ONLINE PAGE PROOFS. Anti-differentiation and introduction to integral calculus

Anti-differentition nd introduction to integrl clculus. Kick off with CAS. Anti-derivtives. Anti-derivtive functions nd grphs. Applictions of nti-differentition.5 The definite integrl.6 Review

. Kick off with CAS Integrl clculus Using CAS technolog, find the derivtive of = using d( ) d. Using CAS technolog, find the templte for the integrl nd clculte d. Compring the results in questions nd, wht do ou notice? Using CAS technolog, find the derivtive of = 5 + using d (5 + ). d 5 Using CAS technolog, clculte (5 + ) d. 6 Compring the results in questions nd 5, wht do ou notice? 7 Using CAS technolog, sketch the grph of f () = +. 8 Drw verticl lines from the -is to the grph t = nd t =. Estimte the re enclosed the -is, the verticl lines nd the grph. 9 Using CAS technolog, clculte (, ) f() = + ( + ) d. Wht do ou notice out the nswers to questions 8 nd 9? Plese refer to the Resources t in the Prelims section of our ebookplus for comprehensive step--step guide on how to use our CAS technolog.

. Units & AOS Topic Concept Anti-derivtives Concept summr Prctice questions Anti-derivtives Clculus is mde up of two prts: differentil clculus nd integrl clculus. Differentil clculus rose from the need to mesure n instntneous rte of chnge, or grdient; integrl clculus rose from the need to mesure the re enclosed etween curves. The fundmentl theorem of clculus which estlished the connection etween the two rnches of clculus is one of the most importnt theorems in mthemtics. Before we cn gin some understnding of this theorem, we must first consider the reverse opertion to differentition. Anti-derivtives of polnomil functions Anti-differentition is the reverse process (or undoing process) to differentition. The derivtive of is ; hence, reversing the process, n nti-derivtive of is. The epression n nti-derivtive is of significnce in the ove emple. For n differentite nti-differentite constnt c, d d ( + c) =, so the nti-derivtive of is + c. The constnt c is referred to s n ritrr constnt. If d d =, then = The fmil of curves = + c + c. In this fmil of prols = + c, ech memer hs identicl shpe nd differs onl the mount of verticl trnsltion ech hs undergone. Ech curve hs the sme grdient function of d =. Some memers of the fmil of d nti-derivtives re shown in the digrm. Given d or f (), the process of otining or f() is d clled nti-differentition. The process is lso referred to s finding the primitive function, given the grdient function. The sic rule for nti-differentition of polnomils B studing the ptterns in the following emples, the sic rule of nti-differentition cn e deduced. The vlues in the tle cn e verified differentiting the polnomil. The rule the tle illustrtes is: If d d = n, n N then = n + n+ + c, where c is n ritrr constnt. d d 5 5 + c + c + c + c + c 7 Mths Quest MATHEMATICAL METHODS VCE Units nd

WOrKeD example think This rule cn e verified differentiting with respect to. The linerit properties llow nti-derivtives of sums nd differences of polnomil terms to e clculted. For emple, if f () = + + + + n n, n N, then f() = + + + + n n n + c = + + + + n n n + c Note tht the degree of the primitive function is one higher thn the degree of the grdient function. The rule for nti-differentition of polnomil shows tht differentition nd nti-differentition re inverse opertions. For n N, the derivtive of n is n n ; the nti-derivtive of n is n+ (n + ) = n+ n + + c. If d d = 68, find. Find n nti-derivtive of 5 +. c Given f () = ( )( + ), find the rule for the primitive function. Stte the reverse process required to find. Use the rule to nti-differentite the polnomil term. Note: It s good ide to mentll differentite the nswer to check its derivtive is 6 8. Appl the nti-differentition rule to ech term of the polnomil. Choose n vlue for the constnt c to otin n nti-derivtive. Note: The conventionl choice is c =. c Epress the rule for the derivtive function in epnded polnomil form. WritE d d = 68 The derivtive of is 6 8. Hence, the nti-derivtive of 6 8 is. = 6 8 + 8+ + c = 6 9 9 + c = 9 + c 5 + = 5 + The nti-derivtive is: 5 + + c = 5 + + c When c =, n nti-derivtive is 5 +. c f () = ( )( + ) = ( ) = Clculte the nti-derivtive nd stte the nswer. f() = + c The rule for the primitive function is f() = + c. Topic AnTI-DIfferenTIATIOn AnD InTrODuCTIOn TO InTegrAL CALCuLus 7

The indefinite integrl Anti-differentition is not onl out undoing the effect of differentition. It is n opertion tht cn e pplied to functions. There re different forms of nottion for the nti-derivtive just s there re for the derivtive. The most common form of nottion uses smolism due to Leiniz. It is customr to write the nti-derivtive of f() s f() d. This is clled the indefinite integrl. Using this smol, we could write the nti-derivtive of with respect to equls + c s d = + c. For n function: f() d = F() + c where F () = f() F() is n nti-derivtive or primitive of f(). It cn lso e sid tht F() is the indefinite integrl, or just integrl, of f(). The ritrr constnt c is clled the constnt of integrtion. Here we tke integrtion to e the process of using the integrl to otin n nti-derivtive. A little more will e sid out integrtion lter. The rule for nti-differentition of polnomil term could e written in this nottion s: For n N, n d = n+ n + + c The linerit properties of nti-differentition, or integrtion, could e epressed s: ( f() ± g()) d = f() d ± g() d kf() d = k f() d Anti-derivtives of power functions The rule for otining the nti-derivtive of polnomil function lso holds for power functions where the inde m e rtionl. This is illustrted ppling the rule to. To verif the result, differentite: d d + d = + + c = + c = + c = + c The rule for nti-differentiting n is: + c = + = n d = n+ n + + c, n R \ 7 Mths Quest MATHEMATICAL METHODS VCE Units nd

