Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic in nture, tht is optimising the vlue of function without ny reference to time. In sttic optimistion nd comprtive sttic nlysis we mke the ssumption tht the process of economic djustment leds to n equilirium, nd we then emine the effect of chnges of the eogenous vriles on the equilirium vlues of the endogenous vriles. With dynmic nlysis, time is eplicitly considered in the nlysis. While we re not covering dynmic nlysis t this point, certin mthemtic tools re required for dynmic nlysis, such s integrtion nd differentil equtions. Without these tools, it ecomes impossile to consider prolems which re not sttic in nture. We will e covering oth of these topics in minly mthemticl wy, leving economic prolems for lter dte. Integrtion is the reverse process of differentition. If function () integrl of f () will yield F (). The nottion to denote integrtion is s follows: F hs first derivtive f () then the f ( ) d, where the integrl sign is n elongted S. f () is referred to s the integrnd, nd the d sign reminds us tht we re integrting with respect to the vrile. We go through the following eption to determine where this nottion comes from. Suppose we re given n ritrry function f () nd sked to find the re of the curve etween points, for emple the re under the curve f() etween nd. Figure. With liner function, this equtes to finding the re of tringle nd rectngle s follows: Figure.
However with non-liner function the prolem ecomes slightly more comple. Wht we cn do, however, is ttempt to find the re under the curve using numer of pproimting rectngles s follows: Figure. We let ech of the rectngles hve equl width nd we cll this width Δ. Ech rectngle hs height equl to the function vlue, for emple the height of the lst rectngle where = is equl to f()=. Thus the re of the lst rectngle is equl to (Δ), s re of rectngle equls length times redth, nd here redth is Δ nd length is f()=. The re of ny of the rectngles is equl to length times redth, which equls f() times Δ, s ll rectngles hve equl redth equl to Δ. To find the re under the curve we dd up the re of ech of the rectngles. This gives us the epression: n A f ( ) Where n = numer of rectngles i = the vlue of t ech point = the sum of ll the res, strting from the first one (i = ) nd ending t the nth one (i = ). i Oviously this sum will not e very ccurte representtion of the re. But perhps if we mke our Δ smller, then this epression will ecome more ccurte representtion of the re under the curve, s there will e less overshooting y ech rectngle. If initilly we hd ten rectngles, the re given y the sum of these rectngles res would oviously e more of n over-estimte (or mye underestimte) thn if we douled the numer of rectngles, nd then summed their re. The more rectngles we use in this pproimting process the etter our estimte for the re under the curve. For emple, imgine we wish to find the re under the curve f() = etween nd. i.8.6.....6.8 Figure.
We could tke four rectngles, ech with redth Δ equl to., nd tke the heights from the right hnd side of ech rectngle. Hence the height for ech rectngle will e: (.) (.) (.7) Therefore the entire re equls to.(.) +.(.) +.(.7) +.() = /=.687 If we doule the mount of rectngles from four to eight, we will use Δ of., nd the following right hnd heights (rememer the height of the rectngle is given y the function vlue f()). (.) (.) (.7) (.) (.6) (.7) (.87) The corresponding totl re is given y the sum of ech of the res which is Δ multiplied y ech function vlue: The finl vlue we get is.987 As cn e seen in figure., using right end points for the rectngles for n incresing function will give n over-estimte, while using right end points for decresing function will yield n over-estimte. Thus douling the numer of rectngles while trying to estimte the re under the grph f()= will egin to ring our estimte down to its true vlue. It ppers tht s the numer of rectngles increses, our estimtions ecome etter nd etter pproimtions of the re. If we let the numer of rectngles tend to infinity, we will otin perfectly ccurte estimte for the re under our grph. Our epression for the re under the curve now ecomes: A lim n n i f ( ) This gives us the epression for the definite integrl, which gives us wy of finding the re under the continuous function f() etween = nd =: An eption of the terminology: f ( ) d lim n n i i f ( ) The integrtion sign is n elongted S, nd ws so chosen ecuse n integrl is limit of sums., re the limits of integrtion, f () is known s the integrnd. d is the lower limit of integrtion is the upper limit of integrtion. hs no mening y itself, ut merely reminds us tht we re integrting with respect to the vrile. Fortuntely when we wnt to find the re under curve, we do not hve to go into the long process of finding n epression for the sum of the re of n rectngles: numer of theorems mke the process esier. i
Before we set out the properties of the definite integrl, some rules of integrtion re s follows: (see pge 9 nd onwrds in Ching for emples).. The Power Rule ( ) n n n d c n. The Eponentil Rule e d e c. The Logrithmic Rule d c ( ) Properties of the definite integrl:. cd c( ). [ f ( ) g( )] d f ( ) d g( ) d. cf ( ) d c f ( ) d. [ f ( ) g( )] d f ( ) d g( ) d c c. f ( ) d f ( ) d f ( ) d Property sttes tht the integrl of constnt function y=c is the constnt times the length of the intervl, s seen in figure. Figure. Property sys tht the integrl of sum is the sum of the integrls. The re under f+g is the re under f plus the re under g. This property follows from the property of limits nd sums.
