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# Calculus I MTH 132

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This 4 page Class Notes was uploaded by Donny Graham on Saturday September 19, 2015. The Class Notes belongs to MTH 132 at Michigan State University taught by John McCarthy in Fall. Since its upload, it has received 24 views. For similar materials see /class/207307/mth-132-michigan-state-university in Mathematics (M) at Michigan State University.

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Date Created: 09/19/15

Supplemental Material for Section 53 The De nite Integral We now turn our attention to the second aspect of Integration Theory the De nite Integral The motivation for this part of the theory is the area problem However the applications of the de nite integral are numerous and go well beyond nding areas as will be seen from subsequent study The area problem can be stated as follows Let f be a function de ned on a closed interval a7 12 with 2 0 for all z E a z Let R be the region in dicated in the picture to the left Formally R is de ned by R 0 I I6 Iaybl andOSy fz and is called the region under the graph of The area problem is to de ne the area of R and to discover an easy method to compute it in a large number of cases We begin with several examp es Example 1 Let p be any positive real number and let a7 12 be any interval Set p for all z E a z Then the region under the graph of f is a rectangle of length I 7 a and of height p Consequently the area of the region R is 101 7 a As simple as this example is7 it is of fundamental importance in dealing with general area Example 2 Let 21 for z E 02 Then the region under the graph of f is a triangle with base equal to the length of the interval 07 2 and height 4 So the area of the region R in this case is 4 4 Example 3 Let 4 for z E 14 In this case the region under the graph of f can be seen to be composed of a rectangle surmounted by a triangle So the area can be computed as E The details are left 4 to the student Example 4 Let for z E 724 Here the the region under the graph of f is composed of two triangles It is easy to see that the area of the region is 2 8 10 Example 5 Let l for z E 732 Here the the region under the graph of f is composed of two triangles and a rectangle Clearly the area of the region is Example 6 Let V9 7 12 for z E 73 3 Then the graph of f is a semicircle and the region under the graph of f is the region inside of this semicircle Thus the area of this region is 32K 3 For all of these examples the area under the graph of the given function could by found using known geometric results However for the region under the graph of 12 for z E 02 no such simple solution is possible Instead we approximate the region by a sequence of regions whose areas we can compute by geometric methods and investigate what happens as the sequence of approximations gets better and better Speci cally we will approximate the region by rectangles We will rst describe the process for an arbitrary function f de ned on an arbitrary interval a7 12 with 2 0 for all z E a z We begin by dividing the given interval into two subintervals of equal length In the rst select any number7 517 at random and in the second select a second number7 52 at random Then the area of the region R under the graph is approximately I 7 a b 7 a b 7 a 2 f 52 2 2 f61 f61 f62A This approximation isnlt particularly good but it can be improved by partitioning the interval a 12 into three subintervals of equal length and selecting a number at random in each of the subintervalsi Label these three numbers as cl Cg and Cg respectively Then the area of R is approximated by b7a b7a b7a7b7a 01 3 f02 3 f53 3 i 3 f61 fC2 fCsA This approximation is a slight improvement over the previous one but still needs to be re ned So next the interval is divided into four subintervals of equal length and a number selected at random from each of the 7 a 4 g og fltCggt fC4 1s a better approx1matlon of the area of Rf Each time the number of subintervals is increased the resulting approximation to the area of R improvesi So to investigate the general situation let n be any positive integer and imagine partitioning the as The left four intervalsi Then the corresponding sum interval ab into n subintervals of equal length The common length of each of these is then n i The left endpoint of the second n b 7 b 7 b 7 b 7 a and its right endpoint is a a 7a a 2 ai Continuing in this fashion n b 7 b 7 a and right endpoint a k a b 7 endpomt of the rst 1nterval is of course a while 1ts r1ght endpomt 1s a a subinterval is a we see that the kth subinterval has left endpoint a k 7 l n element is then selected from each of these subintervalsi The one selected from the kth interval is denoted by ckiThusckE ak7lb b a a 7 a k7 i The area of R can be approximated very well if n is very large by i An arbitrary TL TL b afltc2gtb afltCngtb ab a TL TL TL TL f61 f61 f62 f6n To shorten the amount of writing we introduce the following notation Zfck 7 fc1 fc2 an 161 As n gets larger and larger this approximation will get closer and closer to the area of the region under the graph Consequently