Numerical Integration by using Straight Line Interpolation Formula

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1 Glol Jourl of Pure d Applied Mthemtics. ISSN Volume 13, Numer 6 (2017), pp Reserch Idi Pulictios Numericl Itegrtio y usig Stright Lie Iterpoltio Formul Mhesh Chlpuri* d J Suchrith 1,2 Deprtmet of Mthemtics, Osmi Uiversity, Hyderd-07, Idi Astrct The prolems of umericl itegrtio is to fid pproximte vlue of the Defiite Itegrl. I this pper we fid pproximte vlue of the Defiite Itegrl y usig Stright lie Iterpoltio formul. This formuls useful for equl d uequl spce poits. Keywords:Numericl Itegrtio,Numericl methods,averge rule 1 INTRODUCTION With the dvet of the moder high speed electroic digitl computer, the Numericl Itegrtio hve ee successfully pplied to study prolems i Mthemtics, Egieerig, Computer Sciece d Physicl Sciece. Numericl itegrtio, lso clled Qudrture, is the study of how the umericl vlue of defiite itegrl c e foud. The purpose of this pper is to give ew method to fid pproximte clcultio of defiite itegrl (1) I = f(x)dx where f(x) is itegrle, i the Riem sese o [ ]. The limit of the itegrtio my e fiite or semi-fiite. The meig of the itegrtio (1) is the Are of fig. (1). It's ouded re y three stright lies y=0, x=, x= d oe curve f(x).

2 2124 Mhesh Chlpuri d J Suchrith Figure 1: Vlue of Defiite itegrl I this pper, we develop stright lie equtio y(x) = x such tht (2) I = f(x)dx y(x)dx The grphicl meig of ove itegrl is, there exist res A d B such tht A=B, see fig: 1 2 PRELIMINARIES Defitio2. 1: The Itegrtio (1) is pproximted y fiite lier comitio of vlue of f(x) i the form (3) I = f(x)dx = λ k f k k=0 where x k, k=0,1,..., re clled the scisss or odes distriuted withi the limits of itegrtio [ ] d λ k, k=0,1,..., re clled weight of the itegrtio rule of the qudrture formul. The error costet of (3) is give s (4) E = f(x)dx λ k f k k=0 Defitio2. 2: Order of ccurcy, or precisio, of Qudrture formul is the lrgest positive iteger such tht the formul is exct for x k, for ech k = 0,1,...,. Defitio2. 3: The smllest polyomil is lier polyomil, the stright lie. Lest squre stright lie pproximtios re extremely useful d commo pproximte fit. Give +1

3 Numericl Itegrtio y usig Stright Lie Iterpoltio Formul 2125 dt poits (x i, f i ),1,..., the the est stright lie through the set of dt poits is y = x, where 1 d 2 re stisfy the equtios (5) 1 ( + 1) + 2 x i = f(x i ), 1 x i + 2 x 2 i = x i f(x i ). The equtio (5) is clled orml-equtio of lest squre fit. They c solve for 1 d 2 y Guss Elimitio method. Theorem 1 y(x) is stright lie iterpoltio equtio of the cotiuous d itegrle fuctio f(x) o[,].the (6) y(x i ) f(x i ) where =, x = Proof. :-Let y(x) = x where 1, 2 re costts the y( ) = d y(x 1 ) = x 1 d so o y(x ) = x. ddig ll we get y(x i ) = ( + 1) pply Norml Equtio (4) we get x i y(x i ) f(x i ) Theorem 2 Let y(x) is iterpoltio polyomil equtio of the cotiuous d itegrle fuctio f(x) o[,] d y(x i ) f(x i ), where = >, x = < if d oly if x x ydx fdx. As, itegrtio of f o [ ] is pproch to itegrtio of y(x) o [ ]. Proof. Multiplyig with h d limit h 0 i (6), we get It gives theorem lim h h 0 =0 y(x ) = lim h 0 h =0 f(x ).

4 2126 Mhesh Chlpuri d J Suchrith 3 MAIN RESULTS The odes x i,,1,2..., re equispced with =, x = with spcig h=(-)/.we fit stright lie y(x) = x with +1 dt poits (, f 0 ), (x 1, f 1 ),..., (x, f ). But we kow x i = + x x = + + h h = ( + 1) + h( + 1), 2 x i 2 = ( + 1) + h( + 1)(2 + 1) /6 + h 2 2 ( + 1) 2 /4, x i f i = x o f i + h if i. Solve the Norml equtio for 1 d 2, we get 1 = f i x i 2 x i x i f i ( + 1) x i 2 ( x i) 2 2 = ( + 1) x i f i x i f i ( + 1) x i 2 ( x i) 2 Sustitutig ove vlues i 1, 2 we get 1 = 2 h( + 1)( + 2) [(3 + h( + 1)) f i 3( + x ) if i ] 2 = 6 h( + 1)( + 2) [2 if i i f i ] y = x is the est stright lie pproximte fit to f(x) o [ ]. (7) I = f(x)dx ydx = ( x)dx = 1 h + 2h( + ) 2 = 1 h + 2 h(2 + h)/2 sustitutig 1 d 2, we get umericl solutio of give itegrtio, (8) we lso write x f(x)dx = f(x)dx h + 1 f(x i ) x f(x)dx h(f i s verge)

