C1M14. Integrals as Area Accumulators


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1 CM Integrls s Are Accumultors Most tetbooks do good job of developing the integrl nd this is not the plce to provide tht development. We will show how Mple presents Riemnn Sums nd the ccompnying digrms nd then focus on integrls from to. We her bout moving the golpost when stndrds of performnce re rised,but here tht is ectly wht is hppening becuse is vrible. We will be focusing on functions tht re bounded on their domin,which will be n intervl [, b],nd re continuous t ll ecept possibly finite number of points. For this reson,when we brek [, b] upinton subintervls we my ssume tht the subintervls ll hve the sme length, b =. In more generl situtions,we llow the n subintervls to be of rndom length nd then force the lrgest intervl to get smll s mens of controlling the pproimtion in the limit which we will cll definite integrl. Let s brek ll this down into smll prts nd then ssemble them in useful wy.. Suppose tht f is bounded function on [, b], f is continuous t ll but t most finite number of points,nd n is positive integer.. Let = b n. Then there re n + points { } n i determined by = = +, = +, = +,..., i = + i,..., n = + n = b 3. In ech subintervl [ i, i ], vlue t i is selected. Obviously i t i i.. Intuitively,we think of f(t i s the re of rectngle of height f(t i nd bse with width plced bove the intervl [ i, i ]. 5. The vlue f(t i is clled Riemnn Sum nd its vlue provides n pproimtion to the signed n= re between y = f( nd the is.. We write lim f(t i = f( d n n= 7. On subintervls where f( <,there will be negtive contribution,so we must be creful how we use the phrse the integrl of f is the re beneth the curve. It mkes sense for positive f only. Mple Emples: Mple hs three plotting commnds leftbo, middlebo, rightbo tht illustrte Riemnn sums. It lso hs three numericl commnds tht compute Riemnn sums, leftsum, middlesum, rightsum. We will illustrte their use for on the intervl [, 3] using 3 subintervls. These commnds re ll in the pckge student. > restrt: with(plots: with(student: > leftsum(ˆ,=..3,3; ( ( i > vlue(leftsum(ˆ,=..3,3; > evlf(leftsum(ˆ,=..3,3; > rightsum(ˆ,=..3,3; 3 i= ( 3 i= ( + 3 i > vlue(rightsum(ˆ,=..3,3; 57 > evlf(rightsum(ˆ,=..3,3; Leftbo is on the left nd rightbo is on the right. > leftbo(ˆ,=..3,3;
2 Mple Animtion Emple: We will use the function f( = sin(π on[, 3] nd set up n nimtion 3 of the pproimtion of sin(π d using middlebo. The reder is urged to type in the commnds below or to copy nd pste,nd to wtch the nimtion. The vlue of ech Riemnn sum using the midpoint of the subintervl is shown ner the top of the disply. > restrt: with(plots: > f:=>*sin(pi*; f := sin(π > nstrt:=5; frmeno:=5; nstrt := 5 frmeno := 5 > frmenumbers:=[seq(nstrt+i,i=..(frmeno]: > A:=disply(seq(middlebo(f(,=..3,i,i=frmenumbers,insequence=true: > B:=nimte(f(,=..3,y=..3,color=red, frmes=frmeno: > C:=disply(seq(tetplot([.3,.,evlf(middlesum(f(,=..3,i],i=5..(frmeno+, insequence=true: > disply(a,b,c;