the formal de nition of this area is as follows De nition 7 Let f be a function de ned on a closed interval ab with 2 0 for all z E a 12 Then the region under the graph of f has area means there is a number A such that lim bia W n gm 7 where A doesn7t depend on the choice of ck E a k 7 lb 7 aa kbiia i n n The following gures demonstrate how the approximation to the area improves as n increases a actanglashaw signed area 517969 3 actanglashaw signed area 5232a2 d d 3 3 2 2 72 715 71 705 05 1 72 715 71 705 05 1 12 actanglashav signed area 52a219 16 actanglashav signed area 52561 To illustrate the de nition we will show that for the case of a triangle it gives us the expected answeri Example 8 Let mm for z 6 012 where 0 lt b and m gt 0 Let n be any positive integer In this case for each 16 1 2 n we have that ck 6 12 Because f is increasing on 012 for each 16 1 2 n we have that 721 S g milk Hence b n 1671 b n b n 16 me n b2 n me n 72 lt72 lt72 7 lt lt n m n b7 nk1f6ki nk1mnb or n n 161 Di n k1fCk7 n n k k 1 161 1 1 To compute the two sums 221k 7 1 and 221 16 we use formula 221 16 See page 311 of the text for this formula and others of a similar nature By direct substitution the right hand side of equation 1 is meLQMi To compute the left hand side note that 22116 7 1 16 Thus the left hand side n 2 n 71n 2 1 of equation 1 is 12272172 1 Consequently 1 can be rewritten as n 1n71n b n 1nn1 2 2 b 57 S 21100 9 57 161 Computing the limits as n tends to 00 in the same way that we did for functions and z 7gt 00 we see that n 7 n nn 1 b n 2m hmnH00 72 1 and llmnH00 72 1 So by the Sandw1ch Theorem llmnH00 7 2161 n n n b2 Hence the area under the graph is exactly what we expected namely We will now use essentially the same technique to nd an area that can t be computed by simple geometryi Example 9 Let 12 for z E a b where for ease of calculation let 0 S a Let n be any positive integer For each 16 12iiin let ck E a 16 71 previous example we have aa16 n a Because f 1s 1ncreas1ng on ab as 1n the biaiak71b77a2gbgaifckgbaiak 2 161 161 TL 161 We will compute the sum on the right hand side of equation 2 and leave the computation of the left hand side as an exercise for the reader n b7a27n 2 b7a 2127a 7 2 2ab7a n b7a2n 2 2ak7 a 2ak7k 7 am kaq7 2k 161 161 161 161 To complete this calculation we will once again use the formula used in Example 8 but we also clearly need a formula for 221 162 It is 221 162 Thus b7a and evaluate the limit of each term on the 7 a2ab7 a nn 1 n 2 TL To nish this part of the calculation we multiply equation 3 by n b 7 right hand side First we have limnn00 anaQ b7aa21 Next limnn00 ab7a2 n b 7 b 7 2 1 2 1 b 7 3 and lastly limnn00 7a 7a w 1 Now adding these three and doing the algebra n b 7 b 7 2 1 a 7a 023 7 as Doing a similar calculation you will get that n TL n 21lta 1 b 7 2 1 limnn00 1 77 221 lta k 7 1 g b3 7 as Thus again by the Sandwich Theorem the area under the n we get that limnn00 1 graph of z2 on the interval a b is 023 7 a3 A moment7s thought will convince you that the only place in the above presentation where the assumption of 2 0 was used was in the motivation If we ignore that we started out to de ne area and drop the assumption 2 0 we see that it is still possible to write down the same de nition The resulting concept is called the De nite 1ntegral or the Riemann integral of the function The number that results from the de nition is called the Riemann integral of f from a to b and is denoted by f dzi The sign f is an elongated letter S and is to remind us that the number is obtained by computing a sum and then taking a limit So to be speci c here is the de nition De nition 10 Let f be a function de ned on a closed interval abi Then f is Riemann integrable on a b means there is a number denoted by f dz such that b b 7 a n mm 7 hm EM 1 k1 n7gtoo n b7a7akb 04 TL TL regardless of the choice of ck E a k 7 1 i The number f dz is called the De nite or Riemann integral of f from a to 12 It has a geometric interpretation in terms of the region between the zaxis and the graph of f namely f dz is the area of that part of the region lying above the zaxis minus the area of that part of the region lying below the zaxis To compute the approximations to f dz you use exactly the same method as you did above In particular for 0 S a lt b we have f z2 dz b3 7 as To interpret f dz geometrically rst visualize the region between the graph and the z axis Then f dz is the area of that part of the region above the z axis minus the area of that part of the region below the z axis The following properties of de nite integrals follow directly from the de nition I fltzgt dz 7 7 ff mm 7 f M dz 7 0 ff km dz 7 k ff 161 dz for any number k ff at 7 91 dz 7 ff mm 391 dz ff mm f fzdz 7 f fawn Ifm fz g M for all z in M then mb 7 a g ff 161 dz 3 M0 7 a H to 9 hu 5 on 7 1fgz S for all z in ab then fgz dz S f dz Exercises 1 Let Compute 1 77 221 for each of the following choices a ab 15 n 4 and ck is the left endpoint of each subintervali b ab 15 n 4 and ck is the right endpoint of each subintervali c ab 15 n 4 and ck is the midpoint of each subintervali

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