Property tells us tht constnt (ut only constnt) cn e tken in front of n integrl sign. This lso follows from the properties of limits nd sums. Property follows from property nd, using c=-. Property tells us we cn find the re under the grph etween nd c, y splitting it up into two res, etween nd, nd etween nd c. We now find our rule for evluting the definite integrl: f ( ) d F( ) F( ) where the derivtive of F() is f(), i.e. F is ny nti-derivtive of f. For emple, if we differentite F( ) we otin Thus d F() F(). f ( ), so F() is n nti-derivtive of f(). Therefore the re under the curve f() = etween nd, is equl to third, or. recurring. Incidentlly this nswers our previous question which we ttempted using the sum of the res of n rectngles. The fundmentl theorem of clculus motivtes this use of the evlution theorem. In short, it sttes tht differentition nd integrtion re opposite processes. Thus, if we strt with function F(), nd differentite it to otin f(), [ i.e. F ( ) f ( ) ], if we then integrte the function f(), the result will e the initil function F(). Similrly if we integrte f() to otin F(), [ i.e. f ) d F( ) C ( where C is n ritrry constnt ]. Thus to find the integrl of function f(), we must find the function which when differentited yields f(). This theorem is very useful to us, s otherwise whenever we wish to find the vlue of the re tht lies underneth curve, we hve to go through the entire process of finding the limit of the sum of the res of n pproimting rectngles, which is time consuming process! Prior to the discovery of the fundmentl theorem, finding res, volumes nd other similr types of prolems were nigh on impossile. For completeness, the fundmentl theorem is presented elow: The fundmentl theorem of clculus: Suppose f() is continuous function on the closed intervl [,] d g ( f ( t) dt then g ( ) f ( ) i.e. g( ) f ( t) dt f ( ) d. If ). f ( ) d F( ) F( ) where the derivtive of F() is f(), i.e. F is ny nti-derivtive of f. Wht it sys, roughly speking, is tht if you integrte function nd then differentite the result, you retrieve the originl function. More out the ritrry constnt little lter
We now need to discuss two different types of integrls definite nd indefinite. A definite integrl involves finding the integrl of function etween two numer limits i.e. f ( ) d. The nswer to definite integrl is numer, s we know ccording to the evlution rule the nswer to this is just the ntiderivtive F() evluted etween nd, i.e. F()-F(). An indefinite integrl yields function of s its nswer (if we re integrting with respect to ). An indefinite integrl is n integrl of the form f ( ) d (i.e without upper nd lower limits) nd the solution is f ( ) d F( ) C where C is n ritrry constnt which cn tke on ny vlue. The reson we include the ritrry constnt is illustrted in the following emple. Given the prolem: d potentil solution is we otin. However s this is n ntiderivtive of the cuic function (If we differentite is lso solution to this prolem, s is. This is ecuse when differentiting these epressions, the constnt differentited moves to zero. So it would pper tht the most generl form to give the nswer to this prolem would e s follows: d C, where C is n ritrry constnt. Just smll note on ritrry constnts when we dd two together, we otin third one which hs ggregted the first two, when multiplying, dividing dding or sutrcting numer y/to/from n ritrry constnt, the result is just the ritrry constnt. However the ritrry constnt when multiplied y function of, will sty s just tht: f ) g( ) d F( ) C G( ) C F( ) G( ) C ( where C nd C re two ritrry constnts, nd F() nd G() re two nti-derivtes of f() nd g() respectively. f ( ) d [ F( ) C] F( ) C F( ) C However: f ( ) d [ F( ) C] F( ) C We now turn to some rules of integrtion (definite nd indefinite) nd then some emples.. ( ) d c cf f ( ) d. ( f ( ) g( )) d f ( ) d g( ) d n n n. d C. d C (n cnnot equl ) 6
. e d e C sin d cos 6. d C 7. C 8. cos d sin C Rememer to check the nswer to ny integrtion sum just differentite it nd you should rrive ck t the originl function. Some emples:. d C. d C. d d. d d 7 ( ). [ sin ] d cos C cos C C 6. ( 6) d 6 (9) 6. 7 7. 8. 9. 9 t t t t dt 8 () 9 7.6 ( t 9 (t t t t t 6) dt 6t 8 8 c c t t 9 ) dt 8 9 8 (t t t ) dt t t (Note the trick here) t 9 7
