31.2. Numerical Integration. Introduction. Prerequisites. Learning Outcomes

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1 Numericl Integrtion 3. Introduction In this Section we will present some methods tht cn be used to pproximte integrls. Attention will be pid to how we ensure tht such pproximtions cn be gurnteed to be of certin level of ccurcy. Prerequisites Before strting this Section you should... Lerning Outcomes On completion you should be ble to... review previous mteril on integrls nd integrtion pproximte certin integrls be ble to ensure tht these pproximtions re of some desired ccurcy 8 HELM (8): Workbook 3: Numericl Methods of Approximtion

2 . Numericl integrtion The im in this Section is to describe numericl methods for pproximting integrls of the form f(x) dx One motivtion for this is in the mteril on probbility tht ppers in cn be nlysed by working out 39. Norml distributions π e x / dx for certin vlues of nd b. It turns out tht it is not possible, using the kinds of functions most engineers would cre to know bout, to write down function with derivtive equl to π e x / so vlues of the integrl re pproximted insted. Tbles of numbers giving the vlue of this integrl for different intervl widths ppered t the end of 39, nd it is known tht these tbles re ccurte to the number of deciml plces given. How cn this be known? One im of this Section is to give possible nswer to tht question. It is cler tht, not only do we need wy of pproximting integrls, but we lso need wy of working out the ccurcy of the pproximtions if we re to be sure tht our tbles of numbers re to be relied on. In this Section we ddress both of these points, begining with simple pproximtion method.. The simple trpezium rule The first pproximtion we shll look t involves finding the re under stright line, rther thn the re under curve f. Figure 6 shows it best. f(x) f(b) f() b b x Figure 6 HELM (8): Section 3.: Numericl Integrtion 9

3 We pproximte s follows f(x) dx = grey shded re re of the trpezium surrounding the shded region = width of trpezium verge height of the two sides = ) (f() (b ) + f(b) The simple trpezium rule for pproximting the grph of f by the re of trpezium. Key Point 5 Simple Trpezium Rule f(x) dx is given by pproximting the re under The formul is: f(x) dx (b ) ( f() + f(b) ) Or, to put it nother wy tht my prove helpful little lter on, f(x) dx ( (intervl width) f(left-hnd end) + f(right-hnd end) ) Next we show some instnces of implementing this method. Exmple Approximte ech of these integrls using the simple trpezium rule () π/4 sin(x) dx (b) e x / dx (c) cosh(x) dx Solution () (b) π/4 ( π 4 ) ( + ) =.7768, sin(x) dx (b )(sin() + sin(b)) = e x / dx ( ) (b ) e / + e b / = ( ) ( e / + e ) =.3793, (c) cosh(x) dx (b ) (cosh() + cosh(b)) = ( ) ( + cosh()) = 4.76, where ll three nswers re given to 5 deciml plces. 3 HELM (8): Workbook 3: Numericl Methods of Approximtion

4 It is importnt to note tht, lthough we hve given these integrl pproximtions to 5 deciml plces, this does not men tht they re ccurte to tht mny plces. We will del with the ccurcy of our pproximtions lter in this Section. Next re some Tsks for you to try. Tsk Approximte the following integrls using the simple trpezium method () 5 x dx (b) ln(x) dx Your solution Answer 5 ( () x dx (b ) ) + b = ( (5 ) + ) 5 = (b) ln(x) dx (b )(ln() + ln(b)) = ( ) ( + ln()) = The nswer you obtin for this next Tsk cn be checked ginst the tble of results in 39 concerning the Norml distribution or in stndrd sttistics textbook. Tsk Use the simple trpezium method to pproximte π e x / dx Your solution Answer We find tht π e x / dx ( ) π ( + e / ) =.346 to 5 deciml plces. So we hve mens of pproximting pproximtion. f(x) dx. The question remins whether or not it is good HELM (8): Section 3.: Numericl Integrtion 3

