# Numerical Methods of Approximating Definite Integrals

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1 6 C H A P T E R Numericl Methods o Approimting Deinite Integrls 6. APPROXIMATING SUMS: L n, R n, T n, AND M n Introduction Not only cn we dierentite ll the bsic unctions we ve encountered, polynomils, eponentil nd logrithmic unctions, trigonometric nd inverse trigonometric unctions, nd rtionl unctions, but rmed with the Product nd Chin Rules, we cn hppily dierentite ny new unction constructed by multipliction, division, ddition, subtrction, or composition o these unctions. This gives us sense o competence nd stisction. Although t this point we cn integrte mny unctions, there re bsic unctions such s ln nd sec ) tht we hve not yet tckled. From the Chin Rule or dierentition we get the technique o substitution or ntidierentition; rom the Product Rule or dierentition we get technique o ntidierentition known s Integrtion by Prts. The ltter is something to look orwrd to lerning in Chpter 9.) Lerning more sophisticted methods o substitution nd lgebric mnipultion will enlrge the collection o unctions we cn ntidierentite. Tbles o integrls nd high-powered computer pckges cn provide ssistnce i we re in dire strits. Nevertheless, there re some very innocent-looking unctions tht cnnot be delt with esily. For instnce, ll the technicl skill in the world won t help us ind n ntiderivtive or e, or sin ), or sin ). Knowing tht there is no gurntee tht we cn ntidierentite cn be unnerving. This chpter will restore our sense o hving things under control when we re ced with deinite integrl. Suppose we re interested in evluting deinite integrl nd we hve not ound n ntiderivtive or the integrnd. In ny prcticl sitution we ll need to evlute the deinite ln d = ln + C veriy this or yoursel). This cn be ound either by some serious guess-work, methods given in Section 7., or using the technique clled Integrtion by Prts. With some work we cn integrte sec. sec d = sec +tn sec sec +tn d = sec +sec tn sec +tn d = ln sec + tn +C. Tht is, i we wnt to obtin n ntiderivtive tht is inite sum, product, or composition o the elementry unctions. 805

2 806 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls integrl with certin degree o ccurcy. Provided tht n pproimtion is stisctory, it is not necessry to be ble to ind n ntiderivtive o the integrnd. Insted, we return to the bsic ides tht led us to the limit deinition o the deinite integrl. The theoreticl underpinnings o clculus involve the method o successive pproimtions ollowed by limiting process. Dierentil Clculus Let be dierentible unction. We obtin numericl pproimtions o the slope o the tngent to the grph o t point P by looking t the slope o secnt lines through P nd Q, where Q is point on the grph o very close to P. By tking the limit s Q pproches P we determine the ect slope o the tngent line t P. y = ) Q P = = b Figure 6. Integrl Clculus Let be n integrble unction. We obtin numericl pproimtions o the signed re between the grph o nd the -is on [, b] by prtitioning the intervl [, b] into mny equl subintervls, treting s i it is constnt on ech tiny subintervl, nd pproimting the signed re with Riemnn sum. By tking the limit s the number o subintervls increses without bound we determine the deinite integrl. In this chpter we return to Riemnn sums to obtin pproimtions o deinite integrls. The numericl methods discussed here re oten used in prctice. Computers or progrmmble clcultors re idel or perorming the otherwise tedious clcultions. Approimting Sums: L n, R n, T n, nd M n In the contet o the ollowing emple we ll discuss let-hnd, right-hnd, midpoint, nd trpezoidl sums, sums tht cn be used to pproimte deinite integrl. In order to be ble to compre our pproimtions with the ctul vlue, we ll look t n integrl we cn evlute ectly. EXAMPLE 6. Approimte 5 d. Keep improving upon the pproimtion until you know the vlue to our deciml plces. SOLUTION i) To pproimte the integrl we chop up the intervl o integrtion into n equl subintervls, ii) pproimte the re under the curve on ech subintervl by the re o rectngle, nd iii) sum the res o these rectngles.

