y = f(x) $ # Area & " % ! $ b f(x) g(x) dx = [F (x) G(x)] b a

MthsTrck (NOTE Feb 23: This is the old version of MthsTrck. New books will be creted during 23 nd 24) Topic 9 Module 9 Introduction Integrtion to Mtrices y = f(x) Income = Tickets! Price = =! 25 $! 25 3 35$ # Are &# & " 35 5% " 2 5 %! 8,25 9, 9,75b $ # & ",75 y = 2,75 g(x) 3,75% x Are = b f(x) g(x) dx = [F (x) G(x)] b MATHEMATICS LEARNING SERVICE Centre for Lerning nd Professionl Development MATHS LEARNING CENTRE Level, Schulz Building (G3 on cmpus mp) TEL 833 5862 FAX 833 3553 mls@delide.edu.u www.delide.edu.u/clpd/mths/ Level 3, Hub Centrl, North Terrce Cmpus, The University of Adelide TEL 833 5862 FAX 833 734 mthslerning@delide.edu.u www.delide.edu.u/mthslerning/

This Topic... The topic hs 2 chpters: Chpter introduces integrtion which together with differentition re prts of the brnch of mthemtics clled Clculus. The chpter begins by sking the question: Given the rte of chnge of quntity, how cn we find the quntity? This question is relted to the problem of finding the re of the region between curve nd the horizontl xis. Upper nd lower rectngles re used to pproximte this region. The definite integrl is introduced together with its properties. Chpter 2 introduces the Fundmentl Theorem of Clculus. This centrl theorem links integrtion to differentition nd enbles integrls to be evluted by reverse differentition. Antiderivtives nd indefinite integrls re introduced. Stndrd integrls re used to used to integrte more complex functions. The substitution method is used to simplify the integrtion of composite functions. Selected pplictions include clcultion of the exct re between two curves nd of net chnge in quntities. The topic uses the stndrd derivtives nd methods of differentition introduced in Topic 6. Auhor: Dr Pul Andrew Jnury 28 Edited by: Geoff Cotes Printed: Februry 24, 23 i

Contents Integrls. Introduction.................................2 Are under curve............................ 6.3 The definite integrl............................4 Properties of the definite integrl.................... 4.4. Additive Properties........................ 4.4.2 Liner Properties......................... 5 2 Integrtion 7 2. Fundmentl Theorem of Clculus................... 7 2.2 Integrtion................................. 2 2.2. Antiderivtives nd indefinite integrls............. 2 2.2.2 Methods.............................. 23 2.2.3 Selected Applictions....................... 36 A Summtion Nottion 47 B Justifiction for the Fundmentl Theorem 49 C Answers 52 ii

Chpter Integrls. Introduction Differentition ws concerned with the question: Given quntity, how cn we find its rte of chnge? If we know the volume of wter entering dm s function of time, we cn use differentition to find the rte t which the dm is filled. This topic introduces integrtion, which is concerned with the question: Given the rte of chnge of quntity, how cn we find the quntity? If we know the rte t which dm is filled, we use integrtion to find the volume of wter entering the dm s function of time. Differentition nd integrtion re prts of Clculus. i constnt flow Exmple A dm is filled from creek with constnt flow of 2 litres/min. This is shown on the flow-time grph below. 2, flow (litres/min) time (min) i Clculus is brnch of mthemtics tht includes the study of limits, differentition, integrtion nd infinite series, nd hs widespred pplictions in science nd engineering. The word clculus ws introduced in the mid 7th century from Ltin, nd mens smll stone used for counting.

2 CHAPTER. INTEGRALS As the flow is constnt, the volume of wter flowing into the dm fter t minutes is: volume = flow time = 2 t = 2 t litres You cn see tht the volume of wter entering the dm is the re under the flow-time grph from to t minutes: 2, flow (litres/min) t time (min) chnging flow Exmple Suppose insted tht the flow of wter entering the dm chnged t the rte 2 t litres/min fter t minutes. The flow-time grph would then be: flow (litres/min) 2, 2 time (min) We cn estimte the volume of wter entering the dm in 2 minutes by subdividing the intervl [, 2] on the x-xis into four equl prts: flow (litres/min) 2, 5 5 2 time (min)

.. INTRODUCTION 3 You cn see tht the re of the upper rectngle bove intervl [, 5 ] height width = 2 5 =, is n upper estimte of the volume of wter flowing into the dm for t 5, nd tht the re of the lower rectngle height width = (2 5) 5 = 75, is lower estimte of the volume of wter flowing into the dm for t 5. The sum of the res of the four upper rectngles gives n upper estimte of the totl volume wter flowing into the dm for the 2 minutes: 2 5 + (2 5) 5 + (2 ) 5 + (2 5) 5 = 25, litres nd the sum of the res of the lower rectngles give lower estimte of the totl volume: (2 5) 5 + (2 ) 5 + (2 5) 5 + = 5, litres... the volume of wter entering the dm is between 5, nd 25, litres. This estimte cn be improved by tking finer subdivisions of intervl [, 2 ]. flow (litres/min) 2, 5 5 2 time (min) With eight subdivisions, the upper nd lower estimtes re 7, 5 nd 22, 5. As smller subdivisions re tken, you cn see tht the sum of the res of the upper nd lower rectngles become closer to ech other nd tht both sums become very close to the re under the flow-time curve between t = nd t = 2.

4 CHAPTER. INTEGRALS flow (litres/min) 2, 2 time (min)... in this cse the totl volume of wter entering the dm (ie. the shded re under the flow-time grph) is esy to find: 2 2 = 2, litres. 2 rte of chnge totl chnge Exmple The resoning in the previous exmple cn be used to show tht if wter flows into the dm t the rte of 2 t litres/min, then the chnge in the volume of wter in the dm fter t minutes is equl to the re under the flow-time grph from to t, when t 2. flow (litres/min) 2, (t, 2 t) t 2 time (min) This re cn be found by dding the res of the rectngle nd tringle below. t 2, t { { } {{ } t... the volume of wter in the dm increses by fter t minutes. ii t (2 t) + 2 t t = 2t 5t2 litres ii Differentiting this result confirms tht the rte of chnge of in volume of wter in the dm is dv dt = 2 t.