5 How good is the simple trpezium rule? We define e T, the error in the simple trpezium rule to be the difference between the ctul vlue of the integrl nd our pproximtion to it, tht is e T = f(x) dx ) (f() (b ) + f(b) It is enough for our purposes here to omit some theory nd skip stright to the result of interest. In mny different textbooks on the subject it is shown tht e T = (b )3 f (c) where c is some number between nd b. (The principl drwbck with this expression for e T is tht we do not know wht c is, but we will find wy to work round tht difficulty lter.) It is worth pusing to sk wht mening we cn ttch to this expression for e T. There re two fctors which cn influence e T :. If b is smll then, clerly, e T will most probbly lso be smll. This seems sensible enough - if the integrtion intervl is smll one then there is less room to ccumulte lrge error. (This observtion forms prt of the motivtion for the composite trpezium rule discussed lter in this Section.). If f is smll everywhere in < x < b then e T will be smll. This reflects the fct tht we worked out the integrl of stright line function, insted of the integrl of f. If f is long wy from being stright line then f will be lrge nd so we must expect the error e T to be lrge too. We noted bove tht the expression for e T is less useful thn it might be becuse it involves the unknown quntity c. We perform trde-off to get round this problem. The expression bove gives n exct vlue for e T, but we do not know enough to evlute it. So we replce the expression with one we cn evlute, but it will not be exct. We replce f (c) with worst cse vlue to obtin n upper bound on e T. This worst cse vlue is the lrgest (positive or negtive) vlue tht f (x) chieves for x b. This leds to e T mx f (b )3 (x). x b We summrise this in Key Point 6. Key Point 6 Error in the Simple Trpezium Rule The error, e T, in the simple trpezium pproximtion to mx f (x) x b (b )3 f(x) dx is bounded bove by 3 HELM (8): Workbook 3: Numericl Methods of Approximtion

6 Exmple Work out the error bound (to 6 deciml plces) for the simple trpezium method pproximtions to () π/4 sin(x) dx (b) cosh(x) dx Solution In ech cse the trickiest prt is working out the mximum vlue of f (x). () Here f(x) = sin(x), therefore f (x) = cos(x) nd f (x) = sin(x). The function sin(x) tkes vlues between nd when x vries between nd π/4. Hence e T < (π/4)3 =.8548 to 6 deciml plces. (b) If f(x) = cosh(x) then f (x) = cosh(x) too. The mximum vlue of cosh(x) for x between nd will be cosh() = , to 6 deciml plces. Hence, in this cse, e T < (3.7696) ( )3 =.583 to 6 deciml plces. (In Exmple we used rounded vlue of cosh(). To be on the sfe side, it is best to round this number up to mke sure tht we still hve n upper bound on e T. In this cse, of course, rounding up is wht we would nturlly do, becuse the seventh deciml plce ws 6.) Tsk Work out the error bound (to 5 significnt figures) for the simple trpezium method pproximtions to () 5 x dx (b) ln(x) dx Your solution () HELM (8): Section 3.: Numericl Integrtion 33

7 Answer If f(x) = x = x / then f (x) = x / nd f (x) = 4 x 3/. The negtive power here mens tht f tkes its biggest vlue in mgnitude t the left-hnd end of the intervl [, 5] nd we see tht mx x 5 f (x) = f () = 4. Therefore e T < 4 43 =.3333 to 5 s.f. Your solution (b) Answer Here f(x) = ln(x) hence f (x) = /x nd f (x) = /x. It follows then tht mx x f (x) = nd we conclude tht e T < 3 = to 5 s.f. One deficiency in the simple trpezium rule is tht there is nothing we cn do to improve it. Hving computed n error bound to mesure the qulity of the pproximtion we hve no wy to go bck nd work out better pproximtion to the integrl. It would be preferble if there were prmeter we could lter to tune the ccurcy of the method. The following pproch uses the simple trpezium method in wy tht llows us to improve the ccurcy of the nswer we obtin. 34 HELM (8): Workbook 3: Numericl Methods of Approximtion