3 6. Approimting Sums: L n, R n, T n, nd M n 807 We ll construct three Riemnn sums: the let- hnd sum, denoted by L n ; the right-hnd sum, denoted by R n ; nd the midpoint sum, denoted by M n. We re lredy milir with the irst two.) We describe these sums s ollows. L n : The height o the rectngle on ech subintervl is given by the vlue o t the let-hnd endpoint o the subintervl. R n : The height o the rectngle on ech subintervl is given by the vlue o t the right-hnd endpoint o the subintervl. M n : The height o the rectngle on ech subintervl is given by the vlue o t the midpoint o the subintervl. On the i th intervl, [ i, i ], the height o the rectngle is c), where c is midwy between i nd i ; c = i + i. 4 Subdivisions Let n = 4; we chop [, 5] into 4 equl subintervls ech o length = 5 4 =. = 4 5 Below re the let- hnd sum, the right-hnd sum, nd the midpoint sum. 4 5 L 4 = = = R 4 = = = M 4 = = = Figure 6. By... we men there re more deciml plces but we hve stopped recording them.

4 808 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls The unction ) = is decresing; thereore the let-hnd sums provide n upper bound nd the right-hnd sums lower bound or 5 d = R 4 < d < L 4 = Suppose we tke the verge o R 4 nd L 4, L 4+R 4. This is closer to the vlue o the integrl thn either the right- or let-hnd sums. Geometriclly this verge is equivlent to pproimting the re on ech intervl by trpezoid insted o rectngle, s shown below in Figure 6.. E B F C A i ) D i 4 5 b) Figure 6. Averging the re o rectngles o ABCD nd AEFD in Figure 6.) gives the re o the trpezoid AECD. We reer to the verge o the let- nd right- hnd sums, L n+r n, s the trpezoidl sum or the Trpezoidl rule) nd denote it by T n. T 4 = = Becuse is concve up 4 on [, 5] we know tht on ech subintervl the re under the trpezoid is lrger thn tht under the curve. Thus T 4 gives n upper bound or the integrl, better upper bound thn tht provided by L 4. 5 R 4 < d < T 4 <L < d <.68...< At this point the mind o the criticl reder should be buzzing with questions. Perhps they include the ollowing. Wht re the conditions under which T n will be lrger thn the vlue o the integrl? Smller? Where does the midpoint sum it into the picture? Let s investigte the irst question by looking t some grphs. 4 t) = t, so t) = t nd t) = t. The ltter is positive on [, 5], so is concve up on the intervl under discussion.

5 6. Approimting Sums: L n, R n, T n, nd M n 809 Figure 6.4 I is concve up on [, b], then every secnt line joining two points on the grph o on [, b] lies bove the grph. I is concve down on [, b], then every secnt line joining two points on the grph o on [, b] lies below the grph. We conclude tht b t) dt < T n i is concve up on [, b], nd b T n < t) dt i is concve down on [, b]. Where does the midpoint sum it into the picture? It turns out tht T n nd M n re complementry pir. We will show tht b i is concve up on [, b], then M n < t) dt < T n b i is concve down on [, b], then T n < t) dt < M n. while To do this we ll give n lterntive grphicl interprettion o the midpoint sum. Figure 6.5i) illustrtes the midpoint pproimtion, pproimting the re under the grph o on [ i, i+ ] by the re o rectngle whose height is the vlue o t the midpoint o the intervl. In Figure 6.5ii) we pproimte the re under by the re o the trpezoid ormed by the tngent line to the grph o t the midpoint o the intervl. We clim tht these res re identicl; pivoting the line through A, B, nd C bout the midpoint C does not chnge the re bounded below. Look t Figure 6.5iii). The tringles ACD nd BCE re congruent. We rgue this s ollows. Angles CAD nd CBE re right ngles. Angles ACD nd BCE re equl. Thereore the two tringles in question re similr. But AC = CB becuse C is the midpoint o the intervl [ i, i+ ]. We conclude tht tringles ACD nd BCE re congruent nd hence hve the sme re. Thereore, rectngle i AB i+ nd trpezoid i DE i+ Figures 6.5i) nd ii), respectively) hve the sme re; we cn interpret the midpoint sum s the midpoint tngent sum.

6 80 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls A C B C E A C E B E i midpt. i + i) D i midpt. i + ii) D i midpt. i + iii) D A C B congruent tringles s long s the oblique line psses through midpoint C. Figure 6.5 Below is picture o M 4 using the midpoint tngent line interprettion. is concve up, so the tngent lines lie below the curve. We know tht on ech subintervl the re under the midpoint tngent line is less thn the re under the curve; thereore, M 4 gives lower bound or the integrl Figure 6.6 Where unction is concve up its tngent line lies below the curve; where unction is concve down its tngent line lies bove the curve. It ollows tht i is concve up on [, b], then M n < b t) dt; i is concve down on [, b], then b t) dt < M n. Figure 6.7 How do L n, R n, T n, nd M n improve s we increse n? 8 Subdivisions Let n = 8; we chop [, 5] into 8 equl pieces ech o length = 5 8 =.