A few more useful properties:.. f ( ) d f ( ) d f ( ) d We re going to e looking t two very useful techniques used in integrtion: use of sustitution, nd integrtion y prts. There re whole host of other techniques which cn e useful, however it is these two which re most useful to us in economics. Integrtion using Sustitution We use sustitution, when the integrnd contins function nd its own derivtive. i.e: f ( ) f ( ) d For emple: ( )( ) d If this is the cse, we cn mke use of the following sustitution: We let u equl to the function whose derivtive we cn spot (or crete, using constnt: more out this lter). Let u = Then we know tht du ( ) d When we then sustitute the vlues of u nd du into the integrl, we otin the following integrl: udu which hs the nswer u udu C ut: u = Therefore our finl nswer is ( ) ( )( ) d C Thus the generl rule solution for the prolem is s follows: [ f ( )] f ( ) f ( ) d C or more simply f ( f. f ) d C However the sustitution rule cn e used for more complicted emples, when our function f() whose derivtive we cn spot occurs inside nother function: 8
For emple: f g ( f ) d or (6 8) ( 8 ) d In these cses the procedure does not chnge t ll we still mke the sustitution s follows: Let u = f() Therefore du=f`()d And proceed s usul: Some emples:. (6 8) ( 8 ) d Let u ( 8 ) Therefore du ( 6 8) d Therefore our trnsformed integrl is s follows: u u ( 8 ) du u du C C. Sometimes we cn find function whose derivtive we cn crete s follows: However only if we introduce constnt function. For emple: e ( ) d If we let u = (+), we know du = d. However while we hve the d, we do not hve. This is esily solved however through the following mnipultion: ( e ) d This integrl now contins function nd its derivtive, thus sustitution cn e used: Therefore let u = (+), nd du = d. The integrl ecomes: u e du e u C e () C Rememer, the sustitution of u into this function is device tht we employ. Therefore our finl nswer ought not to contin u, s the originl prolem does not contin it. Alwys rememer to sustitute ck for u. Note: Unlike differentition, there eists no generl formul giving the integrl of product of two function i.t.o. the seprte integrls of those functions. There is lso no generl formul giving the integrl of quotient of functions in terms of their seprte integrls. As result, integrtion is trickier thn differentition, on the whole. 9
Integrtion y Prts We use this technique when we hve to integrte product: Eg. f ( ) g'( ) d When we re given this type of emple, we mke use of the following formul: ( ) g'( ) d f ( ) g( ) f f '( ) g( ) d For emple: e d We pick f s the function which is esy to differentite, nd Often picking squred or cuic term for your f is good ide, s g ' s the function which is esy to integrte. f ' will hve power tht is then one f ', g', f nd g, to lower, nd hence simpler. It is very good ide to mke yourself mini tle with keep things stright. Also, note tht when finding g, we do not other with the ritrry constnt. e d Therefore: f f ' g' e g e e d e () e d e e C Thus we hve mnged to use the formul to integrte our originl question. ] [Cn lso use the lterntive nottion used in Ching: where vdu uv udv
Another emple: sin d Therefore: f f ' g' sin g cos sin d cos cos d cos sin C An emple with trick: d Oviously we do not know the integrl of, tht is why we re using this method. So we mke equl to f, nd we cn then find its derivtive. But then wht will our g ' e? Simple, mke it. f f ' g' g d d C This is hndy trick which cn lso e used to find the integrls of some of the trigonometric functions. Some more emples: e d Therefore: f f ' g' e e g
C e C e e d e e d e Another emple: d ' ' g f g f C d d d Another emple: d g f g f ' '
d d d C The lst line uses result tht we proved few emples go. Now let s try definite integrl using integrtion y prts: te t dt f t f ' g' e t g e t We first clculte the indefinite integrl, then go ck nd sustitute in the limits. te t dt te te t t e e t t dt C Therefore: te t dt te t e e e e e e t You should now mke sure you cn do the integrtion prctice questions. An emple of n economic ppliction of integrls: One simple ppliction is to find totl quntity from mrginl quntity. Suppose firm hs C '( ) e where denotes output. Then totl cost is: mrginl cost C( ) ( e ) d 6e B, where B is the constnt of integrtion.