.. INTRODUCTION 5 Exercise.. Rinwter flowed into, litre tnk t constnt rte of litres/min. () Drw grph of the constnt flow f(t) into the tnk for t minutes. (b) Clculte the totl volume of the wter flowing into the tnk fter t minutes. (c) Drw grph of the volume V (t) of wter in the tnk for t 6, if the tnk contined litres before the rin begn. (d) Interprete the grdient of the line in (c). 2. At time t =, cr begn trvelling up hill with velocity iii of where t is mesured in seconds. v(t) = 3.5t m/s, () Drw grph of the velocity v(t) of the cr for t 6 seconds. (b) Clculte the distnce trvelled by the cr fter 6 seconds. (c) Clculte the distnce s(t) trvelled by the cr fter t seconds. 3. Sketch the prbol y = x 2 for x. Estimte the re between the prbol nd the x-xis for x by: i. subdividing the intervl [, ] into five equl prts. ii. constructing upper nd lower rectngles on ech subintervl to obtin upper nd lower estimtes of the re under the prbol nd bove ech subintervl. iii. summing of the res of the upper nd lower rectngles. iii velocity = rte of chnge of distnce with time.

6 CHAPTER. INTEGRALS.2 Are under curve In the previous section we discovered tht the nswer to the question: Given the rte of chnge of quntity, how cn we find the quntity? ws found by exmining the re under the rte of chnge grph. We cn clculte the exct re under some curves, but for others we cnnot nd in these cses we need to estimte the re. iv If f(x) is positive continuous function on [, b ], then we cn estimte the re between the curve nd the x-xis from x = to x = b by using upper nd lower rectngles. upper rectngles lower rectngles Exmple Consider the function f(x) = x 2 nd the region between the grph of f(x) nd the x-xis, bounded by the verticl lines x = nd x = 4. 25 2 3 4 5 We cn estimte the re A of this region by subdividing the intervl [, 4 ] into three equl intervls of length, nd then using upper rectngles on ech subintervl to estimte the re under the curve. Upper rectngles re rectngles with height equl to the mximum vlue of function on subintervl. 25 2 3 4 5 iv This will be discussed in more detil lter.

.2. AREA UNDER A CURVE 7 The sum A U of the res of the upper rectngles A U = f(2) + f(3) + f(4) = 4 + 9 + 6 = 29 is n upper estimte of re A. We cn find lower estimte for A by using lower rectngles. Lower rectngles re rectngles with height equl to the minimum vlue of function on subintervl. 25 2 3 4 5 The sum A L of the res of the lower rectngles A L = f() + f(2) + f(3) = + 2 + 9 = 2 is n lower estimte of re A. This shows tht re A is between 2 nd 29 unit 2. We cn get better estimte of A by tking smller subintervls. For exmple, if [, 4 ] ws divided into 6 equl prts of length.5, then the sum of the res of the upper rectngles would be: A U =.5 f(.5) +.5 f(2) +.5 f(2.5) +.5 f(3) +.5 f(3.5) +.5 f(4) =.5.5 2 +.5 2 2 +.5 2.5 2 +.5 3 2 +.5 3.5 2 +.5 4 2 = 24.875, nd the sum of the res of the lower rectngles would be: A L =.5 f() +.5 f(.5) +.5 f(2) +.5 f(2.5) +.5 f(3) +.5 f(3.5) =.5 2 +.5.5 2 +.5 2 2 +.5 2.5 2 +.5 3 2 +.5 3.5 2 = 7.375.

8 CHAPTER. INTEGRALS... showing 7.375 A 24.875. As further subdivisions re tken, the difference between A U nd A L becomes smller nd ech become closer to the re A (= 2 unit 2 ). It is esy to estimte the difference A U A L when f(x) is either n incresing function or decresing function. A U A L Exmple (continued) As f(x) = x 2 is n incresing function, the digrm below shows tht the difference in res A U A L is equl to the re of rectngle with: width = width of subintervl height = height of lrgest upper rectngle height of smllest lower rectngle 25 2 3 4 5 This observtion cn be used to clculte how smll subintervls need be in order to estimte re A with predetermined precision. For exmple, if we wish to estimte A to within. unit 2, then we need to use subintervls of width w where w (6 ).

.2. AREA UNDER A CURVE 9 Exercise.2. Sketch the prbol y = x 2 for x. () Estimte the re between the prbol nd the x-xis for x by: i. subdividing the intervl [, ] into two equl prts. ii. constructing upper nd lower rectngles for ech subintervl. iii. summing the res of the upper nd lower rectngles. (b) How mny subintervls do you need to estimte the re to within. unit 2? 2. Sketch the prbol y = x 2 for x. Estimte the re between the prbol nd the x-xis by: i. subdividing the intervl [, ] into four equl prts. ii. constructing upper nd lower rectngles for ech subintervl. iii. summing the res of the upper nd lower rectngles.

CHAPTER. INTEGRALS.3 The definite integrl Suppose tht f(x) is positive continuous function for x b, nd tht the intervl [, b ] is divided into n equl prts by the points x, x,..., x n, with = x nd b = x n. The re A between the curve y = f(x) nd the x-xis from x = to x = b cn be estimted by constructing rectngles of heights f(x ), f(x ),..., f(x n ) on ech of the intervls [ x, x ], [ x, x 2 ],..., [ x n, x n ] s in the digrm below. y y = f(x)...... x x 2 x 3 x n 2 x n b x x x n The sum of the res of these rectngles is equl to v n f(x ) x + f(x ) x + f(x 2 ) x + + f(x n ) x = f(x i ) x where x = b n. As ech rectngle is between the upper nd lower rectngles on the sme subintervl, you cn see tht n f(x i ) x A s n. The limit is represented by i= n lim n i= f(x) x b f(x) dx which is red loud s the integrl from to b of f(x) dee x. vi v See Appendix A. vi The verb to integrte mens to form into one whole, nd integrl is the whole obtined fter integrtion. i=

.3. THE DEFINITE INTEGRAL In this form : the limit replces the with n elongted S, symbol., clled the integrl the x is replced by dx the vlues t the top nd bottom of the integrl symbol re the boundries of the region between the curve y = f(x) nd the x-xis. They re clled the upper nd lower limits of the integrl. This integrl is clled definite integrl s its upper nd lower limits re given. We will consider indefinite integrls in the next chpter. If f(x) is positive continuous function for x b, then the re between the curve y = f(x) nd the x-xis from x = to x = b is represented by the definite integrl: b f(x) dx Note: It is esier to work with sum like n f(x ) x + f(x ) x + f(x 2 ) x + + f(x n ) x = f(x i ) x, thn it is with sums of res of upper nd lower rectngles, s the heights of the rectngles hve cler pttern. We need to investigte the definite integrl further.... Suppose tht f(x) is negtive continuous function for x b, nd tht the intervl [, b ] is divided into n equl prts by the points x, x,..., x n, with = x nd b = x n. The re A between the curve y = f(x) nd the x-xis from x = to x = b cn be estimted using rectngles of heights f(x ), f(x ),..., f(x n ) on ech of the intervls [ x, x ], [ x, x 2 ],..., [ x n, x n ], s in the digrm below. i= y x x n x x 2 x 3 x n 2 x n b x height f(x )...... (x, f(x )) y = f(x)

