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# CALCULUS Math 2A

UCI

GPA 3.73

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This 11 page Class Notes was uploaded by Adam Crona on Saturday September 12, 2015. The Class Notes belongs to Math 2A at University of California - Irvine taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/201864/math-2a-university-of-california-irvine in Mathematics (M) at University of California - Irvine.

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Date Created: 09/12/15

Chapter 5 Integrals In Chapter 2 we used the tangent and velocity problems to introduce the derivative which is the central idea in differential calculus In much the same way this chapter starts with the area and distance problems and uses them to formulate the idea of a de nite integral which is the basic concept of integral calculus 51 Areas and Distances I The Area Problem We begin by attempting to solve the area problem Find the area of the region S that lies under the curve y from a to 12 Area The area of a rectangle is de ned as the product of the length and the width the area of a triangle is half the base times the height the area of polygon is found by dividing it into triangles and adding the areas of the triangles Refer Figure 2 page 315 in the textbook To de ne the area of a region with curved sides we will rst approximate the region S by rectangles and then we take the limit of the areas of these rectangles as we increase the number of rectangles Example 1 Use rectangle to estimate the area under the parabola y 12 from 0 and ll Solution Suppose we divide S into four trips 51 5392 5393 and 54 by drawing the vertical lines I ix 2 and z 7 2 We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the trip Each rectangle has width i and the heights are i 2 22 and 12 3 If we let R4 be the sum of the areas of these approximating rectangles we get 1 l l l l 3 l 15 R 7 72 7 72 7 2 71270 46875 4 4442444 32 We see that the area A of S is less than R4 so A lt 0746875 4 Instead of using the rectangles whose heights are the values of f at the right endpoint of the subintervals we could use the smaller rectangles whose heights are the values of f at the left endpoints of the subintervals The sum of the areas of these approximating rectangles is l 2 l l 2 l l 2 l 3 2 7 L4740 474 472 474 73270721875 5 We see that the area of S is larger than L4 so we have lower and upper estimates for A 0721875 lt A lt 0746875 6 We can repeat this procedure with a larger number of strips By computing the sum of the areas of the smaller rectangles L3 and the sum of the ares of the larger rectangles R3 we obtain better lower and upper estimates for A 072734375 lt A lt 073984375 1 7 We could obtain better estimates by increasing the number of strips n Ln Rn 10 012850000 013850000 20 01308 7500 013587500 30 013168519 013501852 50 013234000 013383500 100 03283500 013383500 1000 013328335 013338335 TABLE 1 With 1000 strips the area lies between 03328335 and 03338335 A good esti mate is obtained by averaging these numbers A m 0133333351 D From the values in the table 1 it looks as if Rn is approaching as n increasei Example 2 For the region S in Example 1 show that the sum of the areas of the upper approximating rectangles approaches that is Solution Rn is the sum of the areas of the n rectanglesi Width 1n and the heights are the values of the function 1 7352 Rn g Each rectangle has I at the points 1n 2n3n nn that is the heights are 1n2 2702 3702 us Rn llt1gt TL TL 2 1 2 2 1 3 2 1 n 2 7 7 7 7 7lt7 TL TL TL TL TL TL 122232n2 1 122232n2 i nn12n1 39 6 Here we use the formula for the sum of the squares of the rst n positive integers 122232n2 Thus we have nn12n1 6 lim 1 I nn1 2n1 namn 6 n1 lt2n1gt hm 7 7 naoo n n 1 6 1 1 lt1gtlt21gt nam6 n n It can be shown that the lower approximating sums also approach that is 1 l39 L 7 n13 3 Thus we have 1 A lim Rn lim Ln7 neoo neoo 3 Apply the idea of Examples 1 and 2 to the more general region S that bounded by y and z a and z b we have the following De nition The area A of the region S that lies under the graph of the con tinuous function f is the limit of the sum of the areas of approximating rectangles A lim Rn lim fzlAz fzgAz new new b 7 where AI 7a n It can be proved that the limit in above de nition always exists since we are assuming that f is continuous It can also be shown that we get the same value if we use left endpoints A lim Rn lim fzoAz fzlAz fzn1Az new new In fact instead of using left endpoints or right endpoints we could take the height of the ith rectangle to be the value of f at my number in the ith subinterval zi1zi We call the numbers xix 1 the sample points So a more general expression for the area S is A liirgoRn We often use sigma notation to write sums with many terms more compactly For instance fzlAz fzgAz i1 So the expressions for area can be written as follows A lim ZfxiAI i1 A lim Z z mz i1 A lim Z zgmz