3 Let s look t the lst pproimtion nd the ctul nswer in floting point. By using uppercse I in Int on the left we obtin n inert integrl,while the lowercse i yields n ctive commnd integrte it on the right in int. > evlf(middlesum(f(,=..3,frmeno+;.9577 > Int(f(,=..3=evlf(int(f(,=..3; 3 sin(π d = Ply the nimtion! Click on the disply. A bo will pper round the figure nd tpe plyer window will pper in the contet br. Click on the Ply button nd wtch! It is importnt to note tht Mple dels with definite integrls by simply including the rnge. For emple,the vlue of f( d results from int(f(,=..b. Now we will turn our ttention to f( nd A( = re s moves from left to right. f(t dt. This integrl intuitively ccumultes A( 8 height = re ccumulted f( A( 3 We will stte the first prt of the Fundmentl Theorem of Clculus nd then illustrte it using function defined in piecewise mnner. The Fundmentl Theorem of Clculus (Prt One: Assume tht f is continuous on [, b] nd tht the function A is defined by A( = f(t dt for t b. Then, A ( =f( for ll in (, b. In other words, A is n ntiderivtive for f. Sometimes we write Mple Emple: D ( f(t dt = f( Suppose tht f is defined by:, if f( = ln(, if <<e ( e, if e We will plot the grph of f( nd of A( = f(t dt,which of course is the re ccumultor function for f(. We will note tht f is continuous,but is not differentible t = nd = e. It will be importnt to observe tht the grph of A( is smooth t ll points,indicting tht A is differentible everywhere,s it should be. > restrt: with(plots: with(student: with(plottools: > e:=ep(; 3
4 e := e > f:=>piecewise(<=,,> nd <e,ln(,>=e,(eˆ; f := piecewise(,, <nd <e,ln(,e, ( e > A:=>int(f(t,t=..; A := f(t dt > A:=plot(f(,=..(e+,color=red: > A:=plot(A(,=..(e+,color=blue: > A3:=plot(f(,=..3,color=yellow,filled=true: > A:=line([3,],[3,A(3],color=green: > A5:=tetplot([3.,,"A("]: > disply(a,a,a3,a,a5; A(.5 height = re ccumulted.5 f( A( 3 Without fnfre we will stte the rest of the Fundmentl Theorem of Integrl Clculus. The sme hypotheses still pply. Fundmentl Theorem of Clculus (Prt Two: If F is ny ntiderivtive of f,then f( d = F (b F ( Suppose tht f is continuous on [, b] nd tht F ( = to F on [, b],then there is point c in (, b for which F (b F ( = F (c b f(t dt f(t dt = f(c b b f(t dt = f(c b f(t dt. If we pply the Men Vlue Theorem This lst vlue is sometimes clled the verge vlue of f( over [, b]. We know tht F = f nd g( d = for ny function g nd these fcts were used bove. We summrize this s the Men Vlue Theorem for Integrls: If f is continuous on [, b],then there is number c in [, b] such tht f( d = f(c(b. Mple Emple: Find vlue c tht stisfies the MVT for integrls for f( =3e + sin(π on the intervl [, 3] nd disply grph tht illustrtes this theorem.
5 > restrt: with(plots: with(plottools: > f:=>3*ep(+*sin(pi*; f := 3 e + sin(π > A:=int(f(,=..3; A := 3 (e( 3 π π π > c:=fsolve(f(=a/3,,..3; c :=.98 > A:=plot(f(,=..3,color=red: > A:=plot(f(c,=..3,color=blue: > A3:=plot(f(c,=..3,color=yellow,filled=true: > A:=line([c,],[c,f(c],color=green: > A5:=tetplot([.5,.,"f("]: > A:=line([3,],[3,f(c],color=blck: > A7:=rrow([.,.8],[,.],.5,.3,.3,color=khki: > A8:=tetplot([.5,,"(c,f(c"]: > A9:=tetplot([.5,.7,"f(c(b"]: > disply(a,a,a3,a,a5,a,a7,a8,a9; 3 f( (c,f(c f(c(b CM Problems: Use Mple to solve the problems nd plot the grphs.. For f( =3e sin( on[,π],evlute (use evlf nd disply grphiclly the left,right,nd middle sums with 7 subintervls. Remember tht the commnds you will need re in student.. Define g( = for in [/, 5]. Then define G( = g(t dt. This is how ln( is defined in some tetbooks when eponentils nd logrithms re delyed until fter the integrl hs been developed. Disply g nd G on the sme grph nd fill the grph below g( fromto3. 3. For f( = sin( on[,π],use Mple to find the verge vlue of f on this intervl nd disply grph tht illustrtes the Men Vlue Theorem for Integrls. 5
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