ONLINE PAGE PROOFS. Integration. 8.1 Kick off with CAS

8. Kick off with CAS Integrtion 8. The fundmentl theorem of integrl clculus 8. Ares under curves 8. Applictions 8.5 Review 8

8. Kick off with CAS Are under curves Using the grph ppliction on CAS, sketch the grph of f() = +. Estimte the re enclosed the -is, the verticl lines = nd =, nd the curve. (, ) = + c Using the clcultion ppliction, find the integrl templte nd clculte ( + )d. d Wht do ou notice out the nswers to prts nd c? Sketch the grph of f() =. Using one clcultion, find the re enclosed the -is, the verticl lines = nd =, nd the curve. c How do ou eplin our nswer? Is there method ou cn use to otin the correct re under the curve? d Using the clcultion ppliction, how could ou ppl the integrl templte to clculte the re? 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.

8. Units & AOS Topic Concept Approimting res under curves Concept summr Prctice questions Interctivit Estimtion of re under curve int 6 The fundmentl theorem of integrl clculus Estimtion of the re under curve There re severl different ws to pproimte or estimte the re etween curve nd the -is. This section will cover the left end-point rectngle rule nd the right end-point rectngle rule. The left end-point rectngle rule Consider the curve defined the rule f : R R, f() = +. Suppose we wish to know the re etween this curve nd the -is from = to =. This cn e chieved constructing rectngles of width unit nd = f() height such tht the top left corner touches the curve. The height of the first rectngle is otined evluting f(), the height of the second rectngle is otined evluting f(), nd the height of the (, f()) (, f()) third rectngle is otined evluting f(). Ech (, f()) rectngle hs width of unit. Approimte re: A = f() + f() + f() = f() + f() + f() = + + 6 = The theoreticl eplntion of this method cn e eplined for the generl function = f() s follows. R R R R = f() R n n n n R n The height of R is f(). The height of R is f(). The height of R is f(). The height of R is f(). The height of R n is f(n ). The height of R n is f(n ). Approimte re = f() + f() + f() + f() +... + f(n ) + f(n ) = f() + f() + f() + f() +... + f(n ) + f(n ) 8 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

If the sme re ws pproimted using rectngles of width.5 units, there would e twice s mn rectngles, so A.5 f() +.5 f(.5) +.5 f() +... +.5 f(n.5) +.5 f(n ) +.5 f(n.5). The nrrower the rectngles re, the closer the pproimtion is to the ctul re under the curve. In the emple shown, the pproimte re is less thn the ctul re. However, if the function ws decresing function, for emple =, then the re of the left end-point rectngles would e greter thn the ctul re under the curve. The right end-point rectngle rule We will now pproimte the re etween the curve f() = +, the -is, = nd = (, f()) constructing rectngles of width unit nd height such tht the top right corner touches the curve. The height of the first rectngle is otined evluting f(), the height of the second rectngle is (, f()) otined evluting f(), nd the height of the third rectngle is otined evluting f(). Ech rectngle hs width of unit. (, f()) Approimte re, A = f() + f() + f() = f() + f() + f() = + 6 + = The theoreticl eplntion of this method cn e eplined for the generl function = f() s follows. = f() R n R n = f() The height of R is f(). The height of R is f(). The height of R is f(). The height of R is f(). R R R R n n n The height of R n is f(n ). The height of R n is f(n). Approimte re, A = f() + f() + f() + f() +... + f(n ) + f(n) = f() + f() + f() + f() +... + f(n ) + f(n) Topic 8 Integrtion 9

The nrrower the rectngles re, the closer the pproimtion is to the ctul re under the curve. In the emple shown, the pproimte re is greter thn the ctul re. However, if the function ws decresing function, s =, then the re of the right end-point rectngles would e less thn the ctul re under the curve. WOrKED EXAMPLE The grph of the function defined the rule f() = e is shown. Give our nswers to the following correct to deciml plces. Use the left end-point rectngle method with rectngles of width.5 units to estimte the re etween the curve nd the -is from = to =.5. Use the right end-point rectngle method with rectngles of width.5 units to estimte the re etween the curve nd the -is from = to =.5. think Drw the left end-point rectngles on the grph. Stte the widths nd heights of the rectngles. Find the pproimte re dding the res of ll the rectngles. WRitE/dRAW.5 (, ).5 (, ).5.5.5 = e.5.5 = Rectngle widths =.5 units Rectngle heights re f(), f(.5), f(), f(.5) nd f()..5 A =.5f() +.5f(.5) +.5f() +.5f(.5) +.5f() =.5 f() + f(.5) + f() + f(.5) + f() =.5 e + e.5 + e + e.5 + e =.5 7.77 8.6 The re is pproimtel 8.6 units. = e = MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Drw the right end-point rectngles on the grph. Stte the widths nd heights of the rectngles. = e Find the pproimte re dding the res of ll the rectngles. The definite integrl The definite integrl,.5 (, ).5.5.5 = Rectngle widths =.5 units Rectngle heights re f(.5), f(), f(.5), f() nd f(.5). A =.5f(.5) +.5f() +.5f(.5) +.5f() +.5f(.5) =.5 f(.5) + f() + f(.5) + f() + f(.5) =.5 e.5 + e + e.5 + e + e.5 =.5 8.. The re is pproimtel. units. f()d, is similr to the indefinite integrl, f()d, ecept tht it hs end points, or terminls, nd. The indefinite integrl involves finding onl n ntiderivtive of f, ut the presence of the end points mens tht the definite integrl requires further clcultion involving these vlues. In fct, the end points nd indicte the rnge of the vlues of over which the integrl is tken. Consider ( )d = = () ( ) = + = Note: For the definite integrl, no ritrr constnt is required for the ntidifferentition, s this would onl e eliminted once the end points were used in the clcultion. Topic 8 Integrtion