WOrKeD example think This rule must eclude n = to prevent the denomintor n + ecoming zero. This eception to the rule will e studied in Mthemticl Methods Units nd. Use the linerit properties to clculte ( + ) d. 7 Clculte 5 8 7 d. 5 c Given f() = form F(), where F () = f()., d Find d. Epress the integrl s the sum or difference of integrls of ech term. Anti-differentite term term. Note: In prctice, the clcultion usull omits the steps illustrting the linerit properties. Epress the term to e nti-differentited in polnomil form. WritE ( + ) d = d + d d = d + d d = + + c = + + c 7 5 8 7 d 5 = 7 5 5 87 5 d = 7 5 8 5 6 d Clculte the nti-derivtive. = 7 5 5 5 8 5 7 7 + c c Epress the given function with power of in its numertor. Use the rule to otin the nti-derivtive of the power function. = 7 5 5 8 5 7 + c c f() = = F() is the nti-derivtive function. F() = + ( + ) + c = + c = + c Topic AnTI-DIfferenTIATIOn AnD InTrODuCTIOn TO InTegrAL CALCuLus 7

Rewrite the nti-derivtive function with positive indices. d Prepre the epression to e nti-differentited epnding the perfect squre nd simplifing. d F() = + c d = + d Eercise. PRctise Work without CAS Consolidte Appl the most pproprite mthemticl processes nd tools Anti-derivtives WE If d d = 5, find. Find n nti-derivtive of + 5. c Given f () = ( )( + 8), otin the rule for the primitive function. Given the grdient function d d =,, otin the primitive function. WE Use the linerit properties to clculte ( 7 + 6) d. Clculte 5 8 + d. c Given f() =, form F(), where F () = f(). d Otin + d. Given f() =, otin F(), where F () = f(). Clculte ech epression to show tht (t + ) dt is the sme s (t + ) dt. 5 Given d, otin in terms of for ech of the following. d d d = 59 d = + 7 d c d d = ( 6 + 7) = ( + ) d Clculte the nti-derivtive. = + + = + c = + c + c d d = (8 )( + 5) d 7 Mths Quest MATHEMATICAL METHODS VCE Units nd

6 For ech f () epression, otin n epression for f(). f () = 5 + 7 6 c f () = 6 8 f () = 6 + 5, d f () = ( ) 7 Clculte the following indefinite integrls. 8 d d 5 c ( 5 7 ) d 8 Clculte the primitive function of +, for, R. Clculte the nti-derivtive of.5 99. c Clculte n nti-derivtive of ( + ). d Clculte the primitive of 7 (5 8). 9 Find F(), where F () = f() for the following: d (9 + 6 5.5 ) d f() = f() = ( ) + c f() =.5( + 5 ) d f() = (5 ) () Given d, otin in terms of for ech of the following. d d d = d d = For ech f () epression, otin n epression for f(). f () = 5 Clculte the following indefinite integrls. f () = 6 + 6 5 + 7 5 d ( + ) d c 5 ( + ) d d ( + )( ) For the following, find F(), where F () = f(), given, R: f() = () + f() = + Clculte the primitive of ( )( + )( + ). Clculte the nti-derivtive of 5 5 c Clculte n nti-derivtive of p q, stting the restriction on the vlues of p nd q. d Clculte n nti-derivtive of 5.. d e Clculte d d ( + 7) d. Topic Anti-differentition nd introduction to integrl clculus 75

MstEr 5 The clcultor enles the nti-derivtive to e clculted through numer of menus. Fmilirise ourself with these clculting the nti-derivtive of using ech of these methods. Rememer to insert the constnt ourself if the clcultor does not. 6 Write down the nswers from the clcultor to the following. d Given d d = ( + ), otin.. Units & AOS Topic Concept Anti-derivtive functions nd grphs Concept summr Prctice questions Interctivit Sketching the nti-derivtive grph int 5965 WOrKeD example c If f(t) = t 5 t, otin F(t). d (5 5 ) d e u + 5 du f Choose our own function nd use CAS technolog to otin its nti-derivtive. Anti-derivtive functions nd grphs If some chrcteristic informtion is known out the nti-derivtive function, such s point on its curve, then this informtion llows the constnt of integrtion to e clculted. Determining the constnt Once the constnt of integrtion is known, we cn determine more thn just fmil of nti-derivtives. A specific nti-derivtive function cn e found since onl one memer of the fmil will possess the given chrcteristic informtion. While there is fmil of nti-derivtive functions = + c for which d =, if it is known the d nti-derivtive must contin the point (, ), sustituting (, ) in = + c gives = + c nd therefore c =. Onl the specific nti-derivtive function = stisfies this condition. The fmil of curves = + c The grdient of curve is given d = + where is constnt. d Given the curve hs sttionr point t (, 5), determine its eqution. If f () = nd f() =, otin the vlue of f(). think Use the given informtion to determine the vlue of. WritE d d = + At the sttionr point (, 5), d d =. = + = 6 76 MAThs QuesT MATheMATICAL MeThODs VCe units nd

Clculte the nti-derivtive. Use the given informtion to clculte the constnt of integrtion. d = 6 + d = 6 + + c = 6 + + c Sustitute the point (, 5): 5 = 6() + () 5 = + 6 + c c = Stte the nswer. = 6 + + Epress f () in the form in which it cn e nti-differentited. Sketching the nti-derivtive grph Given the grph of function = f() whose rule is not known, the shpe of the grph of the nti-derivtive function = F() m e le to e deduced from the reltionship F () = f(). This mens interpreting the grph of = f() s the grdient grph of = F(). + c The eqution of the curve is = 6 +. f () = = = Clculte the nti-derivtive. f() = + c Use the given informtion to clculte the constnt of integrtion. = + + c Sustitute f() = to clculte c. = + + c = + + c c = Clculte the required vlue. f() = + + f() = + + = Topic Anti-differentition nd introduction to integrl clculus 77

