The Traditional Definition of the Integral

Transcription

1 The Trditionl Definition of the Integrl y 2 y = f ( ) f ( ) d = lim n n f( i) i=1 where = 5 2 n nd i lies somewhere between 2+(i 1) nd 2 + i

2 1 Wht Is n Integrl? An integrl is process ( blck bo) tht tkes n input consisting of function f nd closed intervl [, b] nd produces n output which is numericl vlue [, b] f b f() d (For mthemticins, the most importnt issue here is: Wht sort of function cn f() be? We will ssume here tht we re working with functions which re continuous or brek up into finite number of continuous pieces pieced together This is dequte generlity for most pplictions of integrtion in the physicl sciences nd engineering) The integrl is chrcterized by three importnt properties

3 Aioms for the Integrl 1 Cumultive c f() d = b f() d + c b f() d 2 Quntizble If we replce the function f() by function by one which insted of incresing continuously increses by n incredibly lrge number of incredibly smll quntum jumps, then we will get very good pproimtion of the integrl More precisely, if we tke sequence of step functions which pproches f() s the limit, then b f() d will be the limit of the integrls of these step functions 3 Constnt functions If f() =C, where C is constnt, then b f() d =(b )C Properties 1 nd 3 imply tht if we hve two pprently different blck boes with these sme properties, then they would produce the sme vlue whenever f() is step function Since every Riemnn-integrble function is limit of step functions, Property 2 then implies tht the two blck boes ctully produce the sme vlue for ll functions

4 Mny blck boes in the Rel World, stisfy the three properties which chrcterize the integrl Rel World Reltionships With These Properties 1 The re under curve 2 The work done by force pplied over certin distnce 3 The distnce trveled by n object moving with given velocity function over given time intervl 4 The volume of solid obtined by revolving the grph of function y = f() round the -is 5 The force creted by given pressure over n re (This will correspond to double integrl) 6 The mss of solid corresponding to given density function (This will correspond to triple integrl) Since the three properties uniquely chrcterize the integrl, we cn then conclude tht the bove reltionships re given by integrls We don t need to re-invent the integrl for ech new ppliction s is commonly done in clculus books 2 Work = b 4 Volume = 5 Force = b Ω Force() d πf() 2 d Pressure(, y) d dy etc

5 HOWEVER Mny importnt pplictions of integrtion stisfy the first nd second conditions tht chrcterize the integrl but do not stisfy the third Nmely, when the function involved is constnt, the quntity to be clculted is not given simply by multipliction by b For instnce, let V be the volume of the solid of revolution obtined by revolving tht portion of the grph of function y = f() between = nd = b (ssuming 0 <<b)round the y-is y y =8 2 A cone, seen s volume of revolution obtined by revolving the line y =8 2 round the y-is

6 Then when f() isconstnth, we get hollow cylinder nd the volume is V = π(b 2 2 )H y A hollow cylinder, obtined by revolving the constnt function f() =H between = nd = b round the y-is H b Volume = πb 2 H π 2 H = πh(b 2 2 ) The multiplier b 2 2 doesn t fit our prdigm for the integrl If we try to force it into the previous prdigm, we might try this: Volume = b 2 2 πf() d( 2 ) (??) Now s good clculus students, we do in fct know tht the volume of the solid of revolution obtined from revolving curce round the y-is is given by n integrl, nmely V =2π b f() d This is different sort of integrl thn the ones we hve been considering, since the integrl contins not only the function f() but lso the independent vrible (In the lnguge of integrtion theory, for this ppliction of integrtion we need to use new mesure, nmely 2π d insted of d)

7 The Structure of n Appliction of Integrtion In generl, prticulr ppliction of integrtion will hve the following structure [, b] f Appliction Numericl Vlue All pplictions of integrtion will stisfy Property 1 of the integrl 1 Cumultive If <b<c then the vlue determined by looking t the function over the intervl [, c] will be the sum of the vlue determined over [, b] nd the vlue determined over [b, c] Most pplictions will lso stisfy Property 2 2 Quntizble If we replce the function f() by one which insted of incresing continuously increses by n incredibly lrge number of incredibly smll quntum jumps, then we will get very good pproimtion More precisely, if we tke sequence of step functions which pproches f() s the limit, then the vlue corresponding to f() will be the limit of the vlues corresponding to the step functions Mny importnt pplictions, however, will not stisfy the previously stted Property 3 When f() is constnt, the vlue of the quntity in question will not simply be given by multipliction by b

8 However we ssume tht it is known wht the correct formul is when the function f() is constnt then we will be ble to figure out the correct integrl formul The Golden Rule of Thumb for Integrls If resonble formul given by n integrl gives the correct nswers when pplied to constnt functions, nd if the prticulr ppliction in question stisfies Property 2, i e is quntizble, then the formul is in fct correct * The fine print Bsiclly, this will be good for formuls of the form b Φ(, f()) d, where Φ is continuous function of two vribles The only esy wy to screw up is to include the derivtive f () (or higher derivtives) in the integrnd For instnce, if v(t) is velocity, then the formul Distnce = t1 t 0 v(t)(1+v (t)) dt (??) gives the correct nswer when the velocity v(t) is constnt, or for tht mtter step function, but gives the wrong nswer in ll other cses