= + c = () + c ( ) + c = + c + c = = Units & AOS Topic Concept 7 Properties of definite integrls Concept summr Prctice questions WOrKED EXAMPLE think Properties of the defi nite integrl If f nd g re continuous functions on n intervl where < < nd k is constnt, then the following rules ppl. Evlute: cos()d f()d = f()d = kf()d = k f()d f()d ( f() ± g())d = f()d = c f()d + WRitE c f()d ± g()d f()d, providing < c <. (e + )d. Antidifferentite the given function nd specif the end points for the clcultion using squre rckets. cos()d = sin() MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Sustitute the upper nd lower end points into the ntiderivtive nd clculte the difference etween the two vlues. = sin sin() = = Antidifferentite the given function nd specif the end points for the clcultion using squre rckets. Sustitute the upper nd lower end points into the ntiderivtive nd clculte the difference etween the two vlues. WOrKED EXAMPLE think (e + )d = e + = ( e + ()) ( e + ()) = e + + = e + 5 Sometimes, definite integrl questions tke more of theoreticl pproch to prolem solving. Even if the function is unknown, we cn use the properties of definite integrls to find the vlues of relted integrls. Given tht i iii f ()d = 8, find: f ()d ii f()d Find k if i Appl the definite integrl propert kf()d = k k f()d. ( + )d =. ii Appl the definite integrl propert: ( f() ± g())d = f()d ± g()d. WRitE i ii f()d = iv = 8 = 6 ( f() + )d = f()d ( f() + )d ( f() )d. f()d + d Topic 8 InTEgrATIOn

Integrte the second function nd evlute. = 8 + = 8 + ( ) = iii Appl the definite integrl propert: f()d = f()d iv Appl the definite integrl propert: f() ± g())d = f()d ± g()d iii iv f()d = = 8 ( f() )d = f()d f()d Integrte the second function nd evlute. = 8 Antidifferentite nd sustitute the vlues of nd k. = k ( + )d d = 8 () () = 8 9 = 8 = = + k = k + k () + () Simplif nd solve for k. = k + k 5 = k + k 5 = (k + 5)(k ) k = 5 or k = Write the nswer. k =, 5 Interctivit The fundmentl theorem of integrl clculus int 6 The fundmentl theorem of integrl clculus In this section, the vrile t is used nd the function f is defined s continuous function on the intervl [, ] where [, ]. A() is defined s A() = f(t)dt where A() is the re etween the curve = f() nd the t-is from t = to t =. A( + δ) represents the re etween the curve = f() nd the t-is from t = to t = + δ. MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Units & AOS Topic Concept 5 The fundmentl theorem of clculus Concept summr Prctice questions (, f()) The difference etween the res is therefore A( + δ) A(). As we hve lred seen, this re lies etween the res of the left end-point rectngle nd the right end-point rectngle. = f() Left end-point rectngle method + δ t + δ t Therefore, f()δ A( + δ) A() f( + δ)δ. If we divide δ, we hve A( + δ) A() f() f( + δ). δ Of course, the width of ech rectngulr strip ecomes smller nd smller s δ, which results in n incresingl ccurte re clcultion etween the curve nd the t-is. This concept is limiting sitution. B definition, lim δ nd s δ, then f( + δ) f(). A( + δ) A() = d δ d (A()) Consequentl, we cn s tht d (A()) = f(). d If we then integrte oth sides with respect to, we hve d (A())d = f()d or A() = f()d. d To further investigte this theorem, we will let F e n ntiderivtive of f, nd A e the specil ntiderivtive defined s A() = f() Are etween the curve, the t-is nd the lines t = nd t = = f() Actul re under the curve t A( + δ) A() f(t)dt. Are etween the curve, the t-is nd the lines t = nd t = + δ = f() A( + δ) + δ = f() (, f( + δ)) + δ Right end-point rectngle method t t Topic 8 Integrtion 5