WOrKeD example The -intercepts of the grph of = f() identif the -coordintes of the sttionr points of = F(). The nture of n sttionr point on = F() is determined the w the sign of the grph of = f() chnges out its -intercepts. If f() is polnomil of degree n then F() will e polnomil of degree n +. If the rule for = f() cn e formed from its grph, then the eqution of possile nti-derivtive functions cn lso e formed. Additionl informtion to determine the specific function would e needed, s in the previous Worked emple. Consider the grph of the qudrtic function = f() shown. Descrie the position nd nture of n sttionr point on the grph of = F() where f() = F (). Drw three possile grphs for which = F(). c Otin the rule for f(). think Identif the position of n sttionr point on the grph of = F(). d Given F() =, determine the rule for F(), nd sketch the grph of = F(). Determine the nture of the sttionr point. Stte the degree of F(). Drw three possile grphs for the nti-derivtive function. WritE The -intercept on the given grph identifies sttionr point on = F(). Therefore the grph of = F() hs sttionr point t the point where =. As increses, the sign of the grph of = f() chnges from positive to zero to positive out its -intercept. Therefore, there is sttionr point of inflection t the point where = on the grph of = F(). Since f() is qudrtic, its nti-derivtive F() will e polnomil of degree. Other thn t =, the grph of = f() is positive so the grdient of the grph of = F() is positive other thn t =. The -coordinte of the sttionr point on = F() is not known. Three grphs of n incresing cuic function with sttionr point of inflection t the point where = re shown for = F(). = f() (, ) = f() (, ) 78 MAThs QuesT MATheMATICAL MeThODs VCe units nd

c Determine the eqution of the qudrtic function using the informtion on its grph. d Use nti-differentition to otin the rule for F(). Determine the vlue of the constnt of integrtion using the given informtion nd stte the rule for F(). Sketch the grph of = F. Eercise. PRctise Work without CAS c The grph of = f() hs minimum turning point t (, ) nd contins the point (, ). Let f() = ( h) + k. The turning point is (, ). f() = Sustitute the point (, ): = () = The rule for the qudrtic function is f() =. d f() = F() = + c Since F() =, = () + c c = F() = The point (, ) is sttionr point of inflection nd the -intercept. -intercept: let = = = =, is the -intercept. (, ) = (, ) Anti-derivtive functions nd grphs WE The grdient of curve is given d = 6 where is constnt. d Given the curve hs sttionr point t (, ), determine its eqution. If f () = + 9 nd f() =, find the vlue of f( ). If d = nd = when =, find when =. d Topic Anti-differentition nd introduction to integrl clculus 79

Consolidte Appl the most pproprite mthemticl processes nd tools WE Consider the grph of the liner function = f() shown. Descrie the position nd nture of n sttionr point on the grph of = F() where f() = F (). Drw three possile grphs for which = F(). c Otin the rule for f(). d Given F() =, determine the rule for F() nd sketch the grph of = F(). The grph of the grdient, d, of prticulr curve is shown. d Given tht (, ) lies on the curve with this grdient, determine the eqution of the curve nd deduce the coordintes of, nd the nture of, n sttionr point on the curve. = f() (, ) 5 The grdient of curve is given f () = +. Find the eqution of the curve if it psses through the point (, ). 6 The grdient of curve is given d d =. It is lso known tht the curve psses through the point (5, ). 5 Determine the eqution of the curve. Find its -intercepts. 7 The grdient of curve is directl proportionl to nd t the point (, 5) on the curve the grdient is. Determine the constnt of proportionlit. Find the eqution of the curve. ( )(5 ) 8 A function is defined f() =. Determine its primitive function if the point (, ) lies on the primitive function. 9 Given d = ( ) nd = when =, otin the vlue of when =. d Given dz d = ( ) nd z = when =, otin the vlue of z when =. c Given da dt = nd A = when t = 6, otin the vlue of A when t = 6. t d Given d d = nd = when =, otin the vlue of when = 75. At n point (, ) on curve, the grdient of the tngent to the curve is given d d =. The curve hs sttionr point t (8, ). Find the vlue of. Determine the eqution of the curve. c Clculte the eqution of the tngent to the curve t the point where =. d d (, ) (, ) 7 Mths Quest MATHEMATICAL METHODS VCE Units nd

A curve with horizontl smptote with eqution = psses through the point (, ). If the grdient of the curve is given F () = : Find the vlue of. Determine the eqution of the curve. c Clculte, to the nerest degree, the ngle t which the curve cuts the -is. The grdient function of curve is illustrted in the digrm. Descrie the position nd nture of the sttionr points of the curve with this grdient function. Sketch possile curve which hs this grdient function. c Determine the rule for the grdient grph nd hence otin the eqution which descries the fmil of curves with this grdient function. d One of the curves elonging to the fmil of curves cuts the -is t =. Otin the eqution of this prticulr curve nd clculte the slope with which it cuts the -is t =. For ech of the following grphs of = f(), drw sketch of possile curve for = F() given F () = f(). c (, ) = f() (, ) (, ) d (, ) (, ) = f() (, ) d d = f() (, ) (, ) (, ) (, ) = f() Topic Anti-differentition nd introduction to integrl clculus 7

