9 Integrtion 9A Antidifferentition 9B Integrtion of e, sin ( ) nd os ( ) 9C Integrtion reognition 9D Approimting res enlosed funtions 9E The fundmentl theorem of integrl lulus 9F Signed res 9G Further res 9H Ares etween two urves 9I Averge vlue of funtion 9J Further pplitions of integrtion res of STud Antiderivtives of polnomil funtions nd of f ( + ), where f is n for n Q, e, sin (), os () or liner omintions of these Definition of the definite integrl s the limiting vlue of sum n δ i = f( ) d = lim f( * ) δ, where the intervl [, ] is prtitioned into n suintervls, with the ith suintervl of length δ i nd ontining * i, nd δδ = m{δ δ i : i =,, n} nd evlution of numeril pproimtions sed on this definition Informl pproimtion using res under urves left nd right retngles Emples of the definite integrl s limiting vlue of sum involving quntities suh s re under urve, distne trvelled in stright line nd umultive effets of growth suh s infltion Antidifferentition reognition tht F ( ) = f( ) implies f( ) d = F ( ) + i i Informl tretment of the fundmentl theorem of lulus, f( ) d = F ( ) F () Properties of ntiderivtives nd definite integrls: [ f ( ) ± g g( ( )] d = f ( d ) ) d ± + f ( ) d f ( )d = f( ) d f ( ) d = f( ) d = f( ) d g ( ) d Applition of integrtion to prolems involving lultion of the re of region under urve nd simple ses of res etween urves, suh s distne trvelled in stright line; verge vlue of funtion; other situtions modelled the use of the definite integrl s limiting vlue of sum over n intervl; nd finding funtion from known rte of hnge ebookplus 9 ntidifferentition Digitl do Quik Questions As we hve seen, the proess of differentition enles us to find the grdient of funtion. The reverse proess, ntidifferentition (or integrtion), will find the funtion for prtiulr grdient. Integrtion hs wider pplitions inluding lultion of res, volumes, energ, proilit nd mn more quntities in siene nd usiness. Note tht d d f( ) mens differentite f () with respet to ; tht is, d d f( ) = f ( ). 6 mths Quest mthemtil methods CS for the Csio ClssPd
So f () is the ntiderivtive of f '(), denoted s f( ) = f ' ( ) d where mens ntidifferentite, or integrte, or find n indefinite integrl nd d indites tht the integrtion of the funtion is with respet to. Sine d d + ) =, where nd re onstnts then d = + Sine then Worked Emple d n+ d n + = n+ n d = n n + +, In the epression ove, the term is used to denote onstnt. In the ntiderivtive of funtion, there re n infinite numer of possiilities for. However, when we re finding n ntiderivtive, we set to zero. Tht is, finding n ntiderivtive mens let =, or do not dd on the. For emple, the ntiderivtive of + + is + + +. An ntiderivtive of + + is + +. Properties of integrls Sine d d is liner opertor, so too is. Therefore, n [ f( ) ± g ( )] d = f( ) d ± g ( ) d Tht is, eh term n e integrted seprtel, nd kf( ) d= k f( ) d Tht is, onstnt ftor of the funtion n e tken to the front of the integrl. So [ f ( ) ± g ( )] d = f( d ) ± gd ( ) Antidifferentite eh of the following, epressing nswers with positive indies. 7 Think Integrte rule; tht is, dd to the inde nd divide the new inde. Write 8 7 d = + 8 Simplif. 8 = + Integrte rule. d = + Simplif. = + Epress the nswer with positive inde. = When squre root is involved, reple it with frtionl inde. d = d Chpter 9 Integrtion 7
Bring the to the numertor nd hnge the sign of the inde. Integrte rule. = + Simplif. = 6 Write the nswer in the form it ws given. = 6 + = + d Worked Emple Find the following indefinite integrl. ( - )( + ) d Think Write Epnd the epression. ( )( + ) d = ( + ) d Collet like terms. = ( + ) d Integrte eh term seprtel. = + + Simplif eh term. = + + Integrtion of ( + ) n where n B ppling the hin rule for differentition: d ( + ) n+ = ( n+ )( + ) d so n ( + )( + ) n d= ( + ) n+ + or n ( + ) ( + ) n d= ( + ) n+ + or ( ) ( ) n + n + d = n ( + ) + n + Worked Emple Antidifferentite ( ) using ( ) ( ) n + n + d = n ( + ) Think + Write +. Epress s n integrl nd tke out s ftor. ( ) d= ( ) d ( ) Appl the rule where = nd n =. = + ( ) Simplif the ntiderivtive nelling the frtion. = ( ) + 8 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Worked Emple Integrtion of Sine d log e() = d then d = log e() +,where > or d= log e +. Antidifferentite 7. Think Write Tke out s ftor. 7 7 d = 7 d = 7 d Integrte rule. = log + 7 e Integrtion of ( + ) - This n e done ppling the hin rule for differentition: Multipling oth sides gives So d log e ( + ) =, where nd re onstnts. d + d log e ( + ) = d + = + d d = loge + d + d so ( + ) d = loge + + Note tht the in the frtion is the derivtive of the liner funtion +. Worked Emple Antidifferentite Think +. Write Method : Tehnolog-free Epress s n integrl nd tke out s ftor. d = + d + Integrte rule where =. = log + + e Chpter 9 Integrtion 9
Method : Tehnolog-enled On the Min pge, using the soft keord, tp: ) - P Complete the entr line s: ( ) d + Press E. Write the nswer nd put in the onstnt of integrtion, s the CAS lultor omits it. Also note tht the CAS uses ln insted of log e. d + = log e + + Worked Emple 6 6 + Find d. Think Write Epress s seprte frtions. 6 + 6 d = d + Simplif eh frtion. = ( + ) d Integrte eh term seprtel rule. = 6log e + + Simplif leving the nswer with positive indies. = 6log e + = 6log e + Worked Emple 7 Find the eqution of the urve g() given tht g () = + nd the urve psses through (, ). Think Write Write the rule for g (). g () = + Rewrite g () in inde form. g () = + Epress g() in integrl nottion. g() = + d Antidifferentite to otin generl rule for g(). = + + Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Simplif. = + + g() = + + 6 Sustitute oordintes of the given point into g(). As g() =, () + () + = 7 Find the onstnt of ntidifferentition,. + + = so = 8 Stte the rule for g() in the form tht it is given. g() = + = + Worked Emple 8 If urve hs sttionr point (, ), nd grdient of k, where k is onstnt, find: the vlue of k when =. Think Write The grdient is d so write the rule for the d grdient. Let d d = (s sttionr points our when the derivtive is zero) nd sustitute the vlue of into this eqution. d d = k For sttionr points, d =, so k = d () k = s = Solve for k. k = so k = Rewrite the rule for the grdient funtion, using the vlue of k found in ove. d = d Integrte to otin the generl rule for. = ( ) d Sustitute the oordintes of the given point on the urve to find the vlue of. = + Sine urve psses through (, ), = () + = 8 + = 7 Stte the rule for. So = + 7 Sustitute the given vlue of nd lulte. When =, = () () + 7 = The reltionship etween the grph of n ntiderivtive funtion nd the grph of the originl funtion f () is the ntiderivtive of f () nd is written s f( ) = f ( ) d. f () is the grdient funtion of the ntiderivtive funtion f (). Chpter 9 Integrtion
The grph of the ntiderivtive funtion f () n e derived from the grph of f () nd results in fmil of urves. For emple if f () = then the ntiderivtive funtion is f () = +, where n tke n rel vlue. Skething the ntiderivtive funtion from the grph of the originl funtion. The generl shpe of the grph of the ntiderivtive funtion n e determined from the grph of polnomil funtion inresing the degree one. For emple, if f () is qudrti funtion, then f () is ui funtion.. The -interepts of f () eome the turning points on the grph of f ().. When f () is ove the -is, the grdient of f () is positive.. When f () is elow the -is, the grdient of f () is negtive. Worked Emple 9 Sketh the grph of the ntiderivtive funtion from the grph of the grdient funtion f () shown. f () (, ) (, ) (, ) Think Write Stte the shpe of the ntiderivtive funtion. The ntiderivtive funtion will e positive ui funtion. Find the -interepts of the grdient funtion, f (), nd hene find the -oordintes of the turning points. Find when the given grph, f (), is ove the -is nd hene find when f () hs positive grdient. Find when the given grph, f (), is elow the -is nd hene find when f () hs negtive grdient. There re -interepts when = nd when =, so f () hs turning points when = nd =. f () is ove the -is when < nd when >, so f () hs positive grdient when < nd when >. f () is elow the -is when < <, so f () hs negtive grdient when < <. Sketh the urve. The grph ould e n of the fmil of grphs formed vertil trnsltions of the grph shown. f() Mths Quest Mthemtil Methods CAS for the Csio ClssPd
REMEMBER. d d f( ) = f ( ). f( ) = f ( ) d. d = + n + n. d = n +. 6. 7. 8. 9. + n, [ f( ) ± g ( )] d = f( ) d ± g ( ) d kf( ) d= k f( ) d ( ) ( ) n + n + + d = +, n n ( + ) d = loge +, or d = loge + d = loge + + or + d = e + + ( ) log +. Write the nswer in the sme form s the question unless otherwise stted.. To drw the grph of the ntiderivtive funtion f () from the grph of the grdient funtion f (), use the -interepts s the -oordintes of the turning points, nd mke the grdient positive when f () is ove the -is nd negtive when f () is elow the -is.. For polnomil funtion, the grph of f () is one degree greter thn the grph of f (). Eerise 9 Antidifferentition WE Antidifferentite eh of the following, giving nswers with positive indies. 7 d e f g 6 h i j k l m n o 7 p q 9 r 6 s 8 t 6 ( ) WE Find the following indefinite integrls. ( + ) d ( + ) d ( + 6 + ) d d ( + 6 + ) d e ( + ) d f ( + )( 7) d g ( + ) d h ( + )( ) d i )( + ) d Chpter 9 Integrtion