Eercise 8. PRctise Work without CAS From our knowledge of ntidifferentition Therefore, If we let =, then Therefore, If we now let =, then A() F() = c where c is numer. f(t)dt F() = c. f(t)dt =, so F() = c or F() = c. f(t)dt F() = F() f(t)dt F() = F() or f(t)dt = F() F(). It is customr tht F() F() is represented F(). Therefore, f (t)dt = F() = F() F() Finding the re etween the curve nd the -is from to is the sme s finding the definite integrl with the end-point terminls nd. The fundmentl theorem of integrl clculus WE The left end-point rectngle method nd the right end-point rectngle method re shown for the clcultion of the pproimte re etween the curve f() =, >, nd the -is from =.5 to =.5. = = =.5.5.5 = =.5.5.5 = 6 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Consolidte Appl the most pproprite mthemticl processes nd tools Clculte the pproimte re under the curve: using the left end-point rectngle rule using the right end-point rectngle rule. Consider the function defined the rule f : R R, f() =. ( 5)( + 5),. The grph of the function is shown. Use the left end-point rule with rectngles unit wide to pproimte the re ound the curve nd the -is. WE Evlute: Evlute: ( + + + )d (cos() + sin())d. ( + ) d 5 WE Given tht c 5 5 m()d 6 Find k if (m() + )d k 5 m()d = 7 nd ( + )d =. 5 n()d =, find: d 5 5 (e + e ) d. (m() )d (m() + n() )d. 7 The grph of f : [, ] R, f() = ( ) + 9 is shown. (, 8) = ( ) + 9 (, ) (, 8) =. ( 5)( + 5) (5, ) 5 6 = ( ) + 9 (, ) (, ) Topic 8 Integrtion 7

Use the left end point rule with rectngles unit wide to estimte the re etween the curve nd the -is from = to =. Use the right end point rule with rectngles unit wide to estimte the re etween the curve nd the -is from = to =. 8 The grph of the function f() = is shown. (, ) f() = (, ) 5 (, ) 5 Estimte the re ound the curve nd the -is using the right end point method nd using rectngles of width. 9 The grph of f() = ( ) for [, ] is shown. The grph intersects the -is t the point (, ) s shown. Find the vlue of the constnt. Use oth the left end point nd the right end point rules to determine the pproimte re etween the curve nd the -is from = to =. Use rectngle width of nd give our nswers correct to deciml plces. Evlute the following. d ( + )d sin d Given tht 5 f()d = 7.5 nd e 5 5 + d c f() = ( ) (, ) (e e )d d f cos() sin d. g()d =.5, find: 5 f()d g()d c ( f () + )d 5 5 e d (g() + f())d 5 (8g() f())d f 5 g()d + g()d. 8 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Mster Determine h if Determine if h d = 5. e d = e 8. The grph of the function f : R R, f() = 8 + is shown (, ) f() = 8 + The grph cuts the -is t the point (, ). Find the vlue of the constnt. Evlute 5 ( 8 + )d 5 If = sin(), find d d. Hence, find the vlue of 6 If = e +, find d d. 7 If Hence, find the vlue of k 8 Find cos()d. ( )e d. ( )d = 7 5, find k, given k >. + e e (e + ) d, correct to deciml plces. Topic 8 Integrtion 9

8. Units & AOS Topic Concept 9 Are under curve Concept summr Prctice questions Interctivit Are under curve int 5966 Ares under curves If we re interested in the re etween curve tht is continuous function, = f(), nd the -is etween = nd =, then the following grph shows us ectl wht we require. As we hve lred seen, this re cn e pproimted dividing it into series of thin verticl strips or rectngles. The pproimte vlue of the re is the sum of the res of ll the rectngles, whether the re left end-point or right end-point rectngles. Suppose A represents the sum of the res of ll the rectngulr strips etween = nd =, where ech strip hs width of δ. Providing there is ver lrge numer of rectngulr strips so tht δ is etremel smll, where A = lim = δ = = = δ () mens the sum from = to =. Also, since ech strip cn hve its re defined s δa δ, = δ = δa δ. Therefore, if the re under the curve is divided into ver lrge numer of strips, then δa lim δ δ = This leds to the sttement tht da d = A = d. = = f() = = f() But since = nd = re the oundr points or end points, then A = d () MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Equting () nd (), we hve = A = lim δ = d. δ = WOrKED EXAMPLE This sttement llows us to clculte the re etween curve nd the -is from = to =. Consider the function defined the rule f : [, ] R, f() = ( ) + 9. A = ( ( ) + 9)d ( ) = + 9 = () + 9() ( ) + 9() = 9 + 6 + = 7 = 6 units (, 8) f() = ( ) + 9 Determine the re ound the curve defined the rule = e + nd the -is from = to =. think Sketch the grph of the given function nd shde the required re. WRitE/dRAW = e + = (, ) = (, ) Write the integrl needed to find the re. A = (e + )d Topic 8 InTEgrATIOn