7 6. Approimting Sums: L n, R n, T n, nd M n L 8 = = R 8 = = T 8 = L 8 + R 8 = midpoint rectngles OR midpoint tngent trpezoids M 8 = = We know tht.48...= R 8 <M 8 = < 5 d < T 8 =.68...<L 8 = ) = is decresing nd concve up on [, 5]. Thereore, we know tht or ny n 5 5 R n < d < L n nd M n < d < T n. I we re interested in more deciml plces, we cn simply choose lrger vlues o n. 50 Subdivisions Suppose n = 50; we chop [, 5] into 50 equl pieces ech o length = 5 50 = We don t ctully wnt to sum up 50 terms by hnd. Work like this is pinul to do by hnd but it s child s ply or progrmmble clcultor or computer. Get out your progrmmed clcultor or computer nd check the igures given below. 5 R 50 = M 50 = T 50 = L 50 = Subdivisions Suppose n = 400; we chop [, 5] into 400 equl pieces ech o length = = 0.0. We obtin R 400 = M 400 = T 400 = L 400 = Using T 400 s n upper bound nd M 400 s lower bound, we ve niled down the vlue o this integrl to 4 deciml plces. 5 You ll hve to enter the ollowing inormtion: the unction oten s Y ), the endpoints o integrtion, nd the number o pieces into which you d like to prtition the intervl. And, i you re using clcultor without L n, R n, T n, nd M n progrmmed, you ll hve to enter the progrm.)

8 8 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls REMARK M 50 nd T 50 give better pproimtions o 5 d thn do R 400 nd L 400. Summry o the Underlying Principles I is incresing on [, b], then L n < b t) dt < R n. L n R n I is decresing on [, b], then R n < b t) dt < L n. L n R n I is concve up on [, b], then M n < b t) dt < T n. T n M n I is concve down on [, b], then T n < b t) dt < M n. T n M n For ny given n, generlly the trpezoidl nd midpoint sums re much closer to the ctul vlue o the deinite integrl thn re the let- nd right-hnd sums. EXAMPLE 6. Approimte 0 e d by inding upper nd lower bounds diering by no more thn SOLUTION This is gret problem on which to prctice, becuse it is impossible to ind n ntiderivtive or ) = e unless we resort to n ininite sum o terms. Look t the grph o ). It is decresing on the intervl [0, ] nd ppers to hve point o inlection somewhere on this intervl. 7 ) = e Figure 6.8 Suppose we were plnning to use let- nd right-hnd sums only. As shown in Section., the dierence between the let- nd right-hnd sums is given by 6 The unction e k is o prcticl importnce becuse, or the pproprite k, its grph gives the bell-shped norml distribution curve. The re under the norml distribution curve over some given intervl is o vitl importnce to probbilists nd sttisticins. 7 How cn we be sure is decresing? ) =. As increses rom 0 to, increses, so e is positive nd incresing. e d Thereore, its reciprocl is decresing. Alterntively, d e = e 4) = 4 0 or 0. e

9 6. Approimting Sums: L n, R n, T n, nd M n 8 R n L n = b) ) = b) ) b n. Thereore, in this emple R n L n = e 8 e 0 0 n = e 8 n < n I we wnt R n L n < 0.00, then we cn solve.9994 n = 0.00 or n nd choose ny integer lrger thn this. n = = 999.4, so we cn choose n = 000. I your clcultor will ccept number this lrge, you re in good shpe; the let-hnd sum will be n upper bound nd the right-hnd sum will be lower bound, becuse is decresing on [0, ]. The computtion my tke some time or the mchine to perorm, depending upon its power. You should get L 000 = nd R 000 = REMARK Suppose is monotonic 8 over the intervl [, b], so the ctul vlue o b ) d is between L n nd R n. Although it is possible simply to try lrger nd lrger vlues o n until L n nd R n re within the desired distnce rom one nother, it is more eicient, i high degree o ccurcy is demnded, to use to ind n pproprite vlue o n. R n L n = b) ) Generlly, the midpoint nd trpezoidl sums give us much better bounds or prticulr n. I we ind the point o inlection, we cn use these sums to orm sndwich round the vlue o the integrl. ) = e ) = 4e ) = 4[e + 4e )] = 4e + 4 ) ) = 0, where + 4 = 0, tht is, where = 4. On [0, ] ) = 0 t =. Looking bck t ) = 4e + 4 ), we cn see tht chnges sign rom negtive to positive) t =, so = 0.5 is point o inlection. is concve down on [0, ] nd concve up on [, ]. Thereore, 8 is monotonic on [, b] i it is either lwys incresing on [, b] or is lwys decresing on [, b].