2 CHAPTER. INTEGRALS The sum of the res of these rectngles is equl to n f(x ) x f(x ) x f(x 2 ) x f(x n ) x = f(x i ) x where x = b n. As n, you cn see tht If f(x) is negtive continuous function for x b, the integrl b n f(x) dx = lim f(x i ) x n is equl to the negtive of the re between the curve y = f(x) nd the x-xis from x = to x = b. i= i=... nd wht if f(x) is continuous function on [, b ] with sections bove nd below the x-xis? How should we interpret the integrl b f(x) dx in this cse? Once more we subdivide the intervl [, b ] into n equl prts using the points x, x,..., x n, with = x nd b = x n, nd then consider the limit ( b ) f(x i ) x = f(x) dx. n lim n i= y........ b x Using the digrm bove s guide, you cn see tht s n, the re of ech rectngle, nd n f(x i ) x (re below curve nd bove x-xis minus i= re bove curve nd below x-xis)

.3. THE DEFINITE INTEGRAL 3 In generl, If f(x) is continuous function for x b, then b is equl to the difference between n f(x) dx = lim f(x i ) x n i= () (b) the sum of the res under f(x) nd bove the x-xis nd the sum of the res bove f(x) nd below the x-xis for x b. Exercise.3. It is known tht π/2 sin(x) dx = Use the grph of sin x below to evlute () (b) (c) π π 3π/2 sin(x) dx sin(x) dx sin(x) dx y π π 2π x 2. Drw n pproprite grph nd use it to evlute π ( + sin(x)) dx

4 CHAPTER. INTEGRALS.4 Properties of the definite integrl.4. Additive Properties The following properties re useful when evluting integrls. Property (Additivity) If f(x) is continuous on [, b ] nd < c < b, then c f(x) dx + b c f(x) dx = b f(x) dx This property is clerly true when f(x) is positive continuous function on [, b ], s the re between the curve y = f(x) nd the x-xis from x = to x = b is equl to the sum of res from x = to x = c, nd from x = c to x = b. It cn lso be confirmed directly when f(x) tkes negtive vlues on [, b ]. Property 2 f(x) dx def = The originl definition of the integrl b f(x) dx ssumed tht < b. Property 2 is n extension of this definition to the cse = b. vii It is intuitively vlid s rectngle with zero width hs zero re. Property 3 If f(x) is continuous on [, b ], then b f(x) dx def = b f(x) dx The originl definition of the integrl b f(x) dx ssumed tht < b. Property 3 is n extension of this definition to the cse > b. It is consistent with properties nd 2 s b f(x) dx + b f(x) dx = f(x) dx = Observe tht properties 2 nd 3 show tht there is no restriction on the numbers tht cn be used s upper nd lower limits in integrls. vii The symbol def = mens is defined s.

.4. PROPERTIES OF THE DEFINITE INTEGRAL 5.4.2 Liner Properties The following properties re used to rewrite integrls of complex functions in terms of integrls of simpler functions. Property If f(x) is continuous on [, b ] nd k is constnt, then b kf(x) dx = k b f(x) dx This follows directly from the definition of n integrl s ( n n ) kf(x) x = k f(x) x Property 2 i= If f(x) nd g(x) re continuous on [, b ], then b (f(x) + g(x)) dx = b i= f(x) dx + This follows directly from the definition of n integrl s b g(x) dx n n n (f(x) + g(x)) x = f(x) x + g(x) x i= i= i= Note: Properties nd 2 cn be extended to ny liner combintion viii of functions. If f(x), g(x), h(x),... re continuous on [, b ] nd h, k, l,... re constnts, then b (hf(x) + kg(x) + lh(x) +...) dx = h b f(x) dx + k b g(x) dx + l b h(x) dx +... viii A liner combintion of functions is sum of multiples of the functions. Mny mthemticl functions re constructed from liner combintions of simpler functions. For exmple, polynomils re liner combintions of powers.

6 CHAPTER. INTEGRALS Exercise.4. It is known tht for integers n. Use this to evlute x n dx = n + () (b) (c) x 4 dx (x 2 + x + ) dx (x + 2)(x ) dx 2. If f(x) dx =, 3 f(x) dx = b nd 3 2 2 f(x) dx. f(x) dx = c, find

Chpter 2 Integrtion 2. Fundmentl Theorem of Clculus The most importnt ide in clculus is tht it is possible to clculte definite integrl without needing to use limits or to evlute the re under curve. This is clled the Fundmentl Theorem of Clculus nd ws discovered by Newton nd Leibnitz. i Fundmentl Theorem of Clculus ii Let f(x) be continuous function on the intervl [, b ]. If F (x) is solution of F (x) = f(x), then b f(x) dx = F (b) F () The difference F (b) F () is written s [F (x)] b. region bove x-xis Exmple Wht is the re of the region enclosed by the prbol y = x 2 nd the x-xis from x = to x =? y y = x 2 x i Sir Isc Newton (643-727), Gottfried Wilhelm von Leibniz (646-76) ii See Appendix B for justifiction of the fundmentl theorem. 7

8 CHAPTER 2. INTEGRATION Answer As x 2 when x, the enclosed re is One solution of F (x) = x 2 is F (x) = x 3 x3, so ] ( x 2 ) dx = [x x3 3 = 2 3 ( 2 3 ) ( x 2 ) dx. = 4 3 unit2 region strddles x-xis Exmple Wht is the re of the region enclosed by the prbol y = x 2 nd the x-xis from x = to x =.5? Answer The region cn be split into two prts: x, where x 2. x.5, where x 2. The re of the region is ( x 2 ) dx.5 ( x 2 ) dx. One solution of F (x) = x 2 is F (x) = x 3 x3, so.5 ] ].5 ( x 2 ) dx ( x 2 ) dx = [x x3 [x x3 3 3 ( ) 2 = 3 (.5 (.5)3 3 =.958 3 unit 2 2 3 ) Integrtion ws initilly described s being concerned with the question: Given the rte of chnge of quntity, how cn we find the quntity? This question is centrl to the Fundmentl Theorem of Clculus. In order to evlute the definite integrl b f(x) dx, we need to nswer: Given the rte of chnge of quntity f(x), how do we find the quntity F (x)? This is explored in Section 2.2.