Mster. Units & AOS Topic Concept Applictions of nti-differentition Concept summr Prctice questions The digrm shows the grph of cuic function = f() with sttionr point of inflection t (, 8) nd pssing through the point (, 9). Find f() d. = f() A prticulr nti-derivtive (, 8) function of = f() psses through the origin. For this prticulr function: (, 9) i Determine its eqution. ii Find the coordintes of its other -intercept nd determine t which of its -intercepts the grph of this function is steeper. iii Deduce the ect -coordinte of its turning point nd its nture. iv Hence show the -coordinte of its turning point cn e epressed s p q, specifing the vlues of p nd q. v Drw sketch grph of this prticulr nti-derivtive function. 5 Use the CAS clcultor s keord templte to sketch the shpe of the grphs of the nti-derivtive functions given : = = c = ( )( + ) d ( ) ( + ) d ( ) ( + ) d. d Comment on the effect of chnging the multiplicit of ech fctor. 6 Define f() = 6 nd use the clcultor to sketch on the sme set of es the grphs of = f(), = f () nd = etween the three grphs. f() d. Comment on the connections Applictions of nti-differentition In this section, nti-differentition is pplied in clcultions involving functions defined their rtes of chnge. The nti-derivtive in kinemtics Reclling tht velocit, v, is the rte of chnge of displcement,, mens tht we cn now interpret displcement s the nti-derivtive of velocit. v = d dt = v dt 7 Mths Quest MATHEMATICAL METHODS VCE Units nd

WOrKeD example 5 think Further, since ccelertion is the rte of chnge of velocit, it follows tht velocit is the nti-derivtive of ccelertion. = dv dt v = dt The role of differentition nd nti-differentition in kinemtics is displed in the digrm elow. Velocit is of prticulr interest for two resons: The nti-derivtive of velocit differentite differentite with respect to time gives displcement. The derivtive of velocit with respect to time gives displcement velocit v ccelertion ccelertion. nti-differentite nti-differentite A prticle moves in stright line so tht its velocit t time t seconds is given v = t + t, t. Initill the prticle is 8 metres to the left of fied origin. After how mn seconds does the prticle rech the origin? Clculte the prticle s ccelertion when its velocit is zero. Clculte the displcement function from the velocit function. Use the initil conditions to clculte the constnt of integrtion. Set up nd solve the eqution which gives the time the prticle is t the required position. WritE v = t + t Anti-differentite the velocit to otin displcement. = t + t t + c = t + t t + c When t =, = 8. 8 = c = t + t t 8 When the prticle is t the origin, =. t + t t 8 = t (t + ) (t + ) = (t + )(t ) = (t + ) (t ) = t =, t = As t, reject t =. t = The prticle reches the origin fter seconds. Topic AnTI-DIfferenTIATIOn AnD InTrODuCTIOn TO InTegrAL CALCuLus 7

Clculte the ccelertion function from the velocit function. Clculte the time when the velocit is zero. v = t + t Accelertion = dv dt = 6t +. When velocit is zero, t + t = (t )(t + ) = Clculte the ccelertion t the time required. WOrKeD example 6 t, so t = When t =, = 6 + t =, t = = 8 The ccelertion is 8 m/s when the velocit is zero. Accelertion nd displcement To otin the displcement from n epression for ccelertion will require ntidifferentiting twice in order to proceed from to v to. The first nti-differentition opertion will give velocit nd the second will otin displcement from velocit. This will introduce two constnts of integrtion, one for ech step. Where possile, evlute the first constnt using given v t informtion efore commencing the second nti-differentition opertion. To evlute the second constnt, t informtion will e needed. The nottion used for ech constnt should distinguish etween them. For emple, the first constnt could e written s c nd the second s c. A prticle moves in stright line so tht its ccelertion t time t seconds is given = + 6t, t. If v = nd = when t =, clculte the prticle s velocit nd position when t =. think Clculte the velocit function from the ccelertion function. Evlute the first constnt of integrtion using the given v t informtion. Clculte the displcement function from the velocit function. Evlute the second constnt of integrtion using the given t informtion. WritE = + 6t Anti-differentite the ccelertion to otin velocit. v = t + t + c = t + t + c When t =, v = = c v = t + t + Anti-differentite the velocit to otin displcement. = t + t + t + c When t =, = = c = t + t + t + 7 MAThs QuesT MATheMATICAL MeThODs VCe units nd

5 Clculte the velocit t the given time. v = t + t + When t =, v = + + = 7 6 Clculte the displcement t the given time. = t + t + t + When t =, = + + + = 5 7 Stte the nswer. The prticle hs velocit of 7 m/s nd its position is 5 metres to the right of the origin when t =. WOrKeD example 7 think Other rtes of chnge The process of nti-differentition cn e used to solve prolems involving other rtes of chnge, not onl those involved in kinemtics. For emple, if the rte t which the re of n oil spill is chnging with respect to t is given da = f(t), then the re of dt the oil spill t time t is given A = f(t) dt, the nti-derivtive with respect to t. An ice lock with initil volume 6π cm strts to melt. The rte of chnge of its volume V cm fter t seconds is given dv dt =.. Use nti-differentition to epress V in terms of t. Clculte, to deciml plce, the numer of minutes it tkes for ll of the ice lock to melt. Clculte the volume function from the given derivtive function. Use the initil conditions to evlute the constnt of integrtion. Identif the vlue of V nd clculte the corresponding vlue of t. Epress the time in the required units nd stte the nswer to the required degree of ccurc. Note: The rte of decrese of the volume is constnt, so this prolem could hve een solved without clculus. WritE dv dt =. Anti-differentite with respect to t: V =.t + c When t =, V = 6π 6π = c Therefore, V =.t + 6π When the ice lock hs melted, V =. =.t + 6π t = 6π. = 8π The ice lock melts fter 8π seconds. Divide this 6 to convert to minutes: 8π seconds is equl to π minutes. To deciml plce, the time for the ice lock to melt is 9. minutes. Topic AnTI-DIfferenTIATIOn AnD InTrODuCTIOn TO InTegrAL CALCuLus 75

