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All three l abs are to be used as a supplement by you to better internalize the \" big picture\" and " }{TEXT 289 11 "should not" }{TEXT -1 466 " replac e the formal approach of the proofs and theory used to establish the p rocesses in general which are being rigorously presented in your text \+ and in class. Refer back to all three modules as you read the text and listen to your teacher to help solidify the notational and conceptual aspects of this very important section of calculus. As you work throu gh this module you will need to \"scroll\" down the pages using the co ntrol bar on the far right of the screen." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Introduction:" }}{PARA 0 "" 0 "" {TEXT -1 42 "In the firs t two modules you saw that the " }{TEXT 282 13 "area question" }{TEXT -1 27 " can be approached through " }{TEXT 283 32 "the summing of rect angular areas" }{TEXT -1 60 " based on various selection schemes or by defining an area " }{TEXT 284 43 "accumulation function (the antider ivative)." }{TEXT -1 2 " " }{TEXT 285 102 "It is the goal of this thi rd module to connect the two approaches in order to obtain a unified s ystem." }{TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 19 "Objecti ves Defined:" }}{PARA 0 "" 0 "" {TEXT 286 195 "The goal of this third \+ module is to connect the two approaches in determining the area beneat h a curve and to establish the proper notation for generalized results . We shall achieve this goal by:" }}{PARA 0 "" 0 "" {TEXT -1 57 "1. Re visiting the Riemann Rectangle Approximation Method." }}{PARA 0 "" 0 " " {TEXT -1 47 "2. Revisiting the Accumulation Function Method." }} {PARA 0 "" 0 "" {TEXT -1 46 "3. Examining their connections to each ot her. " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 45 "The Rectangle Approxima tion Method Revisited:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 300 "We first revisit the Riemann Rectangle Approximatio n Method from module 1. It seemed reasonable from that investigation t hat the approximation to the area under the curve improved as we incre ased the number of rectangles regardless of the scheme used. This supp osition can be written symbolically as:" }}{PARA 0 "" 0 "" {TEXT -1 26 " " }{OLE 1 4640 1 "[xm]Br=WfoRrB:::wk;nyy I;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::fyyyyyy:nY^::yAJyyyywY::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::JcvGYM t>^:fBWMtNHm=;:::::::n:;`:Z@[::ZOPyQaaf^r;V:>j=B:<:El rfH=MtFGYMq>>Wlj:gmlJ::::::>>?jyyiy=J:B:::::::;a::=ja^GE=;:::::::::N;? 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0 "" 0 "" {TEXT -1 65 " \+ " }{OLE 1 4128 1 "[xm]Br=WfoRr B:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::fyyyyyqyyyYJ^:fBWMtNHm=;:::::::n:;`:Z@[::JJTM;\\WXlL \\Aj;JZY:<:ElrfH=MtFGYMq>>Wlj:gmlJ::::::>^:N:yyyxI:;Z:::::::^h;Z:j:vCS mlJ::::::::::OJ;Zy=J:B::::::n:^<>:N:Y<>D_mlVH[KR<:;B:::::::JFNZ;N: :yayA:<::::::I:[X;:W:YJ:><:i:kJ:F=N=V=^=f=n=nYvY:::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::^=N:kwokT:F:WdG_dnB:arOMeU=DUSe J=uVMuRAtUCUS[TRIUSamBPJd`ppPpsErJYUW_U TEeV;cMEM:@K\\jS:DZ=pF?J;]kkB:=B:bUf[<^Z@H\\\\fVOb:Cb[cv:Cd:HRv>:;::F:;b:JEQZ:F:MZ=Vs<>:?jUC:;JZ:FZ\\:B:;xyvoyyyyArEjqI:QB:DZywaZ:JBAZ:B:DJ:Z:^:;jP@:MB:;B:_cl;Z:b::::^:f?=JX=j@B:_;Z:>Z<:VH::C:Uk:^:>X;j>>:_;l;J:>Z<:U=::C:UK:^:>X=j>>:_;: ;b::::^:f?=JX=jA>:_; j;Vv>Nf;vw<B:>L;J:DZaTXDpql`;^Z:jPF: C:[Y:;j>>:_c<l;Z:qGK_g;Hj;b:DZJ:Y=<:Uk:^:>X;j>>:_cWDjCZf;H@:JdRSa=:WmBZ\\wG@[=Ks:>:j\\bV=MC>;Y; Bw[j]BN_A[DweZ:^x?;ZJrJ@>YVKd<>=tYKLBFBZYgB@>B:=b:?J;<:=j:V:=J<> ZDFZ:B:DRNj;]Q:DZ:QB::C:[q:V:;B:;:::::::::::::: :::::::::ZcwghK:<:::3:" }{TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The notation " }{OLE 1 4136 1 "[xm]B r=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::fyyyyyq:nY^:f:nYnyyy YyA::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::_lqvGcMJ:::::::JEf:yyyxIN:R<:T><::;c G?BpsCECX:A:;X;:N:Y<> D_mlVH[KR<:;B:::::::JFNZ;F:VZ:jysy;Z::::::j:>:Gv:B:fB]mtFFcmnvGWMJnC== nHEM:^:N:yyyxI:;Z::::::JAJL@:<:wyyN:j>J?>:Q:SJ:f ;;jDjA>:[Z:Fr:GW::CZ:NZ;jI[Z:>rymTyyyyY:L:eY:V[:JMJ@fc[_hb_ds?h_Kbf:DZywaZ:JBAZ:>Z:b:;B>cTTUUSaEBWTSiEB_tUUURW=CZ:f_;j>B:>L=J::C:[q:F;;JSZ:F::GEX?U:HJAKMJC@j`@Pt:eTOGR:ER:LZ>WdG_ dnB:arOMeU=DUSeJ=uVMuRAtUCUS[TRIUSamBPJd`pp::WmBHjc`qpDpql`KfZ=wgoOhcg fl?^HgF;R>CRMcDBEDXceV=MC>[[v_?wk>w_;?dnA\\IvbYGbJOxYn];?dni\\IFb:^x?_ 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Z:>:[B:s:?j^iW>@Z:>bpfC>:m iGILphXvfZ;Z>WDjC:Z];:^DP@BBJq:D`f[=;:;:fBJC:ZnAH^;kA Q^dZID:<[:kZxQ^><=:UTR^DPP JLMQ:ETV:WCKaTMBwmcf]]:>::::::VY[j=J:^q:C:[Y:=j;>Z:>::::::::::::::::::::::::FZ:VBYE:UMrvC? MoB:;::::::::_B:?jysy:>Z:B::::::V\\:>:MM:>:N:YLpJbNHVH>@>Z::::::::kJ:v YxI:;Z::::::JywYB:::::::::::::yay=J:B:::::::::::::::::::jysy:>:<:::::: ::::::::5:" }{TEXT -1 40 " is the height of the ith rectangle and " } {OLE 1 4136 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: fyyyyyi:nY^:n:nYnyyyYyA::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::_lqvGcMJ:::::::JEf:yyyxIN :R<:T><::;cG?BpsCECX:A:;F;B:F:YLpfF>:::::::::J?NZ;vyyyyya:vYx Y:B:::::::c:;:?ja:[LsfFaMR>`:J:<:::::::>=?R:=:AB:vYxY:B::::::F:;JJ?Z:B :fB]mtFFcmnvGWMJnC==nHEM:>?jyyiy=J:B::::::^;>Y>:nyyM;B:A:C:EJ :n:v:>;;j>J?>:Q:wAU:W:YJ:nYF<:yA:::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::j:b:r:aS:N:k=;JZ:f:^[<>r ;>^I[Z:>ryM]yyyyYjAJ:Z=fZ:JDJW><fVUTR:eTOm`;B;LncpNdL]:V\\QGghGbf_gZK^ZG_dZfbr_hlGF_J>@ lQPnAMnQ@NbLYBxj]xoZHni@>YVKciDyDLcy]= D:<[:_xIHby?cnA\\IFbdve>we?