1 Part II: Numerical Integration


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1 Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble nd stisfing, but for most pplictions, numericl nswer with severl digits of ccurc is just s useful. For instnce, suppose ou re utomotive or ircrft designer. You m wish to know how much metl is required to build our design, creted using computerided design grphics progrm. Due to the fluctutions in the price for metl, ou onl need the pproimte cost bsed on piecewiseliner pproimtion to our design. Numericl techniques such s those discussed in this section cn be used for tht. 1.1 Emples of Riemnn sums: five rectngulr rules We now describe five methods of computing Riemnn sums. We will cll these rectngulr rules, nd the correspond to five nturl ws of choosing the point c k t which to compute the height, f(c k ), of the rectngle over the subintervl [ k 1, k ]. (Lter, we will consider two other methods of computing Riemnn sums which re nonrectngulr rules. These methods use shpes other thn rectngles to cheive better pproimtions to the re under curve.) Left endpoint Rule (LRAM) Let c k = k 1, the left endpoint of the kth subintervl [ k 1, k ]. LRAM n = f( k 1 )( k k 1 ). If the subintervls re ll of the sme size = k k 1, then k=1 LRAM n = [f( 0 ) + f( 1 ) + + f( n 1 )]. Right endpoint Rule (RRAM) Let c k = k, the right endpoint of the kth subintervl [ k 1, k ]. RRAM n = f( k )( k k 1 ). If the subintervls re ll of the sme size = k k 1, then k=1 RRAM n = [f( 1 ) + f( ) + + f( n )]. Midpoint Rule (MRAM) Let c k = k 1+ k, the midpoint of the kth subintervl [ k 1, k ]. MRAM n = ( ) k 1 + k f ( k k 1 ). k=1 If the subintervls re ll of the sme size = k k 1, then [ ( ) ( ) ( )] n 1 + n MRAM n = f + f + + f. 1
2 Mth 4 Lb Lower Sum (LS) Let m k be the vlue in the kth subintervl so tht f(m k ) is the minimum vlue of f in tht intervl. LS n = n k=1 f(m k)( k k 1 ). Upper Sum (US): Let M k be the vlue in the kth subintervl so tht f(m k ) is the minimum vlue of f in tht intervl. US n = n k=1 f(m k)( k k 1 ). The figures below give n emple illustrting the rectngles tht result from ppling the first three rules. Regrdless of which of these three rules we ppl, the rectngles m be bove or below the grph of the function f, depending on the shpe of the grph. In contrst, the lower sum lws builds the rectngles below the grph of f. Consequentl, the lower sum is lws less thn or equl to the ect re A: LS A for ever prtition P. Similrl, the upper sum builds rectngles over the grph of f, nd the upper sum is lws greter thn or equl to the ect re A: US A for ever prtition P. The lower nd upper sums provide bounds on the size of the ect re: LS A US. Left Sum = Right Sum = Midpoint Sum = Lower Sum = Upper Sum =
3 Mth 4 Lb Emple 1. Use the wmim softwre to crete function tht ou cn use to pproimte the re under the curve f() = 1/ between the points = 1 nd b = using the first method described bove (LRAM). Solution: First, we tell Mim the intervl nd the function s follows: :1$ b:$ f():=1/$ This defines the intervl strting point = 1, ending point b =, nd the function f() = 1/. We will use n rectngles of equl width, so the width of ech rectngle is = b n no mtter which method we re using. We cll this the step size. There re n + 1 points in the intervl [, b] over which we re integrting: = 0 < 1 < < n = b We need to tell Mim the vlue of ech point i. To get the vlue i, strt t the point nd move i steps to the right. So i is plus i times the step size: ( ) b i = + i. n This is true no mtter which method we re using. In Mim, we define this s follows: 1 (i,n) := + i*(b)/n; This function tkes two input rguments, i nd n, nd outputs the vlue i, the ith element in the prtition = 0 < 1 < < n 1 < n = b. Net, define function to pproimte the integrl using the left hnd rule pproimtion method (LRAM) s follows: LRAM(n) := sum( f((i,n)), i, 0, n1 ) * ( (b)/n ); Notice tht we hve defined LRAM s function of n, so b plugging in vrious vlues for n we cn use this function to pproimte the re under the curve using n number of rectngles. For emple, we find tht with 5 rectngles LRAM(5); returns 1879/50. To show the nswer in deciml nottion, flot(lram(5)); returns Similrl, for 10 rectngles, flot(lram(10)); returns , while for 100 rectngles flot(lram(100)); ields We know tht the integrl 1 1 d is ln(). Checking the deciml representtion of ln() in Mim, we find tht flot(log()); returns So we see tht the pproimtion obtined with the LRAM function is ok for smll n, nd is much improved when n = Mke sure ou hve lred defined nd b before defining (i,n). Recll tht ln() is represented in Mim b log(). 3