2.. FUNDAMENTAL THEOREM OF CALCULUS 9 Exercise 2.. F (x) = x 3 x3 + is solution of F (x) = x 2. Rework the first exmple using this solution of F (x) = x 2. 2. Wht is the re of the region between the prbol y = x 2 nd the x-xis, bounded by the verticl lines x = nd x = 4? 25 y = x 2 2 3 4 5 3. Let f(x) be continuous function on the intervl [, b ]. If G(t) = for t b, show tht G (t) = f(t). t f(x) dx

2 CHAPTER 2. INTEGRATION 2.2 Integrtion 2.2. Antiderivtives nd indefinite integrls Let f(x) be given function. In order to use the fundmentl theorem of clculus we need to find function F (x) for which F (x) = f(x). The function F (x) is clled n ntiderivtive of f(x). iii We often wnt to find the most generl solution for F (x) = f(x), or fmily of functions whose derivtive is f(x). This cn sometimes be done by process of systemtic guessing. systemtic guessing Exmple Consider the function f(x) = x 2. We know tht the wy to get power of x through differentition is to differentite nother power of x nd, s differentition reduces the power of x by, it is nturl to consider F (x) = x 3. This is the first guess. If we differentite F (x) = x 3, then we get F (x) = 3x 2. This gives n x 2, but it is multiplied by 3. If we try F (x) = 3 x3 insted, then will get F (x) = x 2. A better nswer is F (x) = 3 x3 + C, where C is constnt, s the constnt differentites to zero. We write F (x) = 3 x3 + C s the ntiderivtive of f(x) = x 2. We cn think of it s representing fmily of solutions, one for ech specific vlue of C. It is useful to hve compct nottion for n ntiderivtive. We use the sme nottion with the definite integrl but without the limits. Insted of sying the ntiderivtive of f(x) is F (x) + C, we write f(x) dx = F (x) + C. The left side is red loud s the integrl of f(x) dee x, f(x) is referred to s the integrnd nd C is clled the constnt of integrtion or n rbitrry constnt. iv The process of finding n integrl is clled integrtion. Integrtion is typiclly crried out by systemtic guessing nd checking the guess using differentition. This cnnot lwys be done s some ordinry looking functions v hve very complex integrls which re impossible to express in terms of common functions let lone guess! iii ntiderivtive = reverse of differentition iv The phrse rbitrry constnt is commonly used to indicte tht the constnt is yet to be specified nd cn potentilly be given ny vlue. v For exmple e x2.

2.2. INTEGRATION 2 In following exmple the constnts re represented by different letters. becuse they my not ll hve the sme vlue. This is indefinite integrl rbitrry constnt Exmple () x 2 dx = x3 3 + C Check: d dx (x3 3 ) = 3x2 3 = x2 (b) x 2 dx = x x3 3 +D Check: d x3 3x2 (x ) = dx 3 3 = x2 (c) e 2t dt = e2t 2 + E Check: d dt (e2t 2 ) = 2e2t 2 = e2t Notes. In (), 3 x3 + lso hs derivtive x 2. Using this insted of 3 x3 leds to x 2 dx = 3 x3 + + C. However indefinite integrls re trditionlly written compctly with single constnt, so the right side should be rewritten s x 2 dx = 3 x3 + D where D = + C. As C cn be ny constnt, D lso cn be ny constnt. 2. In (c), the integrl is tken with respect to the vrible t. The vrible used in the integrnd is the independent vrible tht the function is expressed in terms of. definite integrl rbitrry constnt Exmple Clculte Answer e 2t dt The function e 2t is continuous on [, ] nd e 2t dt = e2t 2 + C By the fundmentl theorem e 2t dt = = [ e 2t ] 2 + C [ ] [ ] e 2 e 2 + C 2 + C = 2 (e2 )

22 CHAPTER 2. INTEGRATION Notes. The exmple bove shows tht it doesn t mtter which vlue of the rbitry constnt is used when evluting [F (x)] b s the constnt term lwys cncels out. 2. For simplicity, some texts just use C = when evluting [F (x)] b. Exercise 2.2.. Find the generl ntiderivtive of () x 3 by differentiting x 4 (b) x 4 by differentiting x 5 (c) 7x 2 by differentiting x 3 (d) x 2 + 2x + by differentiting x 3, x 2 nd x (e) 4x /2 by differentiting x /2 (f) e 5t by differentiting e 5t 2. Clculte ech of the following indefinite integrls. () x 3 dx (b) 3x 7 dx (c) x 2 + 2x + 3 dx (d) 4r 3 dr (e) t /2 dt (f) w /2 dw 3. Clculte ech of the following definite integrls. () x 2 dx (b) 2 2x 5 dx (c) x 2 + 2 dx (d) /2 4u 2 du (e) 6 4 2v /2 dv (f) 2 8w /2 dw

2.2. INTEGRATION 23 2.2.2 Methods (A) Stndrd Integrls (Prt ) The stndrd integrls covered in this topic re: f(x) f(x) dx k x n, n e x x kx + C n + xn+ + C e x + C ln x + C (k is constnt) The first three integrls cn be checked directly by differentition (do check them). The fourth is less obvious nd hs this form becuse logrithms re only defined for positive numbers. Checking the integrl dx = ln x + C : x If x >, then ln x = ln x nd If x <, then ln x = ln( x) nd So, for x, x d d ln x = dx dx ln x = x d d ln x = ln( x) = dx dx x = x dx = ln x + C. squre root Exmple Clculte Answer x x dx. x x dx = x 3/2 dx write s stndrd integrl = 2 5 x5/2 + C = 2 5 x2 x + C finl nswer presented in the sme form s the integrnd

24 CHAPTER 2. INTEGRATION reciprocl power Exmple Clculte Answer x 4 dx. x 4 dx = x 4 dx = 3 x 3 + C = 3x 3 + C write s stndrd integrl finl nswer presented in the sme form s the integrnd logrithm Exmple Clculte Answer 2 x dx nd 2 x dx The function /x is continuous on [, 2 ] nd [ 2, ], nd By the fundmentl theorem nd 2 2 x dt = ln x + C. x dt = [ ln x + C ] 2 = (ln 2 + C) (ln + C) = ln 2 x dt = [ ln x + D ] 2 = (ln + D) (ln 2 + D) = ln ln 2 = ln 2

2.2. INTEGRATION 25 Exercise 2.2.2. Clculte ech of the following integrls. () (c) (e) (g) dx (b) x dx r 2 r dr t 3 dt (d) dx x dx (f) s 2 s ds (h) u du 3 2. Use the Fundmentl Theorem of Clculus to clculte the following integrls. () e x dx (b) e x dx (c) 2 x dx (d) 2 x dx (e) T e x dx (f) S x dx, S >

26 CHAPTER 2. INTEGRATION (B) Liner Combintions Mny mthemticl functions re constructed from liner combintions of simpler functions. vi For exmple, polynomils re liner combintions of powers. We my be ble to integrte these functions by systemticlly guessing the integrl (or ntiderivtive) of ech term in the liner combintion. This is clled integrting term-by-term. Exmple integrting term-by-term. If f(x) = x 2, then x 2 dx = 3 x3 + C = 3 x3 + C 2. If f(x) = x 2 + 2x + 3, then x 2 + 2x + 3 dx = 3 x3 + 2 2 x2 + 3 x + D = 3 x3 + x 2 + 3x + D 3. If f(x) = (x + 3)(x 7), then (expnding the brckets first) (x + 3)(x 7) dx = x 2 4x 2 dx = 3 x3 4 2 x2 2 x + E = 3 x3 2x 2 2x + E We cn summrise this method by the following rules: Rule (multiples) The integrl of constnt multiple is the multiple of the integrl. cf(x) dx = c f(x) dx Rule 2 (sums of terms) The integrl of sum of terms is the sum of their integrls. f(x) + g(x) +... dx = f(x) dx + g(x) dx +... vi A liner combintion of the functions f(x), g(x), h(x)... is sum of multiples of the functions, e.g. f(x) + bg(x) + ch(x) + for constnts, b, c... For exmple, x 2 4x 2 is liner combintion of x 2, x nd with constnts, 4 nd 2. Its terms re x 2, 4x nd 2.