i1 II The Distance Poblem Now let s consider the distance problem Find the distance travelled by an object during a certain time period if the velocity of the object is known at all times Example 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30second time interval We take speedometer readings every ve seconds and record them in the following table In order to have the time and the velocity in consistent units let s convert the velocity readings to feet lTimes l0l5l10l15l20l25l30l Velocity mih 17 l 21 l 24 l 29 l 32 l 31 l 28 TABLE 2 per second lmih 52803600 ft s During the rst ve seconds the velocity lTimes l0l5l10l15l20l lgol lVelocityftsl25l31l35l43l47l46l41l TABLES doesnlt change very much so we can estimate the distance travelled during that time by assuming that the velocity is constant If we take the velocity during that time interval to be the initial velocity 25ft3 then we obtain the approximate distance travelled during the rst ve seconds 25ft3 X 53 125ft Similarly the distance travelled from t 53 to t 103 is estimated as 31ft3 X 53 155ft If we add similar estimates for the other time intervals we obtain an estimate for the total distance travelled 25x531gtlt535gtlt543gtlt547gtlt546gtlt51135ft We could just as well have used the velocity at the end of each time period instead of the velocity at the beginning as our assumed constant velocity Then our estimate becomes 31gtlt535gtlt543gtlt547gtlt546gtlt541x51215ft The calculations in Example 4 is similar to that we used earlier to estimate the areas In general suppose an object moves with velocity v ft where a S t S b and ft 2 0 We take velocity readings at times to a t1t2 tn 1 so that the velocity is approximately constant on each subintervali If these times are equally spaced then the time between consecutive readings is A b 7 ani During the rst time interval the velocity is approximately ft0 and so the distance travelled is approximately ft0Ati Similarly the distance travelled during the second time interval is about ft1At and the total distance travelled during the time interval a b is approximately ft0At ft1At ftn1At Z mam i1 If we use the velocity at the right endpoints instead of left endpoints our estimate for the total distance becomes ft1At ft2At mnw imam The more frequently we measure the velocity7 the more accurate our estimates become7 so it seems plausible that the exact distance d travelled is the limit of such expressions d lim ZmHmt lim 21mm i1 i1 52 The De nite Integral De nition of a De nite Integral If f is a continuous function de ned for a S I S b we divide the interval ab into n subintervals of equal width AI bianl We let IO a 11 12 zn b be the endpoints of these subintervals and we let Ii 1 I be any sample points in these subintervals so lies in the ith subinterval II1 Then the de nite integral of f from a to b is b n mm niggoz zgmz 039 i1 Remark 01 Because we have assumed that f is continuous7 it can be proved that the limit in above de nition always exists and gives the same value no matter how we choose the sample points Although most of the functions that we encounter are continuous7 the limit in above definition also exists if f has a finite number of removable or jump discontinuouties but not infinite discontinuities So we can also define the definite integral for such functions The symbol f is introduced by Leibniz and is called an integral sign It is an elongated 5 an was chosen because an integral is a limit of sums In the notation ffzdx fx is called the integrand and a and b are called the limits of integration a is the lower limit and b is the upper limit The symbol d1 has no official meaning by itself f is all one symbol The procedure of calculating an integral is called integratingl The definite integral f is a number it does not depend on I In fact7 we could use any letter in place of 1 without changing the value of the integrals 1 bftdt 5mm Emma is called a Riemann sum after the German mathematician Bernhard Riemannl If f is positive7 then the Riemann sun can be interpreted as a sum of areas of approximating rectanglesl If f takes on both positive and negative values a definite integral can be interpreted as a net area Although we have defined f by dividing a z into subintervals of equal width7 there are situations in which it is advantageous to work with subintervals of unequal widthi If the subinterval widths are A11 A12 Azn we have to ensure that all these widths approach 0 in the limiting process This happens if the largest width7 max Ari approaches 0 So in this case the definition of a definite integral becomes The sum 12 n fzdz lim 2 fz Axi 1 m i1 aancIHO Evaluating Integrals Useful formulas to evaluate the integral nnl 2 H nnl 2nl 6 MS N 0 H H MS N w l WT H TLC H H n n E cai 5 al i1 i1 n ZWHHM Zai Zbi i1 i1 i1 2ai 7 12 Zai 7 Zbi Example 2 a Evaluate the Riemann sum for 13 7 61 taking the sample points to be right endpoints and a 07 b 3 and n 6 b Evaluate 