MC ( + + ) d is equl to: A d + + B d + d + d C ( + ) d + D + ( + ) d E + + d MC ( + ) d is equl to: A d ( + ) d B ( + ) d C ( + ) d D d + ( + ) d E ( + ) d ( ) WE Antidifferentite eh of the following using ( ) n + n + d = n ( + ) ( + ) ( ) ( + ) d ( ) e (6 + ) f ( ) g ( ) h (7 ) i (8 ) j (8 9) k ( + ) l (6 + ) m 6( 7) n ( 8) 6 o (6 ) p (7 ) 6 MC ( + ) d is equl to: A + ( + ) d B d + ( + ) d C ( + ) d D d ( + ) d E ( + ) d 7 WE, Antidifferentite the following. + +. d g j d 7 d e d h + d k + 8 d 7 d f d + 8 d l + 6 i 6 d d + 6 d + d m + d n 6+ 7 d o d p 6 d q d r 8 d Mths Quest Mthemtil Methods CAS for the Csio ClssPd
ebookplus Digitl do SkillSHEET 9. Sustitution nd evlution 8 mc 6 d is equl to: + A 6 d B 6d d + D 6d ( + ) d 9 We6 Find ( + 7) d. E + 6 ( + ) d C 6d + d + For the following mied sets, find: + + ( ) + + d d d d e + d f ( ) d 6 g ( + ) ( + ) d h d i j + d k + d l m + d n + d We7 Find the eqution of the urve f () given tht: f () = + nd the urve psses through (, ) f () = nd the urve psses through (, ) f () = + nd the urve psses through (, ) d f () = + nd f () = e f () = + nd f (8) = ( )( + ) d + f f () = nd f () = g f () = ( + ) nd the urve psses through (, ) h f () = 8( ) nd f () = i f () = ( + ) nd the urve psses through (, ) 8 j f () = nd f () = 7 7 We8 If urve hs sttionr point (, ), nd grdient of 8 + k, where k is onstnt, find: the vlue of k when =. A urve g() hs g ( ) = k +, where k is onstnt, nd sttionr point (, ). Find: the vlue of k g() g(). We9 Sketh the grph of the ntiderivtive funtions from eh of the following grphs. f () f () f () d (, ) (, ) (, ) (, ) (, ) (, ) Chpter 9 Integrtion
d f () e f () f f () (, ) (, ) (, ) (, ) (, ) g f () (, ) (, ) (, ) 9B Integrtion of e, sin () nd os () Integrtion of the eponentil funtion e Sine d ( e ) d = e then ed =e e + nd d d ke k ( e ) =, where k is onstnt or kek d = e + k ek d = e + Therefore, e d k = e +. Worked emple Antidifferentite eh of the following. e e (e ) ThInk WrITe Integrte rule where k =. e d e = + Simplif. = e + Rewrite the funtion to e integrted so tht the oeffiient of the e term is ler. Integrte rule where k =. e d = = e e + d 6 mths Quest mthemtil methods CS for the Csio ClssPd
Simplif the ntiderivtive. = e = e + + Epnd the funtion to e integrted. ( e ) d= ( e e + ) d Integrte eh term the rule. = e e + + Integrtion of trigonometri funtions d d Sine [sin ( )] = os ( ) nd [os ( )] = sin ( ) d d it follows tht sin( ) d = os ( ) + os( ) d = sin ( ) + Worked emple Antidifferentite the following. sin (6) 8 os () sin ThInk WrITe Integrte rule. sin( 6d ) = os ( 6) + 6 8 Integrte rule. 8 os( d ) = sin ( ) + Simplif the result. = sin () + Integrte rule. sin os d = + Simplif the result. = 6 os + Worked emple Find [ e sin ( ) + d ]. ThInk WrITe Method : Tehnolog-free Integrte eh term seprtel. [ e sin ( ) + d ] = e os( ) + + Simplif eh term where pproprite. = e + os( ) + + Chpter 9 Integrtion 7
Method : Tehnolog-enled On the Min pge, using the soft keord, tp: ) - P Complete the entr line s: e sin( ) + d Press E. Note: When integrting trigonometri funtions, ensure the CAS is set to Rdin mode. Write the nswer nd put in the onstnt of integrtion. ( e sin ( ) + ) d = e os( ) + + + REMEMBER. ed e = + nd ek d k e k = + sin( ) d =. os ( ) + nd os( ) d sin ( ) = + Eerise 9 Integrtion of e, sin () nd os () WE Antidifferentite eh of the following. e e e d e e e f 7e g e6 e h i e 6 j 8e k e l. e e m e n p e e Find n ntiderivtive of ( + e ). Find n ntiderivtive of (e ). Find n ntiderivtive of + 6e. o e + e MC If f () = e + k nd f () hs sttionr point (, ), where k is onstnt, then: k is equl to: A e B e C D E e f () is equl to: A e B e + C e + D e E 8 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
6 WE Antidifferentite the following. sin () sin () os (7) d os( ) e sin ( ) f os ( ) g sin( 6 ) h 8 os () i 6 sin () j os ( ) k sin l os m sin n sin o os p s 6 os sin q sin r 6 os t 7 os u sin (π) π v os w π os sin 7 WE Find: [sin ( ) + os( )] d [sin ( ) os( )] d [os ( ) + sin( )] d d sin ( os ( ) d ) e [ os ( ) sin( )] d f [ + sin ( )] d g sin π os π ( d ) ( + ) h [ sin ( 8 ) 7] e 6 + d 8 Find the ntiderivtive of e + sin () +. 9 Find n ntiderivtive of os () + 6e. Antidifferentite eh of the following. e + + + os () e sin e ( ) + d e sin os e e + + os π f + sin π + In eh of the following, find f () if: π f () = os () nd f = f () = sin () nd f () = f ( ) = os nd f ( π ) = 9 d = f ( ) os sin nd f (π) =. If d π = sin k d +, where k is onstnt, nd hs sttionr point (, ), find: 6 the vlue of k the eqution of the urve when = 6. A urve hs grdient funtion f () = os () + ke, where k is onstnt, nd sttionr point (, ). Find: the vlue of k the eqution of the urve f () π f 6 orret to deiml ples. Chpter 9 Integrtion 9
9C Integrtion reognition As we hve seen, if d d [ f( )] = g ( ) then gd ( ) = f( ) +, where g( ) = f' ( ). This result n e used to determine integrls of funtions tht re too diffiult to ntidifferentite, vi differentition of relted funtion. Worked emple Find the derivtive of the funtion = ( + ). Use this result to dedue the ntiderivtive of ( + ). ThInk WrITe ebookplus Tutoril int-6 Worked emple Write the funtion nd reognise tht the hin rule n e used. Let u equl the funtion inside the rkets. Find du d. = ( + ) Let u = + du d = Epress in terms of u. = u Find d du. 6 Write the hin rule. 7 Find d d using the hin rule. d 8 Reple u with the epression inside the rkets nd simplif where pplile. du Sine d d, epress the reltionship in integrl nottion. Remove ftor from d so tht it resemles d the integrl required. d = u du d d = d du du d d = u = ( + ) ( + ) d = = ( +) ) + ( + ) d = ( + ) + Divide oth sides the ftor in order to otin the integrl required. ( +) d = = ( + ) +, where = Therefore, the ntiderivtive of ( + ) is ( + ) +. Note tht the shorter form of the hin rule elow n e used to differentite. If f( ) = [ g] ( )] n then f ( ) = ng ( ) ) [g g( )] n. mths Quest mthemtil methods CS for the Csio ClssPd
Worked emple Differentite e. Hene, ntidifferentite 6 e. ThInk WrITe Write the eqution nd ppl the hin rule to differentite. = e Let u equl the inde of e. Let u = Find du d. du = d Epress in terms of u. = e u Find d du. 6 Find d using the hin rule nd d reple u. d = e u du d = e u d = e Epress d d in integrl nottion. e d = e + Multipl oth sides onstnt to otin the integrl required. e d =e e + 6e d =e e +, where =. Therefore, the ntiderivtive of 6 e is e +. Worked emple Note tht the shorter form of the hin rule elow n e used to differentite. If = e f ) ( then d f e f ( = ( ) ). d Find the derivtive of sin ( + ) nd use this result to dedue the ntiderivtive of 8 os ( + ). Differentite log e ( ) nd hene ntidifferentite. ThInk WrITe Define f (). f () = sin ( + ) Differentite using f () = g () os [g()] where f () = sin [g()]. f () = os ( + ) Epress f () using integrl nottion. os( +) )d d = sin ( + ) + Multipl oth sides whtever is neessr for it to resemle the integrl required. Write the integrl in the form in whih the question is sked. os( +)d ) d = si in ( + ) + 8 os( +) )d d = si in ( + ) + The ntiderivtive of 8 os ( + ) is sin ( + ) +. Define f (). f () = log e ( ) Chpter 9 Integrtion
g Differentite using f = ( ) () g ( ) where f () = log e [g()]. Epress f () using integrl nottion. Tke out ftor so tht f () resemles the integrl required. Divide oth sides the ftor to otin the required integrl. f ( ) = d = loge + d = log e + d e = log + The ntiderivtive of is log e +. Worked emple 6 Differentite os () nd hene find n ntiderivtive of sin (). ThInk WrITe Write the rule. Let = os () Appl the produt rule to differentite os (). Epress the result in integrl nottion. (Do not dd, s n ntiderivtive is required.) d d = [ sin ()] + [os ()]() = sin () + os () = os () sin () [os ( ) ) sin( )] d = os( ) Epress the integrl s two seprte integrls. os( )d d sin ( )d ) d = os( ) Simplif integrting. (Do not dd.) sin( ) ) sin ( )d ) d = os( ) 6 Mke the epression to e integrted the sujet of the eqution. sin( ) d = os ( ) sin( ) 7 Simplif. sin( ) d =sin s () os( ) Therefore, n ntiderivtive of sin () is sin () os (). Worked emple 7 Show tht + = + +. Hene, find + d. + ThInk Use lgeri long division to divide the numertor into the denomintor. WrITe + ) ) ) + + So + + = + mths Quest mthemtil methods CS for the Csio ClssPd
Write the epression using integrl nottion. + d = + = + Epress s two seprte integrls. = d d + Antidifferentite eh prt. = log + + e remember. To differentite using the hin rule, use one of the following rules. () If f () = [g()] n then f () = ng ()[g()] n () If = e f (), d d f e f ( = ( ) ) () d d du = d du d (d) f () = g () os [g()] where f () = sin [g()] f () = g () sin [g()] where f () = os [g()] g (e) f = ( ) () g ( ) where f( ) = log e g( ( ).. To ntidifferentite, use g ( ) d = f ( ) + where g() = f (). eercise 9C Integrtion reognition We For eh of the following find the derivtive of the funtion in i nd use this result to dedue the ntiderivtive of the funtion in ii. i ( ) 8 ii ( ) 7 i ( + ) ii ( + ) i ii d i + ii + e i ( + 7) ii ( + )( + 7) f i ii ( ) mc The derivtive of ( + 7) is ( + 7). Therefore, the ntiderivtive of ( + 7) is: A ( + 7) + B ( + 7) + C ( + 7) + D ( + 7) + E ( + 7) + The ntiderivtive of ( + 7) is: A ( + 7) + B ( + 7) + C ( + 7) + D ( + 7) + E ( + 7) + Chpter 9 Integrtion