10 84 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls on [0, ], T n gives lower bound or the integrl nd M n gives n upper bound; on [, ], T n gives n upper bound or the integrl nd M n gives lower bound. ) = e Figure 6.9 The combined error on [0, ] nd [, ] must be no more thn We cn split up the llowble error however we wnt. We cn ply round with the clcultor or computer until the sum o T n M n on [0, ] nd T m M m on [, ] is less thn Here is the result o some plying round. I we try n = 50 on both intervls, we obtin the ollowing. On [0, ] T 50 = M 50 = On [, ] T 50 = M 50 = On ech intervl the dierence between T 50 nd M 50 is substntilly less thn To get the inl nswer we need to dd the lower bounds T 50 on [0, ] nd M 50 on [, ]) to obtin lower bound or the integrl, nd dd the upper bounds to obtin n upper bound. 0 e d =.5 0 e d + lower bound: = upper bound: = e d In the net section we will show n lterntive nd more eicient method o using the trpezoidl nd midpoint sums to pproimte this integrl. Numericl methods o integrtion, such s let- nd right-hnd sums nd trpezoidl nd midpoint sums, re very useul not only when we re trying to pproimte b ) d nd cn t ind n ntiderivtive or, but lso in situtions in which we don t even hve ormul or. A scientist my be tking periodic dt redings, or surveyor my be tking mesurements t preset intervls, nd these dt sets my constitute the only inormtion we hve bout. Numericl methods o integrtion cn be pplied directly to the dt sets. Eercise 6. below dels with inormtion rom dt set, nd the results o Eercise 6. nd Eercise 6. cn be esily pplied to dt sets in which dt hve been collected t eqully spced intervls. EXERCISE 6. Between noon nd 5:00 p.m. wter hs been leving reservoir t n incresing rte. We do not hve rte unction t our disposl, but we do hve some mesurements indicting the rte tht wter hs been leving t vrious times. The inormtion is given below. These mesurements were not tken t eqully spced time intervls.

11 6. Approimting Sums: L n, R n, T n, nd M n 85 Time noon :00 :0 :00 :00 :45 4:5 5:00 Rte out in gl/hr) EXERCISE 6. ) Find good upper nd lower bounds on the mount o wter tht hs let the reservoir between noon nd 5:00 p.m. b) Use trpezoidl sum to pproimte the mount o wter tht hs let the reservoir between noon nd 5:00 p.m. Answers re provided t the end o the section. Our im is to pproimte b ) d when we do not hve ormul or ). Our dt consist o collection o mesurements o ) tken t eqully spced intervls. Use trpezoidl sum to pproimte the deinite integrl. Prtition the intervl [, b] into n equl subintervls, ech o length = b n. 0,,,..., re s indicted. 0 = nd k = + k or k =,..., n. y k = k ) or k = 0,..., n. Show tht the trpezoidl sum cn be computed s ollows: ) T n = [y 0 + y + y + +y n + y n ]. y 0 y y n y n 0 = 4 n n Figure 6.0 = b Answers to Selected Eercises Eercise The nswer is provided t the end o the section. ) The rte is incresing, so the lower bound is given by the let-hnd sum nd the upper bound by the right-hnd sum. Lower bound: 40) ) ) + 00) ) ) ) = Upper bound: 60) ) ) + 50) ) ) ) = 0.5 Between nd 0.5 gllons o wter hve let the reservoir.