8 3. The composite trpezium rule The generl ide here is to split the intervl [, b] into sequence of N smller subintervls of equl width h = (b )/N. Then we pply the simple trpezium rule to ech of the subintervls. Figure 7 below shows the cse where N = (nd h = (b )). To simplify nottion lter on we let f = f(), f = f( + h) nd f = f( + h) = f(b). f f(x) f f h Figure 7 b x Applying the simple trpezium rule to ech subintervl we get f(x) dx (re of first trpezium) + (re of second trpezium) = h(f + f ) + h(f + f ) = h ( f + f + f ) where we remember tht the width of ech of the subintervls is h, rther thn the b we hd in the simple trpezium rule. The next improvement will come from tking N = 3 subintervls (Figure 8). Here h = (b ) 3 is smller thn in Figure 7 bove nd we denote f = f(), f = f( + h), f = f( + h) nd f 3 = f( + 3h) = f(b). (Notice tht f nd f men something different from wht they did in the N = cse.) f 3 f(x) f f f h Figure 8 b x HELM (8): Section 3.: Numericl Integrtion 35

9 As Figure 8 shows, the pproximtion is getting closer to the grey shded re nd in this cse we hve f(x) dx h(f + f ) + h(f + f ) + h(f + f 3 ) = h ( f + {f + f } + f 3 ). The pttern is probbly becoming cler by now, but here is one more improvement. In Figure 9 N = 4, h = 4 (b ) nd we denote f = f(), f = f( + h), f = f( + h), f 3 = f( + 3h) nd f 4 = f( + 4h) = f(b). f 4 f(x) f 3 f f f h Figure 9 b x This leds to f(x) dx h(f + f ) + h(f + f ) + h(f + f 3 ) + + h(f 3 + f 4 ) = h ( f + {f + f + f 3 } + f 4 ). We generlise this ide into the following Key Point. 36 HELM (8): Workbook 3: Numericl Methods of Approximtion

10 The composite trpezium rule for pproximting. Choose N, the number of subintervls,. where Key Point 7 Composite Trpezium Rule f(x) dx h ( f + {f + f + + f N } + f N ), f(x) dx is crried out s follows: h = b N, f = f(), f = f( + h),..., f n = f( + nh),..., nd f N = f( + Nh) = f(b). Exmple Using 4 subintervls in the composite trpezium rule, nd working to 6 deciml plces, pproximte cosh(x) dx Solution In this cse h = ( )/4 =.5. We require cosh(x) evluted t five x-vlues nd the results re tbulted below to 6 d.p. x n f n = cosh(x n ) It follows tht cosh(x) dx h (f + f 4 + {f + f + f 3 }) = (.5) ( { }) = HELM (8): Section 3.: Numericl Integrtion 37

11 Tsk Using 4 subintervls in the composite trpezium rule pproximte ln(x) dx Your solution Answer In this cse h = ( )/4 =.5. We require ln(x) evluted t five x-vlues nd the results re tbulted below t 6 d.p. x n f n = ln(x n ) It follows tht ln(x) dx h (f + f 4 + {f + f + f 3 }) = (.5) ( { }) = HELM (8): Workbook 3: Numericl Methods of Approximtion

12 How good is the composite trpezium rule? We cn work out n upper bound on the error incurred by the composite trpezium method. Fortuntely, ll we hve to do here is pply the method for the error in the simple rule over nd over gin. Let e N T denote the error in the composite trpezium rule with N subintervls. Then e N T mx = h3 st subintervl f (x) h3 + mx nd subintervl f (x) h mx lst subintervl f (x) ( ) mx f (x) + mx f (x) mx f (x). st subintervl nd subintervl lst subintervl }{{} N terms This is ll very well s piece of theory, but it is wkwrd to use in prctice. The process of working out the mximum vlue of f seprtely in ech subintervl is very time-consuming. We cn obtin more user-friendly, if less ccurte, error bound by replcing ech term in the lst brcket bove with the biggest one. Hence we obtin e N T h 3 ( N mx f (x) x b ) This upper bound cn be rewritten by reclling tht Nh = b, nd we now summrise the result in Key Point. h3 Key Point 8 Error in the Composite Trpezium Rule The error, b e N, T in the N-subintervl composite trpezium pproximtion to f(x) dx is bounded bove by mx f (x) x b (b )h Note: the specil cse when N = is the simple trpezium rule, in which cse b = h (refer to Key Point 6 to compre). The formul in Key Point 8 cn be used to decide how mny subintervls to use to gurntee specific ccurcy. HELM (8): Section 3.: Numericl Integrtion 39