Eercise. PRctise Work without CAS Consolidte Appl the most pproprite mthemticl processes nd tools Applictions of nti-differentition WE5 A prticle moves in stright line so tht its velocit t time t seconds is given v = t t 8, t. Initill the prticle is metres to the right of fied origin. After how mn seconds does the prticle first rech the origin? Clculte the prticle s ccelertion when its velocit is zero. A prticle moves in stright line so tht its velocit t time t seconds is given v = +, t >. When t =, the prticle is metre to the left of fied origin. t Otin epressions for the prticle s displcement nd ccelertion fter t seconds. WE6 A prticle moves in stright line so tht its ccelertion t time t seconds is given = 8 8t, t. If v = nd = when t =, clculte the prticle s velocit nd position when t =. A prticle moves in stright line so tht its ccelertion, in m/s, t time t seconds is given = 9.8, t. If v = = when t =, clculte the prticle s displcement when t = 5. 5 WE7 An ice lock with initil volume.5π cm strts to melt. The rte of chnge of its volume, V cm, fter t seconds is given dv dt =.5. Use nti-differentition to epress V in terms of t. Clculte, to deciml plce, the numer of seconds it tkes for ll of the ice lock to melt. 6 When first purchsed, the height of smll ruer plnt ws 5 cm. The rte of growth of its height over the first er fter its plnting is mesured h (t) =.t, t, where h is its height in cm t months fter eing plnted. Clculte its height t the end of the first er fter its plnting. 7 The velocit, v m/s, of prticle moving in stright line t time t seconds is given v = 8t t, t. Initill the prticle is 5 metres to the right of fied origin. Otin n epression for the prticle s displcement t time t seconds. How fr from its initil position is the prticle fter the first second? c Determine the position of the prticle when its velocit is zero. 8 A prticle strts from rest nd moves in stright line so tht its velocit t time t is given v = t, t. When its position is metre to the right of fied origin, its velocit is m/s. Wht ws the initil position of the prticle reltive to the fied origin? 9 Strting from point 9 metres to the right of fied origin, prticle moves in stright line in such w tht its velocit fter t seconds is v = 6 6t, t. Show tht the prticle moves with constnt ccelertion. Determine when nd where its velocit is zero. c How fr does the prticle trvel efore it reches the origin? d Wht is the verge speed of this prticle over the first three seconds? e Wht is the verge velocit of this prticle over the first three seconds? 76 Mths Quest MATHEMATICAL METHODS VCE Units nd

After t seconds the velocit v m/s of prticle moving in stright line is given v(t) = (t )(t ), t. Clculte the prticle s initil velocit nd initil ccelertion. Otin n epression for (t), the prticle s displcement from fied origin fter time t seconds, given tht the prticle ws initill t this origin. c Clculte (5). d How fr does the prticle trvel during the first 5 seconds? A prticle moving in stright line hs velocit of m/s nd displcement of metres from fied origin fter second. If its ccelertion fter time t seconds is = 8 + 6t, otin epressions in terms of t for: its velocit its displcement. The ccelertion of moving oject is given =. Initill, the oject ws t the origin nd its initil velocit ws m/s. Determine when nd where its velocit is zero. After how mn seconds will the oject return to its strting point? The digrm shows velocit time grph. v Drw possile displcement time grph. (, ) Drw the ccelertion time grph. c If it is known tht the initil displcement is, (, ) determine the rule for the displcement time t grph. d Sketch the displcement time grph with the rule otined in prt c. Pouring oiling wter on weeds is mens of keeping unwnted weeds under control. Along the crcks in n sphlt drivew, weed hd grown to cover n re of 9 cm, so it ws given the oiling wter tretment. The rte of chnge of the re, A cm, of the weed t ds fter the oiling wter is poured on it is given da dt = 8t. Epress the re s function of t. According to this model, how mn whole ds will it tke for the weed to e completel removed? c For wht ect vlues of t is the model da = 8t vlid? dt 5 Rinwter collects in puddle on prt of the surfce of n uneven pth. For the time period for which the rin flls, the rte t which the re, A m, grows is modelled A (t) = t, where t is the numer of ds of rin. After how mn ds does the re of the puddle stop incresing? Epress the re function A(t) in terms of t nd stte its domin. c Wht is the gretest re the puddle grows to? d On the sme set of es, sketch the grphs of A (t) versus t nd A(t) versus t over n pproprite domin. Topic Anti-differentition nd introduction to integrl clculus 77

Mster.5 Units & AOS Topic Concept The definite integrl Concept summr Prctice questions 6 The rte of growth of colon of microes in lortor cn e modelled dm = k t where m is the numer of microes fter t ds. Initill there were dt microes nd fter ds the popultion ws growing t microes per d. Determine the vlue of k. Epress m s function of t. c Hence clculte the numer of ds it tkes for the popultion size to rech 6. 7 The velocit of n oject which moves in stright line is v = (t + ), t. Use CAS technolog to: stte the ccelertion clculte the displcement, given tht initill the oject s displcement ws. metres from fied origin c find the time nd the velocit, to deciml plces, when the displcement is 8. metres. 8 With the id of CAS technolog, otin n epression for in terms of t if = nd = v = when t =. (t + ) The definite integrl The indefinite integrl hs een used s smol for the nti-derivtive. In this section we will look t the definite integrl. We shll lern how to evlute definite integrl, ut onl riefl eplore its mening, since this will form prt of the Mthemticl Methods Units nd course. The definite integrl The definite integrl is of the form f() d; it is quite similr to the indefinite integrl ut hs the numers nd plced on the integrl smol. These numers re clled terminls. The define the endpoints of the intervl or the limits on the vlues of the vrile over which the integrtion tkes plce. Due to their reltive positions, is referred to s the lower terminl nd s the upper terminl. For emple, d is definite integrl with lower terminl nd upper terminl, wheres d without terminls is n indefinite integrl. In the definite integrl the term f() is clled the integrnd. It is the epression eing integrted with respect to. Clcultion of the definite integrl The definite integrl results in numericl vlue when it is evluted. It is evluted the clcultion: f() d = F() F(), where F() is n nti-derivtive of f(). The clcultion tkes two steps: Anti-differentite f() to otin F(). Sustitute the terminls in F() nd crr out the sutrction clcultion. 78 Mths Quest MATHEMATICAL METHODS VCE Units nd