Antidifferentite the function nd evlute. A = e + = ( e + ()) ( e + ()) = e + 9 + = e + Write the nswer. The re is e + squre units. Signed res When we clculte the re etween grph = f() nd the -is from = to = using the definite integrl or negtive. f()d, the re cn either e positive Consider the function defined the rule f : R R, f() = ( )( ), which is shown. If we evlute ( )( + )d = (6 + )d = + = ( ) + ( ) ( ) = 8 = 8 8 = 6 This re is negtive ecuse the region lies elow the -is. Wheres ( )( + )d f() = ( )( + ) (, ) (, ) (, ) = (6 + )d = + MATHS QUEST MATHEMATICAL METHODS VCE Units nd

= () + () () = 7 + 9 8 = 6 8 = 6 This re is positive s the region lies ove the -is. If we wnt n ccurte nswer for the re ound the curve from = to =, we counterct the negtive region sutrcting it from the positive region. B sutrcting the negtive re, we re ctull dding the re. A = ( )( + )d = 6 6 = 5 = units ( )( + )d The totl re ound the curve, the -is nd the lines = nd = is squre units. This confirms the theor tht if f() >, then the region ove the -is hs positive re, ut if f() <, then the region elow the -is hs negtive re. Hd we not roken up the intervl nd clculted would hve een (6 + )d (6 + )d, the result = + = () + () () ( ) + ( ) ( ) = 6 6 = 5 This result would not hve given us the required re. The vlue of 5 is the vlue of the definite integrl, ut not the re under the curve. This shows tht it is impertive to hve picture of the function to determine when f() > nd when f() < ; otherwise, we re just evluting the definite integrl rther thn finding the necessr re. Topic 8 Integrtion

The totl re etween the function = f() nd the -is from = to = is given c WOrKED EXAMPLE 5 think A totl = c f()d = A A The other method to ccount for the negtive re is to switch the terminls within the integrl for the negtive region. So, A totl = c f()d + = A + A c f()d f()d Find the re ound the curve = ( )( ) nd the -is from = to =. Mke creful sketch of the given function. Shde the required region. WRitE/dRAW The grph cuts the -is where =. the -intercept is (, ). The grph cuts the -is where = : ( )( ) = ( )( + )( )( + ) = = ±, = ± (, ) (, ) (, ) (, ) (, ) = ( )( ) A c A = f() MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Epress the re using definite integrls. Account for the negtive regions sutrcting these from the positive res. Note tht the region from = to = is the sme s the region from = to = due to the smmetr of the grph. A = = ( )( )d ( 5 + 5)d ( )( )d ( 5 + 5)d Antidifferentite nd evlute. = 5 5 5 + 5 5 5 5 + 5 = 5 ()5 5 () + 5() 5 ( )5 5 ( ) + 5( ) 5 ()5 5 () + 5() 5 ()5 5 () + 5() = 5 5 + 5 5 + 5 5 5 + 5 5 + 5 = 5 5 + 5 + 5 5 + 5 = 6 + 5 = + = 8 Write the nswer. The re is 8 units. Eercise 8. PRctise Work without CAS Consolidte Appl the most pproprite mthemticl processes nd tools Ares under curves 5 + 8 5 + + 5 + 5 + 5 5 WE Determine the re ound the curve defined the rule =, nd the -is from = to = 5. Find the re ounded the curve = sin() +, the -is nd the lines = nd =. WE5 Sketch the grph of = e nd hence find the ect re etween the curve nd the -is from = to =. Sketch the grph of = nd hence find the re etween the curve nd the -is from = 8 to = 8. = 5 The grph of =, < is shown. Find the re of the shded region (i.e. for.5.5)..5.5.5 Topic 8 Integrtion 5

6 Consider the function defined the rule f : R \{} R, f() =. Sketch the grph of f for >. Using clculus, find the re enclosed the function, the lines = nd =, nd the -is. 7 The grph of = sin() + cos() for is shown. The grph intersects the is t (m, ). Find the vlue of the constnt m, correct to deciml plces. Find m ( sin() + cos())d, correct to deciml plces. 8 The grph of = e is shown. Find d d e. Hence, find the ect re etween the curve = e nd the -is from = to =. 9 Consider the function f : R R, f() = ( )( 9). Sketch the grph of f, showing the is intercepts nd turning points. Using clculus, find the re enclosed the function, the lines = nd =, nd the -is, correct to deciml plces. The grph of the function = sin() + cos() is shown. Using clculus, find the re etween the curve nd the -is from = to =. (, ).5.5 = sin() + cos() (, ) 5 7 = sin() + cos() (, ) (m, ) = e (, ).5.5 6 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