2.2. INTEGRATION 27 Exercise 2.2.2 3. Clculte ech of the following integrls. () x 2 + 4x + 8 dx (b) 25 6x 3 dx (c) x 4 x dx (d) x 2 x dx (e) (t + )(t + 2) dt (f) (2 u) 2 du (v + ) 2 e w e w (g) v dv (h) 2 dw 4. Wht is the re of the region enclosed by y = (x + )(x 3) nd the x-xis?

28 CHAPTER 2. INTEGRATION (C) Stndrd Integrls (Prt 2) The stndrd integrls introduced in (A) cn be extended to include : f(x) (x + b) n, n f(x) dx (n + ) (x + b)n+ + C e x+b ex+b + C x + b ln x + b + C for constnts nd b. Ech integrl in the tble cn be checked directly by differentiting, using the chin rule. (Do this.) squre root Exmple Clculte 2 2x + 3 dx. Answer 2 2x + 3 dx = 2 (2x + 3) /2 dx = 2 2 3 (2x + 3) 3/2 + C check here vii 2 = 4(2x + 3) 2x + 3 + C logrithm Exmple Clculte Answer 2 3x + 5 dx. 2 3x + 5 dx = 2 3x + 5 dx = 2 ln 3x + 5 + C 3 check here = 2 ln 3x + 5 + C 3 vii See the note on rbitry constnts on pge 2.

2.2. INTEGRATION 29 Some functions my need to be rewritten s function in the form f(x + b) before they cn be integrted. squre root Exmple Clculte Answer As 2x 2x + 3 dx. 2x 2x + 3 = (2x + 8 8) 2x + 3 = 6(2x + 3) 2x + 3 8 2x + 3 we hve... 2x 2x + 3 dx = 6(2x + 3) 3/2 8(2x + 3) /2 dx = 6 2 5 (2x + 3) 5/2 8 2 3 (2x + 3) 3/2 + C 2 2 = 6 5 (2x + 3)5/2 6(2x + 3) 3/2 + C = ( ) 6 (2x + 3) 6 (2x + 3) 3/2 + C 5 = 6 5 ((2x + 3) 5) (2x + 3)3/2 + C = 2 5 (x )(2x + 3) 2x + 3 + C

3 CHAPTER 2. INTEGRATION Exercise 2.2.2 5. Clculte the following integrls. () (3x + ) dx (b) 6( 2x) 3 dx (c) 2 x + 5 dx (d) 2 p + 5 dp (e) 2 3q + 7 dq (f) e w/2 e w/2 2 dw 6. Rewrite ech integrnd in n pproprite form nd then clculte the integrl. () 3x(3x + ) dx (b) 6x( 2x) 3 dx (c) 2x x + 5 dx (d) 2q 3q + 7 dq

2.2. INTEGRATION 3 (D) Composite Functions The stndrd integrls on pge 28 were obtined by pplying the chin rule to simple functions of the form f(x + b). We cn extend our integrting skills by mking use of the generl chin rule for composite functions : The derivtive of composite function is the derivtive of the outside function multiplied by the derivtive of the inside function. In symbols... The Chin Rule If f(u) nd g(x) re given functions, then F (x) = f(g(x)) = F (x) = f (g(x))g (x)... giving the indefinite integrl : f (g(x))g (x) dx = f(g(x)) + C Integrtion involves systemtic guessing followed by checking using differentition. The most efficient wy to decide if n integrl hs the form f (g(x))g (x) dx is to () guess which function is the inside function g(x) (b) confirm the presence of g (x) (c) verify tht the integrnd hs the form f (g(x))g (x). first g(x) then g (x) Exmple Clculte Answer x x 2 + dx. If g(x) = x 2 +, then g (x) = 2x. The integrnd hs x s fctor rther thn g (x) = 2x, but this shouldn t be problem s x = 2 2x. x x 2 + dx = 2 x2 + 2x dx = 2 3/2 (x2 + ) 3/2 + C = 3 (x2 + ) x 2 + + C

32 CHAPTER 2. INTEGRATION first g(x) then g (x) Exmple Clculte Answer x 3 x 4 + dx. If g(x) = x 4 +, then g (x) = 4x 3. The integrnd hs x 3 s fctor rther thn g (x) = 4x 3, but this isn t problem s it is constnt multiple of g (x). x 3 x 4 + dx = 4 x 4 + 4x3 dx = 4 ln x4 + + C It ws difficult to jump from x 4 + 4x3 to ln x 4 + + C in this exmple. The substitution method below mkes this esier. The substitution or chnge of vrible method mkes the integrl f (g(x))g (x) dx esier nd more strightforwrd to clculte. The ide is to simplify the integrl by using the new vrible u insted of x, where u = g(x). viii This is done by replcing g(x) by u g (x) dx by du... s u = g(x)... s du dx = g (x) ix When this is done f (g(x))g (x) dx is trnsformed to f (u) du with integrl f(u) + C = f(g(x)) + C. Observe the difference when the substitution method is pplied to the previous two exmples (next pge). viii Any letter cn be used to represent new vrible, not just u. ix du While it doesn t mke sense to seprte the top nd bottom prts of the symbol dx, the procedure lwys leds to correct outcome. It s best to think of this s working with the nottion in suggestive wy.