12 86 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls b) The trpezoidl pproimtion is the verge o the let- nd right-hnd sums given bove. The dierent lengths o the intervls do not chnge this bsic principle. Convince yoursel o this.) Trpezoidl sum = ) = 955 Eercise. T n = [y 0 + y + y y n ) + y + y + y y n ) ] = y 0 + y + y y n + y n ) P R O B L E M S F O R S E C T I O N 6. Some o the problems in this problem set require the use o progrmmble clcultor or computer. These re not highlighted in ny wy, but common sense should tell you tht summing couple o hundred terms is not good use o your time. On the other hnd, in order to mke sure tht you understnd the numericl methods, some problems eplicitly sk you to rerin rom using progrm to eecute the computtion.. Suppose we use right- nd let-hnd sums to pproimte b t) dt. We prtition the intervl [, b] into n equl pieces ech o length t. Let R n be the right-hnd sum using n subdivisions nd L n be the let-hnd sum using n subdivisions. ) Show tht R n = L n + b) t ) t. b) Conclude tht R n L n = [ b) )] t = [ b) )] b n.. ) Find t dt, where >0. b) ln cn be deined to be t dt. Sketch the grph o /t nd pproimte ln numericlly by prtitioning the intervl [, ] into 4 equl pieces nd computing L 4 nd R 4. Write out the sum long-hnd, without using clcultor progrm. c) Into how mny equl pieces would you hve to subdivide the intervl [, ] in order to pproimte ln so tht R n nd L n give upper nd lower bounds tht dier by no more thn 0.0? Cll this number p. d) Compute M p nd T p. e) Into how mny equl pieces would you hve to subdivide the intervl [, ] in order to pproimte ln so tht R n nd L n give upper nd lower bounds tht dier by no more thn 0.00? Tht dier by no more thn D?. Compute e ln d with error less thn Try, by eperimenttion, to see how mny subdivisions re required or M n nd T n to dier by less thn How mny re required or L n nd R n to dier by less thn 0.00?

13 6. Approimting Sums: L n, R n, T n, nd M n ) Let be the unction grphed below. is incresing nd concve down. b Let A = b ) d. Suppose tht estimtes o ) d re computed using the let, right, nd trpezoid rules, ech with the sme number o subintervls. We ll denote these estimtes by L n, R n, nd T n, respectively. Put the numbers A, L n, R n, nd T n in scending order. Justiy your nswer nd eplin why your nswer is independent o the vlue o n used. b) Answer the sme question s bove i the grph o is the one given below. is decresing nd concve up on [, b]. b 5. Consider 4 d. ) Find vlue o n or which L n b ) d 0.0. b) Use the vlue o n rom prt ) to ind L n nd R n. c) Is the verge o the let- nd right-hnd sums lrger thn the integrl, or smller? d) Compre your numericl pproimtions to the nswer you get using the Fundmentl Theorem o Clculus. 6. Approimte d with error less thn Give upper nd lower bounds or 0 0 d such tht the upper nd lower bounds + 5 dier by less thn Give upper nd lower bounds or ln d such tht the two bounds dier rom one nother by less thn Eplin how you know tht the upper bound is indeed n upper bound nd the lower bound is indeed lower bound. 9. Consider 9 4 d. ) Find vlue o n or which L n R n 0.0. b) Use tht vlue o n to ind L n nd R n.

14 88 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls c) I you tke the verge o the let- nd right-hnd sums, will your pproimtion be lrger thn the integrl, or smller? d) Compre your numericl pproimtions to the nswers you get using the Fundmentl Theorem o Clculus. 0. One o your riends is doing his mthemtics homework on the run. He mnged to squeeze in one problem between lunch nd his epository writing clss. He used his clcultor to ind L n, R n, T n, nd M n to pproimte deinite integrl nd jotted down the results on npkin. They were , 0.885, 0.489, nd He didn t write down the problem, nor did he record the vlue o n he used. But he did sketch the unction over the relevnt intervl) on the npkin; it looked like this. He wnts your help in lbeling the dt he recorded s let-hnd sum, right-hnd sum, midpoint sum, nd trpezoidl sum. Which is which? Eplin your nswer.. Mesurements o the width o pond re tken every 0 yrds long its length. The mesurements re: 0 yrds, 60 yrds, 50 yrds, 70 yrds, 50 yrds, nd 0 yrds. Approimte the surce re o the pond using the Trpezoidl rule Approimte ech o the ollowing integrls with error less thn /00. Note: I you look t ll the questions nd think bout strtegy in dvnce, you will only hve to compute two integrls in order to nswer ll our questions with the desired degree o ccurcy. There is no problem with being more ccurte thn is requested.) Briely eplin wht you hve done nd how mny subdivisions you used. i) 0 e ) d ii) e ) d iii) 0 e ) d iv) e ) d