The grph of =.5( + )( + )( )( ) is shown. (, ) =.5( + )( + )( )( ) (, ) (c, ) (d, ) (, 6) The grph intersects the -is t (, ), (, ), (c, ) nd (d, ). Find the vlues of the constnts,, c nd d. Find the re etween the curve nd the -is from = to = d, correct to deciml plces. The Octgon Digitl store on the corner of two min rods in the north-estern suurs of lrge Austrlin cit hs two ver distinctive prolic windows, ech one fcing one of the min rods. In the erl hours of Sund morning, motorist smshed through one of the windows. The owner decided it would e eneficil to replce oth windows with strongl reinforced nd quite hevil tinted glss. Ech window hs the dimensions shown in the digrm. Find the eqution of the prol tht defines the shpe of ech window. Find the re of glss required to replce ech window. c If the cost per squre metre of the replcement reinforced nd tinted glss is $55, find the cost of replcing the two windows. The grph of f : R R, f() = is shown. Find the re ounded the curve nd the -is from = to =. Hence, or otherwise, find the re of the shded region. The grph of = sin(), is shown. Clculte sin()d. Hence, or otherwise, find the re of the shded region. metres metres f() = = sin() Topic 8 Integrtion 7

Mster 5 The Red Fish Resturnt is new resturnt out to open. The owners commissioned grphic rtist to design logo tht will e seen on the menus nd on dvertisements for the resturnt, nd will lso e etched into the front window of the resturnt. The logo is shown in Figure. As the logo is to pper in numer of different scenrios, the owners need to know the re of the originl to llow for enlrgement or diminishing processes. Figure The grphic rtist formed the shpe using the rule = ( ), for the upper prt of the fish nd = ( ), for the lower prt of the fish. The originl outline is shown in Figure. Clculte the re etween the upper curve nd the -is from = to =. Clculte the re of the entire fish logo. (All mesurements re in centimetres.) Figure Give our nswer correct to deciml plce. c The etched fish on the front window of the resturnt hs n re of.875 m. Wht ws the scle fctor, to the nerest integer, used to enlrge the fish motif? 6 The grph of = e is shown. Find the re etween the curve nd the -is from = to =, giving our nswer correct to deciml plces. The grph of the function (, ) = e = + is shown. + Clculte the re, correct to deciml plces, etween the curve nd the -is from = to =. (, ) (, ) (, ) (, ) (, ) = = + + (, ) 8 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

8. Units & AOS Topic Concept Ares etween curves Concept summr Prctice questions Interctivit Ares etween curves int 65 WOrKED EXAMPLE 6 Applictions The definite integrl is etremel useful in solving lrge vriet of pplied prolems including the verge vlue or men vlue of function, the re etween curves, totl chnge s the integrl of instntneous chnge, nd kinemtic prolems. Ares etween curves Consider the functions f nd g, which re oth continuous on the = f() intervl [, e]. Sometimes f > g nd on other occsions f < g. It is solutel criticl to know when f > g or f < g, = g() so grphic representtion of the c d e sitution is essentil, prticulrl to show the points of intersection of the grphs. We cn find the re etween the curves, providing we tke ech section one t time. Within ech section, the re is found sutrcting the lower function from the higher function. As we re finding the re etween two curves, we don t need to worr out whether the region is ove or elow the -is. Are etween the curves: A = ( g() f()) d + c ( f() g()) d + d c ( g() f()) d + e d ( f() g()) d f < g f > g f < g f > g The functions f : R R, f() = ( + )( ) nd g : R R, g() = + re shown. The grphs intersect t (m, ) nd (p, q). Find the vlues of the constnts m, p nd q. Find the re ound the two curves. think Points of intersection re found solving the equtions simultneousl, so equte the equtions nd solve for. (m, ) WRitE + = ( + )( ) + = + = ( + )( ) = =, = (p, q) = g() = f() Topic 8 InTEgrATIOn 9

Find the corresponding -vlues. When =, = + =. When =, = + =. Stte the solution. m =, p =, q = Determine whether f > g or f < g. Epress the re in definite integrl nottion nd simplif the epression within the integrl. As f() = lies ove g() = +, f > g. A = = = ( f() g())d ( ( + ))d ( + )d Antidifferentite nd evlute. = + = () () + () ( ) ( ) + ( ) = + 8 + + = + 8 = Write the nswer. The re is.5 units. WOrKED EXAMPLE 7 The grphs of f() = sin() nd g() = cos() re shown for [, ]. Find the coordintes of the point(s) of intersection of f nd g for the intervl,. Using clculus, determine the re enclosed etween the curves on the intervl,. think WRitE Use simultneous equtions sin() = cos() to find where the grphs sin() intersect, nd equte the two cos() =, equtions. tn() =, (, ) = g() = f() Solve for for,. = = 8 5 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Find the corresponding -vlue. f 8 = sin = Write the solution. The coordintes re Determine when f > g nd f < g. Epress ech re individull in definite integrl nottion. Use clculus to ntidifferentite nd evlute. When < < 8, g > f. When 8 < <, f > g. The re is equl to: A = 8 8,. ( cos() sin())d + = sin() + cos() 8 = sin + cos 8 ( sin() cos())d + cos() sin() 8 sin() + cos() + cos() sin() cos sin = + + + + = + + + = Write the nswer. The re is squre units. Units & AOS Topic Concept 8 Averge vlue of function Concept summr Prctice questions Interctivit Averge vlue of function int 6 The verge or men vlue of function The verge or men vlue of function f() over the intervl [, ] is given Averge = f()d. Geometricll, the verge vlue of function is the height of rectngle, vg, with width of ( ), tht hs the sme re s the re under the curve = f() for the intervl [, ]. = f() vg Topic 8 Integrtion 5