mc If the derivtive of ( ) 6 is ( ), then 6 ( ) d is: A ( ) 6 + B ( ) 6 + C ( ) 6 + D 6( ) 6 + E )6 + We For eh of the following differentite i nd hene ntidifferentite ii. i e ii e i e 6 ii e 6 i e ii e d i e ii ( e ) We For eh of the following find the derivtive of the funtion in i nd use this result to dedue the ntiderivtive of the funtion in ii. i sin ( ) ii os ( ) i sin ( + ) ii 6 os ( + ) i os ( 7) ii sin ( 7) d i os (6 ) ii sin (6 ) e i sin ( ) ii os ( ) f i os ( ) ii sin ( ) g i log e ( + ) ii h i log e ( + ) ii + + i i log e ( ) ii 6 We6 Differentite i nd hene find n ntiderivtive of ii. i os () + sin () ii sin () i sin( ) [ os ( ) sin( )] ii i e sin () ii e [sin () + os ()] d i sin () ii os () e i e ii e 7 For eh of the following differentite i nd use this result to ntidifferentite ii. i ( ) 6 ii 6 ( )( ) i + ii + + 8 We7 Show tht = +. Hene, find d. 9 Show tht + 8 = + +. Hene, find + 8 d. + Show tht 8 7 = +. Hene, find 8 7 d. Show tht 6 = +. Hene, find 6 d. If = log e [os ()]: find d d. Hene, find tn( )d d. Differentite os( ) sin( ) nd hene find n ntiderivtive of sin ( ). Differentite log e ( ) nd hene find n ntiderivtive of. Differentite sin ( + ) nd hene find n ntiderivtive of os ( + ). (Here, nd re onstnts.) 6 Differentite os ( + ) nd hene find n ntiderivtive of sin ( + ). (Here, nd re onstnts.) mths Quest mthemtil methods CS for the Csio ClssPd
7 Differentite e + nd hene find n ntiderivtive of e +. (Here, nd re onstnts.) 8 Antidifferentite eh of the following. sin (π + ) os ( π ) e π + π d sin + π e os + f os ()esin () ebookplus Digitl do WorkSHEET 9. 9d pproimting res enlosed funtions There re severl ws of finding n pproimtion to the re etween grph nd the -is. We will look t two methods:. the left retngle method. the right retngle method ebookplus Intertivit int- Approimting res enlosed funtions The left retngle method Consider the re etween the urve f () shown t right, the -is nd the lines = nd =. If the re is pproimted left retngles of width unit then the top left orner of eh retngle touhes the urve t one point. So, the height of retngle R is f () units nd the re of R = f () squre units (re of retngle = height width). Similrl, the re of R = f () squre units, the re of R = f () squre units, the re of R = f () squre units. Therefore, the pproimte re under the grph etween the urve f (), the -is nd the lines = to = is [ f () + f () + f () + f ()] squre units, (the sum of the re of the four retngles). If the sme re ws pproimted using retngle widths of. there would e 8 retngles nd the sum of their res would e:.[ f () + f (.) + f () + f (.) + f () + f (.) + f () + f (.)] squre units. From the digrm it n e seen tht the left retngle pproimtion is less thn the tul re under the urve. f() R R R R Worked emple 8 Find n pproimtion for the re etween the urve f () shown nd the -is from = to = using left retngles of width. units. The grph shown hs the eqution f () =. +. f() ebookplus Tutoril int-6 Worked emple 8... ThInk Method : Tehnolog-free Write the numer of retngles nd their width. WrITe There re retngles of width. units. Chpter 9 Integrtion
Find the height of eh retngle (left) sustituting the pproprite -vlue into the f () eqution. Are equls the width multiplied the sum of the heights. Clulte this re. = 7. h = f () =.() + =. h = f (.) =.(.) + =. h = f () =.() + =.8 h = f (.) =.(.) + =. Are = width (sum of heights of retngles) =.(. +. +.8 +.) =.(.7) Stte the solution. The pproimte re is 7. squre units. Method : Tehnolog-enled On the Min pge tpe:. + Highlight the eqution nd tp: Intertive Define OK To find the pproimte re, omplete the entr line s:.( f() + f(.) + f() + f(.) Press E. Write the nswer with the orret units for re. The pproimte re is 7. squre units. The right retngle method Consider the re etween the urve f () shown t right, the -is nd the lines = nd =. If the re is pproimted right retngles of width unit then the top right orner of eh retngle touhes the urve t one point. So, the height of R is f () units nd the re of R is f () squre units. Similrl, the re of R = f () squre units, the re of R = f () squre units the re of R = f () squre units. Therefore, the pproimte re etween the urve f (), the -is R R R R nd the lines = to = is (R + R + R + R ) = [ f () + f () + f () + f ()] squre units. If the sme re ws pproimted with upper retngle widths of. units, the sum of their res would equl:.[ f (.) + f () + f (.) + f () + f (.) + f () + f (.) + f ()] squre units. From the digrm it n e seen tht the right retngle pproimtion is greter thn the tul re under the urve. For n inresing funtion, left retngle pproimtion tul re right retngle pproimtion. For deresing funtion, left retngle pproimtion tul re right retngle pproimtion. f() 6 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Worked Emple 9 Find n pproimtion for the re in the digrm in worked emple 8 using right retngles of width. units. f () =. + Think Write Find the numer of retngles nd the height of eh one (from left to right). Are is the width of the intervl multiplied the sum of the heights. Clulte the re. =.(6.) = 8. There re retngles: h = f (.) =.(.) + =. h = f () =.() + =.8 h = f (.) =.(.) + =. h = f () =.() + =.8 Are =.(. +.8 +. +.8) Stte the solution. The pproimte re is 8. squre units. It n e seen tht the left retngle pproimtion (7. units) is less thn the right retngle pproimtion (8. units). If the re is divided into nrrower strips, the estimte of the re would e loser to the true vlue. Worked Emple With width intervls of unit, lulte n pproimtion for the re etween the grph of f () = + nd the -is from = to = using: left retngles right retngles verging of the left nd right retngle res. Think Write Sketh the grph of f () over domin whih eeeds the width of the required re Drw the left nd right retngles. = + = Left retngles = Right retngles Clulte the height of the left retngles sustituting the pproprite vlues of into the eqution for f (). Note the two retngles to the right nd left of the origin hve the sme height nd re equl in re. Find the re multipling the width the sum of the heights. Left retngle heights: f ( ) = ( ) + = 6 f ( ) = ( ) + = f () = + = f () = + = f () = + = 6 Are = (6 + + + + 6) = Using left retngles, the pproimte re is squre units. Chpter 9 Integrtion 7
Clulte the height of the right retngles sustituting the pproprite vlues of into the eqution for f (). Find the re multipling the width the sum of the heights. Find the verge dding the re of the left retngles nd right retngles nd dividing. Right retngle heights: f ( ) = (from ove) f () = f () = f () = 6 f () = + = Are = ( + + + 6 + ) = Using right retngles, the pproimte re is squre units. + Averge of the res = =. The pproimte re is. squre units when verging the left nd right retngle res nd using widths of unit. Note tht this verge is etween the re of the left retngles nd the re of the right retngles nd is loser to the tul re. remember. An pproimtion to the re etween urve nd the -is n e found dividing the re into series of retngles tht re ll the sme width. The pproimtion is found finding the sum of ll the res of the retngles.. For n inresing funtion, left retngle pproimtion tul re right retngle pproimtion.. For deresing funtion, left retngle pproimtion tul re right retngle pproimtion. eercise 9d pproimting res enlosed funtions We8 Find n pproimtion for the re etween the urve f () t right nd the -is from = to = using left retngles of width units. Find n pproimtion for the re etween the urves elow nd the -is, from = to =, lulting the re of the shded retngles. f() (, ) (, ) f() 9 7 (, ) (, 7) (, ) (, 9) (, ) (, ) f() 8 mths Quest mthemtil methods CS for the Csio ClssPd
MC Consider the grph of = from = to = t right. The width of eh retngle is: A unit B units C units D units E vring The height of the right-most retngle is: A 9 units B units C 6 units D units E unit The re etween the urve = nd the -is from = to = n e pproimted the re of the left retngles s: A sq. units B sq. units C 8 sq. units D sq. units E sq. units WE 9 Find n pproimtion for the re in the digrm t right using right retngles of width unit. A etter pproimtion for the re under this urve n e found verging the right nd left retngle res. Stte this pproimte vlue. Find n pproimtion for the re etween the urves elow nd the -is, from = to =, lulting the re of the shded retngles. (, 8) (, 8) (, 8) 8 8 7 6 8 7 = = (, ) (, ) (, 7) f() f() f() d g 7 (, ) (, ) f() (, 7) (, ) (, 7) e 9 6 (, 6) (, ) (, 9) (, ) f() f (, ) f() (, ) f() = + 6 WE With width intervls of unit lulte n pproimtion for the re etween the grph of f () = + nd the -is from = to = using: left retngles right retngles verging of the left nd right retngle res. Chpter 9 Integrtion 9