2.2. INTEGRATION 33 first guess g(x) Exmple Clculte Answer x x 2 + dx. If g(x) = x 2 +, then g (x) = 2x. The integrnd hs x s fctor rther thn g (x) = 2x, but this shouldn t be problem s x = 2 2x. Put u = x 2 +, then du = 2x dx, nd x x 2 + dx = 2 x2 + 2x dx = 2 u du = 2 3/2 u3/2 + C = 3 u u + C Rewriting the nswer in terms of x gives x x 2 + dx = 3 (x2 + ) x 2 + + C first g(x) then g (x) Exmple Clculte Answer x 3 x 4 + dx. If g(x) = x 4 +, then g (x) = 4x 3. The integrnd hs x 3 s fctor rther thn g (x) = 4x 3, but this isn t problem s it is constnt multiple of g (x). Put u = x 4 +, then du = 4x 3 dx, nd x 3 x 4 + dx = 4 = 4 x 4 + 4x3 dx u du = 4 ln u + C Rewriting the nswer in terms of x gives x 3 x 4 + dx = 4 ln x4 + + C

34 CHAPTER 2. INTEGRATION Exercise 2.2.2 7. Clculte ech of the following integrls. () (c) (e) (g) 2x(x 2 + ) 3 dx (b) (x 2 + 4)(x 3 + 2x + ) 5 dx (d) (2t + ) t 2 + t dt (f) 2v dv (h) (v 2 + ) 3 8. Clculte ech of the following integrls. () (c) (e) (g) 2x e x2 2 dx (b) (x + )e x2 +2x 3 dx (d) e t (e t + ) 5 dt (f) e v dv (h) (e v + ) 3 (2x + 4)(x 2 + 4x) 4 dx 2x x 2 + dx u 3 + u4 + 4u du w + 2 (w 2 + 4w + 3) 5 dw x 2 e x3 + dx e /x x 2 dx e 2u e 2u + du e 2w + 2e w (e 2w + 4e w + 3) 5 dw 9. Clculte ech of the following integrls. () (c) (e) 3x 2 x 3 + dx x 2 + 4 x 3 + 2x + dx e 2t + 3e t e 2t + 6e t + dt (b) (d) (f) 2x + 4 x 2 + 4x dx e x e x + dx e u e u du e u + e u

2.2. INTEGRATION 35 (E) Numericl Methods (optionl) There re mny ntiderivtives tht cn not be found explicitly, for exmple e x2 dx. In these cses we use numericl methods to estimte their definite integrls. The method of using upper nd lower methods to estimte n integrl is one such method, but there re other methods tht provide closer closer pproximtions nd re esier to progrm. See http://en.wikipedi.org/wiki/numericl integrtion http://en.wikibooks.org/wiki/numericl Methods/Numericl Integrtion nd http://people.hofstr.edu/stefn Wner/relworld/integrl/numint.html http://people.hofstr.edu/stefn wner/relworld/integrl/integrl.html [All pges ccessed 8/3/8]

36 CHAPTER 2. INTEGRATION 2.2.3 Selected Applictions Integrtion hs widespred pplictions in science nd engineering including : clcultion of res clcultion of volumes clcultion of centres of mss of solids clcultion of the work done by force problems concerned with rte of chnge problems in economics sttisticl problems This topic only considers problems ssocited with res nd chnge in quntities over time. (I) Clcultion of the re between two curves We hve seen in section.3 tht if f(x) is continuous function for x b, then b f(x) dx is equl to the the difference between the sum of the res under f(x) nd bove the x-xis the sum of the res bove f(x) nd below the x-xis for x b. Let f(x) nd g(x) be positive continuous functions for x b. Suppose lso tht f(x) g(x) on [, b ]. This is shown in the digrm below. Wht is the re of the shded region between f(x) nd g(x) tht is bounded by the lines x = nd x = b? y = f(x) b x y = g(x)

2.2. INTEGRATION 37 The re remins the sme if ech curve is trnslted verticlly by the sme distnce k, with k tken so tht both curves re bove the x-xis. y = f(x) + k y = g(x) + k b x You cn now be seen tht the re of the shded region is the difference between the res below: y = f(x) + k Are = b (f(x) + k) dx y = g(x) + k b x y = f(x) + k Are2 = b (g(x) + k) dx y = g(x) + k b x

38 CHAPTER 2. INTEGRATION... nd tht the difference in res is A = b (f(x) + k) (g(x) + k) dx = b f(x) g(x) dx In generl, If f(x) nd g(x) re continuous functions for x b, then b is equl to the the difference between f(x) g(x) dx () (b) the re under f(x) nd bove g(x) when f(x) g(x) the re under g(x) nd bove f(x) when g(x) f(x) for x b. re between curves Exmple Wht is the re of the region enclosed by the prbol y = x 2 nd the line y = x? y x Answer The prbol nd line intersect when x 2 = x x 2 x = x(x ) = x = &

2.2. INTEGRATION 39 The enclosed re is x x 2 dx = = [ ] x 2 2 x3 3 + C [ 2 ] 3 = 6

4 CHAPTER 2. INTEGRATION Exercise 2.2.3. Clculte the re between the curves below over the given intervl () f(x) = x 2 4 nd g(x) = x 2 + 9 x (b) f(x) = x 2 nd g(x) = x 3 x (c) f(x) = 2x nd g(x) = x 2 + 3 3 x (e) f(x) = e x nd g(x) = x x 3 2. Clculte the re enclosed by the curves below () f(x) = x nd g(x) = x 2 (b) f(x) = x nd g(x) = x 2 (c) f(x) = x 4 nd g(x) = 3x 2 (d) f(x) = x 4 nd g(x) = 2x 2 + 3 3. The cble used to construct new rts centre will hve the cross-section shown below. y y = (x.2) 2 x () Express the cross-sectionl re s n integrl. (b) Clculte the re. (c) If 3 m of cble re requied, clculte the volume of the cble. 4. Consider the grphs of the form y = x n for positive integers n. () At wht points do these grphs intersect? (b) Does the re between the grphs nd the x-xis, from x = to x = increse or decrese s n increses? (c) Wht hppens if the intervl in (b) ws chnged to [, 2 ]? (d) Wht hppens if n is llowed to be negtive integer?

2.2. INTEGRATION 4 (II) Chnge in Quntities We discovered in section.3 tht if the rte of chnge of quntity is known nd is positive, then the net chnge in the quntity is equl to the re under the grph of the rte of chnge nd bove the horizontl xis. The Fundmentl Theorem in section 2. extends this discovery: Net Chnge If the rte of chnge f(x) with respect to x is known for quntity F (x), then the net chnge in the quntity from x = to x = b is F (b) F () = b f(x) dx totl chnge Exmple The mintennce costs M(x) for building increse with ge x yers. Records for certin building show tht the rte of increse in costs is pproximtely Wht is the totl mintennce cost for () the first 5 yers? (b) the first t yers? Answer dm dx = 6x2 + 4 dollrs/yer. () The totl cost for the first 5 yers is 5 6x 2 + 4 dx = [ 2x 3 + 4x + C ] 5 (b) The totl cost for the first t yers is t = [ 2 5 3 + 4 5 ] = $ 45 6x 2 + 4 dx = [ 2x 3 + 4x + D ] t = 2t 3 + 4t dollrs

42 CHAPTER 2. INTEGRATION definite integrl vs indefinite integrl Exmple The mrginl cost x of mnufcturing x rdios per week is dc dx = 25.x dollrs/rdio when x 2. The fixed costs per week before production begins re $. xi Wht is the totl cost of producing rdios per week? Answer (definite integrl) As we hve Totl Cost = Fixed Costs + Cost of Production Totl cost ( rdios) = + 25.x dx = + [ 25x.5x 2 + C ] = + [ 25.5 2] = $ 3 Answer (indefinite integrl) As the mrginl cost is dc = 25.x, the cost of production is dx C(x) = 25.x dx = 25x.5x 2 + C for some constnt C nd, s C() = C =, So = 25x.5x 2 Totl cost ( rdios) = + C() = + (25.5 2 ) = $ 3 x The mrginl cost of production is the rte of chnge of cost of production reltive to output. xi The fixed costs re necessry costs tht re independent of the number of rdios produced. They might include lese, wges, insurnce, etc.