WOrKED EXAMPLE 8 Find the verge vlue for the function defined f() = sin() for the intervl 8, 8. think Write the rule for the verge or men vlue of function. Sustitute the pproprite vlues into the rule. WRitE Averge = = = 8 8 8 8 f()d 8 8 sin()d Antidifferentite nd evlute. = cos() 8 = cos 8 sin()d 8 + cos 8 = + = = = + Totl chnge s the integrl of instntneous chnge If we re given the eqution for the rte of chnge nd we wnt to find the mount tht hs chnged over prticulr time period, we would integrte the rte of chnge eqution using the strting nd finishing times s the terminls. For emple, if we know the rte of wter flowing, dv in L/min, nd we wnt to find the mount of dt dv liquid tht hs flowed in the first minutes, we would evlute dt dt. 5 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

WOrKED EXAMPLE 9 think It is common prctice to include heting in concrete sls when new residentil homes or units re eing constructed, ecuse it is more economicl thn instlling heting lter. A tpicl reinforced concrete sl, 5 centimetres thick, hs tuing instlled on top of the reinforcement, then concrete is poured on top. When the sstem is complete, hot wter runs through the tuing. The concrete sl sors the het from the wter nd releses it into the re ove. The numer of litres/minute of wter flowing through the tuing over t minutes cn e modelled the rule dv t = cos dt The grph of this function is shown. dv dt (, 8) dv dt = 8 t t [cos( ) + sin( 9 ) + ] + sin t 9 6 8 +. Wht is the rte of flow of wter, correct to deciml plces, t: i minutes ii 8 minutes? Stte the period of the given function. c Find the volume of wter tht flows through the tuing during the time period for one whole ccle. i Sustitute t = into the given eqution nd evlute. ii Sustitute t = 8 into the given eqution nd evlute. WRitE t i dv t = cos + sin t dt 9 + When t =, dv dt = cos + sin 9 + = 6.97 The rte t minutes is 6.97 litres/minute. ii When t = 8, dv dt = cos 8 + sin 8 9 + = 5.68 The rte t 8 minutes is 5.68 litres/minute. Topic 8 InTEgrATIOn 5

Determine the ccle for the function nlsing the shpe of the grph. A complete ccle for the function occurs etween t = 6 nd t =, so the period is 6 = 8 minutes. c The re under the curve of the eqution of the rte of flow gives the totl volume tht hs flowed through the tuing. c A = = cos t 6 6 cos t = t sin + sin t 9 + sin t 9 9 t cos 9 Antidifferentite nd evlute. = sin(8) 9 cos 8 sin() 9 cos + dt + dt = 9 cos + 7 + 9 cos 8 = 5 = 8 Write the nswer. The volume of wter tht psses through the tuing during one ccle is 8 litres. WOrKED EXAMPLE Kinemtics You re lred wre of the reltionships etween displcement, velocit nd ccelertion. However, our knowledge out the definite integrl nd the re under curves now gives us dditionl skills for the clcultion of fcts relted to kinemtics. + t + 7 + 8 A prticle strting from rest ccelertes ccording to the rule = t( t). Find reltionship etween the velocit of the prticle, v metres/second, nd the time, t seconds. Find the displcement of the prticle fter seconds. c Sketch the grph of velocit versus time for the first seconds of the motion. d Clculte the distnce trvelled the prticle in the first seconds. d dt dv dt 6 Displcement (t) Velocit v(t) v(t)dt (t)dt Accelertion (t) 5 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