7 Find the pproimte re etween the urves elow nd the -is, over the intervl indited, lulting the re of the shded retngles. Give et nswers. = e = + + 8 = to = = to = = log e () d = ( ) = to = 6 = to = 6 e f() = + 8 f f() = g = to = = to = 6 6.. = to = = 6 8 Clulte n pproimtion for the re etween the grph of = ( ), the -is nd the lines = nd =, using intervl widths of unit nd: left retngles right retngles verging the left nd right retngle res. 9 Clulte n pproimtion for the re under the grph of = + to the -is etween = nd =, using intervl widths of. units nd: left retngles right retngles verging the left nd right retngle res. Find n pproimtion for the re under the grph of = etween = nd =, using intervl widths of unit, verging the left nd right retngle res. Mths Quest Mthemtil Methods CAS for the Csio ClssPd
9e The fundmentl theorem of integrl lulus Consider the region under the urve f () etween = nd =, where f () nd is ontinuous for ll [, ]. Let F () e the funtion tht is the mesure of the re under the urve etween nd. F ( + h) is the re under the urve etween nd + h nd F ( + h) F () is the re of the strip indited on the grph. The re of the strip is etween the res of the left nd right retngles; tht is, f ()h < F ( + h) F () < F ( + h)h F ( +h) F( ) or f( ) < <f f ( + h) ), h (dividing h). h As h, f ( + h) f () = f() + h F() F( + h) F() F( or lim ( + h) ) F ( ) = f( ) h h tht is, F () = f () (differentition from first priniples). Therefore, F ( ) = f( ) d tht is, F () is n ntiderivtive of f () or f( ) d = F ( ) + ut when =, f( ) d = F ( ) + or = F( ). Therefore, f( ) d = F ( ) F () nd when =, f( ) d = F ( ) F (. ) = (s the re defined is zero t = ) Tht is, the re under the grph of f () etween = nd = is F () F (). f( ) d is the indefinite integrl, whih represents the generl ntiderivtive of the funtion eing integrted. This is the fundmentl theorem of integrl lulus nd it enles res under grphs to e lulted etl. It pplies onl to funtions tht re smooth nd ontinuous over the intervl [, ]. It n e stted s re = f( ) d = [ F( ()] [do not dd s F () is n ntiderivtive of f ()] = F () F () nd re lled the terminls of this definite integrl nd indite the domin over whih the integrl is tken. Chpter 9 Integrtion
f( ) d is lled the defi nite integrl euse it n e epressed in terms of its terminls nd, whih re usull rel numers. In this se the vlue of the definite integrl evlutes sis rel numer nd not funtion. The funtion eing integrted, f (), is lled the integrnd. Properties of definite integrls Definite integrls hve the following five properties.. f( ) d =. f( ) d = f( ) d + f( ) d, < <. kf( ) d= k f( ) d. [ f( ) + g ( )] d = f( ) d + g ( ) d. f( ) d = f( ) d ebookplus Digitl do Investigtion Definite integrls Worked emple Evlute the following definite integrls. ( + ) d d ( + ) ebookplus Tutoril int-66 Worked emple ThInk WrITe Method : Tehnolog-free Antidifferentite eh term of the integrnd nd write in the form [ F ( )]. Sustitute vlues of nd into F () F (). ( + ) d = [ + ] Evlute the integrl. = = Epress the integrnd with negtive power. = [ + ( ) ] [ + ( ) ] d ( + ) = ( + ) d Antidifferentite rule. ( + ) = = ( + ) Epress the integrl with positive power. = ( + ) mths Quest mthemtil methods CS for the Csio ClssPd
Sustitute the vlues of nd = into F () F () where = nd ( + ) =. = Evlute the definite integrl. = + 9 = 6 Method : Tehnolog-enled On the Min pge, using the soft keord, tp: ) - P Complete the entr lines s: ( + ) d d ( + ) Press E fter eh entr. Write the nswers. ( + ) d = d ( + ) = 6 Worked Emple Find the et vlue of eh of the following definite integrls. π sin d π 6 ( ) e e d Think Antidifferentite the integrnd, writing it in the form [ F ( )]. Sustitute vlues of nd into F () F (). Write π sin d 6 os 6 = π 6 = = π π π 6 os 6 π 6 os π 6 os 6 π 6 os 6 Chpter 9 Integrtion
Evlute the integrl. = Antidifferentite the integrnd, using [ F ( )]. Sustitute vlues of nd into F () F (). ( e e ) d = = e + Evlute. = e + e 6 6 = [ ] [ ] = + e e + e e e + e = ( e + ) Worked Emple k If 8d = 6, find k. Think Method : Tehnolog-free Antidifferentite the integrnd, using [ F ( )]. Write k 8d= [ ] k So [ ] k = 6 Sustitute the vlues of nd into [k ] [() ] = 6 F () F (). Simplif the integrl. k = 6 k = 6 k = 9 k = ± 9 Solve the eqution. k = or Method : Tehnolog-enled On the Min pge, omplete the entr line s: k 8d = 6 Highlight the integrl nd tp: Intertive Eqution/Inequlit solve Chnge the vrile to k. OK Write the solutions. Solving ( 8d ) = 6 for k implies k = or k =. k Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Sigm nottion An lterntive nottion for the definite integrl of f () for [, ] is the sigm nottion (mening the sum of ). ( i, f( i)) = f() Divide the intervl [, ] into n equl suintervls with the ith suintervl of width δ i nd height f ( i ), i [, n]. The re of the retngle formed the ith suintervl is δa i = f ( i ) δ i. n As n, δ i nd δ A i f( ) d. So n i f( ) d = lim f ( i ) δ i δ = Worked Emple The intervl [, ] is divided into n equl suintervls the points,,... n, n, where = < <... < n < n =. Let δ = i i for i =,,... n. Rewrite s definite integrl nd thus evlute lim ( 6 δ ). n δ i = i Think Reognise the lternte nottion for the definite integrl. Write n i δ i = lim ( 6 δ ) = 6 d Evlute. 6 d = = 7 = = REMEMBER. The fundmentl theorem of lulus is f( ) d = [ ( )] where F () is n ntiderivtive of f (). F = F () F ().. The epression f( ) d is lled the definite integrl where nd re the terminls nd represent the upper nd lower vlues of.. Properties of definite integrls () f( ) d = () f( ) d = f( ) d + f( ) d, < < Chpter 9 Integrtion
() kf( d ) = k f( ) d (d) [ f( ) + g ( )] d = f( ) d + gd ( ) (e) f( ) d = f( ) d. f( ) d = lim f( ) δ n δ i i = i i eercise 9e The fundmentl theorem of integrl lulus We Evlute the following definite integrls. ebookplus Digitl do SkillSHEET 9. Sutrting funtion vlues d d d ( ) d 6 d e ( + ) d f ( + ) d g ( 6 + ) d h ( + ) d i 9 d j ( + 6) d k ( ) + d l ( ) d m p s ( ) d n + d q 6 8 d d o ( 7) d d r 7 d We Find the et vlue of eh of the following definite integrls. e d e d e d ( e 6 + ) d e π + e d f ( e e ) d g sin( ) d h π sin( ) d i sin d j π sin d π ( ) π m os d π 6 p π π π π ( ) k os( d ) l 8os( ) d π n 7 π os d π π sin d q [ sin ( )] d + 6 r + ( ) o d π π π os( ) d + os ( d ) 6 mths Quest mthemtil methods CS for the Csio ClssPd
If f( ) d = 6, find the vlue of f( ) d. MC Given tht f( ) d = 6, [ f( ) + ] d is equl to: A 6 B C D 9 E f ( ) d is equl to: A 6 B C D 6 E Evlute the following. ( t t) dt os( t ) dt π d sin d π g π π e e d os( ) dt h [ sin ( ) ] t t e d k 6 WE If ( + ) d =, find k. k 7 If d = 8, find k. k 8 If 9 d = log e(), find k. 9 If e d =, find the vlue of. π If os( ) d =, find the vlue of k given tht k 6 Given tht f( ) d = 8, evlute the following. < k < π. f 7 ( p ) dp 8 dm m 6 f( ) d f( ) d [ f( ) ] d d [ f( ) + ] d 6 WE MC The intervl [, ] is divided into n equl suintervls the points,,... n, n, where = < <... < n < n =. Let δ = i i for i =,,... n. Then lim ( i δ ) is equl to: δ = i A d n d 6 D E 6 MC The intervl [, ] is divided into n equl suintervls the points,,... n, n, where = < <... < n < n =. Let δ = i i for i =,,... n. Then lim ( i δ ) is equl to: δ = A 9 i n d D d E 7 6 Chpter 9 Integrtion 7
9f Signed res When lulting res etween the grph of funtion f () nd the -is using the definite integrl f( ) d, the re is signed; tht is, it is positive or negtive. If f () >, the region is ove the -is; if f () < it is elow the is. We shll now emine these two situtions nd look t how we lulte the re of regions tht inlude oth. Region ove is If f () >, tht is, the region is ove the -is, then f( ) d >, so the vlue of the definite integrl is positive. For emple, if f () >, then the re = f( ) d. Region elow is If f () <, tht is, the region is elow the -is, then f( ) d <, so the vlue of the definite integrl is negtive. For emple, if f () <, then the re = f ( ) d or re = f( ) d, s the region is elow the -is or f( ) d (reversing the terminls hnges the sign). Therefore, for res elow the -is, ensure tht the re hs positive vlue. (Ares nnot e negtive.) = f() = f() Comining regions For regions tht re omintions of res ove nd elow the -is, eh re hs to e lulted seprte integrls one for eh re ove nd one for eh re elow the -is. For emple, from the digrm, Are = A + A However, f( ) d = A A, euse the res re signed. A = f() A To overome this diffiult we find the orret re s: Are = f( ) d f( ) d ( = A A = A + A ) or = f( ) d + f( ) d or = f( ) d + f( ) d Note: When lulting the re etween urve nd the -is it is essentil tht the -interepts re determined nd grph of the urve is skethed over the intervl required. The term mens tht we should mke the vlue of positive even if it is negtive. 8 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Worked Emple Epress the shded re s definite integrl. Evlute the definite integrl to find the shded re, giving our nswer s n et vlue. = Think Epress the re in definite integrl nottion where = nd =. Write Are = d Antidifferentite the integrnd. Are = log e Evlute. = [log e ()] [log e ()] = log e () = log e () Stte the solution s n et nswer. The re is log e () squre units. Worked Emple 6 Clulte the shded re. = Think Epress the re in definite integrl nottion, with negtive sign in front of the integrl s the region is elow the -is. Write Are = ( ) d Antidifferentite the integrnd. = [ ] Evlute. = [( () ()) ( () ())] = [( 98) ( )] [ 9 ( )] = = [ 9+ ] = ( 7 ) = 7 Stte the solution. The re is 7 squre units. Chpter 9 Integrtion 9