13 Exmple 3 The function f is known to hve second derivtive with the property tht f (x) < for x between nd 4. Using the error bound given in Key Point 8 determine how mny subintervls re required so tht the composite trpezium rule used to pproximte 4 f(x) dx cn be gurnteed to be in error by less thn 3. Solution We require tht tht is (b )h 4h <.5. <.5 This implies tht h <.5 nd therefore h <.83. Now N = (b )/h = 4/h nd it follows tht N > Clerly, N must be whole number nd we conclude tht the smllest number of subintervls which gurntees n error smller thn.5 is N = 358. It is worth remembering tht the error bound we re using here is pessimistic one. We effectively use the sme (worst cse) vlue for f (x) ll the wy through the integrtion intervl. Odds re tht fewer subintervls will give the required ccurcy, but the vlue for N we found here will gurntee good enough pproximtion. Next re two Tsks for you to try. 4 HELM (8): Workbook 3: Numericl Methods of Approximtion

14 Tsk The function f is known to hve second derivtive with the property tht f (x) < 4 for x between nd 4. Using the error bound given in Key Point 8 determine how mny subintervls re required so tht the composite trpezium rule used to pproximte 4 f(x) dx cn be gurnteed to hve n error less thn.. Your solution Answer We require tht tht is 4 (b )h 7h <. <. This implies tht h <.74 nd therefore h <.444. Now N = (b )/h = 5/h nd it follows tht N > Clerly, N must be whole number nd we conclude tht the smllest number of subintervls which gurntees n error smller thn. is N = 8. HELM (8): Section 3.: Numericl Integrtion 4

15 Tsk It is given tht the function e x / hs second derivtive tht is never greter thn in bsolute vlue. () Use this fct to determine how mny subintervls re required for the composite trpezium method to deliver n pproximtion to π e x / dx tht is gurnteed to hve n error less thn. (b) Find n pproximtion to the integrl tht is in error by less thn. Your solution () Answer We require tht (b )h π <.5. This mens tht h <.5398 nd therefore, since N = /h, it is necessry for N = 3 for the error bound to be less thn ±. Your solution (b) 4 HELM (8): Workbook 3: Numericl Methods of Approximtion

16 Answer To crry out the composite trpezium rule, with h = 3 x =, h, h,. This evlution gives we need to evlute f(x) = π e x / t f() = f =.39894, f(h) = f =.37738, f(h) = f =.3945 nd f() = f 3 =.497, ll to 5 deciml plces. It follows tht π e x / dx h(f + f 3 + {f + f }) =.339 We know from prt () tht this pproximtion is in error by less thn. Exmple 4 Determine the minimum number of steps needed to gurntee n error not exceeding ±., when evluting cosh(x ) dx using the trpezium rule. Solution f(x) = cosh(x ) f (x) = x sinh(x ) f (x) = sinh(x ) + 4x cosh(x ) Using the error formul in Key Point 8 E = h { sinh(x ) + 4x cosh(x )} x [, ] E mx occurs when x =. > h { sinh() + 4 cosh()} h <./{( sinh() + 4 cosh()} h <.48 h <.3753 n 6.65 n = 7 needed HELM (8): Section 3.: Numericl Integrtion 43

17 Tsk Determine the minimum of strips, n, needed to evlute by the trpezium rule: π/4 {3x.5 sin(x)} dx such tht the error is gurnteed not to exceed ±.5. Your solution Answer f(x) = 3x.5 sin(x) f (x) = sin(x) Error will be mximum t x = π 4 so tht sin(x) = E = (b ) h f () (x) x [, π 4 ] E = π 48 h 6{ + sin(x)}, x [, π 4 ] E mx = π 48 h () = πh 4 We need πh 4 <.5 h <. π Now nh = (b ) = π so n = π 4 4h π We need n > = so n = required h < HELM (8): Workbook 3: Numericl Methods of Approximtion