2.2. INTEGRATION 43 When n object trvels long stright line, its displcement s(t) is its position from the origin. A positive displcement corresponds to position on the right of the origin, nd negtive displcement corresponds to position on the left of the origin. If the object moves m wy from the origin then returns to the origin its displcement will be zero even though the distnce trvelled is 2m. Velocity v(t) is the rte of chnge of displcement, s (t). An object trvelling in positive direction hs positive velocity. When it returns towrds the origin its velocity is negtive. velocity displcement distnce Exmple A object P moves in stright line with velocity v(t) = ds = 2 2t m/s. dt for t seconds. Wht is () its chnge in displcement between t = nd t =? (b) the distnce it trvelled between t = nd t =? (c) its chnge in displcement between t = nd t = 2? (d) the distnce it trvelled between t = nd t = 2? Answer () The chnge in displcement between t = nd t = is (b) As v(t) for t, s() s() = 2 2t dt = [ 2t t 2 + C ] = m distnce trvelled = 2 2t dt = [ 2t t 2 + C ] = m (c) The chnge in displcement between t = nd t = 2 is s(2) s() = 2 2 2t dt = [ 2t t 2 + C ] 2 = m

44 CHAPTER 2. INTEGRATION (d) As v(t) for t, nd v(t) for t 2 distnce trvelled = 2 2t dt 2 2 2t dt = [ 2t t 2 + C ] [ 2t t 2 + D ] 2 = ( ) = 2 m

2.2. INTEGRATION 45 Exercise 2.2.3 5. The mrginl cost of mnufcturing nd storing x crdbord crtons per week is C (x) = 3.2 +.4x dollrs per item, nd the fixed costs re $5 per week. Wht is the totl cost of mnufcturing nd storing crtons per week? 6. A new suburb is estimted to grow t the rte of dp dt = 5 + 3 t people per yer. If the current popultion is, estimte () the popultion in 25 yers (b) the popultion P (t) fter t yers 7. The re A(t) covered by n ulcer chnges t rte da dt = 4 (t + ) 2 cm2 /dy s it hels, where t is in dys. If the re is initilly A() = 4 cm 2, wht is () the re of the wound in dys (b) the re A(t) sfter t dys 8. A object P moves in stright line with velocity v(t) = ds dt = 2 t2 m/s. for t seconds. Wht is the () chnge in displcement between t = nd t =? (b) distnce it trvelled between t = nd t =? (c) chnge in displcement between t = nd t = 2? (d) distnce it trvelled between t = nd t = 2? 9. A oil tnker hit reef nd is producing circulr oil slick tht is expnding t n pproximte rte of dr dt = 2 t + where r metres is the rdius of the slick fter t minutes. If r() = 2, estimte () the rdius of the slick fter 3 minutes (b) the rdius of the slick fter t minutes

46 CHAPTER 2. INTEGRATION (c) the rte t which the re of the slick is chnging. A circulr plstic tube, with internl dimeter 4 cm nd externl dimeter 4.5 cm, crries wter t constnt temperture of 7 C. The temperture inside the tube drops off t rte of dt dx = 5 x where x is the distnce from the centre of the tube nd 4 x 4.5. Wht is the temperture on the outside of the tube?

Appendix A Summtion Nottion Summtion nottion i is used in situtions where we need to write down the sum of mny numbers or terms. We could write the sum of the squres of the numbers from to s 2 + 2 2 + 3 2 +... + 2, leving it to the reder to guess the pttern of numbers, but summtion nottion cn be used to express this sum concisely nd without mbiguity s i= i 2. If f(i) represents n expression involving i, then n f(i) hs the following mening : i= n f(i) = f() + f(2) + f(3) +... + f(n). i= This nottion hs number of prts : end summtion sign index n f(i) i= strt generl term i Sometimes sigm nottion becuse the Greek letter sigm Σ is used. 47

48 APPENDIX A. SUMMATION NOTATION The summtion sign Σ Σ is the Greek upper cse letter corresponding to S. It tells us to dd the terms in sum, where the generl term is given to the right of the summtion sign. The index vrible i This vrible is used to number or lbel ech term in the sum. The index is often represented by i. Other common possibilities include j nd k. The strt The i = prt beneth the summtion sign tells shows which index number to begin with. It is usully either zero or one but it cn be nything. You lwys increse the index by one t ech successive step in the sum. The end The number on top of the summtion sign is the finl index number. sum of indexed terms Exmple 5 () i 2 = 2 + 2 + 2 2 +... + 5 2 (b) (c) (d) (e) i= 5 j= 5 k=3 8 t= j 2 = 2 + 2 2 + 3 2 +... + 5 2 k = 3 + 4 + 5 +... + 5 ( ) i 2 + i = 2 + 2 + + 2 + 2... + 2 + 8 n 3 t = 3 + 3 2 + 3 3 +... + 3 n t=

Appendix B Justifiction for the Fundmentl Theorem Let f(x) be positive continuous function on the intervl [, b ]. The re between f(x) nd the x-xis from x = to x = b is equl to the definite integrl b f(x) dx. Define the re function F (x) for x b to be the re between f(x) nd the x-xis on the intervl [, x ] s shown in the digrm below. y y = f(x) Are F (x) x b x The re between f(x) nd the x-xis from x = to x = b is equl to F (b) F () so b f(x) dx = F (b) F (). We now show tht F (x) is solution of the eqution F (x) = f(x). 49

5 APPENDIX B. JUSTIFICATION FOR THE FUNDAMENTAL THEOREM The re between f(x) nd the x-xis on the intervl from to x + h is equl to F (x + h). It cn be seen from the digrm below tht when h is very smll this re is closely pproximted by dding the re between f(x) nd the x-xis from to x the re of the rectngle with height f(x) nd bse the intervl [ x, x + h ]. y y = f(x) Are F (x) x x + h b x In other words F (x + h) F (x) + hf(x) nd F (x + h) F (x) h f(x). We know tht s h becomes very smll, F (x + h) F (x) h F (x) which implies F (x) = f(x). So b f(x) dx = F (b) F (), where F (x) is solution of the eqution F (x) = f(x). The finl step in the justifiction is to show tht this is true for ny function tht stisfies F (x) = f(x)...

