ENGR 1990 Engineering Mathematics The Integral of a Function as a Function


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1 ENGR 1990 Engineering Mhemics The Inegrl of Funcion s Funcion Previously, we lerned how o esime he inegrl of funcion f( ) over some inervl y dding he res of finie se of rpezoids h represen he re under f( ) from o. We now consider he concep of niderivives nd how hey re used o clcule he inegrl of funcion. This ide lso eposes he direc link eween he conceps of differeniion nd inegrion. Aniderivives A funcion G ( ) is clled n niderivive of funcion f( ) over some inervl if nd only if G ( ) is coninuous nd G( ) f ( ) on he inervl. The following le gives niderivives of some common funcions used in engineering. Nme Funcion, ( ) f Aniderivive, G( ) G( ) f ( ) Consn Polynomil erms n n1 n 1 Eponenil e e Sine sin( ) cos( ) Cosine cos( ) sin( ) Noe h he niderivive is no unique, ecuse we cn dd consn o ny known niderivive o produce noher niderivive. Recll h he derivive of consn is zero. Fundmenl Theorem of Inegrl Clculus If f( ) is coninuous funcion over some inervl, nd G ( ) is n niderivive of f( ) on h sme inervl, hen f ( ) d G ( ) d G( ) G( ) G( ) (1) So, if we know he funcion nd n niderivive, hen we cn clcule he inegrl direcly. We do no hve o pproime i y summing res. Kmmn ENGR 1990 Engineering Mhemics pge: 1/6
2 Inegrls of Funcions s Funcions The resul of Eq. (1) is numer h represens he re under funcion f( ) from o. These resuls cn e mde more generl o pply n rirry ending poin of y simply replcing he upper limi of he inegrl wih he vrile. f ( ) d G ( ) d G( ) G( ) G( ) () This resul my e used o clcule he inegrl for sring poin of o ny ending poin. Indefinie Inegrls When we do no specify he inervl over which he inegrl is o e evlued (h is, we do no specify he limis of he inegrl), we cll he inegrl indefinie. Effecively, we re using rirry upper nd rirry lower limis. We wrie, f ( ) d G( ) D (3) Here, D is n rirry consn. If G ( ) is n niderivive of f( ) on he inervl, we cn evlue he resul of Eq. (3) over h inervl. As epeced, we ge he sme resul s in Eq. (1). f ( ) d G( ) D G( ) D G( ) D G( ) G( ) (4) If we use Eq. (3) nd evlue over n rirry upper limi, we ge he sme resul s in Eq. (). f ( ) d G( ) D G( ) D G( ) D G( ) G( ) Kmmn ENGR 1990 Engineering Mhemics pge: /6
3 Emple 1: Given: The displcemen of cr s i moves wih velociy from ime 1 o is he inegrl of over h period of ime. s v() d 1 The displcemen cn e posiive or negive depending on wheher is posiive or negive. Find: Assuming he cr hs velociy v ( ) 7. (f/s ), () find he displcemen of he cr from o seconds; () find he ol disnce rveled from o seconds. Soluion: () An niderivive of v( ) 7. is G( ) , so s v( ) d 7. d (f) () As efore, ecuse is posiive in he rnge, he ol disnce rveled is d s 78.7 (f). Emple : Given: The velociy of ll for cerin period of ime fer i is hrown upwrd is v( ) (f/s) Find: () he vericl displcemen of he ll from 0 o seconds; nd () he ol disnce rveled y he ll from 0 o seconds. 3 Kmmn ENGR 1990 Engineering Mhemics pge: 3/6
4 Soluion: () An niderivive of v( ) is G( ) , so s d (f) () To find he ol disnce rveled, we mus o rek he inervl ino wo segmens: 0 3, nd s d d (f) Emple 3: Given: The velociy of cr over he ime inervl from 0 o seconds is v ( ) (f/s). Find: The disnce rveled y he cr from o seconds. Soluion: Since he funcion is posiive hroughou he enire rnge of, he ol disnce rveled is equl o he displcemen. We cn clcule he displcemen y noing h n niderivive of v( ) G(). So, 3 3 is d s d (f) Kmmn ENGR 1990 Engineering Mhemics pge: 4/6
5 Emple 4: Given: The velociy of ll for cerin period of ime fer i is hrown upwrd is v( ) (f/s) Find: () s (), he vericl displcemen of he ll s funcion of ime; () he ime required for he ll o is sring poin; nd (c) he mimum heigh of he ll. 3 Soluion: () The displcemen funcion relive o he sring poin my e found y inegring o n rirry upper limi. () s( ) v( ) d d In his resul, he displcemen funcion is zero 0. If we wn o specify nonzero displcemen 0, sy s(0) 0, hen we ke slighly differen pproch. Firs, we se s( ) v( ) d D Then, we solve for D o sisfy he iniil condiion. In his cse, D 0. () To find he ime required for he ll o reurn o is sring poin, se s () from Eq. () o zero, nd solve for. s( ) or 0 0 (sec) 6 (sec) (c) Recll h mim or minim of funcions occur when heir derivives re zero. d s d ( ) (16.1 ) or 3 (sec) m s (96.6 3) (f) 3 This corresponds physiclly o he condiion h he ll s velociy e zero. Kmmn ENGR 1990 Engineering Mhemics pge: /6
6 Recll lso h for he funcion o hve mimum, he second derivive should e negive. d s( ) (checks) d Emple : Given: A ll is hrown upwrd wih n iniil velociy of 0 (f/s) from n iniil heigh of 6 (f). The ll hs downwrd ccelerion of 3. (f/s ). v( ) ( ) d s( ) v( ) d Find: () he velociy funcion ; nd () he displcemen funcion s (). Soluion: () Using he indefinie form of inegrion nd he iniil condiion v(0) 0 (f/s) v( ) ( ) d 3.d 3. D v( ) 0 3. (f/s) () Using he indefinie form of inegrion nd he iniil condiion s(0) 6 (f) s( ) v( ) d 0 3. d D s( ) (f) Kmmn ENGR 1990 Engineering Mhemics pge: 6/6
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