_xYgB@nAlj:F:=b:?b:?B:?j:@j:V:=J<>ZD><:C:[q:Z:>:::::::::^=;J;FsUaU>bpfCYVKd<>=tYKLBFBZYgB@> B:=b:DJ;<:=j:j:^:;b:<::::::[B:;B:MM:>:N:YLpJbNHEmsJR<:;B:::::::JF>:yay=J:B::::::nYZ :::::::::::::vYxI:;Z::::::::::::::::::::yay=J:B::::::::F:wyyAbr:B:bKgJ:fc[KdRS;CGWCF;SK]UW=EWMuUWM@B::P`?oBYJ:fbk;n_>WdG[nIalHH> :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: BK>Z::::::5:" }{TEXT -1 43 "is the width of each rectangle. The produc t" }{OLE 1 2600 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyy yyy::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::fyyyyyqyyyY:vY:::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::gj^pNZRaaf^r;V:>j;Z:j:vCSmlJ::::::::::OJ;@jyyyyyY;jysy;Z: ::::::^<>:fB]mtFFcmnvGWMJnC==nHE=;:::::JJ:<::::::=J:>`;Z:::: :::::::::yay=J:B::::::::nyyM;B:AB:;J:wAyA::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::j:b:CRMcDBEDXce V=MC>[[v_?wk>w_;?dnA\\IvbYGbJOxYn];?dni\\IFb:^x?_x_^xIPbYGB<[=we;kAQNt eli\\NtyrZ:B>N:F:nyyyyy]::yyyyyy::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::fyyyyyq:nY^:f: nYnyyyYyA::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::_lqvGcMJ:::::::JEf:yyyxIV:VZHQ:R<:T ><::;cG?BpsCECX:A:;b;B:F:YLpfF>:::::::::J?NZ;vyyyyyQ:vYxY:B:::::::c:;: =j[vGUMrvC?MoJ::::::::JCN:yyyxI:;Z::::::j:>:M=;:?ja^G>D_mlVH[KR<:;B::: ::::JFNZ;F:^:vYxY:B::::::VZ:JoI:<:wyyN::C:EJ::[Z:n<;JC>:a:c:e:wAi:jy;::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::F:DZ:B::::::::::F:wyyAbR<:TNEj`@Pt \\Pd`QrPPJLMQp=j:F;HJfA:?JFK:^:NZ;j<Z:>DEZ:F[l;B:;B:DJ:Z:FZ:B:OKXAjn;Ve;fl;Z:b::::^:f?=J< JrA:O:_[:B:?J:>_cVd<>b=fZ:JDFp;N_=f<;jwOZy]:JBG:;B:;b:<;VH::C:UK:^:>X= j>>:_;NInDjw?FB:>l;J:>Z<:::JX;J@b=fZ:JDjZLJBB:qAB:>L;J:DZaTXDpql`;^Z:>:UK:^:>X=j>>:_cG;ES:UTRc=PpdPoF@j<@Z>BKaTMB:arOMeU=DUSeJ=uVMuRAtUCUS[TRIUSamB PJd`ppPps:JqF\\=Vdsgg\\wG@kZK^ZG_dZfbr_hlGF_J>@lQPnAM nQ@NbLYBxj]xoZHni@>YVKciDyDLcy]=D:<[:_ xIHby?cnA\\IFbdve>we?_xYgB@nAlj:@JJAZZWDjC:Z];:^DP@BBJq:D`fS>H>:>:j\\:: ZnAH^;kAQ^dZID:<[:kZxQ^><=Z:fX:M:HRmJ;]Z:>:[B:;B:qQBv:>:sg:B:=b:Dlc`qsLqlp@CZ:f?=JD_=aMR>@>Z::::::::kJ:vYxI:;Z::::::JywYB:::::::::::::yay=J: B:::::::::::::::::::jysy:>:<::::::::::::::5:" }{TEXT -1 183 " is simpl y the area of the ith rectangle and sigma notation is used to designat e their sum for a given number, n, of them. This summation is a simpli fied version of a more generalized" }{TEXT 309 12 " Riemann Sum" } {TEXT -1 159 " named after Bernhard Riemann, but will serve us adequat ely for now. If we let the number of rectangles increase without bound we have the following limit: " }{OLE 1 5152 1 "[xm]Br=WfoRrB:::wk; nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::fyyyyyA;nY^:f:n:v:nYnyyyYyA::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :JcvGYMt>^:fBWMtNHm=;:::::::n:;`:Z@[::JJTM;\\WXlL\\Aj ;J:`:<:ElrfH=MtFGYMq>>Wlj:gmlJ::::::>^:N:yyyxI:;Z:::::::^o;>:<:=ja^GE= ;:::::::::N;?R:yyyyyyC:yayA:<::::::I:c:;:?ja:[LsfFaMR>`:J:<:::::::>=?R :?Z:F:;jysy;Z::::::J>JyT:<:=:?J:VZ:JZ:n:nyyMyk>>:OJ:V;^;f;;JAjA>:[ Z:F:m:o:q:s:u:w:y:;[:V>N>nYnYvY:::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::s:?JVYWBA:;dGU<;jvUm]^wdAIER:LZ >WdG_dnB:arOMeU=DUSeJ=uVMuRAtUCUS[TRIUSamBPJd`ppPpsErJYUW_UTEeV;cMEM:@KZ;b:?bZSM;RTDJ:fGhj:NZ:Vjhc:CJCALQEJZDV:[BV@:nF?J;]<;rZkm;^=hZ;F:?J;eA]:FZ:ZhC:Ub:Cb[]bBEQ?DJ:=b:yyyyI: C:WS:mjI[B:^;UTRcETcTX[USDZyc:E:cb:cX:ErcTTUUSnC<Z:N`DZ:FZ:B:OKXAjj<^a;fl;Z:b::::^:f?;J_c^gcggbF?[T:qb:GMpFb>f`B:>L=J::;b::nC:J:C:[q:F;;JSZ:F::;b::::^:f?=JX=j>>:_;JklDjw?JB:>l;J:>Z<:::J: _;j;^]ANf;vw<;B:qQKZ:JB?:;b:DZJVdscRYEU@J<<: UK:^:>x;F:OB:N`DZ:N:;JQQTk_q;:cb:SQoXj>JSZ:F:<l; ZL>ZvS;JMJDAj`@Pt:eTOKV:ER:LZ>WDjCZf;H@:JdRS a=:WmBZ\\wG@[=wG:>:j\\bV=MC>;Y;Bw[j]BN_A[DweZ:^x?;ZJrJ@>YVKd<>=t YKLBFBZYgB@>B:=b:?J;<:=j:V:=J<>ZDFZ:B:\\jS:JmJ;]<;rZkm;ZER:N:?fDZ:V[:B:G;Sjyw yyI:S\\:>:[B:;B:qQBv:=Jyyy;d:yayYZHQ:>:;H:<:TNE>:UTR^DPPJLMQLyVDurJ:::::::sg:B:=b:Dlc`qsLqlp@CZ:f?=JZ:>:N:YLpfF[LsjsJRJ:<:::: :::>=vYxI:;Z::::::JywYB:::::::::::::yay=J:B:::::::::::::::::::jysy:>:< :::::::::::::3:" }{TEXT 280 0 "" }}{PARA 256 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 28 "This equation says that the " }{TEXT 266 24 "exact area under a curve" }{TEXT -1 308 " is the infinite sum of a pproximating rectangles provided that our function is non-negative ove r a specified interval. If our function takes on both positive and neg ative values, then we have a net change/signed area value. We utilize \+ Leibniz's notation for this limit and write the following definition. \+ " }{OLE 1 4128 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::y yyyyy::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::fyyyyyi:nY^:n:nYnyyyYyA::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::_lqvGcMJ:::::::JEf: yyyxI^:VZHQ:R<:T><::;cG?BpsCECX:A:;F;B:fB]mtFFcmnvGWMJnC==nHE=;::: ::JJ:<::::::JjC:j:vCSmlJ::::::::::OJ;@jyyyyyy;jysy;Z::::::j= JDJ:j:VBYmp>HYLkNG>::::::::N:;:=J:N:;B:AB:;JyyIE:jy Kyky;:S:UJ:n;v;;JBB:]:wAyA:::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::JHJ;>qQaR>bpfC>:miGILphXvfZ;B;LncpNdL]:V\\ QGghGbf_gCRMcDBEDXceV=MC>[[v_?wk>w_;?dnA\\IvbYGbJOxYn];?dni \\IFb:^x?_x_^xIPbYGB<[=we;kAQNteli\\NtyrZ^?GhoGfnGgioGSZ::RSPN]v;fbk;n_>WdG[nIalHH>::::::]bBE Q?DJnjwRIAjA@a;>ZZ=PFpJ;LJ=iLKDi:>Z::::::::=Jy;ryA:>:;P:<:T>fC^DP` nZ?;B:B::P`?kAj\\:n?:Z:>::::::yayYZ:JZ:>:C:?R:=j<>:=b:yyyyI:C:WS:mjr;>>a:Gc;YJHvyy uy;B:kMBryMMyyyyYjC:GU::>:j\\bVNdni<D_mlVH[KR< :;B:::::::JFNZ;N:;B:=B:;B:yayA:;B::::::V;^p;Z:J;<:Jb:a=[;;B:::::::JF>Z :vYxI:;Z::::::JywYB:::::::::::::yay=J:B:::::::::::::::::::jysy:>:<:::: ::::cI>H^;CwYVKd<>=tYKLBFBZYgB@>j:DJ;NZ:j:F:Aj:ZDj?JMJ@vYxy: ^C:;:::JrAj:B:_c<>Z:F:;Jk^q;f:Z< B>cTTUUSZZ:WM:^:X;j>>:_;l;J:>Z<:::J< jP>:C:[q:F;;JSZ:F::;b::::^:f?