18 4. Other methods for pproximting integrls Here we briefly describe other methods tht you my hve herd, or get to her, bout. In the end they ll mount to the sme sort of thing, tht is we smple the integrnd f t few points in the integrtion intervl nd then tke weighted verge of ll these f vlues. All tht is needed to implement ny of these methods is the list of smpling points nd the weight tht should be ttched to ech evlution. Lists of these points nd weights cn be found in mny books on the subject. Simpson s rule This is bsed on pssing qudrtic through three eqully spced points, rther thn pssing stright line through two points s we did for the simple trpezium rule. The composite Simpson s rule is given in the following Key Point. The composite Simpson s rule for pproximting Key Point 9 Composite Simpson s Rule. Choose N, which must be n even number of subintervls,. Clculte f(x) dx f(x) dx is crried out s follows: 3 h ( f + 4{f + f 3 + f f N } + {f + f 4 + f f N } + f N ) where h = b N, f = f(), f = f( + h),..., f n = f( + nh),..., nd f N = f( + Nh) = f(b). The formul in Key Point 9 is slightly more complicted thn the corresponding one for composite trpezium rule. One wy of remembering the rule is the lern the pttern which show tht the end point vlues re multiplied by, the vlues with odd-numbered subscripts re multiplied by 4 nd the interior vlues with even subscripts re multiplied by. HELM (8): Section 3.: Numericl Integrtion 45

19 Exmple 5 Using 4 subintervls in the composite Simpson s rule pproximte cosh(x) dx. Solution In this cse h = ( )/4 =.5. We require cosh(x) evluted t five x-vlues nd the results re tbulted below to 6 d.p. x n f n = cosh(x n ) It follows tht cosh(x) dx 3 h (f + 4f + f + 4f 3 + f 4 ) = (.5) ( ) 3 = , where this pproximtion is given to 6 deciml plces. This pproximtion to cosh(x) dx is closer to the true vlue of sinh() (which is to 6 d.p.) thn we obtined when using the composite trpezium rule with the sme number of subintervls. Tsk Using 4 subintervls in the composite Simpson s rule pproximte ln(x) dx. Your solution 46 HELM (8): Workbook 3: Numericl Methods of Approximtion

20 Answer In this cse h = ( )/4 =.5. There will be five x-vlues nd the results re tbulted below to 6 d.p. x n f n = ln(x n ) It follows tht ln(x) dx 3 h (f + 4f + f + 4f 3 + f 4 ) = (.5) ( ) 3 =.3866 to 6 d.p. How good is the composite Simpson s rule? On pge 39 (Key Point 8) we sw formul for n upper bound on the error in the composite trpezium method. A corresponding result for the composite Simpson s rule exists nd is given in the following Key Point. Key Point Error in Composite Simpson s Rule The error in the N-subintervl composite Simpson s rule pproximtion to bove by mx f (iv) (x) x b (b )h4 8 f(x) dx is bounded (Here f (iv) is the fourth derivtive of f nd h is the subintervl width, so N h = b.) The formul in Key Point cn be used to decide how mny subintervls to use to gurntee specific ccurcy. HELM (8): Section 3.: Numericl Integrtion 47

21 Exmple 6 The function f is known to hve fourth derivtive with the property tht f (iv) (x) < 5 for x between nd 5. Determine how mny subintervls re required so tht the composite Simpson s rule used to pproximte 5 f(x) dx incurs n error tht is gurnteed less thn.5. Solution We require tht 5 4h4 8 <.5 This implies tht h 4 <.45 nd therefore h < Now N = 4/h nd it follows tht N > For the composite Simpson s rule N must be n even whole number nd we conclude tht the smllest number of subintervls which gurntees n error smller thn.5 is N =. Tsk The function f is known to hve fourth derivtive with the property tht f (iv) (x) < for x between nd 6. Determine how mny subintervls re required so tht the composite Simpson s rule used to pproximte 6 f(x) dx incurs n error tht is gurnteed less thn.5. Your solution 48 HELM (8): Workbook 3: Numericl Methods of Approximtion

22 Answer We require tht 4h4 8 <.5 This implies tht h 4 <.875 nd therefore h <.89. Now N = 4/h nd it follows tht N > 9.49 N must be n even whole number nd we conclude tht the smllest number of subintervls which gurntees n error smller thn.5 is N =. The following Tsk is similr to one tht we sw erlier in this Section (pge 4). Using the composite Simpson s rule we cn chieve greter ccurcy, for similr mount of effort, thn we mnged using the composite trpezium rule. Tsk It is given tht the function e x / hs fourth derivtive tht is never greter thn 3 in bsolute vlue. () Use this fct to determine how mny subintervls re required for the composite Simpson s rule to deliver n pproximtion to π e x / dx tht is gurnteed to hve n error less thn 4. Your solution Answer We require tht 3 π (b )h 4 8 <.5. This mens tht h 4 < nd therefore h < Since N = /h it is necessry for N = 4 for the error bound to be gurnteed to be less thn ± 4. HELM (8): Section 3.: Numericl Integrtion 49

