instructions a Skim the section of the textbook before class a Read the section of the textbook after class a Graded homework will be continuously assigned on MyMathLab lwill not announce this every time in class There will be one week of time for every homework assignment Check the due date in MyMathLab Grading based on a 20 Homework MyMathLab a 30 Midterms there will be 2 midterms I drop the lower grade 9 50 Final Exam Schedule to be announced Two similar examples Suppose we know the velocity vt given in mph say of a truck driving on a highway without changing direction Because the odometer is broken we would like to compute the distance traveled between times t a and t b given in hours say The Trigonometric Truck An Ordinary Truck The truck driver obsessed with The truck driver adapts the velocity trigonometric functions maintains according to traffic conditions on an otherwise empty road an The velocity vt is a function of exact velocity given by the formula time t but there is no formula for this function known However we can use the speedometer to determine the velocity vt at times t t1t2t vt 4010sint The Trigonometric Truck as an Antiderivative Problem 0 The velocity function vt is the derivative of the position function st 0 Thus 5t is an antiderivative of vt 0 We can also say that 5t is a solution of the differential equation 0 7s t v t d 0 Let s compute it from vt 4010sint 5t 4Ot710cost C This is the general solution of the diff equation Which constant C The distance traveled is the change in position D 51 7 5a 0 So it turns out that the constant C does not matter D 51 7 5a 40b710cosb C 7 40a710cosa C 4Ob7a710cosb7cosa a Heye hemhcuch m s hmgweh by accymma so we cahthyue down ah ant devwa ve a Let s subdmdeheume mevva an We h submtevva s 0Q equa ength A hem each Wena aha p okaume IZVWIM ch a We mad the speedomem a 01252 umes We assumetha the Ve oony mm b m 01 esscchaahwduhhg each submtevva a 80 we can ccmpme the amahce Uave ed ch each submtevva as Dmahce Ve oony gtlt Thhe The Ordinary Truck Approximating the Distance 0 In total we get an approximation of the distance D x vt1At vt2At vtnAt a We hope that by making the subintervals short enough the approximation will be good Ham s a mum x h loemno on ooovdma omo aveg on a What 5 n5 avea o m sweveamang e ova 0 wow w ma mm a magva oa ou us anudenva ves wmeu ustheansw v We cannot compute the area because of the curved boundary so let39s do something else that is simpler Let us approximate the area of the region by summing up areas of rectangular strips 9 Divide the interval into some subintervals XoyXllx Xszly Xzstly X3X4 not necessarily of equal length 0 Pick some point or from X0Xl pick some point 02 from page etc not necessarily the midpoint a We use a rectangle of height fci whose base is the interval X0Xi etc quot 05 U75 35 375 75 75 o The area of each rectangle is height x width so fci Xi 7X0 etc Then we hope Area H Area fci Xi 7 X0 fcz X2 7 Xi 03 39 X3 Xz 04 39 X4 Xa sue Riemann sums as approximations o The sum Area fc1x1i x0 f02 xz 7 x1 03 39 X3 X2 04 39 X4 X3 is called a Riemann sum 0 lntuitively clear is this we will make it precise later If we divide our interval into more and smaller subintervals then the approximation Area m Area should get better W o I o l V 0 o The textbook contains numerical examples that illustrate this and MyMathLab Figure 51 Animation lets you experiment with the subdivisions and will show you the value of the resulting Riemann sum g Lower and upper sums o A problem with these Riemann sums We don t seem to have control over the error we make in the approximation 9 The area of some of the rectangles overestimates the true area a the area of other rectangles underestimates the true area In total for now we can just hope that the approximation is close to the true area of the region but we don t know whether is above or below and what the error is 0 Here s a new idea After choosing the subdivision of the interval we make a specific choice for all the points 0 in the respective subintervals x1x a We can choose all c such that fc is the maximum function value on the interval 4 The resulting Riemann sum is called an upper sum a Or we can choose all c such that fc is the minimum function value on the interval x1x The resulting Riemann sum is called a lower sum 0 Then we have inequalities estimates Lower Sum g Any Riemann Sum g Upper Sum and Lower Sum g True Area of the Region g Upper Sum 5H0 Bounding the error If we have computed both a lower and an upper sum we have an upper bound for the error that we make by using the approximation instead of the true area So now the error is under control Error magnitude lTrue Area of the Region 7 Riemann Suml g Upper Sum 7 Lower Sum Note that we have an upper bound on the error magnitude without having to know the true area 0 However lower and upper sums may be hard to compute We need to find the minimum and the maximum of the function over all the chosen subintervals Still


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