0313 7 6zdzi Solution a With n 6 the interval Width is b7a 370 1 A 77 I n 6 2 and the right endpoints are 11 0512 1013 1514 2015 257 and 15 30 So the Riemann sum is 5 R5 7 Z xi Ax i1 f0i5 A1 fli0 A1 fli5Az f2i0 A1 f2i5 A1 f3i0 AI 1 7 E72875 7 5 7 5625 7 4 0625 9 7 739375 b With n subintervals we have 8 Thus 10 011 3nzg 6nzg 9n and in general I Sin1 Since we are using right endpoints we have 3 n n 3239 3 3 7 6 d l39 A l39 7 7 0ltI I I 22ng I 2012 n gt71 7 ltgt376ltgt 81 l 2 l 7 73133010757 7 271 81 27 7727776175 4 4 Properties of the de nite Integral When we de ned the de nite integral ffzdz we implicitly assumed that a lt 121 But the de nition as a limit of Riemann sums makes sense even if a gt 121 Notice that if we reverse a and b then AI changes from b 7 an to a 7 Therefore a b b fltzgtdz77 fltzgtdz If a b then AI 0 and so a 0 Properties of the integral 1 f cdz 01 7 a where c is any constant 2 ffl z gltzgt1dz 7 1 MW 7 15mm 3 f cfzdz cf fzdz where c is any constant 4 ffl z 7 wow 7 1 mm 7 15mm 5 1 mm ff mm 7 1 mm Example 6 Use the properties of integrals to evaluate 014 312dz Solution Using Properties 2 and 3 of integrals we have 1 1 1 1 int 4 312dz 4dr 312dz 4dr 3 IQdI 0 0 0 0 We know from Property 1 that 1 44141704 0 and we had evaluate that fol IQdI so 1 1 1 4312dz 4dz3 12dz4315 0 0 0 3 D Example 7 If it is known that 630 17 and f fzdz 12 nd 810 f Solution By property 5 we have 08fzdz810fzdz 0101ng 010 010 7 08fzdz 17712 5 Comparison properties of the integral 6 1f 2 0 for a S I S b then 2 01 7 1f 2 91 for a S I S b then f 2 fgzdzi 81fmS SMforaSzS b then b mbia S SMbia Example 8 Use property 8 to estimate 14 dzi Solution Since is an increasing function its absolute minimum on 1 4 is m 1 and its absolute maximum on 14 is M 21 Thus Property 8 gives 1471S4Esz2471 4 SS szG 1 53 The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus is appropriately named because it es tablishes a connection between the two branches of calculus differential calculus and integral calculusi Differential calculus arose from the tangent problem whereas integral calculus arose from a seemingly unrelated problem the area probleml New ton7s teacher at Cambridge lsaac Barrow discovered that these two problems are actually closely related The Fundamental Theorem of Calculus Part I If f is continuous on ab then the function 9 de ned by 91Eftdt agng us continuous on ab and differentiable on ab and g z Example 2 Find the derivative of the function 91 Eg xl thti Solution Since xl t2 is continuous Part I of the Fundamental Theorem of Calculus gives gz l 12 4 Example 4 Find 57 sec tdt Solution Here we have to be careful to use the Chain Rule in conjunction with FTCll Let u 14 Then 4 d E d u l sec tdt sec tdt d u du 7 sectdt du secui dz secz4 413 The Fundamental Theorem of Calculus Part 2 If f is continuous on ab then b Fb 7 Fa where F is any antiderivative of f that is a function such that F Part 2 of the Fundamental Theorem states that if we know an antiderivative F of f then we can evaluate fa simply by subtracting the values of F at the endpoints of the interval abi Its very surprising that f fzdz which was de ned by a complicated procedure involving all of the values of for a S I S b can be found by knowing the values of Fz at only two points a and 12 Although the theorem may be surprising at rst glance it becomes plausible if we interpret it in physical terms If vt is the velocity of an object and 3t is its position at time t then vt s t so 8 is an antiderivative of vi In Section 51 we considered an object that always moves in the positive direction and made the 11 guess that the area under the velocity curve is equal to the distance travelled ln symbols 1 vtdt 7 312 7 3a That is exactly what FTC2 says in this context We often use the notation FWl Fa FW So the equation of FTC2 can be written as b Fz where F f Other common notations are Fzl2 and Example 5 Evaluate the integral 612 Sdzi Solution The function 13 is continuous on 721 and we know from Section 410 that an antiderivative is Fz Z 14 so Part 2 of the Fundamental Theorem gives 21 7 F1 7 F72i14 7 i 724 7 7 Example 8 What is wrong with the following calculations 3 71 3 l I l 4 7d 7 7 7 1 7 1 12 I 71l1 3 3 Solution To start7 we notice that this calculation must be wrong because the answer is negative but 112 2 0 and Property 6 of integrals says that f 2 0 when f 2 0 The fundamental Theorem of Calculus applies to con tinuous functions It can t be applied here because 112 is not continuous n 71 3 ln fact7 f has an in nite discontinuity at z 07 so 3 l 7dr does not exist 1 z Differentiation and Integration as Inverse Processes The Fundamental Theorem of Calculus Suppose f is continuous on m 1 If 91 f fmdty then MI f1 2 f Fb7Fa7 where F is any antiderivative of f7 that is7 F We noted that Part 1 can be rewritten as d as g NW 7 fltzgt and Part 2 can be rewritten as b Fzdz F02 7 Fa

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