5 Numericl Itegrtio y usig Stright Lie Iterpoltio Formul 2127 or f(x)dx ( )(f i s verge) I this pper ove formul is clled Averge rule of umericl itegrtio of f(x). This method is exct for polyomil of degree 1, hece is of order 1. Error estimte The error of the this rule is x E = x 2 dx (x ) + 1 ( 2 + x x x 2 ) = h 3 2 /6 there fore the (9) R = E 2! f" (ξ) = h3 2 ( )3 f"(ξ) = f"(ξ), where ξ [ ]. So we kow from (9), the error of (9) is iverse proportiol to umer of suitervls(i.e R 1/). Other proof of Averge rule,we fit lest squre stright lie y = l for the discrete dt (x i, f i ),,1,...,, where x i s re eed't e equispced poits, we hve i.e or so I = I(l) = (f i y i ) 2 = miimum x f(x)dx = di dl = 2(f i l).1 = 0 ( ) = + 1 f(x i ) l = f i. x f(x) x ydx = ldx = l(x ) Note:-This rule does ot depedig o h, so this rule c e use for uequispced odes x i s. The grphiclly meig of Averge rule is, suppose the crossig pits of f(x) d y re P[l 1, m 1 ] d Q[l 2, m 2 ]. See grph fig: 2 A,B d C re res of itegrtio (f(x) y) from limits to l 1, l 1 to l 2 d l 2 to respectively. The the vlues of B, A+C re pproximtely sme. i.e B A+C.

6 2128 Mhesh Chlpuri d J Suchrith Figure 2: Grph of Averge rule The ifiite series i=1 f(x i ) is covergece if x 1 f(x) is fiite. We hve Averge rule x f(x)dx x 1 Write d h=1 we get lim ( 1)h i=1 ( 1)h f(x i ). i=1 f(x i ) f(x)dx x 1 f(x i ) f(x)dx. i=1 x 1 This is sme s itegrl test of ifiite series. Hece the theorem. Note:-Let, (, + ), d or h 0 the f(x)dx = lim [ h + 1 f(x i )] = lim h 0 [h f(x i )]. 3.1 Composite Formule To void the use of higher order methods d still oti ccurte results, we use the composite itegrtio methods. We divide the itervl [, ] ito umer of suitervls d evlute the itegrl i ech suitervl y prticulr method. We c derive composite formule from Averge rule x f(x)dx = f(x)dx h + 1 f(x i )

7 Numericl Itegrtio y usig Stright Lie Iterpoltio Formul Trpezoidl rule We divide the itervl [ ] ito suitervl, ech of legth h=(-)/. we deote the suitervl s (, x 1 ), (x 1, x 2 ),..., (x 1, x ) where =, x = d x i = + ih, i=1,2,...,. We write x f(x)dx f(x)dx x 1 x 1 x 2 I = f(x)dx = f(x)dx + x 1 write =1 i verge rule d pply to ech itegrtio we get (10) f(x)dx h 2 (f 0 + f + 2 f i ) i=1 this formul is clled composite Trpezoidl rule or 1/2 Averge rule /3-Averge rule I usig the 1/3 Averge rule, we eed three scisss. We divide the itervl [ ] i eve umer of suitervls of equl legth givig odd umer of scisss. If we divide the itervl [ ] ito 2N suitervl ech of legth h=(-)/(2n), the we get 2N+1 scisss, x 1,..., x 2N, x = + ih i=1,2...,2n-1.we write I = x f(x)dx = x 2 f(x)dx + x 4 f(x)dx x 2N f(x)dx. x 2 x 2N 2 We write =2 i Avrge we get I = x f(x)dx = 2h [(f f 1 + f 2 ) + (f 2 + f 3 + f 4 )+... +(f 2N 2 + f 2N 1 + f 2N )] (11) I = x f(x)dx 1 = 2h 3 (f 0 + f 1 + f 3 + f f 2N 1 + f 2N + 2(f 2 + f f 2N 2 )) This formul is clled 1/3-Averge rule Geerl form of 1/(m+1)-Averge rule I usig the 1/(m+1)-Averge rule, we eed mn+1 scisss. We divide the itervl [ ] i mn umer of suitervls of equl legth givig mn+1 umer of scisss. If we divide the itervl [ ] ito mn suitervl ech of legth h=(-)/(mn), the we get mn+1 scisss, x 1,..., x mn, x = + ih i=1,2...,mn-1.we write I = x f(x)dx = x m We write =m i Avrge we get f(x)dx + x 2m x m f(x)dx x mn x m(n 1) f(x)dx. I = x f(x)dx = mh x [(f 0 m f 1 + f f m ) + (f m + f m+1 + f m f 2m ) (f mn m + f mn (m 1) + f mn 1 + f 2N )]