Worked emple 7 Epress the shded re s definite integrl. Clulte the re. Clulte the re using CAS lultor. = ThInk WrITe Epress the re ove the -is s n integrl nd the re elow the -is s n integrl. For the re elow the -is, tke the negtive of the integrl from to. Antidifferentite the integrnds. = Are = d d ( ) ( ) ( ) ( ) ( ) Evlute. = ( ) ( ) ( ) ( ) ( ( ) ( ) ) ( ) ( ) Simplif. = [ ( 8)] [( 8) ] = ( ) = 8 Stte the solution. The re is 8 squre units. On the Min pge, using the soft keord, tp: ) - P Complete the entr line s: ( ) d ( ) d Press E. Alterntivel, tke the solute vlue for the re elow the -is s shown. Write the re with the orret units. The re is 8 squre units. Worked emple 8 Sketh the grph of = e showing ll interepts nd using et vlues for ll ke fetures. Find the re etween the urve nd the -is from = to =. ebookplus Tutoril int-67 Worked emple 8 ThInk Find the -interept letting = nd solving for. WrITe When =, e = e = mths Quest mthemtil methods CS for the Csio ClssPd
Tke log e of oth sides. log e (e ) = log e () = log e () (or pproimtel.69) so the -interept is log e (). Find the -interept letting =. Note the vertil trnsltion nd hene sketh the grph showing the pproprite horizontl smptote nd interept. Shde the region required. Epress the re ove the -is s n integrl nd the re elow the -is s n integrl. Sutrt the re elow the -is from the re ove the -is. When =, = e = = so the -interept is. = e log e = loge ( ) Are = ( e ) d - ( e ) d log ( ) Antidifferentite the integrnds. = [ e ] [ e ] Evlute. (Rememer: e log e () = ) e loge ( ) loge ( ) = [e ()] [e log e () log e ()] [[e log e () (log e ()] [e ()]] = [e ] [ log e ()] [[ log e ()] [ ]] Simplif. = e + log e () + log e () + = e 7 + log e () Stte the solution. The re is [e 7 + log e ()] or pproimtel.6 squre units. REMEMBER. If f () >, re = f( ) d.. If f () <, re = f( ) d, s the region is elow the -is or = f( ) d, or = f( ) d, (reversing the terminls hnges the sign).. Chek if the required re lies ove nd elow the -is. = f(). Are = f( ) d f( ) d (= A A = A + A ) or = f( ) d + f( ) d A A. e log e ( ) = Chpter 9 Integrtion
Eerise 9f Signed res Find the re of the tringle t right. geometrill using integrtion = Find the re of the tringle t right. geometrill using integrtion = WE Epress the following shded res s definite integrls. = = = d = e = 9 + f = g = e h i = sin () j = os ( ) = e π π WE Evlute eh of the definite integrls in question to find the shded re. Give our nswer s n et vlue. For eh of the following, sketh grph to illustrte the region for whih the definite integrl gives the re. d ( 6 ) d d d ( ) d e d f e d g log e d π ( ) h sin( ) d Mths Quest Mthemtil Methods CAS for the Csio ClssPd
6 WE6 Clulte eh of the shded res elow. = = = d e = f = = + g h i = sin () = e = e π π π j π π = os ( ) 7 MC The re etween the grph, the -is nd the lines = nd = is equl to: A f ( ) d B f( ) d C E f( ) d f( ) d D f( ) d The re etween the grph, the -is nd the lines = nd = is equl to: A f( ) d + f( ) d B f ( ) d = f() C f( ) d + f( ) d E f( ) d f( ) d D f( ) d 8 WE7 Epress the following shded res s definite integrls whih give the orret re. g() f() h() Chpter 9 Integrtion
d e g() f() 9 mc Emine the grph. = + 6 The re etween the urve nd the -is from = nd = is equl to: A 7 sq. units B sq. units C 7 sq. units D sq. units E sq. units The re etween the urve nd the -is from = nd = is equl to: A 6 sq. units B sq. units C sq. units D sq. units E 6 sq. units The re etween the urve nd the -is from = nd = is equl to: A B sq. units C sq. units D sq. units E sq. units We8 Sketh the grph of the urve =, showing ll interepts nd using et vlues for ll ke fetures. Find the re etween the urve nd the -is: from = to = from = to = from = to =. Sketh the grph of the urve = +, showing ll interepts. Find the re etween the urve nd the -is etween the lines: = nd = = nd = = nd =. Sketh the grph of the urve = + os () over [, π]. Find the et re etween the urve nd the -is from: = to = π = π to = π. ebookplus Digitl do WorkSHEET 9. Sketh the grph of f () = nd find the re etween the urve nd the -is nd the lines = nd =. Give oth n et nswer nd n pproimtion to deiml ples. Find the et re etween the urve =, the -is nd the lines = nd =. Find the et re ounded the urve g() = e +, the -is nd the lines = nd =. mths Quest mthemtil methods CS for the Csio ClssPd
9g Further res Ares ound urve nd the -is For grphs with two or more -interepts, there is n enlosed region (or regions) etween the grph nd the -is. The re ound the grph of f () nd the -is is: f ( ) d (negtive euse the re is elow the -is) or f( ) d. The re ound the grph of g() nd the -is is: g ( ) d g ( ) d or g ( ) d + g ( ) d Tht is, if the grph hs two -interepts then one integrnd is = g() required. If the grph hs three -interepts then two integrnds re required, nd so on. Note: Wherever possile it is good prtie to use sketh grphs to ssist in n prolems involving the lultion of res under urves. = f() Worked Emple 9 Sketh the grph of the funtion g() = ( )( + ). Find the re ound the -is nd the grph of the funtion. Think Determine the tpe of grph looking t the numer of rkets nd the sign of the -terms. Solve g() = to find the -interepts. Sketh the grph. Shde the region ound g() nd the -is. Write g() = ( )( + ) is n inverted prol. For -interepts, g() = ( )( + ) = = nd =. The -interepts re nd. = g() Epress the re s n integrl. Are = ( 6 + ) d Evlute. = [ 6 + ] = [6() + () () ] [6( ) + ( ) ( ) ] Chpter 9 Integrtion
= (8 + 9 9) ( + + 8 ) = ( 7 ) = + 7 = 6 Stte the solution. The re ound g() nd the -is is 6 squre units. Finding res without sketh grphs When finding res under urves tht involve funtions whose grphs re not esil skethed, the re n e lulted providing the -interepts n e determined. Note: The smol f () is known s the solute vlue of f (), whih mens tht we must mke the f () funtion or vlue positive whenever it is negtive. For emple, = = = As res nnot e negtive, tking the solute vlue of the integrnds involved in prolem will ensure tht ll res re mde positive. For emple, The shded re t right = f( ) d + f( ) d or = f( ) d + f( ) d = f() Worked Emple Find the -interepts of = sin () over the domin [, π]. Clulte the re etween the urve, the -is nd = nd = π. Think Write To find the -interepts, let =. For -interepts, = Solve for over the given domin. sin () = =, π, π, π, π, et. =, π,, π π, π Pik the -interepts tht re etween the given end points of the re. Stte the regions for whih it is neessr to lulte the re. = π is the onl -interept etween nd π. π Are = sin(d ) + sin () d π π 6 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Evlute the solute vlue of the integrl for eh region. π = [ os ( )] + [ os ()] = [ os ( π) ( os ())] π π + [ os ( π) ( os ( π))] = [ ( )] + [ ( )] = + Add the result to give the totl re. = + = Stte the solution. The re is squre units. Worked Emple Differentite log e ( ). Hene, find n ntiderivtive of. Find the re etween the grph of, the -is, = nd =, giving our nswer orret to deiml ples. Think Write Let equl the epression to e differentited. Epress u s funtion of in order to ppl the hin rule for differentition. (Let u equl the funtion inside the rkets.) Let = log e ( ) Let u = Find du d. du d = Write in terms of u. = log e (u) Find d du. d du = u 6 Find d d using the hin rule. So d d = u d Sine d d = +, epress the reltionship in integrl nottion. Remove ftor from d so tht it d resemles the integrl required. = d = log e + d = log e + Chpter 9 Integrtion 7
Divide oth sides the ftor in order to otin the required integrl. Find the -interepts. (For =, the numertor =.) If the -interepts re not etween the terminls of the re, find the re evluting the integrnd. d = loge + An ntiderivtive of is log. For -interepts, = = Are = = d log e = = = = log ( ) ( ) e log e log () 8 log () e 8 log e ( ) 8 log e ( ) ( ) Stte the solution. The re is log e squre units. 8 e or pproimtel.9 REMEMBER. For grphs with two or more interepts, there is n enlosed region (or regions) etween the grph nd the -is.. The numer of regions is one less thn the numer of interepts.. Where possile, sketh grphs to mke it esier to lulte the res under urves.. As res nnot e negtive, tke the solute vlues of the integrls.. When grphs re not esil drwn, res n e lulted finding the -interepts nd determining whether the re within the ounds of the required re. Eerise 9G Further res In the following eerise give ll nswers orret to deiml ples where pproprite, unless otherwise stted. WE9 i Sketh the grph of eh of the following funtions. ii Find the re ound the -is nd the grph of eh funtion. f () = g() = ( )( + ) Find the re ound the -is nd the grph of eh of the following funtions. h() = ( + )( ) h() = + 6 g() = 8 d g() = e f () = ( )( ) f f () = + 6 g g() = + h h() = ( )( + )( + ) MC The re ound the urve with eqution = 6 + 8 nd the -is is equl to: A squre units B 6 squre units C squre units D squre units E squre units 8 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
