Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to solve emples o pproimte integrls. Wht is integrtion? Integrtion is the proess o mesuring the re under untion plotted on grph. Why would we wnt to integrte untion? Among the most ommon emples re inding the veloity o ody rom n elertion untion, nd displement o ody rom veloity untion. Throughout mny engineering ields, there re wht sometimes seems like ountless pplitions or integrl lulus. You n red out some o these pplitions in Chpters 7.A-7.G. Sometimes, the evlution o epressions involving these integrls n eome dunting, i not indeterminte. For this reson, wide vriety o numeril methods hs een developed to simpliy the integrl. Here, we will disuss the Guss qudrture rule o pproimting integrls o the orm where I d is lled the integrnd, lower limit o integrtion upper limit o integrtion 7..
7.. Chpter 7. Figure Integrtion o untion. Guss Qudrture Rule Bkground: To derive the trpezoidl rule rom the method o undetermined oeiients, we pproimted d Let the right hnd side e et or integrls o stright line, tht is, or n integrted orm o d So d But rom Eqution, we wnt d. d to give the sme result s Eqution or Hene rom Equtions nd,
Guss Qudrture Rule 7.. Sine nd re ritrry onstnts or generl stright line Multiplying Eqution y nd sutrting rom Eqution gives 6 Sustituting the ove ound vlue o in Eqution gives 6 Thereore d 7 Derivtion o two-point Guss qudrture rule The two-point Guss qudrture rule is n etension o the trpezoidl rule pproimtion where the rguments o the untion re not predetermined s Method : nd, ut s unknowns nd. So in the two-point Guss qudrture rule, the integrl is pproimted s d I There re our unknowns,, nd. These re ound y ssuming tht the ormul gives et results or integrting generl third order polynomil,. Hene d d 8 The ormul would then give d 9
7.. Chpter 7. Equting Equtions 8 nd 9 gives Sine in Eqution, the onstnts,,, nd re ritrry, the oeiients o,,, nd re equl. This gives us our equtions s ollows. Without proo see Emple or proo o relted prolem, we n ind tht the ove our simultneous nonliner equtions hve only one eptle solution Hene d We n derive the sme ormul y ssuming tht the epression gives et vlues or the individul integrls o Method :, d, d, d nd d. The reson the ormul n lso e
Guss Qudrture Rule 7.. derived using this method is tht the liner omintion o the ove integrnds is generl third order polynomil given y. These will give our equtions s ollows d d d d These our simultneous nonliner equtions n e solved to give single eptle solution Hene d 6 Sine two points re hosen, it is lled the two-point Guss qudrture rule. Higher point versions n lso e developed. Higher point Guss qudrture ormuls For emple d 7 is lled the three-point Guss qudrture rule. The oeiients, nd, nd the untion rguments, nd re lulted y ssuming the ormul gives et epressions or integrting ith order polynomil d. Generl n -point rules would pproimte the integrl
7..6 Chpter 7. d....... n n 8 Arguments nd weighing tors or n-point Guss qudrture rules In hndooks see Tle, oeiients nd rguments given or n -point Guss qudrture rule re given or integrls o the orm n g d g 9 i i i Tle Weighting tors nd untion rguments used in Guss qudrture ormuls Points Weighting Ftors Funtion Arguments...7769.7769.6.888888889.6.7796669..7796669.788.6.6.788.866.998.998.866.69688.786867.68888889.786867.69688.967986.869..869.967986 6.79.6767.6799.6799.969.66986.86986.86986