4 Mth 4 Lb Problem 1: Use wmim to crete function to pproimte the re under the curve f() = 1 between the points = 0 nd b = 1 using the second method described bove (RRAM). Write our nswers below. (Hint: For LRAM we evlute the function nd sum over the points { 0, 1,..., n 1 }, while for RRAM we will evlute the function nd sum over the points { 1,,..., n }.) Answers: flot( RRAM(5) ) = flot( RRAM(10) )= Problem : Use wmim to crete function to pproimte the re under the curve f() = 1/ between the points = 1 nd b = using the third method described bove (MRAM). (Hint: ou will need function to compute the midpoints, which ou cn define in terms of the function (i,n) tht ou hve bove. For emple, (i,n) := ( (i,n) + (i+1,n) ) / ; Your definition of the function MRAM(n) will involve f((i,n)) insted of f((i,n)).) Answers: flot( MRAM(5) ) = flot( MRAM(10) )= Compre our nswers with those in the emple on the previous pge. Which is more ccurte in this cse? (circle one) LRAM MRAM 4
5 Mth 4 Lb 1. The trpezoid rule nd Simpson s rule The methods in this section pproimte the definite integrl of function f b building es functions close to f nd then ectl evluting the definite integrls of the es functions. If the es functions re close enough to f, then the sum of the definite integrls of the es functions will be close to the definite integrl of f. Above, we sw tht the left, right nd midpoint pproimtions fit horizontl lines to f. There, the es functions were piecewise constnt functions, nd the pproimting regions re rectngles. The Trpezoidl Rule fits slnted lines to f, the es functions re piecewise liner, nd the pproimting regions re trpezoids. Finll, Simpson s Rule fits prbols to f, nd the es functions re piecewise qudrtic polnomils. For simplicit, we will divide the intervl [, b] into n equll long subintervls. Ech subintervl hs length h = i = b n, nd the points of the prtition re 0 =, 1 = + h, = + h,..., i = + ih,..., n = + nh = + n( b n ) = b. The re of trpezoid is (bse) (verge height), so the re of the trpezoid with coordintes ( 0, 0), ( 0, 0 ), ( 1, 0), ( 1, 1 ), is ( 1 0 ) ( ). (See Figure 1.) ( 1, 1) ( 0, 1) ( 0, 0) ( 1, 0) Figure 1: Trpezoid Theorem 1. ( Trpezoidl Approimtion Rule ) If f is integrble on [, b], nd [, b] is prtitioned into n subintervls of length h = b n, then the Trpezoidl pproimtion of b f() d is T n = h [f( 0) + f( 1 ) + f( ) f( n 1 ) + f( n )]. Proof: The re of the trpezoid with coordintes ( i, 0), ( i, i ), ( i+1, 0), ( i+1, i+1 ), where i = f( i ), is ( i+1 i ) i+i+1 = h ( i + i+1 ). Therefore, the sum of the trpezoidl res pproimting b f() d is s desired. i= ( i+1 i ) i + i+1 = h ( i + i+1 ) = h ( n 1 + n ), i= 5
6 Mth 4 Lb f() Figure : Tble for Emple. Clculte T 4, the Trpezoidl pproimtion of 3 f() d, for the function vlues in Figure. 1 Solution: The step size is h = (b )/n = (3 1)/4 = 1/. Therefore, T 4 = h [f( 0)+f( 1 )+f( )+f( 3 )+f( 4 )] = 1 [4.+(3.4)+(.8)+(3.6)+(3.)] = (0.5)(7) = Problem 3: Let s see how well the trpezoidl rule pproimtes n integrl whose ect vlue we know, 3 1 d = 6 3. Clculte T 4, the Trpezoidl pproimtion of 3 1 d. Answer: 1.3 Simpson s rule If the grph of f is curved, even the slnted lines m not fit the grph of f s closel s we would like, nd lrge number of subintervls m still be needed with the Trpezoidl rule to get good pproimtion of the definite integrl. Curves tpicll fit the grph of f better thn stright lines, nd the esiest nonliner curves re prbols. Theorem. Three points ( 0, 0 ), ( 1, 1 ), (, ) re needed to determine the eqution of prbol, nd the re under prbolic region with evenl spced i vlues is 3 ( ). Theorem 3. ( Simpson s Rule ) If f is integrble on [, b], nd [, b] is prtitioned into n even number n of subintervls of length h = = b n, then the Prbolic pproimtion of b f() d is S n = 3 [f( 0) + 4f( 1 ) + f( ) + 4f( 3 ) + f( 4 ) f( n 1 ) + f( n )]. Emple 3. Clculte S 4, the Simpson s rule pproimtion of 3 f() d, for the function vlues tbulted 1 in Figure. Solution: The step size is h = (b )/n = (3 1)/4 = 1/. Then S 4 = h 3 [f( 0) + 4f( 1 ) + f( ) + 4f( 3 ) + f( 4 )] = 1 6 [4. + 4(3.4) + (.8) + 4(3.6) + (3.)] = 41 6 =
7 Mth 4 Lb 1.4 Trpezoidl vs. Simpson: Which Method Is Best? The hrdest nd slowest prt of these pproimtions, whether b hnd or b computer, is the evlution of the function t the i vlues. For n subintervls, ll of the methods require bout the sme number of function evlutions. The rest of this section discusses error bounds of the pproimtions so we cn know how close our pproimtion is to the ect vlue of the integrl even if we don t know the ect vlue. Theorem 4. (Error Bound for Trpezoidl Approimtion) If the second derivtive of f is continuous on [, b] nd M m [,b] f (), then the error of the T n pproimtion is b f() d T n (b )3 1n M. The error bound formul (b )3 M 1n for the Trpezoidl pproimtion is gurntee : the ctul error is gurnteed to be no lrger thn the error bound. In fct, the ctul error is usull much smller thn the error bound. The word error does not indicte mistke, it mens the devition or distnce from the ect nswer. Emple 4. How lrge must n be to be certin tht T n is within of 1 sin() d? 0 Solution: We wnt to pick n so tht (b )3 1n M 1/1000. We m tke M = 1, so (b )3 1n M = 1 1n 1/1000, or n 1000/1. Tking n = 10 will work. Theorem 5. (Error Bound for Simpson s Rule Approimtion) If the fourth derivtive of f is continuous on [, b] nd M 4 m [,b] f (4) (), then the error of the S n pproimtion is b f() d S n (b )5 180n 4 M 4. Problem 4: How lrge must n be to be certin tht S n is within of 1 0 sin() d? Answer: 7
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