15 6. Approimting Sums: L n, R n, T n, nd M n 89. Let ) =. Let nd b be positive constnts, 0 < < b. Below re two prtitions + o the intervl [, b]. One given with w s) prtitions [, b] into 8 equl subintervls, ech o length w; the other given with t s) prtitions the intervl into equl pieces ech o length t. w i = + i w, i = 0,,...,8 t i = + i t, i = 0,,..., Put the ollowing epressions in scending order, with < or = signs between them. Identiy the sum with the letter preceding it.) A = 7 w i ) w B = i=0 C = t i ) t D = E = i=0 b 8 w i ) w i= i= t i ) t w) dw F = ) b ) 4. Approimte rctn d using let- nd right-hnd sums to obtin n upper nd lower bound or the integrl with dierence less thn Sve time by grphing y = rctn nd using symmetry to simpliy the problem. 5. The unction ) is decresing nd concve down on the intervl [, 5]. Suppose tht you use right-hnd sum, R 00, let-hnd sum, L 00, trpezoidl sum, T 00, nd midpoint sum, M 00, ll with 00 subdivisions, to estimte 5 ) d. Select ll o the ollowing tht must be true. ) L 00 R 00 b) 5 ) d T 00 c) T 00 R 00 d) T 00 M 00 e) M 00 L 00 ) T 00 = L 00 + R 00 )/ 6. Suppose tht g is dierentible unction whose derivtive is g ) = +. Prtition [0, ] into n equl pieces ech o length nd let k = k, where k = 0,,..., n. Put the ollowing epressions in scending order with < or = signs between them). A = n n g i ) B = g i ) i= n C = lim g i ) n i=0 i=0 D = lim n n g i ) i=

16 80 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls 7. Suppose L 0 < b ) d < R 0 nd R 0 L 0 < 0.0. Which o the ollowing sttements must be true? Circle ll such sttements. ) R 0 b ) d < 0.0 b) c) T 0 b ) d < d) T 0 b ) d < 0.0 L 0 b ) d < Two o the ollowing three integrls you cn evlute ectly. One you cnnot, until lerning integrtion by prts. Identiy the one you cnnot evlute ectly nd pproimte it with n error under Find ect nswers or the other two integrls. ) e ln d b) 0 e d c) 0 e d 6. SIMPSON S RULE AND ERROR ESTIMATES Simpson s Rule We begn our discussion with let- nd right-hnd sums, sums tht provide bounds or b ) d i is monotonic on [, b]. Then we cme up with the clever ide o using the verge o L n nd R n to pproimte deinite integrl. This led us to the trpezoidl sum, which generlly gives much better pproimtion to deinite integrl thn either the let or right sums or the sme number o subdivisions. Encourged by our success, we might strt thinking bout verging the midpoint nd trpezoidl sums. An even better ide would be to tke weighted verge o these two sums. This is becuse the midpoint rule generlly gives us better it thn does the trpezoidl rule. In ct, nlysis o numericl dt indictes tht the error in using the trpezoidl rule is bout two times the size s tht rom using the midpoint rule. From the picture below this seems plusible. The weighting M n+m n +T n = M n+t n gives us method o pproimting deinite integrls known s Simpson s rule. i midpoint i+ = Error using trpezoid = Error using midpoint tngent trpezoid The shded re is bout double the blckened re. Weighted verge: weight midpoint twice s hevily s the trpezoidl sum. M n + T n Figure 6. = Simpson's rule Simpson s rule: S n = M n + T n

