Section 3: Antiderivatives of Formulas

Transcription

1 Chptr Th Intgrl Appli Clculus 96 Sction : Antirivtivs of Formuls Now w cn put th is of rs n ntirivtivs togthr to gt wy of vluting finit intgrls tht is ct n oftn sy. To vlut finit intgrl f(t) t, w cn fin ny ntirivtiv F of f n vlut F() F(). Th prolm of fining th ct vlu of finit intgrl rucs to fining som (ny) ntirivtiv F of th intgrn n thn vluting F() F(). Evn fining on ntirivtiv cn ifficult, n w will stick to functions tht hv sy ntirivtivs. Builing Blocks Antiiffrntition is going ckwrs through th rivtiv procss. So th sist ntirivtiv ruls r simply ckwrs vrsions of th sist rivtiv ruls. Rcll from Chptr : Drivtiv Ruls: Builing Blocks In wht follows, f n g r iffrntil functions of n k n n r constnts. () Constnt Multipl Rul: kf kf ' () Sum (or Diffrnc) Rul: f g f ' g' (or f g f ' g' n n n (c) Powr Rul: Spcil css: k () Eponntil Functions: (cus (cus ln k k ) ) ) ln () Nturl Logrithm: Thinking out ths sic ruls ws how w cm up with th ntirivtivs of n for. This chptr is (c). It ws rmi y Dvi Lippmn from Shn Clwy's rmi of Contmporry Clculus y Dl Hoffmn. It is licns unr th Crtiv Commons Attriution licns.

2 Chptr Th Intgrl Appli Clculus 97 Th corrsponing ruls for ntirivtivs r nt ch of th ntirivtiv ruls is simply rwriting th rivtiv rul. All of ths ntirivtivs cn vrifi y iffrntiting. Thr is on surpris th ntirivtiv of / is ctully not simply ln(), it s ln. This is goo thing th ntirivtiv hs omin tht mtchs th omin of /, which is iggr thn th omin of ln(), so w on t hv to worry out whthr our s r positiv or ngtiv. But you must crful to inclu thos solut vlus othrwis, you coul n up with omin prolms. Antirivtiv Ruls: Builing Blocks In wht follows, f n g r iffrntil functions of n k, n, n C r constnts. () Constnt Multipl Rul: kf ( ) k f () Sum (or Diffrnc) Rul: f ( ) g f f (c) Powr Rul: n n C, provi tht n = n Spcil cs: k k C (cus k k ) () Eponntil Functions: C C ln () Nturl Logrithm: ln C

3 Chptr Th Intgrl Appli Clculus 98 Empl 7 Fin th ntirivtiv of 8 / 7 7 / C 8 / Tht s littl hr to look t, so you might wnt to simplify littl: 8 7 / C. 8 Empl Fin ln C Empl Fin F() so tht F ' n F. This tim w r looking for prticulr ntirivtiv; w n to fin ctly th right constnt. Lt s strt y fining th ntirivtiv: C So w know tht F som constnt; w just n to fin which on. For tht, w ll us th othr pic of informtion (th initil conition): F C F C 9 C C Th prticulr constnt w n is 9; F 9.

4 Chptr Th Intgrl Appli Clculus 99 Th rson w r looking t ntirivtivs right now is so w cn vlut finit intgrls ctly. Rcll th Funmntl Thorm of Clculus: F ' F F If w cn fin n ntirivtiv for th intgrn, w cn us tht to vlut th finit intgrl. Th vlution F() F() is rprsnt y th symol F ] or F. Empl Evlut in two wys: (i) By sktching th grph of y = n gomtriclly fining th r. (ii) By fining n ntirivtiv of F() of th intgrn n vluting F() F(). (i) Th grph of y = is shown to th right, n th sh rgion corrsponing to th intgrl hs r. (ii) On ntirivtiv of is F( ), n 9 ]. Not tht this nswr grs with th nswr w got gomtriclly. If w h us nothr ntirivtiv of, sy F ( ) 7, thn ] Whtvr constnt you choos, it gts sutrct wy uring th vlution; w might s wll lwys choos th sist on, whr th constnt =. Empl Fin th r twn th grph of y = n th horizontl is for twn n. This is 7. ]

5 Chptr Th Intgrl Appli Clculus Empl 6 A root hs n progrmm so tht whn it strts to mov, its vlocity ftr t scons will t ft/scon. () How fr will th root trvl uring its first scons of movmnt? () How fr will th root trvl uring its nt scons of movmnt? () Th istnc uring th first scons will th r unr th grph of vlocity, from t = to t =. Tht r is th finit intgrl t t. An ntirivtiv of t is t, so t t t ft. () t t t ft. Empl 7 Suppos tht t minuts ftr putting ctri on Ptri plt th rt of growth of th popultion is 6t ctri pr minut. () How mny nw ctri r to th popultion uring th first 7 minuts? () Wht is th totl popultion ftr 7 minuts? () Th numr of nw ctri is th r unr th rt of growth grph, n on ntirivtiv of 6t is t. 7 So nw ctri = 6t t = t 7 = (7) () = 7 () Th nw popultion = (ol popultion) + (nw ctri) = + 7 = 7 ctri. Empl 8 A compny trmins thir mrginl cost for prouction, in ollrs pr itm, is MC ( ) whn proucing thousn itms. Fin th cost of incrsing prouction from thousn itms to thousn itms. Rmmr tht mrginl cost is th rt of chng of cost, n so th funmntl thorm tlls us tht MC ( ) C( ) C( ) C( ). In othr wors, th intgrl of mrginl cost will

6 Chptr Th Intgrl Appli Clculus giv us nt chng in cost. To fin th cost of incrsing prouction from thousn itms to thousn itms, w n to intgrt MC ( ). W cn writ th mrginl cost s MC ( ) /. W cn thn us th sic ruls to fin n ntirivtiv: / C( ) 8. Using this, / Nt chng in cost = It will cost.889 thousn ollrs to incrs prouction from thousn itms to thousn itms.

7 Chptr Th Intgrl Appli Clculus. Erciss For prolms -, fin th inict ntirivtiv y. w. P P t t 9.. t t For prolms -8, fin n ntirivtiv of th intgrn n us th Funmntl Thorm to vlut th finit intgrl.... ( + ) For prolms 9 - fin th r shown in th figur. 9...