This clcultion is commonl written s: f() d = F() = F() F() WOrKeD example 8 think For emple, to evlute d, we write: d = = () () = 8 Note tht onl n nti-derivtive is needed in the clcultion of the definite integrl. Hd the constnt of integrtion een included in the clcultion, these terms would cncel out, s illustrted the following: d = + c = (() + c) (() + c) = 9 + c c = 8 The vlue of definite integrl m e positive, zero or negtive rel numer. It is not fmil of functions s otined from n indefinite integrl. Evlute ( + ) d. Clculte n nti-derivtive of the integrnd. Sustitute the upper nd then the lower terminl in plce of nd sutrct the two epressions. WritE ( + ) d = + = + ( ) + ( ) Evlute the epression. = 8 + = () ( ) = Stte the nswer. ( + ) d = Topic AnTI-DIfferenTIATIOn AnD InTrODuCTIOn TO InTegrAL CALCuLus 79

Integrtion nd re Ares of geometric shpes such s rectngles, tringles, nd trpeziums cn e used to pproimte res enclosed edges which re not stright. Leiniz invented the method for clculting the ect mesure of n re enclosed curve, there estlishing the rnch of clculus known s integrl clculus. His method ers some similrit to method used in the geometr studies of Archimedes round BC. Leniz s pproch involved summing lrge numer of ver smll res nd then clculting the limiting sum, process known s integrtion. Like differentil clculus, integrl clculus is sed on the concept of limit. To clculte the re ounded curve = f() 8 6 nd the -is etween = nd =, the re could e pproimted set of rectngles s illustrted in the digrm elow. Here, 8 rectngles, ech of width.5 units, hve een constructed. The lrger the numer of rectngles, the etter the ccurc of the pproimtion to the ctul re; lso, the lrger the numer, the smller the widths of the rectngles. To clculte the re etween = nd =, Leiniz prtitioned (divided) the intervl [, ] into lrge numer of strips of ver smll width. A tpicl strip would hve width δ nd re δa, with the totl re pproimted the sum of the res of such strips. For δ, the pproimte re of tpicl strip is δa δ, so the totl re A is pproimted from the sum of these res. This is written s δ A = δ. The pproimtion improves s δ nd the numer of strips increses. The ctul re is clculted s limit; the limiting sum s δ, nd therefore A = lim δ = δ. Leiniz chose to write the smol for the limiting sum with the long cpitl S used in his time. Tht smol is the now fmilir integrl sign. This mens tht the re ounded the curve, the -is nd =, = is A = d or A = f() d. The process of evluting this integrl or limiting sum epression is clled integrtion. The fundmentl theorem of clculus links together the process of nti-differentition with the process of integrtion. The full proof of this theorem is left to Units nd. = f() = f() 75 Mths Quest MATHEMATICAL METHODS VCE Units nd

signed re As we hve seen, the definite integrl cn result in positive, negtive or zero vlue. Are, however, cn onl e positive. If f() >, the grph of = f() lies ove the -is nd f() d will e positive. Its vlue will e mesure of the re ounded the curve = f() nd the -is etween = nd =. However, if f() <, the grph of = f() is elow the -is nd f() d will e negtive. Its vlue gives signed re. B ignoring the negtive sign, the vlue of the integrl will still mesure the ctul re ounded the curve = f() nd the -is etween = nd =. If over the intervl [, ] f() is prtl positive nd prtl negtive, then f() d could e positive or negtive, or even zero, depending on which of the positive or negtive signed res is the lrger. f() d is the sum of signed re mesures This dds compleit to using the signed re mesure of the definite integrl to clculte re. Perhps ou cn lred think of w round this sitution. However, for now, nother tret witing ou in Units nd will e to eplore signed res. WOrKeD The re ounded the line =, example 9 the -is nd =, = is 8 = illustrted in the digrm. 6 Clculte the re using the formul for the re of tringle. Write down the definite integrl which represents the mesure of this re. c Hence, clculte the re using clculus. think Clculte the re using the formul for the re of tringle. Stte the definite integrl which gives the re. WritE Bse is units; height is 8 units. Are of tringle: A = h = 8 = 6 The re is 6 squre units. The definite integrl f() d gives the re. For this re f() =, = nd =. d gives the re mesure. 5 6 Topic AnTI-DIfferenTIATIOn AnD InTrODuCTIOn TO InTegrAL CALCuLus 75

c Evlute the definite integrl. Note: There re no units, just rel numer for the vlue of definite integrl. c d = = = 6 Stte the re using pproprite units. Therefore, the re is 6 squre units. Interctivit Kinemtics int 596 WOrKeD example The definite integrl in kinemtics A prticle which trvels t constnt velocit v of m/s will trvel distnce of metres in 5 seconds. For this motion, the velocit time grph is horizontl line. Looking t the rectngulr v = re under the velocit time grph ounded the horizontl is etween t = nd t = 5 shows tht its re mesure is lso equl to. 5 The re under the velocit time grph gives the mesure of the distnce trvelled prticle. t For positive signed re, the definite integrl v dt gives the distnce trvelled over the time intervl t [t t, t ]. The velocit time grph is the most importnt of the motion grphs s it gives the velocit t n time; the grdient of its tngent gives the instntneous ccelertion nd the re under its grph gives the distnce trvelled. The velocit of prticle moving in stright line is given v = t +, t. Give the prticle s velocit nd ccelertion when t =. i Sketch the velocit time grph nd shde the re which represents the distnce the prticle trvels over the intervl t [, ]. ii Use definite integrl to clculte the distnce. Assume units for distnce re in metres nd time in seconds. think WritE Clculte the velocit t the given time. v = t + When t =, v = + = The velocit is m/s. Otin the ccelertion from the velocit function nd evlute it t the given time. = dv dt = 6t When t =, = 6 The ccelertion is 6 m/s. t 75 MAThs QuesT MATheMATICAL MeThODs VCe units nd