Guss Qudrture Rule 7..7.6767. 66986.79 6 6.969 So i the tle is given or g d integrls, how does one solve d? The nswer lies in tht ny integrl with limits o [, ] n e onverted into n integrl with limits [, ]. Let mt I, then t I, then t suh tht m m Solving the two Equtions simultneously gives m Hene t d dt Sustituting our vlues o nd d into the integrl gives us d d Emple For n integrl d, show tht the two-point Guss qudrture rule pproimtes to where d
7..8 Chpter 7. Solution Assuming the ormul d gives et vlues or integrls, d d d d d, d d, nd d. Then E. E. E. E. E. Multiplying Eqution E. y nd sutrting rom Eqution E. gives E.6 The solution to the ove eqution is,, or/nd, or/nd. I. is not eptle s Equtions E.-E. redue to,,, nd. But sine, ut onlits with. II. is not eptle s Equtions E.-E. redue to,,, nd. Sine, then or hs to e zero ut this violtes. III. is not eptle s Equtions E.-E. redue to,,, nd. I, then
Guss Qudrture Rule 7..9 gives nd tht violtes. I, then tht violtes. Tht leves the solution o s the only possile eptle solution nd in t, it does not hve violtions see it or yoursel E.7 Sustituting E.7 in Eqution E. gives E.8 From Equtions E. nd E.8, E.9 Equtions E. nd E.9 gives E. Sine Eqution E.7 requires tht the two results e o opposite sign, we get Hene Emple d For n integrl d, derive the one-point Guss qudrture rule. Solution The one-point Guss qudrture rule is E. d E. d nd d Assuming the ormul gives et vlues or integrls, d d E. the other eqution eomes Sine,
7.. Chpter 7. E. Thereore, one-point Guss qudrture rule n e epressed s d E. Emple Wht would e the ormul or d i you wnt the ove ormul to give you et vlues o, d tht is, liner omintion o nd. Solution I the ormul is et or liner omintion o nd, then d d E. Solving the two Equtions E. simultneously gives 6 6 E. So d 6 6 E. Let us see i the ormul works. Evlute d using EqutionE.
Guss Qudrture Rule 7.. d [ ] [ ] 6 6 6. The et vlue o d is given y d 6. Any surprises? Now evlute d using Eqution E. d 6. The et vlue o d is given y d [ ] 6 9 Beuse the ormul will only give et vlues or liner omintions o nd not work etly even or simple integrl o d. Do you see now why we hoose s the integrnd or whih the ormul gives us et vlues? d, it does Emple Use two-point Guss qudrture rule to pproimte the distne overed y roket rom t 8 to t s given y ln 9.8t dt t 8 Also, ind the solute reltive true error.
7.. Chpter 7. Solution First, hnge the limits o integrtion rom [ 8, ] to [, ] 8 8 8 t dt d 8 9 d using Eqution gives Net, get weighting tors nd untion rgument vlues rom Tle or the two point rule,...7769..7769 Now we n use the Guss qudrture ormul 9 d [ 9 9 ] [.77 9.77 9 ] [.69.8] [96.87 78.8] 8. m sine.69 ln 9.8.69.69 96.87.8 ln 9.8.8.8 78.8 The solute reltive true error,, is True vlue 6. m t 6. 8. 6..6% t Emple Use three-point Guss qudrture rule to pproimte the distne overed y roket rom t 8 to t s given y ln 9.8t dt t 8 Also, ind the solute reltive true error.
Guss Qudrture Rule 7.. Solution First, hnge the limits o integrtion rom [ 8, ] to [, ] 8 8 8 t dt d 8 9 d The weighting tors nd untion rgument vlues re.6.7796669.888888889..6.7796669 nd the ormul is sine using Eqution gives 9 d [ 9 9 9 ].6.77967 9.8888889. 9.6.77967 9 [.6.79.88889 9..6 7.6] [.6 9.7.88889 8.7.6 79.69] 6. m.79 ln 9.8.79.79 9.7 9. ln 9.89. 9. 8.7 7.6 ln 9.87.6 7.6 79.69 The solute reltive true error, t, is True vlue 6. m 6. 6. t 6..%
7.. Chpter 7. INTEGRATION Topi Guss qudrture rule Summry These re tetook notes o Guss qudrture rule Mjor Generl Engineering Authors Autr Kw, Mihel Ketelts Dte August, We Site http://numerilmethods.eng.us.edu