9 Why the Golden Rule Works Becuse the integrl is cumultive (i e dditive over disjoint intervls), if formul given by n integrl gives the correct nswer for constnt functions, then it will lso give the correct nswer for step functions But if the ppliction in question is quntizble, then the correct vlue corresponding to the function f() cn be obtined by tking the limit of the vlues for sequence of step functions converging to f() The reson for clling the principle bove Rule of Thumb is tht there re two wys it cn go wrong: First, the ppliction in question might not in fct be quntizble And second, the formul in question might involve f () or otherwise fil to meet the conditions of the footnote A few importnt pplictions of integrtion re not quntizble ccording to our definition nd hence do not stisfy Property 2 These pplictions cnnot be ignored The most obvious emple is the formul for the length of tht portion of the grph offunctiony=f() between points = nd = b If one quntizes function, i e pproimtes it by step function, one will not get good pproimtion to the length of its grph In fct, it is firly cler tht for step functions, the length is lwys b, wheres this is never true for functions whose slope is non-zero in t lest some plces

10 A step-function pproimtion to stright line b Theorem Suppose the quntity being computed by n ppliction hs n incresing reltionship to the function tht determines it (i e mking the function bigger lwys mkes the quntity bigger), nd tht for constnt functions, over given intervl the quntity depends continuously on the (constnt) vlue of the function Then the reltionship is quntizble Emple The volume obtined by revolving tht portion of the grph of positive function y = f() between = nd = b round the y-is (Clerly mking the function bigger will increse the volume) The continuity ssumption in the theorem is included to rule out the possibility of bizrre, contrived counter-emples In most cses, the theorem cn be proved without it in ny cse Most common pplictions of integrtion re incresing The sme thing holds if the reltionship between the function nd quntity being computed is decresing rther thn incresing

11 The ide of the proof of the theorem is tht one brckets function between step function below it nd step function bove it Since by ssumption, the ppliction in question is incresing, the vlue corresponding to the function will lie between the vlues corresponding to these two step functions But, s I will show in moment, by mking the steps smll enough, the vlues corresponding to the step functions cn be mde rbitrrily smll Thus the vlue corresponding to these step functions cn be mde rbitrrily close to the vlue for the function under considertion This is wht we men when we sy tht the ppliction is quntizble y 2 1 Lower step function: f s () f() y Upper step function: f S () f()

12 Proof of my clim tht the vlues for the ppliction in question corresponding to the two step functions brcketing f() will be etremely close together if the steps re mde smll enough: I will prove this only in the specil cse tht the function f() is incresing from to b Then looking t these two step functions together, we cn see tht ecept for the first step, the lower one is the sme s the upper one ecept shifted one step to the right (This works becuse we hve mde ll the steps of the sme width) y Upper step function: f S () f() 2 1 Lower step function: f s () f() As n emple, for the volume of revolution round the -is, using the lower step function gives n pproimtion (ssuming tht the function y = f() is incresing, s in the picture) V s = πf( 0 ) 2 + πf( 1 ) 2 + πf( 2 ) 2 + +πf( n 1 ) 2 nd using the upper step function gives n pproimtion V S = πf( 1 ) 2 + πf( 2 ) 2 + +πf( n 1 ) 2 + πf( n ) 2, where =(b )/n But it is evident tht V S V s =(πf( n ) 2 πf( 0 ) 2 ) =(πf(b) 2 πf() 2 ), nd this cn be mde s smll s we wnt by mking smll enough (i e mking n lrge enough) (Note tht the continuity ssumption in the theorem ws not required for the proof in this cse tht f() is n incresing function A similr proof works if f() is decresing function And lmost ll continuous functions cn be obtined by gluing together pieces of incresing functions with pieces of decresing ones)

13 And now for something different: An importnt bd emple Note tht the length of the grph of function does not hve n incresing reltionship to the function Mking the function f() lrger mkes its grph between = nd = b higher but not longer If we did not lredy know it, this would be hint tht length does not hve quntizble reltionship to the function However if one mkes the derivtive f () lrger (in bsolute vlue), then the curve gets steeper nd does in fct become longer This suggests the the length of the curve might be given by some integrl involving f () We hve seen tht we cnnot find the length of the grph of function f() by pproimting f() by step function However it seems plusible tht one could find the length by pproimting the grph of the function by set of line segments 2 1 y One should be ble to get better nd better pproimtions to the length of curve by pproimting the curve by sequence of line segments

14 y (b, f(b)) f(b) f() b (, f()) If f() =m + c is stright-line function, then the length of this curve from the point (, f()) to the point (b, f(b)) is (b )2 +(f(b) f()) 2 = (b ) 2 + m 2 (b ) 2 =(b ) 1+m 2 =(b ) 1+f () 2 Thus the formul Length = b 1+f () 2 gives the correct nswer for stright line segments d Since the length of ny curve cn be pproimted rbitrrily closely by replcing the curve by union of very short line segments, we see tht this formul gives the correct result for the length of the grph of ny function