i Sketch the v t grph nd shde the re required. i v = t +, t The v t grph is prt of the prol which hs turning point (, ) nd psses through the point (, ). The re under the grph ounded the t-is nd t =, t = represents the distnce. v 5 ii Stte the definite integrl which gives the mesure of the re shded. Note: The integrnd is function of t nd is to e integrted with respect to t. Evlute the definite integrl. Eercise.5 PRctise Work without CAS The definite integrl WE8 Evlute Evlute the following. ( )( + ) d d ( ) d. WE9 The re ounded the line =.75, the -is nd =, = is illustrted in the digrm. Clculte the re using the formul for the re of tringle. 5 6 ii Are mesure is given (t + ) dt = t + t v = t + = ( + ) ( + ) = 8 t (t + ) dt. Stte the distnce trvelled. The distnce trvelled the prticle over the time intervl [, ] is 8 metres. =.75 5 6 7 Topic Anti-differentition nd introduction to integrl clculus 75

Consolidte Appl the most pproprite mthemticl processes nd tools Write down the definite integrl which represents the mesure of this re. c Hence, clculte the re using clculus. The re ounded the curve = 6, the -is nd =, = is illustrted in the digrm. Write down the definite integrl which represents the mesure of this re. Hence, clculte the re. 5 WE The velocit of prticle moving in stright line is given v = t +, t. Give the prticle s velocit nd ccelertion when t =. i Sketch the velocit time grph nd shde the re which represents the distnce the prticle trvels over the intervl t [, ]. ii Use definite integrl to clculte the distnce. Assume units for distnce re in metres nd time is in seconds. 6 The velocit, v m/s, of prticle moving in stright line fter t seconds is given v = t t + 5, t. Show tht velocit is lws positive. Clculte the distnce the prticle trvels in the first seconds using: i definite integrl ii nti-differentition. 7 Evlute ech of the following. c e 5 d (6 + 5 ) d d ( + ) d f (7 ) d ( ) d ( + )( ) d 8 Show clcultion tht the following sttements re true. c e ( + 5) d = 5 d = d = t dt d 9 Evlute ech of the following. d ( + ) d d f ( + ) d = d = 9 d = d 8 (, 6) = 6 (, ) (, ) d + d 75 Mths Quest MATHEMATICAL METHODS VCE Units nd

c e 5 d 5 t( t + ( t) ) dt f Determine the vlue of n so tht Determine the vlue of p so tht n d d = 9. p 6 + + 5 d = 8. u du Epress the res in ech of the following digrms in terms of definite integrl nd clculte the re using oth integrtion nd known formul for the re of the geometric shpe. The tringulr re ounded the line = + 6, the -is nd =, = s illustrted in the digrm = + 6 The rectngulr re ounded the line =, the -is nd = 6, = s illustrted in the digrm = 7 6 5 5 6 7 c The trpezoidl re ounded the line =, the coordinte es nd = s illustrted in the digrm = d 5 Topic Anti-differentition nd introduction to integrl clculus 755

For ech of the following res: i write definite integrl, the vlue of which gives the re mesure ii clculte the re. The re ounded the curve = +, the -is nd =, = s illustrted in the digrm The re ounded the curve =, the -is nd = s illustrted in the digrm = 8 c The re ounded the curve = ( + )( ), the -is nd =.5, =.5 s illustrted in the digrm 6 = ( + )( ).5.5.5.5.5.5 Sketch grph to show ech of the res represented the given definite integrls nd then clculte the re. d 6 ( ) d c Consider the function defined f() =. Sketch the grph of = f(), showing the intercepts with the coordinte es. On the digrm, shde the re which is ounded the curve nd the coordinte es. c Epress the shded re in terms of definite integrl. d Clculte the re. d = + 756 Mths Quest MATHEMATICAL METHODS VCE Units nd

Mster 5 A prticle strts from rest t the origin nd moves in stright line so tht fter t seconds its velocit is given v = t 5t, t. Clculte n epression for the displcement t time t. When the prticle is net t rest, how fr will it hve trvelled from its strting point? c Drw the velocit time grph for t [, ]. d Use definite integrl to clculte the re enclosed the velocit time grph nd the horizontl is for t [, ]. e Wht does the re in prt d mesure? 6 An thlete trining to compete in mrthon rce runs long stright rod with velocit v km/h. The velocit time grph over 5-hour period for this thlete is shown elow. (, ( ), 6 6 ) (, ) ( 5, 9 (, ) ) 5 6 7 Wht were the thlete s highest nd lowest speeds over the 5-hour period? At wht times did the thlete s ccelertion ecome zero? c Given the eqution of the grph shown is v = + 8t 7t + 7t t, use definite integrl to clculte the distnce the thlete rn during this 5-hour trining period. 7 Write down the vlues of: ( + ) d ( + ) d, nd c eplin the reltionship etween prts nd. 8 Use the clcultor to otin ( + )( ) d. Drw the grph of = ( + )( ). c Wht is the re enclosed etween the grph of = ( + )( ) nd the -is? d Wh do the nswers to prt nd prt c differ? Write down definite integrl which does give the re. Topic Anti-differentition nd introduction to integrl clculus 757