=J:;b::::^:f?;J::B>N:F:nyyyyy]::yyyyyy:::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::fyyyyyA;nY^:f:n:v:F;nYnyyyYyA:: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::_lqvGcMJ:::::::JEf:yyyxIV:VZHQ:R<:T><::;cG?BpsCE CX:A:;f:<::::::JpA:<:=ja^G E=;:::::::::N;?R:yyyyyyC:yayA:<::::::KJ:^<>:N:Y<>D_mlVH[KR<:;B:::::::J FNZ;N::yayA:<::::::M:?w;B:F:N:;j;<:C:EJ:=;jFJGjGJHjHJIjIJJB:=C:N>V>^>f>>?v>nYF?:yA::::::::: :::::::::::::::::::::::::::::::::::::::::::::::JHJ;>qQAt;JZTm`J:FypvBS itkm<@Z>BKaTMcDGZK^ZG_dZfbr_hlGF_J>@lQPnAMnQ@NbLY Bxj]xoZHni@>YVKciDyDLcy]=D:<[:_xIHby?c nA\\IFbdve>we?_xYgB@nAlj:F:=b:?b:?B:?j:@j:V:=J<>ZDFZ:>\\:>Z:fx;B:YR:?j ;A^;R:DJ;DRpN:``<>:U]EF:?B:AfUDJA]:FZ:ZhC:Ub:Cb[]bBEQ?DJ:nFNjbJ:HR vJ;a\\<>Z=`F?ZrC:QZZiywi]::::=J:DZ:B::::::::::vYyyyA:CZ :F:r:cw:N:_A;JryMZyyyC?>f>B:QB:DZyc:E:cb:cY:ErcTTUUSnC<x;j>>:_[:B: =Z:F`:^q;:cJJEJX@jDjw?jx]:JBA::C:[q:f\\:JSZ:F;;JN`Lp`pp\\lN \\lN>d;V]Z:>Z<:VH::C:Uk:^:>X; B:MJ:N@B:=Z:NFsW:JDN];VhAfX=j>>:_;fX=j>>:_;l;J:>Z<:::Jl;J:>Z<:::J:C:[q:F;;JSZ:F:NklDjw? JB:>l;J:>Z<:::J<B:>l;>:;b:DZJVdscRYEU@J<<:UK:^:>X=J?N[u:[Z:VY;><:[V:b:DZJ:Y=<:Uk:^:>X;J@gAQIy<;B:qQZqI<:[>;b:DZJ:Y=<:UK:^:>X=j>>:_ cFu;fCc=PpdPo`Aj<@Z>BKaH^;kAQ^dZID :<[:kZxQ^><=DRNj;]Q:DZn]\\fN=R:;bZ=@IAJH>ZER:N:?f DbI?B;p=AjAR:; :G=?FD>Z=@INjcb:;rZC=\\a:V;D:==Jk>ZGZej:R;B:bKgJ:f c[KdRS;CGWCF;SK]UW=EWMuUWM@B::P`?oBYJ:fbk;n_>WdG[nIalHHHYLkNG:<::::::GC:;jn>Z:>:N:YLpfF[LsjsJR<:;B::: ::::JFB:yay=J:B::::::nYyA<::::::::::::jysy:>:<:::::::::::::::::::vYxI: ;Z:::::::::V[:B:Gc;^;yayI:SL:\\:B:qQBv:x;>:AZ:>::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::3:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "Provided this limit exists, " }{TEXT 294 29 "we call this infinite sum the" }{TEXT -1 1 " " }{TEXT 264 29 "definite integral from a to b " }{TEXT 295 1 "." }{TEXT -1 19 " In short, we have" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 260 0 "" }{TEXT 258 35 "The \+ area under the curve on [a,b] =" }{TEXT 290 1 " " }{OLE 1 3664 1 "[xm] Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::fyyyyyqyyyYJ^pNZ Raaf^r;V:>j?B:<:ElrfH=MtFGYMq>>Wlj:gmlJ::::::>^:N:yyyxI:;Z:::::::>^;Z: j:vCSmlJ::::::::::OJ;vYxI:;Z::::::j@[C:>Z::::::::kJ;@ J;B:=J:vYxY:B::::::n:nt=Z:j:J;>:AB:^:nyyMyk=J>>:M:OJ:V;^;f;;JAjA>:[Z:F :=Z:V;>:>RBKaTMcDG ZK^ZG_dZfbr_hlGF_J>@lQPnAMnQ@NbLYBxj]xoZHni@>YVKciDyDLcy]=D:<[:_xIHby?cnA\\IFbdve>we?_xYgB@nAlj:F:=b:?b:?B:? j:@j:V:=J<>ZDFZ:V:YRV@J:DZ=pF?J;]\\<>Z=@I?J;a\\Zr:WX:N:U>;Jn>>r?V< \\:>Z:NmHJZyIoD?>f>j?<:G;Sj`@Pt\\Pd`QrP`:R:E:Mb:b:E: cb:;g:Mc:e:qi:;fy>Z:JBAZ:B:DJ::CZ:f_;J?<:_c_c>d;r: EB:^Z:JBAj:fDe:qAB:>l;J:>Z<:::J:C:[q:F;;JSZ:F:<:[V:>:; b::::^:f?=JB[J: aTXDpql`;^Z:jP>:C:[q:F;;JSd:v?ZnAdni<Z:>:[B:;B:qQBv:Z:>:::::::::::::::::::::::::::::::::::::ZT`_;B:;B:::: :::5:" }{TEXT 261 1 " " }{TEXT 259 46 "if f is continuous and non-nega tive on [a,b]. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 291 0 "" }{TEXT 292 54 "Key Items for You to Note about the Definite Integral." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 293 0 "" }{TEXT -1 43 "1. The def inite integral is a number which " }{TEXT 262 3 "may" }{TEXT -1 44 " b e interpreted as the area beneath a curve." }}{PARA 0 "" 0 "" {TEXT -1 38 "2. The value of the definite integral " }{TEXT 312 6 "always" } {TEXT 296 1 " " }{TEXT -1 25 "represents a net change. " }}{PARA 0 "" 0 "" {TEXT -1 85 "3. \"a\" is the starting interval value and is calle d the \"lower limit of integration\"." }}{PARA 0 "" 0 "" {TEXT -1 83 " 4. \"b\" is the ending interval value and is called the \"upper limit \+ of integration\"." }}{PARA 0 "" 0 "" {TEXT -1 79 "5. \"f(x)\" is the c urve we seek the net change on and is called the \"integrand\"." }} {PARA 0 "" 0 "" {TEXT -1 4 "6. \"" }{OLE 1 3112 1 "[xm]Br=WfoRrB:::wk; nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::fyyyyyqyyyYJ:::::::JEf:yyyxIV:VZHQ:R<:T><::;cG?BpsCECX:A:;x:::::::;C:?jyyiy=J:B:::::::kA;:=ja^GE=;::::::: ::N;?:xI:;Z::::::J@[;;B:::::::JFNZ;N::yayA:<::::: :EJ::W:wAyA::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::JHJ;FsUaUF:V;>: >RBKaTMcDG@jCXontpZlPpDnjHqnHp[xPlPPb@pmPpsF\\?^dcgg_WhZnc_whZNd igg[oG]r:aTXUeRYEU@kZK^ZG_dZfbr_hlGF_J>@lQPnAMnQ@NbLY Bxj]xoZHni@>YVKciDyDLcy]=D:<[:_xIHby?c nA\\IFbdve>we?_xYgB@nAlj:F:=b:?J;N:=Z:V:=J<>ZDFZ:V:YR:@Z:V[;FZ;FZGNZ>n jeN^lE:xxqyH<:F:DZ:B::::::::::vYyyyA:CZ:F:[=JCB:=j>r:MR:N :m]:>:C:?R:=j\\:B:;xy>ryyY@Z:>j<>Z:V[:JMJ@fc[_hb_ds?h_C:;F:EZ:F[<;B:qi:;XyB:>l;Z<>Z<>ZJVdsgGl`;^Z:jP@:MB:;JSd:WD:BU:::c\\_;<v?ZnAdni<YVKdlir]:G;SjywyyI:S\\:>:[J:x;j; " 0 "" {MPLTEXT 1 0 92 "restart:\nwith(student):\nwith(plot s):\nsetoptions(labels=[\"\",\"\"],font=[SYMBOL,20]):\nf:=x->x^2:" } {TEXT -1 12 "/* function." }{MPLTEXT 1 0 17 "\nDigits:=6:\na:=0:" } {TEXT -1 17 "/* left endpoint." }{MPLTEXT 1 0 7 "\nb:= 4:" }{TEXT -1 18 "/* right endpoint." }{MPLTEXT 1 0 7 "\nn:=20:" }{TEXT -1 24 "/* nu mber of rectangles." }{MPLTEXT 1 0 253 "\ndx:=(b-a)/n:\n`Approximate A rea`:=Sum(f(1+k*dx)*dx,k=1..n)=value(rightsum(f(x),x=0..4.0,n))*`sq.un its`;\ndisplay(seq(rightbox(f(x),x=a..b,NumRects),NumRects =5..20),ins equence=true,title=\"Right-Hand Approximating\\nRectangles\",titlefont =[TIMES,BOLD,14]);\n" }}}{PARA 0 "" 0 "" {TEXT 313 102 "Click on the p icture and use the animation toolbar which will appear at the top to v iew the animation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} {PARA 0 "" 0 "" {TEXT -1 94 "The true area can be found by computing t he infinite limit we defined above which we wrote as:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {OLE 1 5664 1 "[xm]Br=WfoRrB:::wk ;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::fyyyyyi:nY^:n:nYv:>;F;nYvY::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :NDYmq^H;C:ELq^H_mvJ::::::::gj^pNZRaaf^r;V:> jDZ:j\\FHemj^HMmqnG;KaFFJufF>::::::;C:?jyyiy=J:B:::::::SY:B:F:YLpfF>:: :::::::J?NZ;vyyyyyY:vYxY:B::::::v:^<>:F:AlqfG[maNFO=;::::::::_J;Zy=J:B ::::::F;FG;J:j:>:?J:>:wyynyyyYO:J@j@>:W:YJ:><:a:c :e:gJ:v<V>^>f>n>v>>?F?nYvY:::::::::::::::::: :::::::::::::::::::::::::::::::::::::::s:?JVYWbA:;dGU<;jvUm]^wdAIER:LZ >WdG_dnB:arOMeU=DUSeJ=uVMuRAtUCUS[TRIUSamBPJd`ppPpsErJYUW_UTEeV;cMEM:@K;R:DJ;DRpN:``<>:U]EF:?B:AfUDJkkB:=B:bUZ=@IFZ:>ZZiywi]::::=Jyyy;d:yqyyy;t V;J:>R;B:bKgJ:fc[_hb_ds?h_?