MC The re etween the urve t right, the -is nd the lines = nd = is equl to: A C f( ) d B f( ) d + f( ) d f ( ) d D f( ) d f( ) d E f( ) d f( ) d MC The re etween the urve = 6, the -is nd the lines = nd = is equl to: A squre units B squre units C squre units 6 = f() D squre units E squre units 6 For eh of the following: i sketh the grph of the urve over n pproprite domin, lerl lelling n -interepts in the intervl required. ii find the re etween the urve, the -is nd the lines indited elow. =, = nd = =, = nd = =, = nd = d =, = nd = e = e, = nd = f = e, = nd = g = sin (), = π 6 nd = π i = sin (), = π nd = π 6 h = os, = π nd = π j =,, = nd = 7 WE For eh of the following funtions: i find the -interepts over the given domin ii lulte the re etween the urve, the -is nd the given lines. Use sketh grphs to ssist our workings. =,, = nd = = sin () os (), [, π], = nd = π = e e, = nd = d =,, = nd = e = e, = nd = f =, = nd = g = ( ), = nd = 8 Find the et re of the region enlosed the -is, = e nd the lines = nd =. 9 Find the et re of the region enlosed the -is, = os () nd the lines = π nd = π. 6 Find the re ound = ( ), the -is nd the -is. Sketh the grph of = showing ll smptotes nd interepts. ( ) Find the re under the urve etween = nd =. Chpter 9 Integrtion 9
Give the eqution of the smptotes for the funtion f () = ( + ). Find the re etween the urve, the -is, = nd =. Find the re ound the urve = e, the -is, = nd =. (Find the -interepts first.) Find the re ound the urve = sin (), the -is, = π nd = π. (Chek the -interepts first.) WE Differentite log e (). ( > ) Hene, find n ntiderivtive of log e (). Find the re ound the grph of log e (), the -is, = nd = giving et nswers. 6 Differentite log e ( + ). Hene, find n ntiderivtive of Find the re etween +, the -is, = nd =. 7 Find the re etween the grph of =, the -is, = nd =. Use this result to lulte the re etween the grph, the -is nd the line =. +. = e 8 Find the et re of the shded region on the grph = e elow. = (, ) 9 Find the shded re elow. (Hint: It is esier if ou use smmetr.) π π = sin () The re of the region ounded the -is, the -is, the urve = e - nd the line = k, where k is positive rel onstnt, is squre units. Find k. The re of the region ounded the -is, the -is, the urve = sin () nd the line = k, where k is positive rel onstnt, is squre unit. Find k. 9H Ares etween two urves We shll now onsider the re etween two funtions, f () nd g(), over n intervl [, ]. Our pproh depends on whether the urves interset or do not interset over this intervl. If the two urves f () nd g() do not interset over the intervl [, ] Here, we m look t three irumstnes: when the region is ove the -is, when it is elow the -is, nd when it rosses the -is. 6 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Region ove -is The rown shded re = f( ) d g ( ) d = [ f( ) g ( )] d. Note: The lower funtion is sutrted from the higher funtion to ensure positive nswer. Region elow -is Agin, the lower funtion is sutrted from the higher funtion to ensure positive nswer. Brown shded re = [ f( ) g ( )] d, s f () is ove g() over the intervl [, ]. Region rosses -is Shded re = [ f( ) g ( )] d f() g() f() g() f() g() Worked Emple Stte the definite integrl tht desries the shded re on the grph t right. Find the re. = + = Think Write Stte the two funtions f () nd g(). f () = + nd g() = Sutrt the eqution of the lower funtion from the eqution of the upper funtion nd simplif. Write s definite integrl etween the given vlues of. f () g() = + = + Are = [ f( ) g ( )] = ( + ) d Antidifferentite. Are = [ + ] Evlute the integrl. = [ ( ) + ] [ ( ) + ] = ( + ) () = Stte the re. The re is squre units. Worked Emple Find the vlues of where the funtions = nd = interset. Sketh the grphs on the sme es. Hene, find the re ound the urves. Think Write/drw Stte the two funtions. = nd = Chpter 9 Integrtion 6
Find where the urves interset. Solve for. For points of intersetion: = = ( )( + ) = = or = Find the ke points of eh For =, funtion nd sketh. when =, = when =, = = when =, = (, ) Line psses through (, ), (, ) = nd (, ) For =, when =, = Hene the -interept is. (, ) Prol lso psses through (, ) nd (, ). Define f () nd g (). Let f () = nd g() = Write the re s definite integrl etween the vlues of t the points of intersetion. Are = [ f( ) g ( )] d = [ ( )] d = ( + ) d Antidifferentite. = [ + ] Evlute the integrl. = [ ( ) ( ) + ( )] [ ( ) ( ) + ( ) = ( 8 + ) ( + ) = ( ) (- 6 ) ] = + 6 = Stte the re. The re is squre units. If the two urves interset over the intervl [, ] Where nd re the vlues of where f () nd g() interset over the intervl [, ], the re is found onsidering the intervls [, ], [, ] nd [, ] seprtel. For eh intervl re must e tken to mke sure the integrnd is the higher funtion. Sutrt the lower funtion. So the shded re equls: [ g ( ) f( )] d + [ f( ) g ( )] d + [ g ( ) f( )] d Therefore, when finding res etween two urves over n intervl, it must e determined whether the urves interset within tht intervl. If the do, the re is roken into su-intervls s shown ove. As with res under urves, sketh grphs should e used to ssist in finding res etween urves. g() f() 6 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
If sketh grphs re not used, the solute vlue of eh integrl, for eh su-intervl, should e tken to ensure the orret vlue is otined. Worked emple ebookplus Find the vlues of where the grph of the funtions f () = nd g() = interset. Sketh the grphs on the sme es. Shde the region etween the two urves nd = nd =. Find the et re etween f () nd g() from = to = using CAS lultor. Tutoril int-8 Worked emple ThInk WrITe/drW Stte the two funtions. f () =, g () = Let f () = g() to find the vlues of where the grphs interset. For points of intersetion, = Solve for. = = ( )( + ) = = nd = Sketh f () nd g() on the sme es nd shde the region etween the two urves from = to =. f() = g() = Stte the re s the sum of two integrls for the two su-intervls. Are = ( ) + ( ) d d On the Min pge, using the soft keord, tp: ) - P Complete the entr line s: ( d ) ( ) + d Press E. Write the re with the orret units. The re is log e + squre units. Chpter 9 Integrtion 6
remember. If two urves f () nd g() do not interset over the intervl [, ] nd f () > g(), then the re enlosed the two urves nd the lines = nd = is found using the formul: f( ) d gd ( ) = [f( f ) g g( ( )] d. If two urves f () nd g() interset over the intervl [, ], it is neessr to find the points of intersetion nd hene find the re of eh setion, euse sometimes f () > g() nd sometimes g() > f ().. If sketh grphs re not used to determine whih is the upper urve, then it is neessr to tke the solute vlue or positive vlue of eh integrl. eercise 9h res etween two urves We Stte the definite integrl tht will find the shded res on eh grph elow. = = = = = + = d = e = f = e = g = 8 = h = = 9 = = e We Find eh of the res in question. mc Whih one of the following does not equl the shded re? A g ( ) d f( ) d B g ( ) d + f( ) d C f( ) d g ( ) d D [ g ( ) f( )] d E [ f( ) g ( )] d mc The re ound the urves f (), g() nd the lines = nd = t right is equl to: A [ f( ) g ( )] d B [ f( ) + g ( )] d f() g() g() f() 6 mths Quest mthemtil methods CS for the Csio ClssPd
C [ g ( ) f( )] d E [ f( ) + g ( )] d MC The shded re t right is equl to: A [ f( ) g ( )] d B [ g ( ) f( )] d + [ f( ) g ( )] d C [ g ( ) f( )] d D [ f( ) g ( )] d E [ f( ) g ( )] d + [ g ( ) f( )] D [ f( ) g ( )] d 6 WE In eh of the following: i find the vlues of where the funtions interset ii sketh the grphs on the sme es iii hene, find the re ound the urves. = nd = = nd = = nd = d = nd = e = ( + ) nd = f = nd = 7 WE i Find the vlues of where the funtions interset. ii Sketh the grphs on the sme es. iii Find the re etween f () nd g() giving n et nswer. = nd = = nd = + 8 Find the re etween the pirs of urves elow, over the given intervl. =, =, [, ] = sin (), = os (), [, π] = ( ), = ( + ), [, ] d =, = 6, [, ] e =, =, [, ] f = e, = e, [, ] g() f() g = os (), = π, [, π ] h = e, = e, [, ] 9 Find the re etween the urve = e nd the lines =, = nd =. Find the re etween the urve = nd the lines = +, = nd =. Clulte the re etween the urves = sin () nd = os () from = to = π. Clulte the re etween the urves = sin () nd = sin () from = to = π. Find the et re ound the urves = e nd = e. The grph t right shows the ross-setion of riked rhw. (All mesurements re in metres.) Find the -interepts of f (). Find the -interepts of g(). Find the ross-setionl re of the rikwork. f() = g() = Chpter 9 Integrtion 6