23 (b) Find n pproximtion to the integrl tht is in error by less thn 4. Your solution Answer In this cse h = ( )/4 =.5. We require results re tbulted below to 6 d.p. x n π e x n / It follows tht π e x / evluted t five x-vlues nd the π e x / dx 3 h (f + 4f + f + 4f 3 + f 4 ) = (.5) ( ) = to 6 d.p. We know from prt () tht this pproximtion is in error by less thn 4 5 HELM (8): Workbook 3: Numericl Methods of Approximtion

24 Exmple 7 Find out how mny strips re needed to be sure tht 4 sinh(t) dt is evluted by Simpson s rule with error less thn ±. Solution E = (b ) 8 h4 (6) sinh(x) < x < 4 E 64h sinh(8) 8. h sinh(8) h.84 4 nh = b n.84 = 9.9 So n = 9 is needed (minimum even number). HELM (8): Section 3.: Numericl Integrtion 5

25 Engineering Exmple Plstic bottle design Introduction Mnufcturing continers is lrge nd vried industry nd optimum pckging cn sve compnies millions of pounds. Although determining the cpcity of continer nd mount of mteril needed cn be done by physicl experiment, mthemticl modelling provides cost-effective nd efficient mens for the designer to experiment. Problem in words A mnufcturer is designing new plstic bottle to contin 9 ml of fbric softener. The bottle is circulr in cross section, with vrying rdius given by r = 4 +.5z.7z +.z 3 where z is the height bove the bse in cm. () Find n expression for the volume of the bottle nd hence show tht the fill level needs to be pproximtely 8 cm. (b) If the wll thickness of the plstic is mm, show tht this is lwys smll compred to the bottle rdius. (c) Hence, find the volume of plstic required to mnufcture bottle which is cm tll (include the plstic in the bse nd side wlls), using numericl method. A grph the rdius ginst z is shown below: 8 6 z Figure Mthemticl sttement of problem Clculte ll lengths in centimetres. () The formul for the volume of solid of revolution, revolved round the z xis between z = nd z = d is d πr dz. We hve to evlute this integrl. (b) To show tht the thickness is smll reltive to the rdius we need to find the minimum rdius. r 5 HELM (8): Workbook 3: Numericl Methods of Approximtion

26 (c) Given tht the thickness is smll compred with the rdius, the volume cn be tken to be the surfce re times the thickness. Now the surfce re of the bse is esy to clculte being π 4, but we lso need to clculte the surfce re for the sides, which is much hrder. For n element of height dz this is πz (the slnt height) of the surfce between z nd z +dz. ( ) The slnt height is, nlyticlly dr + dz, or equivlently the distnce between dz (r(z), z) nd (r(z + dz), z + dz), which is esier to use numericlly. ( ) dr Anlyticlly the surfce re to height is πr + dz; we shll pproximte this dz numericlly. This will give the re of the side surfce. Mthemticl nlysis () We could clculte this integrl exctly, s the volume is d π(4+.5z.7z +.z 3 ) dz but here we do this numericlly (which cn often be simpler pproch nd possibly is so here). To do tht we need to keep n eye on the likely error, nd for this problem we shll ensure the error in the integrls is less thn ml. The formul for the error with the trpezium rule, with step h nd integrted from to (ssuming from the problem tht we shll not integrte over lrger rnge) is h mx f. Doing this crudely with f = πg where g(z) = 4 +.5z.7z +.z 3 we see tht g(z) = 58 (using only positive signs nd z ) nd g (z).5 +.4z +.6z = 5.7 < 6, nd g (z).4 +.z.38. Therefore f = π(gg + (g ) ) ( )π < 7π, so h mx f 63h. We need h < /63, or h <.43. We will use h =., nd the error will be t most.5. The pproximtion to the integrl from to 8 is πg () i= πg (.i). + πg (8). (reclling the multiplying fctor is hlf for the first nd lst entries in the trpezium rule). This yields vlue of 899.7, which is certinly within ml of 9. (b) From the grph the minimum rdius looks to be bout t bout z = 8. Looking more exctly (found by solving the qudrtic to find the plces where the derivtive is zero, or by plotting the vlues nd by inspection), the minimum is t z = 8.93, when r =.948 cm. So the thickness is indeed smll (lwys less thn.6 of the rdius t ll plces.) HELM (8): Section 3.: Numericl Integrtion 53