^nn_j>^?GhoGfnGgioGSZ::RSPN]v;fbk;n_>WdG[nI alHHr:cw::C:?R:=j<>Z:F Zn>R?aZ:n^@v;sjyyiyAZ:>I[B:l;B:ZJ^d cgg_KaBBJq>Z:^:Z:FZ:B:OKXAjFuAf@[C:>Z::::::::kJ;@J;>:=:yayA:<::::::Q:?w;>Z:J;vCJb:a=[;; B:::::::JF>Z:vYxI:;Z::::::JywYB:::::::::::::yay=J:B::::::::::::::::::: jysy:>:<::::::::=J: ::C:UK:^:>X=jD<:_;>:KSDSUTUEDMCDMKb@jGDJm^G=D;ec:]V;[T:Hj<<:cjnOoIf :; b::::^:f?;J:;b::nC:J:;b::::^:f?;JX;j>>:_;^oIfX=j>>:_;X;jDJSZ:F;;jZ\\ Qm\\QvTPvLPv>j;^]ANf;vwR;r:EB:^Z:WWc?QX;cb:SJoXj>JSZ:F:< l;ZgAQIy<jqZ:jP>:C:[q:F;;JSd:Fu;fCc=PpdPo`Aj<@Z>BKa<Hnq:J::E\\lGF _J>v?ZnAdni<:<[=SJvk?cDr]ZDV:[BV@:nF?J;]<;rZkm;^=;r<@:?J;eRIN:>Z:FZ:ZhC:U:]:OJ<^l=b:>I?B;p=AjAR:;:G=?FD>Z=@INjcb:;rZC=\\a:V; D:==Jk>ZGZeZ:>Z:V[:B:G;Sjysy=J`>Z:>:[B:;B:qQBv: " 0 "" {MPLTEXT 1 0 423 "restart:with(plots):dx:=(4-0)/n:\nan1:=plot (x^2,x=0..4,color=red,thickness=3):\nan2:=plot(x^2,x=0..4,y=0..20,fill ed=true,color=aquamarine,view=[0..4,0..20]):Area:=eval(limit(Sum((0+k* dx)^2*dx,k=0..n-1),n=infinity)):\nan3:=display(textplot([3.5,5,convert (Area,string)],color=blue)):\nLimit(Sum((0+k*dx)^2*dx,k=1..n),n=infini ty)=Int(x^2,x=0..4);\ndisplay(an1,an2,an3,textplot(\{[3,5.1,\"Area =\" ]\},color=blue,font=[TIMES,BOLD,14]));\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 " " }{TEXT -1 2 "/*" }{TEXT 297 23 "Press Enter to Continue" }}{PARA 0 "" 0 "" {TEXT -1 449 "The exact area can be deter mined if the limit from above can be evaluated. This is not usually ea sy to do by hand but a powerful software package like Maple can give y ou excellent results. Furthermore, what if the function takes on both \+ positive and negative values? When this occurs then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis and \+ the sum of the areas of the rectangles that lie below the x-axis. Thus , " }{TEXT 302 46 "the definite integral can be interpreted as a " } {TEXT 263 8 "net area" }{TEXT 303 1 "." }{TEXT -1 212 " When a definit e integral is computed one must take care to know whether the function takes on positive and negative values so that if a true area value is desired, the appropriate signed area can be dealt with. " }{TEXT 400 32 "Consider the following example: " }}{PARA 0 "" 0 "" {TEXT -1 4 "Le t " }}{PARA 0 "" 0 "" {XPPEDIT 401 0 "f(x) = x^3;" "6#/-%\"fG6#%\"xG*$ F'\"\"$" }{TEXT 402 4 "- 4x" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 "Calculate the net area and the total area between f(x) and the x-axis on [0, 3]. Place the cursor in the red " }{TEXT 403 7 "restart" }{TEXT -1 27 " command and press \"E nter\"." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 511 "restart:\nwith(plots):\nf:=x->x^3- 4*x:\nArea1:=Int(f(x),x=0..2):\ng:=t->t^3-4*t:\nk:=int(f(x),x=0..2)+in t(g(x),x=2..3):\nArea2:=Int(g(x),x=2..3):\n`Net Area`:=Area1+Area2=k*` sq.units`;\nl:=abs(int(f(x),x=0..2))+int(g(x),x=2..3):\n`Total Area`: =abs(Area1)+Area2=l*`sq.units`;\nA1:=plot(f(x),x=0..2,filled=true,colo r=aquamarine,title=\"Area 1 & Area 2\",titlefont=[TIMES,BOLD,12]):\nA2 :=plot(g(x),x=2..3,filled=true,color=yellow):\ndisplay(A1,A2,textplot( \{[1.1,-2,\"Area 1\"],[2.6,2,\"Area 2\"]\},color=blue,font=[TIMES,BOLD ,14]));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 12 "Conclu sion: " }{TEXT 305 4 "The " }{TEXT 265 14 "exact net area" }{TEXT 306 27 " can be found by computing " }{TEXT 304 81 "the definite integral \+ which is the limit of an infinite sum of rectangular areas " }{TEXT 307 51 "which we call the Riemann Sum. However, this value " }{TEXT 310 22 "can only be determined" }{TEXT -1 1 " " }{TEXT 308 45 "if we c an actually compute the infinite limit" }{TEXT -1 1 "." }{TEXT 311 218 " Otherwise, we are stuck to choosing a finite number of rectangle s and using them to obtain an approximation to the definite integral's value.This is in practice very cumbersome to do by hand for non-trivi al problems. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 44 "The Accumulati on Function Revisited - FTC 1:" }}{PARA 0 "" 0 "" {TEXT -1 14 "Let's l ook at " }{TEXT 314 7 "f(x) = " }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\"# " }{TEXT -1 127 " again. The definite integral computes the signed are a under this function over an interval [a,b]. Place the cursor in the \+ red " }{TEXT 315 7 "restart" }{TEXT -1 58 " Maple command to see the s ymbolic representation of this." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "restart:\nArea:=Int(x^2,x=a..b);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 196 "In Module 2 \+ we were able to determine the area under this curve by defining a func tion which we called The Accumulation Function, A(x), which we also le arned is called an antiderivative of f(x). " }{TEXT 316 76 "It is cus tomary to actually use F(x) to represent the antiderivative of f(x)" } {TEXT -1 182 ".Thus, instead of using A(x) we will now use F(x) in its place. Since the Accumulation function determines the signed area up \+ to a point, x, in an interval we can write this fact as:" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT 337 42 "The Fundamental Theorem of Calculus Part 1" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 100 "If a function, f, is continu ous on a closed interval [a,b] then we can define a function on [a,b] \+ as" }}{PARA 261 "" 0 "" {TEXT -1 0 "" }{XPPEDIT 318 0 "F(x) = int(f(t) ,t = a .. x);" "6#/-%\"FG6#%\"xG-%$intG6$-%\"fG6#%\"tG/F.;%\"aGF'" } {TEXT -1 3 " ." }}{PARA 261 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 " " {TEXT -1 114 " which is continuous on [a,b] and also d ifferentiable on the open interval (a,b) with F'(x) = f(x). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 151 "F(x) yields the net accumulation of the integral fr om an initial starting point, a, up to a variable ending point x. It i s very important to note that " }{TEXT 317 83 "F(x) is a continuous fu nction and not a number unless x is given a specific value. " }{TEXT -1 161 "For example, if the interval is [0,2] and we specify x to be \+ 1, then F(1) yields the net accumulation from 0 to 1. The two diagram s below show the distinction:" }}{PARA 256 "" 0 "" {TEXT -1 0 "" }} {PARA 256 "" 0 "" {TEXT 319 59 "x unspecified \+ x = 1" }{TEXT -1 0 "" }}{PARA 256 "" 0 "" {OLE 1 38944 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>n=F:nyyyyy]::yyyyyy::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::fyyyyy A>:C:E:G:I:K:M:O:Q:S:UJ:n;v;;JBB:]:_:a:c:e:g:i:k:m:o:q:s:u:wAwyyyA FAN>V>^>f>n>v>>?nYN?V?^?f?n?v?>@:y;;\\:F BNBVB^BfBnBvB>CFCNCnYvY:::::::::::::::::::::::::::::::::::_lqvGcMJ:::::::JEf:yyyxI^:vZCN:R<:T><::;cG?BpsCECX:y:;R:B:F@cljNFS mlJ:::::::::n;?R:K:yay=;Z::::::J;>:WuGB:F:AlqfG[maNFO=;::::::::_J;F:;j ;jysy;Z:::::::>G>:V:Y:<::::::?J:^:>:^Ljs\\n>LJ t\\n>LJ:B:::^A=DJ_yy=OX^;gJ\\fFGmjfHSMt>>[D:=MsFFIMsJr>G;K]vGWMt>:;::j ;J=>:M:SJ:F<^<;jEB:o:=C:^>v>N?f?>@:SM:F:;J::yayY\\>Z:j: J::yaya\\>Z:j:J::yayi\\>B:<:=:;:vYxYEL:<:=:;:vYxyEL:<:=:;:vYxYFL:<:=:; :vYxyFL:<:=:;:vYxYGL:<:=:;:vYxyGL:<:=:;:vYxYHL:<:=:;:vYxyHL:<:=:;:vYxY IL:<:=:;:vYxyIL:<:=:;:vYxYJLZ:B:F:>:jysy=C;B:F:>:jysy?C;B:F:>:jysyAC;B :F:>:jysyCC;B:F:>:jysyEC;B:F:>:jysyGC;B:vYxY:B::::A:G:M:S:]:c:i:o:=;C; 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]i:uh;kMV^:;jP@j:f;N@md;Uh:=B:fL:;C::I;jP@J=:f:jG>x@Nr:nf@Vf:^:f?? :J<:w:n;j>Z:_KnTjkF:Cjk`::;jP@j:f;N @Qe;qn:<:A=JJZ:_kKEjvG:Fy:Fn>UC:C J:v>:US:G:j<:q:?W:=g:CT:CJ:f??:J:?R::UK>Z:F;b:UB:;JSVNC T:=:CT:=oJAJOHJtF^:J:f_;F:U:_KMgf;FZ:^:C:?R::Uk>f:Z:_KGEJP@j:<:SS:qf:?V:Os:KC:= ;;jP@j:f;N@Cg:UT:=J<^?:;kP:I;jP@J=:f:jGV`=Vr;Vn:?R::Uk>Z:F;b:UB:;JSNn:I;jP@J=:f:jGV?kc:=g:=O;B: M:DJBB:_kKGn;VZ:^:=W:=g:gc:OKJp]:^=o;CJ:f?=j@JSnNab:=B:FN:Z:_kKgZ:I;jP@J=:f:jGvyZ:_kAEjUDj:jUDJAEj=EJyO`:f>CJ:f_;F:U:_ksDJ hDj:jU:>^:f_:^:;B:I;IB:;Jxky=J:XSJZ:VZ:J::::::::::::::::::::::5:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 " \+ " }{TEXT 320 15 " F(x) = " }{XPPEDIT 321 0 "int(t ^2,t = 0 .. x);" "6#-%$intG6$*$%\"tG\"\"#/F';\"\"!%\"xG" }{TEXT 322 29 " F(1) = " }{XPPEDIT 323 0 "int(x^2,x = 0 .. 1 );" "6#-%$intG6$*$%\"xG\"\"#/F';\"\"!\"\"\"" }{TEXT 324 3 " = " } {XPPEDIT 325 0 "1/3;" "6#*&\"\"\"F$\"\"$!\"\"" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "From Modu le 2 we observed that which we have now formally stated: " }{TEXT 328 12 "F'(x) = f(x)" }{TEXT -1 74 ". If we take the derivative of the integral for F(x) we see the following:" }}{PARA 256 "" 0 "" {XPPEDIT 374 0 "d/dx;" "6#*&%\"dG\"\"\"%#dxG!\"\"" }{TEXT 326 16 "F(x) = F'(x) = " }{XPPEDIT 375 0 "d/dx;" "6#*&%\"dG\"\"\"%#dxG!\"\"" } {TEXT 329 1 " " }{XPPEDIT 376 0 "int(t^2,t = a .. x);" "6#-%$intG6$*$% \"tG\"\"#/F';%\"aG%\"xG" }{TEXT 327 3 " = " }{XPPEDIT 377 0 "x^2;" "6# *$%\"xG\"\"#" }{TEXT 378 7 " = f(x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 15 "This says that " }{TEXT 330 95 "if you t ake the derivative of the antiderivative, F(x), you get back the origi nal function f(x)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 24 "The \+ implication is that " }{TEXT 331 61 "differentiation and antidifferent iation are inverse processes" }{TEXT -1 149 ". It also implies that ev ery continuous function f(x) has an antiderivative. Finding the antide rivative as module 2 showed may not be an easy matter." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 332 0 "" } {TEXT 333 83 "Key Items for You to Note about the Fundamental Theorem \+ of Calculus Part 1 (FTC 1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 334 0 "" }{TEXT -1 61 "1. The FTC 1 def ines the accumulator function as an integral." }}{PARA 0 "" 0 "" {TEXT -1 135 "2. The integral in the FTC 1 equation has its upper limi t of integration as a variable thus making the integral's value depend ent on x." }}{PARA 0 "" 0 "" {TEXT -1 43 "3. \"a\" is a scalar startin g interval value." }}{PARA 0 "" 0 "" {TEXT -1 43 "4. \"x\" is a variab le ending interval value." }}{PARA 0 "" 0 "" {TEXT -1 79 "5. \"f(t)\" \+ is the curve we seek the net change on and is called the \"integrand\" ." }}{PARA 0 "" 0 "" {TEXT -1 4 "6. \"" }{OLE 1 3116 1 "[xm]Br=WfoRrB: ::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::fyyyyyqyyyYJ:::::::JEf:yyyxIV:VZHQ:R<:T><::;cG?BpsCECX:A:; x:::::::;C:?jyyiy=J:B:::::::kA;:=ja^GE=;:: :::::::N;?:xI:;Z::::::J@[;;B:::::::JFNZ;N::yayA:< ::::::EJ::W:wAyA:::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::JHJ;FsUaUF :V;>:>RBKaTMcDG@jCXontpZlPpDnjHqnHp[xPlPPb@pmPpsF\\?^dcgg_WhZnc_ whZNdigg[oG]r:aTXUeRYEU@kZK^ZG_dZfbr_hlGF_J>@lQPnAMnQ 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}{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "Q9. Create a scenario where L'Hospital's Rule may be appl ied to the FTC 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 160 "Q10. Suppose the upper limit of integration in the FTC 1 is a function like sinx. How does that affect the result of the deriv ative of the integral shown below?