THINK Antidifferentite the ccelertion eqution to find the velocit eqution. WRITE/DRAW v = (t)dt = (t( t))dt Appl the initil conditions to find v in terms of t. Integrte v etween t = nd t =. As we re finding displcement, there is no need to sketch the grph. = (6t t )dt = t t + c When t =, v =, so c =. v = t t = (t t )dt = t t = () () = Write the nswer. After seconds the displcement is zero. c Sketch grph of v versus t. c -intercept: (, ) v t-intercepts: (, ) (, ) = t t = t ( t) t =, v = t t When t =, v = = 6. (, 6) d The re under the curve of velocit-time grph gives the distnce covered. Set up the integrls nd sutrct the negtive region. d D = (t t )dt (t t )dt Antidifferentite nd evlute. = t t t t = = 7 8 = 5 6 =.5 6 + 6 + 7 8 t Write the nswer. The distnce trvelled the prticle in seconds is.5 metres. Topic 8 Integrtion 55

Eercise 8. PRctise Work without CAS Questions, 5, 6, 8, 9 Applictions WE6 The grphs of g() = nd the line f() = re shown. Find the coordintes of the point of intersection etween f nd g, nd hence find the re of the shded region. f() = g() = Find the re enclosed etween the curve f() = ( ) nd the line g() = 9. WE7 Using clculus, find the re enclosed etween the curves f() = sin() nd g() = cos() from = to =. The grphs of f() = e nd g() = re shown. Using clculus, find the re enclosed etween = f(), = g() nd the lines (, ) = nd = 5. Give our nswer correct to g() = deciml plces. 5 WE8 Find the verge vlue or men vlue of the function defined the rule f() = e for,. f() = e (.9,.8) 5 6 Find the verge vlue or men vlue of the function defined the rule f() = for [.5, ]. 7 WE9 The verge rte of increse, in cm/ month, in the length of o from irth until ge 6 months is given the rule dl dt = t where t is the time in months since irth nd L is the length in centimetres. Find the verge totl increse in length of o from 6 months of ge until 6 months of ge. Give our nswer correct to deciml plce. 8 A numer of pprentice ricklers re competing in competition in which the re required to uild fence. The competitors must produce fence 56 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

Consolidte Appl the most pproprite mthemticl processes nd tools tht is stright, netl constructed nd level. The winner will lso e judged on how mn ricks the hve lid during -minute period. The winner lid ricks t rte defined the rule dn dt =.8t + where N is the numer of ricks lid fter t minutes. Sketch the grph of the given function for t. Shde the region defined t. c How mn ricks in totl did the winner l in the -minute period defined t? 9 WE A prticle moves in line so tht its velocit v metres/second from fied point, O, is defined v = + t +, where t is the time in seconds. Find the initil velocit of the prticle. Wht is the ccelertion of the prticle when: i t = ii t = 8? c Sketch the grph of v versus t for the first seconds. d Find the distnce covered the prticle in the first 8 seconds. An oject trvels in line so tht its velocit, v metres/second, t time t seconds is given v = cos t, t. Initill the oject is metres from the origin. Find the reltionship etween the displcement of the oject, metres, nd time, t seconds. Wht is the displcement of the oject when time is equl to seconds? c Sketch the grph of v versus t for t. d Find the distnce trvelled the oject fter seconds. Give our nswer in metres, correct to deciml plces. e Find reltionship etween the ccelertion of the oject, metres/second, nd time, t seconds. f Wht is the ccelertion of the oject when t = seconds? The grphs of f() = nd g() = + re shown. f() = The grphs intersect t the points (, ) nd (c, ). Find the constnts, nd c. Find the re enclosed etween the curves from = to = c. c Find the verge vlue or men vlue of the function f() = (, ) for [,.5]. Give our nswer correct to (c, ) deciml plces. g() = + Topic 8 Integrtion 57

The grphs of f() =.5e nd g() = cos() re shown. The grphs intersect t (.5,.) nd (.5,.86). Using clculus, find the re enclosed etween the curves from =.5 to =.5. Give our nswer correct to deciml plces. g() = cos() (,.5) (, ) (.5,.86) (, ) (.5,.) f() =.5e The grphs of = nd = re shown. Find the points of intersection of the two grphs. Find the lue shded re. c Find the pink shded re. d Hence, find the re enclosed etween the two grphs tht is represented the lue nd pink shded regions. The grphs of =.5 ( )( + ) nd =.5 ( )( + ) re shown. =.5 ( )( + ) (, ) (, ) (, ) =.5 ( )( + ) Using clculus, find the re of the region enclosed etween the curves nd the lines = nd =. 5 Sketch the grphs of =.5( + )( )( ) nd = ( )( + ) on the one set of es. Show tht the three coordinte pirs of the points of intersection of the two grphs re (, ), (, ) nd (, ). c Find the re, correct to deciml plces, enclosed etween the curves from = to =. 6 The edge of grden ed cn e modelled the rule =.5 sin +. (, ) = (, ) = 58 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