8 2130 Mhesh Chlpuri d J Suchrith x (12) I = f(x)dx = mh m + 1 [2(f m + f 2m f m(n 1) ) + f 0 + f f m 1 + f m f mn ] this formul is clled 1/(m+1)-Averge rule. 4 PROBLEMS Prolem 1 Fid the pproximte vlue of 1 1 I = x dx, usig Averge rule with differet equl suitervls. Usig the exct solutio, fid the solute errors. Solutio:Results for the Averge rule to estimte the itegrl of f(x) = 1/(2 + x) from x = 0 to 1. The exct vlue is I exct = 1 1/(2 + x)dx = log(3) 0 log(2)= We hve I = f(x)dx = x f(x)dx h +1 f(x i ) I Error=I I exct This rule slowly covergece to exct vlue I d the error is hlf if is dle. So exct vlue come from =124 for four deciml plces. Prolem 2 Fid the pproximte vlue of 3 2 f(x)dx, usig tle x f(x)

9 Numericl Itegrtio y usig Stright Lie Iterpoltio Formul 2131 Solutio I this prolem h is ot equl to ll itervl, But we c pply Averge rule to this prolem.we hve I = 3 (3 2) f(x)dx = (f(x 2 0 ) + f(x 1 )+... +f(x 6 )) 7 = 1 ( ) 7 = = Prolem 3 Fid pproximte vlue of I = 3 1 si(x)ex dx fit stright lie y(x) such tht 3 y(x)dx = I. 1 Solutio:Let f(x) = si(x)e x d y e the stright lie y fit +1 dt poits (x i, f(x i )),,1,2,..,. Now we divide the itervl [1 3] ito two equl suitervl, tht is = 2 or h = 1. the 3 dt poits re (1, f(1)), (2, f(2)) d (3, f(3)). we fit stright lie y 2 y orml equtio (5) we get y 2 = 0.27x followig this we get y 4 = 0.78x y 8 = 1.17x y 16 = 1.39x y 32 = 1.51x d y 64 = 1.57x But we kow if the 3 y 1 (x)dx 3 f(x)dx. there fore I = 1 3 (1.57x )dx = the figure (3) shows tht the covergece of y (x) Figure 3: Covergece of y (x)

10 2132 Mhesh Chlpuri d J Suchrith 5 CONCLUSION We develop this ew method for esy to solve Defiite Itegrl of fiite itervl with equispced odes.if we hve ot equispced dt poit the we c use this Averge rule for itegrtio o give compct itervl (exmple 4.2 is exmple of this cocept). This method is esy to evlute of defiite itegrl. The Error is esy to derive. It's lier se cocept, we re developig this cocept to hight degree polyomils d defiite itegrl of this polyomil is sme s defiite itegrl of give fuctio o compct itervl [ ]. Also we develop this cocept to high dimesio. We re reserched out the odes, there re o fixed odes for give exct vlue of itegrtio for ll itegrle fuctios f(x). ACKNOWLEDGEMENTS We grtefully ckowledge the support of the UGC(Uiversity Grts Commissio) JRF d NET Fellowship,Idi. REFERENCES [1] M. Cocepcio Ausi, 2007, A Itroductio to Qudrture d Other Numericl Itegrtio Techiques, Ecyclopedi of Sttistics i Qulity d reliility. Chichester, Egld. [2] Gordo K. Smith, 2004, Numericl Itegrtio, Ecyclopedi of Biosttistics.2d editio, Vol-6 [3] Rjesh Kumr Sih, Rkesh Kumr,2010, Numericl method for evlutig the itegrle fuctio o fiite itervl, Itertiol Jourl of Egieerig Sciece d Techology.Vol-2(6) [4] Gerry Sozio, 2009, Numericl Itegrtio, Austrli Seior Mthemtics Jourl, Vol-23(1) [5] J. Oliver, 1971, The evlutio of defiite itegrls usig high-order formule, The Computer Jourl, Vol-14(3) [6] S.S Sstry, 2007, Itroductory Method of Numericl Alysis, Fourth Editio, Pretice-hll of Idi Privte Limited. [7] Richrd L. Burde, 2007, Numericl Alysis, Seve Editio, Itertiol Thomso Pulishig Compy. [8] Joh H. Mthew, 2000, Numericl Method for Mthemtics, Sciece d Egieerig, Secod Editio, Pretice Hll of Idi Privte Limited. [9] Dvid Kicid, Wrd Cheey, Numericl Alysis Mthemtics of Scietific Computig, Idi Editio, Americ Mthemticl Society, Third editio, 2010 pge o [10] M.K Ji, S.R.K. Iyegr, R.K. Ji, Numericl Methods for Scietists d Egieers Computtio.2005, pge d [11] Steve C. Chpr.Applied Numericl Methods with MATLABÂ for Egieers d Scietists, Third Editio.2012, pge o