17 6. Simpson s Rule nd Error Estimtes 8 We cll it S n becuse we need n dt points or M n nd n + ) dierent dt points or T n. Sometimes it will be convenient to write S p = M n+t n where p = n. Simpson s rule generlly gives substntilly better numericl estimte thn either the midpoint or the trpezoidl sums. To get some pprecition or this, try it out on some integrls tht you cn compute ectly. Below we do this or the integrl 5 / d. EXAMPLE 6. Compre M n, T n, nd Simpson s rule the weighted verge) or 5 d, where n = 4, 8, 50, nd 00. Note tht 5 d = ln 5 = ln 5 ln = ln 5 = SOLUTION ) n = 4: M 4 = T 4 = Simpson s rule = M 4+T 4 = b) n = 8: M 8 = T 8 = Simpson s rule = M 8+T 8 = c) n = 50: M 50 = T 50 = Simpson s rule = M 50+T 50 = d) n = 00: M 00 = T 00 = Simpson s rule = M 00+T 00 = Like the trpezoidl sum, Simpson s rule is convenient tool or pproimting b ) d even when ormul or is not vilble. For instnce, the vlues o ) reerred to in the discussion below could be mesurements tken by surveyor estimting the re o plot o lnd or body o wter. Suppose our im is to pproimte b ) d. We cn collect vlues o ) t eqully spced intervls nd use Simpson s rule to pproimte the deinite integrl. Prtition the intervl [, b] into n equl subintervls, ech o length = b n. We need to use weighted verge o the midpoint nd trpezoidl sums; thereore, on ech subintervl we need not only the vlue o t ech endpoint or T n ) but lso the vlue o in the middle or M n ). Although we re chopping [, b] into n equl subintervls, we will use n + vlues o n + vlues or T n nd n vlues or M n ). Let y k = k ) or k = 0,...,n, where 0,,,..., n re s indicted below. ) k = + k or k = 0.,...,n 0 = 4... n n n or M n or M n or M n = b Figure 6. In the ollowing eercise you will show tht using this lbeling convention, Simpson s rule cn be used to pproimte the deinite integrl using the ollowing ormul.

18 8 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls Simpson s rule: ) S n = 6 [y 0 + 4y + y + 4y + y 4 + +y n + 4y n + y n ], where = b n. EXERCISE 6. In this eercise use the lbeling system described bove: Prtition [, b] into n equl pieces o length = b n. k = + k ) or k = 0,,...,n, where = b n Let y k = k ) or k = 0,...,n, where 0,,,..., n. ) Show tht using this lbeling convention T n = b) Show tht using this lbeling convention ) y 0 + y + y 4 + y 6 + +y n + y n ). M n = y + y + y 5 + y 7 + +y n ). c) Show tht using this lbeling convention Simpson s rule is given by 6) y 0 + 4y + y + 4y + y 4 + +y n + 4y n + y n ). The nswer to prt c) is given t the end o the section. Notice tht becuse Simpson s rule requires both the midpoint nd trpezoidl rules nd the relevnt dt points or ech o these two rules re dierent, Simpson s rule requires pproimtely double the number o dt points necessry or computing either M n or T n individully. REMARKS Simpson s rule cn be presented without reerence to the trpezoidl nd the midpoint sums. Insted, it cn be viewed s ollows. On ech subintervl [ i, i ] pproimte the re under by the re under the prbol pssing through the three points on the grph o corresponding to the endpoints o the intervl nd the midpoint, or i, i )), i, i )), nd i + i, i + i )).

19 6. Simpson s Rule nd Error Estimtes 8 A B C D E Find prbol through A, B, nd C. Find nother prbol through C, D, nd E. Sum the res under the prbols to pproimte the re under the curve. i i i+ Figure 6. When Simpson s rule is pplied to cubics, it will lwys give n ect nswer, even when using M +T. Error Bounds In Section 6. we used pproimting sums to provide upper nd lower bounds or the vlue o n integrl. When we use Simpson s rule we hve good estimte, but without getting bound on the error involved we cnnot be sure how good the estimte is. While we cnnot epect to know the ect error in using pproimting sums, it would be quite useul to get bound on the error. Consider, or instnce, Emple 6.. It would be useul to be ble to pproimte 0 e d using the trpezoidl rule or Simpson s rule, hving bound or the error but not worrying bout the point o inlection. Let s begin by thinking bout the error involved in using let- nd right-hnd sums. y y y The steeper the slope o the lrger the error in pproimting by rectngles. Figure 6.4 As illustrted in Figure 6.4, the lrger is, the lrger the mgnitude o the error involved in pproimting the re under the curve by the re o rectngle. When pproimting the re using the trpezoidl rule or the midpoint tngent trpezoid, the mgnitude o is not n issue; the rte tht is chnging, mesured by ), is ctor controlling the error. For L n, R n, T n, nd M n, the lrger n is, the smller the epected error. Put nother wy, the smller is, the smller the epected error. Using the Men Vlue Theorem, the ollowing error bounds cn be proven.