The ed hs edges defined =, = nd =. All mesurements re in metres. Sketch the grph of =.5 sin + long with =, = nd = s edges to show the shpe of the grden ed. Clculte the re of the grden ed, correct to the nerest squre metre. c Topsoil is going to e used on the grden ed in preprtion for new plnting for spring. The topsoil is to e spred so tht it is uniforml 5 cm thick. Find the mount of soil, in cuic metres, tht will e needed for the grden ed. 7 A prticle moves in stright line. At time v t seconds its velocit, v metres per second, is defined the rule v = e.5t.5, t. The (,.5) grph of the motion is shown. ( log e (), ) Find the ccelertion of the prticle, m/s, in terms of t. v = e.5t.5 Find the displcement of the prticle, m, if = when t =. c Find the displcement of the prticle fter seconds. d Find the distnce covered the prticle in the fourth second. Give our nswer correct to deciml plces. 8 A stone footridge over creek is shown long with the mthemticl profile of the ridge. 7 5 6 5 5 6 The rch of the footridge cn e modelled qudrtic function for [ 5, 5], with ll mesurements in metres. Find the eqution for the rch of the ridge Find the re etween the curve nd the -is from = 5 to = 5. c Find the re of the side of the ridge represented the shded re. d The width of the footridge is metres. Find the volume of stones used in the construction of the footridge. 9 The rte of growth of moile phone suscriers with prticulr compn in the UK cn e modelled the rule dn dt =.85e.t t Topic 8 Integrtion 59

Mster where N million is the numer of suscriers with the compn since 998 nd t is the numer of ers since 998, the er the compn ws estlished. Find how mn millions of moile phone suscriers hve joined the compn etween 998 nd 5, correct to deciml plce. The mintennce costs for cr increse s the cr gets older. It hs een suggested tht the increse in mintennce costs of dollrs per er could e modelled dc dt = 5t + 5 where t is the ge of the cr in ers nd C is the totl ccumulted cost of mintennce for t ers. Sketch the grph of the given function for t. Find the totl ccumulted cost of mintennce for t = 5 to t = ers. Consider the functions f() = sin () nd g() = cos (). Sketch the grphs on the sme set of es for. Find the re etween the curves for. The cross-section of wterw is prolic. Its depth is metres, nd the width cross the top of the wterw is metres. When the wterw is one-third full, wht is the depth of the wter in metres, to deciml plces? 6 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

ONLINE ONLY 8.5 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. A 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. Topic 8 InTEgrATIOn 6

8 Answers Eercise 8. 5 units 77 6 units units 65 +.5e.5e 5 c 6 d 8 6 k = ± 7 units units 8 units 9 = Left end point: 7.56 units Right end point: 7.56 units 7 55 6 d e c ( ) f 5.5 c.5 d e 5 f.5 h = 5 9 = = units 5 6 7 k = 5 8.96 d = cos() + sin() + d d d = ( )e (e ) Eercise 8. 66 units units = e units 5.6 units 6 = = units 7 m =.6779.6 units 8 9 d = (, ) 8 8 d = e (e ) units (.6, 6) 6 (.6, 6) (, ) (, ) (, ) (, ) (, 9) (, ) e + e units f() = ( )( 9) 5.7 units 6 units =, =, c =, d = 5.68 units = + 6 metres c $ 6 MATHS QUEST MATHEMATICAL METHODS VCE Units nd

units units ( ) units 5 7.77 cm 5.5 cm c Scle fctor = 6.76 units 9.9 units Eercise 8. Point of intersection = (6, ), re = units Are = 5 6 units units.65 units 5 e 6 7 8. cm 8 nd dn dt 6 (, ) c ricks dn dt =.8t + 9 m/s = t + c (, 6) i.5 m/s ii.5 m/s v d 6 m (, ) = + + = 6 sin t m t (,.95) t c (, v d.9 m e = sin t f m/s =, =, c = 9.6 units c 6..76 units ( t v = cos, ( ( ( ( ( 7, (, ), (, ) 7 6 units c units d.5 units 8.6 units 5 =.5( + )( )( ) = ( )( + ) (, ) ( (, t =.5( + )( )( ) () = ( )( + ) () () = ().5( + )( )( ) = ( )( + ).5( + )( )( ) ( )( + ) =.5( + )( )( ) + ( )( + ) = ( )( + )(.5( ) + ) = ( )( + )(.5.5 + ) = ( )( + )(.5 +.5) = =, + = or.5 +.5 = = = = When =, = ( + )( + ) =. When =, = ( + )( + ) =. When =, = ( )( + ) =. Therefore, the coordintes re (, ), (, ) nd (, ). c 9. units (, 6) (, ) (, ) (, ) ( Topic 8 Integrtion 6

6 v dc dt (, ) =.5 sin ( ) + (, ) 75 (, 75) 5 m c.5 m 7 =.5e.5t = e.5t.5t + c.77 m d. m 8 = 5. m c 5 m d 5 m 9 55. million t (, 5) $565 (, ) units. m.5 dc dt = 5t + 5 t = cos () = sin () (, ) 6 MATHS QUEST MATHEMATICAL METHODS VCE Units nd