The digrm elow shows the outline of window frme. If ll mesurements re in metres, wht is the re of glss whih fits into the frme? = = 6 The digrm t right shows the side view of onrete ridge. (All mesurements re in metres.) Find: the -interepts of the urve the length of the ridge the re of the side of the ridge d the volume of onrete used to uild the ridge if the ridge is 9 metres wide. 7 The ross-setion of rod tunnel entrne is shown t right. (All mesurements re in metres.) The shded re is to e onreted. Find: the et re, ove the entrne, whih is to e onreted the et volume of onrete required to uild this tunnel if it is metres long. f() = sin ( π ) = 9I Averge vlue of funtion Consider the funtion = f (). The verge vlue, v, for the funtion = f () over the intervl [, ] is given : v = f d ( ). This n e rerrnged to give v ( ) = f( ) d. = f() Geometrill, the verge vlue of the funtion is the height, v, of the retngle of width ( ) tht hs the sme re s the re under the grph of = f () for the intervl [, ]. v Worked Emple Find the verge vlue of f () = for the intervl [, ]. Find the vlue of tht orresponds to the verge vlue. Think Write Write the reltionship for the verge vlue. v = f( ) d 66 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Identif f () = nd sustitute in the vlues for nd. v = d Antidifferentite. v = Evlute the definite integrl nd multipl v =. Solve f () = to find. = = 8 = 6 = The verge vlue is. = 7 =± 7 6 Choose the vlue of tht stisfies the given intervl. = 7 s [, ]. Worked Emple 6 Find the verge vlue of f () = log e () for the intervl [, ]. Give our nswer in et form. Think Write the reltionship for the verge vlue nd sustitute the vlues for nd. The integrl of log e () is not overed in this ourse. The verge vlue n onl e evluted using CAS lultor. On the Min pge, using the soft keord, tp: ) - P Complete the entr line s: (ln( )) d Press E. Write v = e d log ( ) Write the nswer. The verge vlue f () = log e () for the intervl [, ] is log e (). Chpter 9 Integrtion 67
remember The verge vlue, v, for the funtion = f () over the intervl [, ] is given v = f( ) d. eercise 9I verge vlue of funtion We Find the verge vlue, in et form, of the funtion for the given intervl. =, [, ] = sin (),, π 6 =, [, ] d = e, [, ] We6 Find the verge vlue of the funtion, in et form, for the given intervl. = log e (), [, ] = tn (),, π 6 = e, [, ] d = +, [, 7] We For the funtion = e for [, ], find: the verge vlue of the funtion the orresponding -vlue in et form. mc The verge vlue of the funtion f () = log e ( + ) over the intervl [, ] is: A log e ( 9 ) B log e (9) C [ 9 log e ( ) ] D 9 log e () E 9 log e() + mc The verge vlue of the funtion = sin () over the intervl, π is: A π B π C π 8 D E π 9J further pplitions of integrtion Differentition n e pplied to rtes of hnge nd relted rtes. If the rte of hnge of funtion is known, ntidifferentition llows the originl funtion to e found. So integrtion hs mn prtil pplitions. Worked emple 7 The rte of hnge of position, veloit, of prtile trvelling in stright line is given d. t = e m/s, t, where is mesured in metres nd t in dt seonds. Find the veloit: i initill ii fter seonds, orret to deiml ples. Find the time tken to reh veloit of m/s. Sketh the grph of d ginst t. dt d Find the totl distne trvelled the prtile in the first seonds. ebookplus Tutoril int-69 Worked emple 7 68 mths Quest mthemtil methods CS for the Csio ClssPd
Think i The initil veloit ours when t =. Sustitute t = into d dt. write d dt =. e = e. = e = = t Answer with the orret units. Veloit is initill m/s. ii Sustitute t = into d dt. d =. e dt = e = 9. 8 Answer with the orret units. After seonds, the veloit is 9.8 m/s. Sustitute d dt =. d =. t e dt = e. Solve for t. = e. Answer the question orret to deiml ples. To grph d = e, dt open the Grph & T pge nd omplete the funtion entr line s: = - e -. Tik the o nd tp!. Note the viewing window settings: XMin:, XM:, YMin: nd YM: e. t =.. t = ln(.) ln (. ) t =. =.7 s Time tken is.7 seonds. t t When drwing our grph, lel the es with the given vriles. Note the horizontl smptote d = is not displed. The dt smptoti ehviour of the funtion, however, is lerl visile. d Distne trvelled = re under the grph Stte the distne s definite integrl. d d dt = e.. = ( t e ) dt t Chpter 9 Integrtion 69
Antidifferntite. = t+ e. t Evlute the integrl. = ( + e. ) ( + e. ) Stte the distne trvelled with orret units. = ( + e ) ( e ) = + e = 7 + e = 7.6 The distne trvelled in the first seonds is 7.6 metres. Worked Emple 8 The rte of hnge of pressure, P tmospheres, of given mss of gs with respet to its volume, V m, is given dp k =, k >. dv V If d P = when V =, find k. dv Find the pressure, P, s funtion of V given tht when P = tmospheres, V = m. Find the volume when the pressure is tmospheres. Think Write Sustitute the onditions dp dv = nd V = into the given rte. dp dv = k = V k ( ) Simplif. = Solve for k. k = Write the rte with the vlue of k found previousl. Antidifferentite. P Sustitute the given onditions P = nd V = to find. dp dv Write the reltionship for P. P k = V = V dv = V = + V P = + V = + = = V dv 7 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Sustitute P =. P = V = V Solve for V. V = Stte the nswer with orret units. The volume is m. Eerise 9J Further pplitions of integrtion If f () = ( ) nd the -interept of f () is, find the rule for f (). If d d = os( ) nd the -interept is, find the et vlue of when = π. The rte of defletion from horizontl position of -metre diving ord when n 8-kg person is metres from its fied end is given d = + + d. ( )., where is the defletion in metres. (Metres) (Metres) Bord Defletion Wht is the defletion when =? Determine the eqution tht mesures the defletion. Hene, find the mimum defletion. On n d the ost per item for mhine produing n items is given dc n = e., where n [, ] nd C is the ost in dollrs. dn Use the rte to find the ost of produing the th item. Epress C s funtion of n. Wht is the totl ost of produing the first items? d Find the verge ost of prodution for: i the first items ii the seond items. WE7 The rte of hnge of position (veloit) of ring r trvelling down stright streth of rod is given d = t( 6 t), where is mesured in metres nd t in seonds. dt Find the veloit when: i t = ii t =. Determine: i when the mimum veloit ours ii the mimum veloit. Sketh the grph of d ginst t for t 6. dt Chpter 9 Integrtion 7
d Find the re under the grph etween t = nd t =. e Wht does this re represent? 6 WE8 The rte t whih wter is pumped out of dm, in L/min, t minutes fter the pump is strted strted is dv t = + dt os π. How muh wter is pumped out in the th minute? Find the volume of wter pumped out t n time, t, fter the pump is strted. How muh wter is pumped out fter minutes? d Find the verge rte t whih wter is pumped in the first hour. e How long would it tke to fill tnk holding 6 litres? 7 The rte of flow of wter into hot wter sstem during -hour period is thought to e dv t = + dt os π, where V is in litres nd t is the numer of hours fter 8 m. Sketh the grph of dv ginst t. dt Find the length of time for whih the rte is ove. L/h. Find the volume of wter tht hs flowed into the sstem etween: i 8 m nd pm ii pm nd 8 pm. 8 The roof of stdium hs the shpe given the funtion f : [, ] R, f () =.. The stdium is 7 metres long nd its ross-setion is shown t right. Find the volume of the stdium. The stdium is to hve severl ironditioners strtegill pled round it. Eh n servie volume of m. How mn ironditioners re required? 9 The ross-setion of hnnel is proli. It is metres wide t the top nd metres deep. Find the depth of wter, to the nerest m, when the hnnel is hlf full. For n point P on the urve =, prove tht the re under the urve is one qurter of the re of the retngle. (metres) (metres) P = 7 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
The rh of onrete ridge hs the shpe of prol. It is 6 metres high nd 8 metres long. Find the rule for the funtion orresponding to the rh of the ridge. Find the re of the shded region. If the ridge is metres wide, find the volume of onrete in the ridge. In the figure t right f () nd g() interset t O nd B. Show tht the oordinte of B is (log e (), ). Find the et re of the region ound f () nd g(). Show tht the sum of the res under f () nd g(), from = to = log e (), is equl to the re of the retngle OABC. (metres) 7 6 (metres) f() = e C B g() = e + O A Chpter 9 Integrtion 7
Summr Antidifferentition rules The reltionships etween f () nd f( ) d re: f () n f( ) d + n + n + ( + ) n ( + ) n n ( + ) + e e k k e sin () os () + + log e + + log + + e + k + os () + sin () + [ f( ) ± g ( )] d = f( ) d ± g ( ) d kf ( ) d= k f( ) d gd ( ) = f( ) +, where g() = f () d f( ) d is the indefinite integrl Definite integrls The fundmentl theorem of integrl lulus: f( ) d = [ F ( )] = F ( ) F ( ) where F () is n ntiderivtive of f (). f( ) d is the definite integrl kf ( ) d= k f( ) d f( ) d = f( ) d + f( ) d, < < [ f( ) ± g ( )] d = f( ) d ± g ( ) d f( ) d = f( ) d Grphs of the ntiderivtive funtion For polnomil funtion, the grph of f () is one degree higher thn the grph of f (). The -interepts of f () re the -oordintes of the turning points of f (). When f () is ove the -is, the grdient of f () is positive. When f () is elow the -is, the grdient of f () is negtive. 7 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Approimting res under urves An pproimtion to the re etween urve nd the -is n e found dividing the re into series of retngles tht re ll the sme width. The pproimtion is found finding the sum of ll the res of the retngles. For n inresing funtion, left retngle pproimtion tul re right retngle pproimtion. For deresing funtion, left retngle pproimtion tul re right retngle pproimtion. Are under urves Are = f ( ) d, if f ( ) > for [, ] Are = f( ) d, if f( ) <, or f( ) d, for [, ] = f() = f() Are = f( ) d f( ) d = f() = f( ) d + f( ) d, if f( ) > for [, ] nd f () < for [, ] A A Are etween urves Are = [ f( ) g ( )] d, if f( ) > g ( ) for [, ] f() g() Are = [ g ( ) f( )] d + [ f( ) g ( )] d, if g() > f () for [, ] nd f () > g() for [, ] g() f() Limiting vlue of sum f( ) d = lim f( i ) δ δ i i = Averge vlue of funtion v = f( ) d Chpter 9 Integrtion 7