5 If G(x) is nother function with G (x) = f(x), then d dx (G(x) F (x)) = G (x) F (x) =. This implies tht G(x) F (x) = k for some constnt k. As G(b) G() = (F (b) + k) (F () + k) = F (b) F () you cn see tht b f(x) dx = F (b) F () = G(b) G(),... so it doesn t mtter which solution of F (x) = f(x) is chosen.

Appendix C Answers Exercise. () Horizontl line with verticl intercept (, ). (b) From grph, V (t) = t. (c) Line with grdient nd verticl intercept (, ). (d) Rte of chnge of volume per minute. 2() Stright line with intercepts (6, ) nd (, 3). 2(b) 9 m. 2(c) 3t.25t 2 3(iii) Upper estimte =.76, Lower estimte =.56 Exercise.2 () Upper estimte =.875, Lower estimte =.375. (b) Ten equl intervls of width.. 2() Upper estimte =.75, Lower estimte =.75. Exercise.3 (i) 2 (ii) 2 (iii) 2. The grph of + sin(x) is obtined by trnslting the grph of sin(x) verticlly by one unit. The re is 2 + π = 2 + π 52

53 Exercise.4 (i) 2 (ii) 6 2(b) c b (iii) 7 6 Exercise 2. 2. 2 Exercise 2.2. () 4 x4 + C (b) 2x 5 + D (c) 7 3 x3 + E (d) 3 x3 + x 2 + x + F (e) 8x /2 + G (f) 2e 5t + H 2() 4 x4 + C 2(b) 3 8 x8 + D 2(c) 3 x3 + x 2 + 3x + E 2(d) 2r 2 + F 2(e) 2 3 t3/2 + G 2(f) 2w /2 + H 3() 3(b) 26 3(c) 4 3 3 3(d) 4 3(e) 224 3(f) 6( 2 ) 3 Exercise 2.2.2 () x + C (b) x + D (c) 2 3 x x + E (d) 2 x + F (e) 2 7 r3 r + G (f) 2 3s s + H (g) 4 t4 + I (e) 2 u 2 + J 2() e 2(b) e 2(c) ln 2 2(d) ln 2 2(e) e T 2(f) ln S 3() 3 x3 + 2x 2 + 8x + C 3(b) 25x 4x 4 + D 3(c) 2 3 x x 8 x + E 3(d) 2 x2 2 ln x + F 3(e) 3 t3 + 3 2 t2 + 2t + G 3(f) 4u 2u 2 + 3 u3 + H 3(g) 2 v2 + 2v + ln v + I 3(h) ew + e w + J 2

54 APPENDIX C. ANSWERS 4. 5() 32 3 36 (3x + )2 + C 5(b) 2( 2x) 4 +D 5(c) 4 3 (x + 5) x + 5 + E 5(d) 4 p + 2 + F 5(e) 4 ln 3q + 7 + G 6() 6(b) 6(c) 6(d) 5(f) e w/2 + e w/2 + H (3x + ) 2 (3x + ) dx = 39 (3x + )3 36 (3x + )2 + C 8( 2x) 4 + 8( 2x) 3 dx = 4 5 ( 2x)5 ( 2x) 4 + D 2(x + 5) 3/2 (x + 5) /2 dx 4 5 (x + 5)2 x + 5 2 3 (x + 5) x + 5 + E 4 28 3q + 7 dq = 4q 28 3 ln 3q + 7 + F 7() 4 (x2 + ) 4 + C 7(b) 5 (x2 + 4x) 5 + D 7(c) 8 (x3 + 2x + ) 6 + E 7(d) 2 3 (x2 + ) x 2 + + F 7(e) 2 3 (t2 + t) t 2 + t + G 7(f) 2 u4 + 4u + H 7(g) 2(v 2 + ) + I 7(h) 2 8(w 2 + 4w + 3) + J 4 8() e x2 2 + C 8(b) + 3 ex3 + D 8(c) +2x 3 2 ex2 + E 8(d) e /x + F 8(e) 6 (et + ) 6 + G 8(f) 3 (e2u + ) 3/2 + H 5 8(g) (e v + ) + I 8(h) 2 8(e 2w + 4e w + 3) + J 4 9() ln x 3 + + C 9(b) ln x 2 + 4x + D 9(c) 3 ln x3 + 2x + + E 9(d) ln e x + + F 9(e) 2 ln e2t + 6e t + + G 9(f) ln e u + e u + H

55 Exercise 2.2.3 () 76 3 2() 6 3() 4.2 (b) 2(b) 3 2 (c) 32 3 (x.2) 2 dx 3(b) 4() They ll intersect t (, ) nd (, ). 2(c) 2 3 5 (d) e 3 2 2(d) 64 5 4 375 m2 3(c) 3.2 m 3 4(b) The re between y = x n nd x-xis from x = to x = is which decreses s n increses. n +, 4(c) The re would then be 2 n+ n +, which increses s n increses. You cn check this lst sttement by considering ( 2 n+ n + 2n 2 n = 2n n + ) ( ) n = 2 n n n(n + ) 4(d) The re is infinite when n. You cn check this by clculting the re from x = h to x = for h >. For exmple, for n = it is which becomes infinite s h. h x dx = [ln x ] h = ln h, Note. We couldn t clculte the integrl directly for the intervl [, ] s x is not defined for x =. Check out wht hppens with x = 2! 5. Totl Cost = 5 + 6() P (25) = + 6(b) P (t) = + 25 t 3.2 +.4x dx = $39 5 + 3 t dt = 63, 5 5 + 3 t dt = + 5t + 2t t

56 APPENDIX C. ANSWERS 7() A() = 4 + 7(b) A(t) = 4 + t 8() s() s() = 8(b) distnce trvelled = 8(c) s() s() = 4 (t + ) 2 dt = 4 cm2 4 (t + ) 2 dt = 4 t + cm2 2 2 t 2 dt = 5 3 cm 2 t 2 dt = 5 3 cm 2 t 2 dt = 4 3 cm 8(d) distnce trvelled = 2 2 t 2 dt 2 2 2 t 2 dt = 8 2 4 3 cm 9() r(3) = r() + 9(b) r(t) = r() + 3 t 2 t + dt = + 3 cm 2 t + dt = + t + m 9(c) A(t) = π( + t + ) 2 = π(2 + 2 t + + t) m 2 ( A (t) = π + ) t + m 2 /min. T (4.5) = T (4) + 4.5 4 5 x dx = 7 + 5 ln(4/4.5) C