ONLINE ONLY.6 Review www.jcplus.com.u the Mths Quest review is ville in customisle formt for ou to demonstrte our knowledge of this topic. the review contins: short-nswer questions providing ou with the opportunit to demonstrte the skills ou hve developed to efficientl nswer questions without the use of CAS technolog Multiple-choice questions providing ou with the opportunit to prctise nswering questions using CAS technolog ONLINE ONLY Activities to ccess ebookplus ctivities, log on to www.jcplus.com.u Interctivities A comprehensive set of relevnt interctivities to ring difficult mthemticl concepts to life cn e found in the Resources section of our ebookplus. Etended-response questions providing ou with the opportunit to prctise em-stle questions. summr of the ke points covered in this topic is lso ville s digitl document. REVIEW QUESTIONS Downlod the Review questions document from the links found in the Resources section of our ebookplus. studon is n interctive nd highl visul online tool tht helps ou to clerl identif strengths nd weknesses prior to our ems. You cn then confidentl trget res of gretest need, enling ou to chieve our est results. Units & Anti-differentition nd introduction to integrl clculus Sit topic test 758 MAThs QuesT MATheMATICAL MeThODs VCe units nd

Answers Eercise. = 6 + 5 c f() = + 6 + c = + c 75 5 + + c 5 8 7 + 8 + c F() = + c d + + c F() = + c (t + ) dt = t + t + c = (t + ) dt 5 = + c = + 8 + c c = 6 + + c d = + + + 6 f() = 6 + 7 + c f() = 7 + 5 + c c f() = 6 7 7 + c d f() = 9 + 5 5 + c 7 5 9 + c + c c 58 + c d 9 +.5 + c 8 + + c.5 + c c + + + d 7 5 + + + c 9 F() = ( ) + c = 5 5 + c = + c f() = 5 + c f() = + c + 7 + 5 + c + 8 + + c c 5 6 6 5 + 6 5 + c d + + c F() = + + c F() = + + c + + c c 5 6 5 + 75 + c p q + p q+, p q d 5 5 e + 7 + c 5 5 9 9 + 5 5 + c 5t + 5 t + d e f (u + 5) 6 Answers will vr. F() = + + c c F() = + 5 + c d F() = 9 9 6 + c Eercise. = 6 + 7 f( ) = = Topic Anti-differentition nd introduction to integrl clculus 759

Mimum turning point t = c f() = d F() = + ; = + + 6; minimum turning point t (, ) 5 f() = + + 5 6 = + 5 (5, ), (, ) 7 k = = + 8 8 F() = + 9 6 9 = 9 z = 8 c A = 7 d = = = 5 5 + 96 5 c 5 5 = 98 = 6 f() = 6 + 6 c 8 Minimum turning point t = ; mimum turning point t = fmil of curves = + c (, ) (, ) = + (, ) = c d d = ; = + c d = + ; slope of 5 An line with grdient of (, ) An prol with mimum turning point when = c An inverted cuic with sttionr point of inflection when = = F() d An qurtic with minimum turning points when = ± nd mimum turning point when = = f() ( = ) = F() = F() 76 Mths Quest MATHEMATICAL METHODS VCE Units nd

9 + 8 + c i = 9 + 8 ii (, ); steeper t (, ) iii Mimum turning point when = iv p =, q = v 5,,c Use our CAS technolog to sketch the grphs. d For fctors of multiplicit, the nti-derivtive grph hs turning points t the zeros of ech fctor. For fctors of multiplicit, the nti-derivtive grph hs sttionr points of inflection t the zeros of ech fctor. 6 Use our CAS technolog to sketch the grphs. The -intercept of = f () is connected to the turning point of = f(). The turning points of = -intercepts of = f(). Eercise. seconds m/s = t, = t + t (, ) 6 9 = + 8 9 (, ) (, ), ) f() d re connected to the ( Velocit 9 m/s; position 9 metres to the right of the origin.5 metres 5 V =.5t +.5π 56.5 seconds 6 6. cm 7 = 8t t t + 5 9 metres c At the origin 8 metres to the right of the origin 9 = 6 second; metres to the right of origin c 5 metres d 5 m/s e m/s m/s; 8 m/s (t) = t 9t + t c d 8 metres v = t + 8t 8 = t + t 8t + 5 seconds; metres to right of origin seconds Prt of prol with mimum turning point t t = t Prt of horizontl line elow the t-is through (, ) t = Topic Anti-differentition nd introduction to integrl clculus 76

c = (t ), t d Prt of prol with mimum turning point t (, ) (, ) = (t ) 5 v = m/s; = m/s i v 8 7 6 5 v = t + 5 6 t A = 9t + 9 ds c t 5 ds A(t) = t t ; domin [, 8] c 8 m d 6 k = 5 m = t + c 6 ds 7 = 8(t + ) (t + )5 = +. c.56 seconds;.7 m/s 8 = t + (t + ) Eercise.5 8 6 squre units.75 d c 6 squre units 8 (, ) (, 8) (6 ) d squre units = A(t) 8 = Aʹ (t) (8, ) ii (t + ) dt = 6 so distnce is 6 metres. 6 Velocit is positive definite qudrtic function in t. i metres 7 6 6 ii metres c 7 d 5 8 e f 8 Proofs required check with our techer 9 8 c 6 d 55 6 e 7 f n = 7 p = 9 c i c 6 ( + 6) d; squre units d; squre units ( ) d; squre units ( + ) d ii squre units i ( ) d ii 6 squre units i.5.5 ( + )( ) d ii squre units 76 Mths Quest MATHEMATICAL METHODS VCE Units nd

Are enclosed =, -is nd =, = is 6 squre units. Are enclosed = nd -is is squre units. c Rectngulr re enclosed the line =, -is nd =, = is 6 squre units. Intercepts (, ), (6, ); = = = c Are mesure equls 6 d squre units 5 = 5t 5t 6 metres c (, 5) 5 v = t 5t 6 ( ) d. d 6 e Distnce trvelled in first seconds. 6 7 km/h; km/h hour, hours nd hours c 6 km 7 6 6 c Interchnging the terminls chnges the sign of the integrl. 8 5 6 5 c squre units 6 d Are lies elow -is so signed re is negtive. Possile integrls re (, ) (, 6) ( + )( ) d. (, ) (, ) 5 = ( + )( ) ( + )( ) d or = 8 6 Topic Anti-differentition nd introduction to integrl clculus 76