20 84 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls Let I = b ) d. Let L n, R n, T n, nd M n be let, right, trpezoidl, nd midpoint pproimtions o I, respectively. Then L n I M b ) n R n I M b ) n T n I M b ) n M n I M b ) 4n where M is the mimum vlue o on [, b]. where M is the mimum vlue o on [, b]. You might guess tht the error bound or Simpson s rule involves n n nd the third derivtive o. In ct, Simpson s rule is much more ccurte thn you might epect. It cn be shown tht S n I M 4b ) 5 80n) 4 I p is n even number, then where M 4 is the mimum vlue o 4) on [, b]. S p I M 4b ) 5 80p 4. EXAMPLE 6.4 Find upper bounds or the error involved in using T 0 nd M 0 to pproimte 0 e d. SOLUTION In order to use the error estimtes or T 0 nd M 0 we need to ind n upper bound or on [0, ]. In Emple 6. we computed ). I ) = e, then ) = 4e + 4 ) = ) e. We cn get crude upper bound or on [0, ] by noting tht the numertor is no more thn ) = 4 + 6) = 5 4 = 60. ) 60 = 60e e e on [ 0, ] so ) < 60. NOTE We hven t ound M, but we ound bound or M. T 0 I M b ) n where = 0, b =, n = 0, nd M < 60 T 0 I < M 0 I < = = 0.4 The error using T 0 is gurnteed to be less thn = 0.4 = 0. The error using M 0 is gurnteed to be less thn 0..

21 6. Simpson s Rule nd Error Estimtes 85 EXAMPLE 6.5 How big should n be to esily gurntee tht M n pproimtes 0 e d to within 0.00? SOLUTION From Emple 6.4 we know tht M n I < 480 = 0. We must ind n such tht 4n n 0 < We ll solve 0 = 0.00 nd pick the net higher integer. n n n 0.00) = 0 =>n = 0,000 Then n 4.4. So we choose n = 4. EXAMPLE 6.6 Get n upper bound or the error involved in using S 0 to pproimte ln = d. SOLUTION ) =. Compute,,, nd 4). Conirm tht 4) ) = 4 5. On [, ], 4 4. S p I M 4b ) 5 80p 4 where =, b =, p = 0, nd M 4 = 4. 4 ) < = Answers to Selected Eercises Eercise 6.c) Simpson s rule = M n+t n ) ) [y + y + y 5 + +y n ) + y 0 + y + y 4 + y 6 + +y n + y n ) ] ) = [y + y + y y n )] + ) ) [y 0 + y + y 4 + y 6 + +y n + y n )] ) = [4y + 4y + 4y y n )] 6 + [y 0 + y + y 4 + y 6 + +y n + y n )] 6 = [y 0 + 4y + y + 4y + y 4 + 4y 5 + y 6 + 4y 7 + +y n + 4y n + y n )] 6 ) )

22 86 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls P R O B L E M S F O R S E C T I O N 6.. Approimte e ln d using Simpson s rule with p = 0. S 0 = M 5+T 5. Find n upper bound or the error. At certin point you ll use the ct tht e<.. Approimte the ollowing integrls using the trpezoidl rule, the midpoint rule, nd Simpson s rule or the speciied number o subdivisions nd compute error bounds using the ormuls given in this section. Compre your nswers to the ect nswer. ) 0 d n = 4 b) 0 + d n = 4. Approimte the ollowing integrl using the trpezoidl rule, the midpoint rule, nd Simpson s rule or the speciied number o subdivisions. Look t T n, M n, nd S n. Compute error bounds. ln d ) n = 4 b) n = 8 c) n = 0 4. ) Approimte 0 cos ) d using M 0. Find n upper bound or the error. b) Approimte 0 cos ) d using S 0. S 0 = M 0+T 0. Find n upper bound or the error. 5. A surveyor is mesuring the cross-sectionl re o 0-oot-wide river beneth bridge. He mesures the river s depth every 5 eet. The dt re given below. 5 t 5 t 5 t 5 t 5 t 5 t {}}{ {}}{ {}}{ {}}{ {}}{ {}}{ Depth in eet) ) Use the trpezoidl sum to pproimte the cross-sectionl re o the river. b) Use Simpson s rule to pproimte the cross-sectionl re o the river.

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