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 256 "" 0 "" {XPPEDIT 18 0 "d/dx;" "6#*&%\"dG\"\"\"%#dxG!\"\"" } {XPPEDIT 18 0 "int(t,t = a .. sinx);" "6#-%$intG6$%\"tG/F&;%\"aG%%sinx G" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 267 0 "" }{TEXT -1 0 "" }}}} {SECT 1 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 371 31 "The Establishment of the FTC 2:" }}{PARA 0 "" 0 "" {TEXT -1 192 "In module 2 we explore d how the net accumulation beneath a curve seemed to be related to the value of the Accumulation Function, A(x), evaluated at the interval e ndpoints. We explored this as:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 256 "" 0 "" {TEXT 353 32 "Does A(b) - A(a) = The net area?" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 131 "To see h ow this might be so, consider the interval [a,b]. Using the accumulati on function A(x) and letting x = a and then b we get:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 339 7 "A(a) = " }{XPPEDIT 340 0 "int(f(t),t = a .. a);" "6#-%$intG6$-%\"fG6#%\"tG/F);%\"aGF," } {TEXT 341 16 " and A(b) = " }{XPPEDIT 342 0 "int(f(t),t = a .. b); " "6#-%$intG6$-%\"fG6#%\"tG/F);%\"aG%\"bG" }{TEXT 343 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "Subtracting yields : " }{TEXT 344 32 " " }{TEXT 354 15 "A (b) - A (a) = " }{XPPEDIT 345 0 "int(f(t),t = a .. b);" "6#-%$intG6$-% \"fG6#%\"tG/F);%\"aG%\"bG" }{TEXT 346 10 " - " }{XPPEDIT 347 0 "int(f(t),t = a .. a);" "6#-%$intG6$-%\"fG6#%\"tG/F);%\"aGF," }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 " But, " }{XPPEDIT 348 0 "int(f(t),t = a .. a);" "6#-%$intG6$-%\"fG6#% \"tG/F);%\"aGF," }{TEXT 349 4 " = 0" }{TEXT 355 1 " " }{TEXT -1 62 "be cause there is no accumulation beneath an individual point. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence," }}{PARA 0 "" 0 "" {TEXT 350 15 "A(b) - A (a) = " }{XPPEDIT 351 0 "int(f(t),t = a .. b);" "6#-%$intG6$-%\"fG6#%\"tG/F);%\"aG%\"bG" }{TEXT -1 69 " whi ch is precisely the limit established by the Riemann Sum Method. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT 352 42 "The Fu ndamental Theorem of Calculus Part 2" }{TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 3 "If " }{TEXT 358 1 "f" }{TEXT -1 28 " is continuous on [a,b] and " }{TEXT 359 4 "F(x)" } {TEXT -1 26 " is any antiderivative of " }{TEXT 360 1 "f" }{TEXT -1 8 " , then " }}{PARA 0 "" 0 "" {TEXT 361 35 " \+ " }{TEXT 364 24 " The Net Accumulation = " }{XPPEDIT 363 0 "in t(f(x),x = a .. b);" "6#-%$intG6$-%\"fG6#%\"xG/F);%\"aG%\"bG" }{TEXT 362 37 " = F(b) - F(a), where F'(x) = f(x)." }}{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 356 181 "This means that if we c annot evaluate the infinite limit of the Riemann Sum to get the exact \+ accumulation, we can still find it if we can find the antiderivative o f the given curve." }{TEXT -1 1 " " }{TEXT 357 77 "This is perhaps one of the greatest achievements in the field of mathematics!" }{TEXT 372 0 "" }{TEXT 373 0 "" }{TEXT -1 135 " This provides us with another means to evaluate definite integrals without resorting to messy Riema nn Rectangle Approximation Methods." }}{PARA 0 "" 0 "" {TEXT 338 0 "" }}{PARA 261 "" 0 "" {XPPEDIT 365 0 "int(f(x),x = a .. b);" "6#-%$intG6 $-%\"fG6#%\"xG/F);%\"aG%\"bG" }{TEXT 366 15 " = F(b) - F(a) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "The equation ab ove has tremendeous implications. " }{TEXT 367 22 "In words it says th is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 368 179 " The definite integral over a closed interval can be determined by simp ly evaluating the antiderivative at the endpoints. This is exactly wha t you intuitively explored in Module 2!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "Of course, we still have " }{TEXT 370 17 "the difficulty of" }{TEXT -1 1 " " }{TEXT 369 35 "actually fin ding the antiderivative" }{TEXT -1 134 " but now we know that if we ca n find it, we can compute the definite integral which can in turn be i nterpreted as a net accumulation. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 379 0 "" }{TEXT 380 83 "Key Items for You to Note about the Fundamental Theorem of Calculus Part 2 (FTC 2)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 381 0 "" }{TEXT -1 65 "1. The FTC 2 returns the net accumulatio n over an interval [a,b]." }}{PARA 0 "" 0 "" {TEXT -1 125 "2. The defi nite integral in the FTC 2 may be interpreted as area if the integrand function is non-negative over the interval." }}{PARA 0 "" 0 "" {TEXT -1 110 "3. The FTC 2 holds if the integrand function does not have any infinite discontinuities on the interval [a,b]." }}{PARA 0 "" 0 "" {TEXT -1 43 "3. \"a\" is a scalar starting interval value." }}{PARA 0 "" 0 "" {TEXT -1 41 "4. \"b\" is a scalar ending interval value." }} {PARA 0 "" 0 "" {TEXT -1 79 "5. \"f(x)\" is the curve we seek the net \+ change on and is called the \"integrand\"." }}{PARA 0 "" 0 "" {TEXT -1 4 "6. \"" }{OLE 1 3120 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:ny yyyy]::yyyyyy::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::fyyyyyqyyyYJ::::::: JEf:yyyxIV:VZHQ:R<:T><::;YQQBpsCECX:A:;x:::::::;K;vyyuy:>:<::::::JvK:j:vCSmlJ::::::::::OJ;Zy=J:B::::::^:^<>: N:Y<>D_mlVH[KRJ:<:::::::>=?R:?Z:F:;jysy;Z::::::j<>Z:^c:AB:nyyM yK=j=J>j>J?>:Q:SJ:f;;JA>:wAyA::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::JHJ;FsUaUF:V;>:>RBKaTMcDG@jCXon tpZlPpDnjHqnHp[xPlPPb@pmPpsF\\?^dcgg_WhZnc_whZNdigg[oG]r:aTXUeRYEU@kZK^ZG_dZfbr_hlGF_J>@lQPnAMnQ@NbLYBxj]xoZHni@>YVKciDyDLcy]=D:<[:_xIHby?cnA\\IFbdve>we?