27 ( ) dr (c) For the re of the side surfce we shll clculte πr + dz numericlly, using dz ( ) dr the trpezium rule with step. s before. + dz = (dz) dz + (dr), which we shll pproximte t point z n by (z n+ z n ) + (r n+ r n ), so evluting r(z) t intervls of. gives the pproximtion πr() 999 (.) + (r(.) r()) + πr(.i) (.) + (r(.(i + )) r(.i)) i= +πr() (.) + (r() r(9.98)). Clculting this gives 473 cm. Approximting the nlyticl expression by direct numericl clcultion gives 474 cm. (The nswer is between nd cm, so this vrition is understndble nd does not indicte n error.) The bottom surfce re is 6π = 5.3 cm, so the totl surfce re we my tke to be = 54 cm, nd hence the volume of plstic is 54. = 5.4 cm 3. Mthemticl comment An lterntive to using the trpezium rule is Simpson s rule which will require mny fewer steps. When using computer progrm such s Microsoft Excel hving n efficient method my not be importnt for smll problem but could be significnt when mny clcultions re needed or computtionl power is limited (such s if using progrmmble clcultor). The reder is invited to repet the clcultions for () nd (c) using Simpson s rule. The nlyticl nswer to () is given by 8 π(6 + 4z.3z.54z z 4.8z 5 +.4z 6 ) dz which gives to 4 d.p. 54 HELM (8): Workbook 3: Numericl Methods of Approximtion

28 Exercises. Using 4 subintervls in the composite trpezium rule pproximte 5 x dx.. The function f is known to hve second derivtive with the property tht f (x) < for x between nd 3. Using the error bound given erlier in this Section determine how mny subintervls re required so tht the composite trpezium rule used to pproximte 3 f(x) dx cn be gurnteed to hve n error in it tht is less thn.. 3. Using 4 subintervls in the composite Simpson rule pproximte 5 x dx. 4. The function f is known to hve fourth derivtive with the property tht f (iv) (x) < 6 for x between nd 5. Determine how mny subintervls re required so tht the composite Simpson s rule used to pproximte 5 f(x) dx incurs n error tht is less thn.. 5. Determine the minimum number of steps needed to gurntee n error not exceeding ±. when numericlly evluting 4 ln(x) dx using Simpson s rule. HELM (8): Section 3.: Numericl Integrtion 55

29 Answers. In this cse h = (5 )/4 =. We require x evluted t five x-vlues nd the results re tbulted below x n f n = x n It follows tht 5 x dx h (f + f 4 + {f + f + f 3 }) = ) ( () { } = (b )h. We require tht <.. This implies tht h <.368. Now N = (b )/h = /h nd it follows tht N > 3.68 Clerly, N must be whole number nd we conclude tht the smllest number of subintervls which gurntees n error smller thn. is N = In this cse h = (5 )/4 =. We require x evluted t five x-vlues nd the results re s tbulted in the solution to Exercise. It follows tht 5 x dx 3 h (f + 4f + f + 4f 3 + f 4 ) = () ( ) 3 = We require tht 6 6h4 8 <.. This implies tht h4 <.5 nd therefore h < Now N = 6/h nd it follows tht N > We know tht N must be n even whole number nd we conclude tht the smllest number of subintervls which gurntees n error smller thn. is N = HELM (8): Workbook 3: Numericl Methods of Approximtion

30 Answers 5. f(x) = ln(x) f (4) (x) = 6 x 4 Error = (b )h4 f (4) (x) 8 =, b = 4 E = h4 (6/x 4 ) 8 x [, 4] E mx = h h h.4467 Now nh = (b ) so n.4467 n n = 8 (minimum even number) HELM (8): Section 3.: Numericl Integrtion 57