hpter review Short nswer Find the eqution of the urve f () if it psses through (, ) nd f ( ) =. A prtiulr urve hs d = k d os π +, where k is onstnt, nd it hs sttionr point (, ). Find: the vlue of k the eqution of the urve the vlue of when = 6. If = sin ( + ), find d nd hene d ntidifferentite ( + ) os ( + ). A urve hs grdient funtion f () = e + k. It hs sttionr point t (, ). Find: the vlue of k the eqution of the urve f (). Clulte the et re etween the urve = e nd the lines =, = nd =. (Hint: = e nd = do not interset etween = nd =.) 6 Evlute eh of the following definite integrls. 9 d ( + ) π os( ) π d k 7 Given tht ( ) d =, find two possile vlues for k. 8 Sketh the grph of the funtion f ( ) =. Find the et re etween the grph of f (), the -is nd the lines = nd = 6. 9 Find the re ound the -is nd the urve g() = ( )(6 + ). Clulte the re etween the urve = os () nd the lines =, = nd = π. Use the method of left retngles to pproimte the re under the urve = +, from = to =, using intervl widths of unit. The grph of f: R R, f( )= e + is shown. The norml to the grph of f where it rosses the -is is lso shown. Find the eqution of the norml to the grph of f where it rosses the -is. Find the et re of the shded region. Em tip Students often hd diffiult orretl finding the -interept; (, ) ws ommon nswer, leding to the norml eing = +. Quite few students differentited ut then did not find the vlue of the derivtive t = nd simpl sustituted the epression for the derivtive into the eqution of the norml, whih led to inorret ttempts t lgeri mnipultion nd nonliner normls. Some students found tngent insted. Most students orretl set up n re epression ut some were unle to proeed due to their norml not eing liner epression. Mn hd the orret terminls, lthough insted of ws populr inorret vlue. Arithmeti errors ppered frequentl nd sutrtion mistkes in the integrnd were lso ommon. Some students lso lost from the eqution of the urve. The use of d ws surprisingl good. A few students used the re under the urve minus the tringle with some suess. Multiple hoie The ntiderivtive of A log e ( ) + B + log e ( ) + 6 log e ( ) + D 6 + log e ( ) + E + ( ) is: [Assessment report 7] [ VCAA 7] 76 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
The indefinite integrl ( ) d is equl to: A ( ) + B ( ) + C ( ) + D ( ) + E ( ) + An ntiderivtive of ( + ) is: A ( + ) B ( + ) + C 9 ( + ) D 9 ( + ) E 8( + ) The ntiderivtive of 6e is: A e + B e + C 8e + D e + + E e + The indefinite integrl os sin ( ) d is equl to: A sin ( ) ( ) ( ) ( ) ( ) B sin C sin D sin E sin + os () + + os () + + os () + + os () + os () + 6 An ntiderivtive of + sin () + e is: A [ os () + e ] B + os( ) + e C os () + e D [ os () + e ] E [ os () + e ] 7 If f () hs sttionr point t (, ) nd f () = e + k, where k is onstnt, then f () is: A e + B e + C e + D e + + E e + + 8 If the derivtive of ( ) 8 is 8( )( ) 7 then n ntiderivtive of ( )( ) 7 is: A ( ) 8 B ( ) 8 C ( ) 8 D ( ) 8 E 8( ) 8 9 If the derivtive of e + is ( ) e + +, then the ntiderivtive of ( ) e + + is: A e + + B + ( + ) C e + + D e + + E undefined If the derivtive of log e ( ) is then the ntiderivtive of is: A log e ( ) + B log e ( ) + C log e ( ) + D log e ( ) + E undefined The pproimtion for the re under the grph t (., 9) right from = to =, (, 6) (., ) using the lower (, ) retngles is: A sq. units B sq. units C sq. units D sq. units E sq. units The re under the grph t right from = to = (, 8) n e (, 6) pproimted the re (, ) of the upper retngles (, ) nd is equl to: A sq. units B sq. units C sq. units D sq. units E sq. units The intervl [, ] is divided into n equl suintervls the points,,... n, n, where = < <... < n < n =. Let δ = i i for i =,,... n. n Then lim ( iδ ) is equl to: δ = A i d B d C D E 8 em TIP The definition of the definite integrl s the limiting vlue of sum, nd its orresponding representtion in integrl form, is fundmentl propert of integrl lulus nd must e understood. [Assessment report ] [ VCAA ] Chpter 9 Integrtion 77
The epression ( ) d is equl to: A B 8 C D E 6 The et vlue of the definite integrl ( e e ) d is: A e e B e e C e e D e + e E e e π 6 The et vlue of os ( ) A B C D E 7 The shded re on the grph t right is equl to: A sq. units B 6 sq. units C sq. units D sq. units E 8 sq. units 8 The shded re on the grph t right is: A sq. units B sq. units C 6 sq. units D 6 sq. units E 8 sq. units 9 The re ound the urve on the grph t right nd the -is is equl to: A sq. units B sq. units C sq. units D sq. units E 7 sq. units d is: = ( ) = = ( + )( ) Questions to ppl to the urve with eqution f () = e. The grph of f () is est represented : A f() B f() C E f() D f() f() The re ound the grph of f (), the -is nd the line = is equl to: A e B e C e + D e + E e The re ound the grph of f (), the -is nd the line = e is equl to: A e B e C e + D e + E e Use the grph elow to nswer questions nd. = = + The two grphs interset where is equl to: A nd B nd C nd D nd E nd The re ound the two grphs is equl to: A sq. units B 7 sq. units C 7 sq. units E 6 sq. units D sq. units The verge vlue of the funtion = os () over the intervl, π is: A B π π d E π π 8 78 Mths Quest Mthemtil Methods CAS for the Csio ClssPd
Etended response From pst reords it hs een found tht the ost rte of mintining ertin r is dc = 7t + t+ 8, where C is the umulted ost in dollrs nd t is the time in ers sine the r dt ws first used. Find: the initil mintenne ost C s funtion of t the totl mintenne ost during the first ers of use of the r d the totl mintenne ost from to ers e the mintenne ost for the seond er. Over -hour period on prtiulr utumn d, strting t midnight, the rte of hnge of the temperture for Melourne ws pproimtel dt dt π πt = os, where T is the temperture in C nd t is the numer of hours sine midnight when the temperture ws C. Find: the temperture t n time, t whether the temperture rehes 7 C t n time during the d the mimum temperture nd the time t whih it ours d the minimum temperture nd the time t whih it ours e the temperture t i m i i pm f the time when the temperture first rehes. C. The digrm t right shows prt of the urve with eqution e =. Find: the oordinte of point A the eqution of the norml to the urve t point A the oordinte of point B d the oordinte of point C e the re ound the urve nd the lines AB nd BC. Find the derivtive of log e (). Hene, find n ntiderivtive of log e (). The ross-setion of pltform is shown t right. (All mesurements re in metres.) Find the height of the pltform. d Find the ross-setionl re of the pltform. e Find the volume of onrete required to uild this pltform if it is metres long. Norml C e B A = e f() = log e e Chpter 9 Integrtion 79
A thik metl pipe is filled with oiling wter nd is kept oiling. The temperture, T C, of the metl in the pipe dereses reltive to its distne, m, from the entre of the pipe. It is known tht dt = nd 8. d Find the rte of hnge of the temperture in the metl on the outside of the pipe. Epress T s funtion of. Find the temperture of the metl, orret to deiml ples: i when = 6 m ii on the outside of the pipe. ebookplus Digitl do Test Yourself Chpter 9 8 mths Quest mthemtil methods CS for the Csio ClssPd
ebookplus Ativities Chpter opener Digitl do Quik Questions: Wrm up with ten quik questions on integrtion. (pge 6) 9A Antidifferentition Digitl do SkillSHEET 9.: Prtise sustitution nd evlution. (pge ) 9 Integrtion reognition Tutoril WE int-6: Wth worked emple on performing integrtion reognition. (pge ) Digitl do WorkSHEET 9.: Determine funtions using ntidifferentition. (pge ) 9D Approimting res enlosed funtions Intertivit int- Approimting res enlosed funtions: Consolidte our understnding of vring pproimtions to res enlosed funtions. (pge ) Tutoril WE 8 int-6: Wth worked emple on the pproimtion of the re under urve. (pge ) 9E The fundmentl theorem of integrl lulus Tutoril WE int-66: Wth worked emple on evluting definite integrls. (pge ) Digitl dos Definite integrls: Investigte the properties of definite integrls. (pge ) SkillSHEET 9.: Prtise sutrting funtion vlues. (pge 6) 9F Signed res Tutoril WE 8 int-67: Wth worked emple on finding the re ound n eponentil urve ove nd elow the -is. (pge ) Digitl do WorkSHEET 9.: Approimte res etween urves nd lulte res etween urves using integrls. (pge ) 9H Ares etween two urves Tutoril WE int-8: int-68: Wth worked emple on lulting the re etween two urves using CAS. (pge 6) 9J Further pplitions of integrtion Tutoril WE 7 int-69: Wth worked emple on pplitions of integrtion. (pge 68) Chpter review Digitl do Test Yourself: Tke the end-of-hpter test to test our progress. (pge 8) To ess ebookplus tivities, log on to www.jplus.om.u Chpter 9 Integrtion 8