_xYgB@nA lj:F:=b:?J;N:=Z:V:=J<>ZDFZ:V:YR:@Z:V[;FZ;FZGNZ>njeN^lE:xxqyH<:F:DZ:B:: ::::::::vYyyyA:CZ:F:[=JCB:=j>r:MR:N:m]:>:C:?R:=j\\:B:;xy>ryyY @Z:>j<>Z:V[:JMJ@fc[_hb_ds?h_C:;F:EZ:F[<;B:qi:;XyB:>l;Z< >Z<>ZJVdsgGl`;^Z:jP@:MB:;JSd:WD:BU:::c\\ _;<v?ZnAdni<YVKdlir]:G;SjywyyI:S\\:>:[ J:x;j; 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Create 3 different integrals which all have a net accumulation of 4. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 212 "Q10. Since the derivative of an object's position equation is its velocity , the integral of an object's velocity graph should be its net accumul ated distance. If an object has a velocity equation given by v(t) = " }{XPPEDIT 18 0 "t^2;" "6#*$%\"tG\"\"#" }{TEXT -1 1 "-" }{XPPEDIT 18 0 "3*t;" "6#*&\"\"$\"\"\"%\"tGF%" }{TEXT -1 70 " ft/sec, what is the net displacement over an interval [0,5] seconds? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 56 "Using the Fundam ental Theorem of Calculus: Applications:" }}{PARA 0 "" 0 "" {TEXT -1 74 "There are a wealth of applications for integration (antidifferenti ation). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 384 23 "Things to keep in mind:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 78 "1. The definite integral returns a value \+ that represents the net accumulation." }}{PARA 0 "" 0 "" {TEXT -1 89 " 2. If you evaluate the definite integral of any rate graph you will ge t the total change." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }{TEXT 268 36 "Examples of Integration using Maple:" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "Run the \+ following Maple code examples by always starting with the cursor in th e red " }{TEXT 385 7 "restart" }{TEXT -1 14 " command line." }}{PARA 0 "" 0 "" {TEXT 278 0 "" }}{PARA 0 "" 0 "" {TEXT 279 9 "Example 1" }} {PARA 0 "" 0 "" {TEXT -1 31 "Evaluate the definite integral " } {XPPEDIT 18 0 "int(x*sin(x),x = 0 .. 7);" "6#-%$intG6$*&%\"xG\"\"\"-%$ sinG6#F'F(/F';\"\"!\"\"(" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 280 "restart:\nwith(plots ):\nk:=evalf(Int(x*sin(x),x=0..7)):\nInt(x*sin(x),x=0..7)=k;\na1:=plot (x*sin(x),x=0..7,y=-6..6,color=red,thickness=3):\na2:=plot(x*sin(x),x= 0..7,y=-6..6,color=green, filled = true,view=[0..7,-6..6], title=\"Gra ph of xsinx\",titlefont=[TIMES,BOLD,14]):\ndisplay(a1,a2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 281 13 " \+ To find the " }{TEXT 386 10 "total area" }{TEXT 387 124 " bounded by t he curve above, we need to identify when the curve is above and below \+ the horizontal axis on the interval [0,7]" }{TEXT -1 1 "." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 350 "rest art:with(plots):\n`Total Area`:= Int(x*sin(x),x=0..Pi)+abs(Int(x*sin(x ),x=Pi..2*Pi))+Int(x*sin(x),x=2*Pi..7);`Total Area`:=evalf(%)*`sq.unit s`;\na3:=plot(x*sin(x),x=0..7,y=-6..6,color=red,thickness=3):\na4:=plo t(x*sin(x),x=0..7,y=-6..6,color=green, filled = true,view=[0..7,-6..6] , title=\"Graph of xsinx\",titlefont=[TIMES,BOLD,14]):\ndisplay(a3,a4) ;\n" }}}{EXCHG {PARA 258 "" 0 "" {TEXT -1 55 "Note: We need to conside r the integral as signed area. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 270 0 "" }{TEXT 271 0 "" }{TEXT 272 0 "" }{TEXT 273 0 "" } {TEXT -1 1 " " }{TEXT 276 0 "" }{TEXT 277 10 "Example 2:" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 388 375 "A s stated earlier, it is importatnt for you to understand that if you i ntegrate a rate function you get a net accumulation quantity. For exa mple, if we have a rate graph of water leaking out for 15 minutes, the n the area beneath the rate graph yields the net amount of water that \+ leaked out over the 15 minute interval. Follow the example below close ly to make this clearer." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 376 "One of the earliest pollution problems brought to the attention of the Environmental Protection Agency (EPA) was the ca se of the Sioux Lake in South Dakota. For years a small paper plant lo cated nearby had been discharging waste containing carbon tetrachlorid e into the waters of the lake. At the time the EPA learned of the situ ation, the chemical was entering the lake at a " }{TEXT 389 27 "rate o f 16 cubic yards/year" }{TEXT -1 360 ". The agency immediately ordered the installation of filters designed to slow (and eventually stop) th e flow of carbon tetrachoride from the mill. It took three years to ac tually implement the program of intervention and the filters during wh ich time the rate of inflow was a steady 16 cubic yards per year. Once the filters were installed, the flow declined. " }{TEXT 394 130 "From the time that the filters were installed until the time the flow stop ped, the rate of flow was well modeled by the quadratic " }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 390 29 "Rate (in cubic yards/year) = " } {XPPEDIT 391 0 "t^2;" "6#*$%\"tG\"\"#" }{TEXT 392 5 " - 14" }{TEXT 274 1 "t" }{TEXT 393 5 " + 49" }{TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 6 "where " }{TEXT 275 1 "t" }{TEXT -1 66 " is time measured i n years since the EPA learned of the situation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 183 "A. Plot the curve which \+ models the flow of carbon tetrachloride into the lake beginning at the time the EPA first learned of the situation. Press \"Enter\" to get t he cursor in the red " }{TEXT 395 7 "restart" }{TEXT -1 58 " command a nd then press \"Enter\" agian to obtain the graph." }{MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 410 "restart:\nwith(plots): \nf:=t->t^2-14*t+49:\namount:=t->piecewise(t<3,16,t>3,t^2-14*t+49):\na 1:=plot(amount(t),t=0..10,title=\"Flow Rate Model in Cubic Yards per Y ear\",labels=[years,rate],titlefont=[TIMES,BOLD,14],labelfont=[TIMES,B OLD,12],thickness=3,color=blue,view=[0..7,0..20]):\na2:=plot(amount(t) ,t=0..7,color=plum,filled=true):\ndisplay(a1,a2,textplot([2,7,\"Total \+ Pollutants\"],font=[TIMES,BOLD,14],color=blue));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {TEXT -1 312 "How many yea rs elapsed between the time the EPA learned of the situation and the t ime the pollution flow stopped entirely? Can you write two separate in tegrals to compute the total amount of pollutants which came into the \+ lake? The code below displays the solution. Study it carefully. Press \+ \"Enter\" to continue." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 105 " k:=int(16,t=0..3)+int(f(t),t=3..7):\n`Total Pollutants`:=Int(16,t=0..3 )+Int(f(t),t=3..7)= k*`cubic yards`;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {TEXT 396 10 "Example 3:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 407 "This third example combines both parts of the FTC. As stated earl ier, sometimes we do not have a nice formula for our function and, thu s, the antiderivative cannot be found. But, if we know that we are dea ling with a rate graph we can still approximate the definite integral \+ by creating a grid beneath the graph and calculating the area by count ing the squares. Work through the four parts of this example. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "The graph below displays the velocity of a 6 hour jog by a student.\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 414 "restart:with(plots):\na1:=p lot((2.1*sin(x)-2.2*cos(x/3+2))+1.1,x=0..6,thickness=3,view=[0..6,0..5 ],color=blue,tickmarks=[8,6],title=\"Jogger's Velocity Graph\",titlefo nt=[TIMES,BOLD,14],labels=[hours,`mile/hr`],labelfont=[TIMES,BOLD,12]) :\na2:=plot(\{1,2,3,4,5\},x=0..6,view=[0..6,0..5.1],color=red):\na3:=i mplicitplot(\{x=1,x=2,x=3,x=4,x=5,x=5.99\},x=0..6,y=0..5,color=red,num points=1):\ndisplay(a1,a2,a3,tickmarks=[6,6]);\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {TEXT 397 31 "Answer the following questions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "Q1. What is the approx imate number of miles jogged? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "Q2. When was the jogger accelerating?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 148 "Q3. Assuming the jogger's unknown velocity function is written as v(t), write a definite integral to re present the jogger's total distance traveled." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "Q 4. Try to sketch the displacement graph for this jogger." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} {PARA 0 "" 0 "" {TEXT 398 10 "Example 5:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 164 "As a last example, examine the cartoon from FOXTROT and determine its accuracy from what you have learned from this modul e and explain why it is or is not accurate." }}{PARA 256 "" 0 "" {METAFILE 525 383 383 1 "iXmGh`B:FZMFeG>w=Z:FnDG:MZ=ngjI:?Jgyx:JDJMTjA n=yyyyyyCJDjexPsLPAfccWflwgmwg`_hZvdik>UKBB:G;gjyyyyyY:FZ:Jyyyyyyoh:AB<^: vZ;^:oxuVbI>>S@GW:kd:B:AjmvyyY:K[:>eJ:>\\WY:Iu=VgIJFKFK:jSJmJ: V@a;GmS>:G=GMmJ:FMakF?Z:^SaK`?:yI:aky?ZmB:asFnfmJ:H=asFaKmHmSZm nFasFGMmHMmFmmnFSxFGM:HmFW`mFMGuFmnFImF_smFM;rFSpSHM`?ZmvYasFyQ:TP:bSn f`kFU@ShS>Z`W`:V``W@GeSakFUpSvy`W@;bSGE:acSGMmTPmFm`oFShSGM:TpFW``GMGe SmnFUpF_s`GMyiSmN:TP`W``_SGeSSpFUP`_s`_SyiSSP:Tpywy`wY;R`;jS`CGU`FmSL` aCyY`>ZSX`:V`SX@GU`akFasS^sSX@yY`aK:`SmZSHm:V`SHM 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:::::::::::::::::::::::::::::::::::::::::Jy[CFj Oar=[B>[r=[B>;r=;J:aB:;B:;::::::::e:::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::^Z:B:IR:=B:KB:;B: Gc;YB:GB:yyyyyy=B:CJ:f_;>Z:VZ:>:;5:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 35 " \+ " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Con clusions:" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 399 106 "You have n ow completed all three Labs for Integration. From these labs you shoul d have seen the following:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 352 " 1.The area under a continuous curve can be dete rmined by a limiting process of approximating rectangles (The Riemann Sum). There are many ways in which to define the rectangles but they \+ converge to the same value as the number of rectangles increases witho ut bound. The process is usually quite cumbersome to do by hand and is not a preferred method." }}{PARA 0 "" 0 "" {TEXT -1 304 "2. The area \+ under a continuous curve can be determined by evaluating the antideriv ative of a function at the endpoints over a given interval (Fundamenta l Theorem of Calculus). This is by far the preferred method but its we akness lies in the fact that one must be able to find the antiderivati ve function." }}{PARA 0 "" 0 "" {TEXT -1 286 "3. The Fundamental Theor em of Calculus is composed of two parts which when taken together disp lay the inverse processes of differentiation and antidifferentiation ( integration). We call the FTC 1 and FTC 2 but we refer to both them ta ken together as The Fundamental Theorem of Calculus." }}{PARA 0 "" 0 " " {TEXT -1 167 "4. Definite integrals return net accumulations, thus, \+ if a function takes on both positive and negative values one needs to \+ account for the sign in the negative cases." }}{PARA 0 "" 0 "" {TEXT -1 235 "5. If the curve being integrated is that of a rate function, t hen the value of the definite integral can be interpreted as a total a mount of change (net quantity). This has many applications which will \+ be investigated in later classes." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} }{MARK "1" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }