CHAPTER FIVE. Solutions for Section 5.1

Transcription

1 CHAPTER FIVE 5. SOLUTIONS 87 Soluions for Secion 5.. (a) The velociy is 3 miles/hour for he firs hours, 4 miles/hour for he ne / hour, and miles/hour for he las 4 hours. The enire rip lass + / + 4 = 6.5 hours, so we need a scale on our horizonal (ime) ais running from o 6.5. Beween = and =, he velociy is consan a 3 miles/hour, so he velociy graph is a horizonal line a 3. Likewise, beween = and =.5, he velociy graph is a horizonal line a 4, and beween =.5 and = 6.5, he velociy graph is a horizonal line a. The graph is shown in Figure 5.. (b) How can we visualize disance raveled on he velociy graph given in Figure 5.? The velociy graph looks like he op edges of hree recangles. The disance raveled on he firs leg of he journey is (3 miles/hour)( hours), which is he heigh imes he widh of he firs recangle in he velociy graph. The disance raveled on he firs leg of he rip is equal o he area of he firs recangle. Likewise, he disances raveled during he second and hird legs of he rip are equal o he areas of he second and hird recangles in he velociy graph. I appears ha disance raveled is equal o he area under he velociy graph. In Figure 5., he area under he velociy graph in Figure 5. is shaded. Since his area is hree recangles and he area of each recangle is given by Heigh Widh, we have Toal area = (3)() + (4)(/) + ()(4) = = 6. The area under he velociy graph is equal o disance raveled. velociy (miles/hour) velociy (miles/hour) Area = Disance raveled Figure 5.: Velociy graph ime (hours) ime (hours) Figure 5.: The area under he velociy graph gives disance raveled. Using he daa in Table 5.3 of Eample 4, we consruc Figure 5.3. rae of sales (games/week) 5 Underesimae 5 ime (weeks) Figure 5.3

2 88 Chaper Five /SOLUTIONS 3. (a) Lower esimae = (45)() + (6)() + ()() = fee. Upper esimae = (88)() + (45)() + (6)() = 98 fee. (b) v We use Disance = Rae Time on each subinerval wih = 3. We know ha A beer esimae is he average. We have Underesimae = = 4, Overesimae = = Disance raveled 465. Disance raveled The car ravels abou 35.5 fee during hese seconds = Figure 5.4 shows he graph of f(). The region under he graph of f() from = o = is a riangle of base seconds and heigh 5 meer/sec. Then Thus he disance raveled is 5 meers. Disance raveled = Area of riangle = 5 = 5 meers. 5 f() = 5 Figure Jus couning he squares (each of which has area ), and allowing for he broken squares, we can see ha he area under he curve from o 6 is beween 4 and 5. Hence he disance raveled is beween 4 and 5 meers. 7. By couning squares and fracions of squares, we find ha he area under he graph appears o be around 3 (miles/hour) sec, wihin abou. So he disance raveled was abou 3 ( ) ( fee, wihin abou 58 36) 5 fee. (Noe ha 455 fee is abou.86 miles) 8. The able gives he rae of oil consumpion in billions of barrels per year. To find he oal consumpion, we use lef-hand and righ-hand Riemann sums. We have Lef-hand sum = (.3)(5) + (.3)(5) + (3.9)(5) + (4.9)(5) + (7.)(5) = 597. bn barrels. Righ-hand sum = (.3)(5) + (3.9)(5) + (4.9)(5) + (7.)(5) + (9.3)(5) = 63. bn barrels Average of lef- and righ-hand sums = = 64.5 bn barrels. The consumpion of oil beween 98 and 5 is abou 64.5 billion barrels.

3 5. SOLUTIONS (a) Le s begin by graphing he daa given in he able; see Figure 5.5. The oal amoun of polluion enering he lake during he 3-day period is equal o he shaded area. The shaded area is roughly 4% of he recangle measuring 3 unis by 35 unis. Therefore, he shaded area measures abou (.4)(3)(35) = 4 unis. Since he unis are kilograms, we esimae ha 4 kg of polluion have enered he lake. Polluion Rae (kg/day) ime (days) Figure 5.5 (b) Using lef and righ sums, we have Underesimae = (7)(6) + (8)(6) + ()(6) + (3)(6) + (8)(6) = 336 kg. Overesimae = (8)(6) + ()(6) + (3)(6) + (8)(6) + (35)(6) = 54 kg.. (a) Based on he daa, we will calculae he underesimae and he overesimae of he oal change. A good esimae will be he average of boh resuls. Underesimae of oal change = = was considered wice since we needed o calculae he area under he graph. Overesimae of oal change = = and 86 were considered wice since we needed o calculae he area over he graph. The average is: ( )/ = 34 million people. (b) The oal change in he world s populaion beween 95 and is given by he difference beween he populaions in hose wo years. Tha is, he change in populaion equals 685 (populaion in ) 555 (populaion in 95) = 353 million people. Our esimae of 34 million people and he acual difference of 353 million people are close o each oher, suggesing our esimae was a good one.. Suppose f() is he flow rae in m 3 /hr a ime. We are only given wo values of he flow rae, so in making our esimaes of he flow, we use one subinerval, wih = 3/ = 3: Lef esimae = 3[f(6 am)] = 3 = 3 m 3 Righ esimae = 3[f(9 am)] = 3 8 = 84 m 3 (an underesimae) (an overesimae). The bes esimae is he average of hese wo esimaes, Bes esimae = Lef + Righ = = 57 m 3.. (a) Noe ha 5 minues equals.5 hours. Lower esimae = (.5) + (.5) = 5.5 miles. Upper esimae = (.5) + (.5) = 5.75 miles. (b) Lower esimae = (.5) + (.5) + (.5) + 8(.5) + 7(.5) + (.5) =.5 miles. Upper esimae = (.5) + (.5) + (.5) + (.5) + 8(.5) + 7(.5) = 4.5 miles.

4 9 Chaper Five /SOLUTIONS 3. (a) See Figure 5.6. (b) The disance raveled is he area under he graph of he velociy in Figure 5.6. The region is a riangle of base 5 seconds and aliude 5 f/sec, so he disance raveled is (/)5 5 = 5 fee. (c) The slope of he graph of he velociy funcion is he same, so he riangular region under i has wice he aliude and wice he base (i akes wice as long o sop). See Figure 5.7. Thus, he area is 4 imes as large and he car ravels 4 imes as far. 5 5 v() v() 5 5 Figure 5.6 Figure The rae a which he fish populaion grows varies beween fish per monh and abou fish per monh. If he rae of change were consan a he lower bound of during he enire -monh period, we obain an underesimae for he oal change in he fish populaion of ()() = fish. An overesimae is ()() = 64 fish. The acual increase in he fish populaion is equal o he area under he curve. (Noice ha he unis of he area are heigh unis imes widh unis, or fish per monh imes monhs, as we wan.) We esimae ha he area is abou grid squares. The area of each grid square represens an increase of (5 fish per monh) (4 monhs) = fish. We have Toal change in fish populaion = Area under curve =. We esimae ha he fish populaion grew by fish during his -monh period. 5. (a) Car A has he larges maimum velociy because he peak of car A s velociy curve is higher han he peak of B s. (b) Car A sops firs because he curve represening is velociy his zero (on he -ais) firs. (c) Car B ravels farher because he area under car B s velociy curve is he larger. 6. (a) Since car B sars a =, he ick marks on he horizonal ais (which we assume are equally spaced) are hours apar. Thus car B sops a = 6 and ravels for 4 hours. Car A sars a = and sops a = 8, so i ravels for 8 hours. (b) Car A s maimum velociy is approimaely wice ha of car B, ha is km/hr. (c) The disance raveled is given by he area of under he velociy graph. Using he formula for he area of a riangle, he disances are given approimaely by Car A ravels = Base Heigh = 8 = 4 km Car B ravels = Base Heigh = 4 5 = km. 7. Skech he graph of v(). See Figure 5.8. Adding up he areas using an overesimae wih daa every second, we ge s = 6 m. The acual disance raveled is less han 6 m. (m/sec) (sec) Figure 5.8

5 5. SOLUTIONS 9 8. (a) Noe ha he rae r() someimes increases and someimes decreases in he inerval. We can calculae an upper esimae of he volume by choosing = 5 and hen choosing he highes value of r() on each inerval, and similarly a lower esimae by choosing he lowes value of r() on each inerval: Upper esimae = 5[ ] = 34 liers. Lower esimae = 5[ + + 6] = 4 liers. (b) A graph of r() along wih he areas represened by he choices of r() in calculaing he lower esimae is shown in Figure r() 5 5 Figure (a) See Table 5.. Table R (b) The oal change is he rae of change, in dollars per year, imes he number of years in he ime inerval, summed up over all ime inervals. We find an underesimae and an overesimae and average he wo: Underesimae = = $ Overesimae = = $ A beer esimae of he oal change in he value of he fund is he average of he wo: Toal change in he value of he fund = $65.. The car s speed increases by 6 mph in / hour, ha is a a rae of 6/(/) = mph per hour, or /6 = mph per minue. Thus every 5 minues he speed has increased by mph. A he sar of he firs 5 minues, he speed was mph and a he end, he speed was mph. To find he disance raveled, use Disance = Speed Time. Since 5 min = 5/6 hour, he disance raveled during he firs 5 minues was beween 5 6 mile and 5 6 mile. Since he speed was beween and mph during his five minue period, he fuel efficiency during his period is beween 5 mpg and 8 mpg. So he fuel used during his period is beween gallons and gallons. Thus, an underesimae of he fuel used is ( Fuel = ) =.73 gallons. 6 An overesimae of he fuel used is ( Fuel = ) =.3 gallons. 6

6 9 Chaper Five /SOLUTIONS We can ge beer bounds by using he acual disance raveled during each five minue period. For eample, in he firs five minues he average speed is 5 mph (halfway beween and mph because he speed is increasing a a consan rae). Since 5 minues is 5/6 of an hour, in he firs five minues he car ravels 5(5/6)=5/4 miles. Thus he fuel used during his period was beween and Using his mehod for each five minue period gives a lower esimae of.843 gallons and an upper esimae of.99 gallons. Soluions for Secion 5.. Dividing he inerval from o 6 ino equal subinervals gives = 3. Using f() =, we have Lef-hand sum = f() + f(3) = = 7.. Dividing he inerval from o ino 3 equal subinervals gives = 4. Using f() = /( + ), we have Lef-hand sum = f() + f(4) + f(8) = = Calculaing boh he LHS and RHS and averaging he wo, we ge (5( ) + 5( )) = Since we are given a able of values, we mus use Riemann sums o approimae he inegral. Values are given every. unis, so =. and n = 5. Our bes esimae is obained by calculaing he lef-hand and righ-hand sums, and hen averaging he wo. Lef-hand sum = 5(.) + 3(.) + (.) + 5(.) + 9(.) = 8.4 Righ-hand sum = 3(.) + (.) + 5(.) + 9(.) + (.) = 3.8. We average he wo sums o obain our bes esimae of he inegral: 4 3 W()d = So we have 5 f()d 543. Lef-hand sum = = 66 Righ-hand sum = = Average = = 543

7 5. SOLUTIONS We esimae 4 f()d using lef- and righ-hand sums: Lef sum = = 6,45. Righ sum = = 7, We esimae ha 4 f()d In his esimae, we used n = 4 and = = 7, f() 6 f() Figure 5.: Lef Sum, = 4 (a) Lef-hand sum = = 4. (b) Righ-hand sum = = Figure 5.: Righ Sum, = f() 6 f() Figure 5.: Lef Sum, = Figure 5.3: Righ Sum, = 8. (c) Lef-hand sum = =. (d) Righ-hand sum = = 36. f() d is equal o he area shaded. We can use Riemann sums o esimae he area, or we can coun grid squares in Figure 5.4. There are abou 5 grid squares and each grid square represens 4 square unis, so he area shaded is abou 6. We have f() d f() Figure 5.4

8 94 Chaper Five /SOLUTIONS 9. We know ha 5 f()d = Area under f() beween = and = 5. The area under he curve consiss of approimaely 4 boes, and each bo has area (5)(5) = 5. Thus, he area under he curve is abou (4)(5) = 35, so 5 f()d 35.. To esimae he inegral, we coun he recangles under he curve and above he -ais for he inerval [, 3]. There are approimaely 7 of hese recangles, and each has area, so 3 f()d f()d is equal o he area shaded. We can use Riemann sum o esimae his area, or we can coun grid squares. These are 3 whole grid squares and abou 4 half-grid squares, for a oal of 5 grid squares. Since each grid square represen 4 square unis, our esimaed area is 5(4) =. We have f()d. See Figure f() 4 3 Figure 5.5. (a) See Figure 5.6. y y = 3 Figure 5.6 (b) 3 d = area shaded, which is less han.5. Rough esimae is abou.3. 3 d =.5

9 5. SOLUTIONS (a) See Figure f() = 3 Figure 5.7 (b) 3 A rough esimae of he shaded area is 7% of he area of he recangle 3 by.7. Thus, d = (a) See Figure d 7% of 3.7 = = 3.6. f() 4 3 Figure 5.8 f() = 3 The shaded area is approimaely 6% of he area of he recangle uni by 3 unis. Therefore, (b) 3 d = (a) See Figure d.6 3 = Figure 5.9 f() = (b) The shaded area appears o be approimaely unis, and so d. d =.545

10 96 Chaper Five /SOLUTIONS 6. (a) If = 4, hen n =. We have: Underesimae of oal change = f() + f(4) = = 7. Overesimae of oal change = f(4) + f(8) = = 38. See Figure 5.. f() = + f() = Figure 5. Figure 5. (b) If =, hen n = 4. We have: See Figure 5.. (c) If =, hen n = 8. Underesimae of oal change Underesimae of oal change = f() + f() + f(4) + f(6) = =. Overesimae of oal change = f() + f(4) + f(6) + f(8) = = 48. = f() + f() + f() + f(3) + f(4) + f(5) + f(6) + f(7) = = 48. Overesimae of oal change See Figure 5.. = f() + f() + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) = =. f() = Figure 5.

11 5. SOLUTIONS (a) 6 ( + ) d = 78 4 y y = (b) Using n = 3, we have Lef-hand sum = f() + f() + f(4) = = 46. This sum is an underesimae. See Figure 5.3. y y 4 4 y = + y = Figure 5.3 Figure 5.4 (c) Righ-hand sum = f() + f(4) + f(6) = = 8. This sum is an overesimae. See Figure The inegral represens he area below he graph of f() bu above he -ais. Since each square has area, by couning squares and half-squares we find 6 f()d = The graph given shows ha f is posiive for. Since he graph is conained wihin a recangle of heigh and lengh, he answers and. are boh eiher oo small or oo large o represen f()d. Since he graph of f is above he horizonal line y = 8 for.95, he bes esimae is and no d = 4.7 (3 + ) d = 448. d =. + d =.35 e

12 98 Chaper Five /SOLUTIONS.7 4. (.85) d = d =.9 6. (.3) d = The inegral ln d e ln d = e d 3 e d = (.886) = (a) Wih n = 4, we have =. Then (b) = 5, = 7, = 9, 3 =, 4 = 3 and f( ) =, f( ) = 3, f( ) = 8, f( 3) =, f( 4) = 3 (c) Wih n =,we have = 4. Then (d) 3. (a) Wih n = 4, we have = 4. Then (b) Lef sum = ()() + (3)() + (8)() + ()() = Righ sum = (3)() + (8)() + ()() + (3)() = 6. = 5, = 9, = 3 and f( ) =, f( ) = 8, f( ) = 3 Lef sum = ()(4) + (8)(4) = Righ sum = (8)(4) + (3)(4) = 9. =, = 4, = 8, 3 =, 4 = 6 and f( ) = 5, f( ) = 3, f( ) =, f( 3) =, f( 4) = 7 (c) Wih n =,we have = 8. Then (d) Lef sum = (5)(4) + (3)(4) + ()(4) + ()(4) = 36 Righ sum = (3)(4) + ()(4) + ()(4) + (7)(4) = 38. =, = 8, = 6 and f( ) = 5, f( ) =, f( ) = 7 Lef sum = (5)(8) + ()(8) = 376 Righ sum = ()(8) + (7)(8) = 3. Soluions for Secion 5.3. The area = ( 3 + )d. Using echnology o evaluae he inegral, we see ha Area = ( 3 + )d = 8. See Figure 5.5.

13 5.3 SOLUTIONS 99 y 5 y = 3 + Figure 5.5. See Figure Area = (.6) d 9.47 P P = (.6) Figure On he inerval, we see ha + 5 is greaer han +. We have Area = (( + 5) ( + )) d = The wo funcions inersec a = and = 3. Beween hese values, 3 is greaer han. We have Area = 3 (3 ) d = (a) The oal area beween f() and he -ais is he sum of he wo given areas, so Area = = 3. (b) To find he inegral, we noe ha from = 3 o = 5, he funcion lies below he -ais, and hence makes a negaive conribuion o he inegral. So 5 f() d = 3 f()d f()d = 7 6 =. 6. The area below he -ais is greaer han he area above he -ais, so he inegral is negaive. 7. The enire graph of he funcion lies above he -ais, so he inegral is posiive. 8. The areas above and below he -ais are approimaely equal in size, so he inegral is approimaely zero.

14 3 Chaper Five /SOLUTIONS 9. The area above he -ais is larger han he area below he -ais, so he inegral is posiive.. (a) Couning he squares yields an esimae of 6.5, each wih area =, so he oal shaded area is approimaely 6.5. (b) 8 f()d = (shaded area above -ais) (shaded area below -ais) 6.5 = 3.5 (c) The answers in (a) and (b) are differen because he shaded area below he -ais is subraced in order o find he value of he inegral in (b).. We know ha 5 f()d = Area above he ais Area below he ais. 3 The area above he ais is abou 3 boes. Since each bo has area ()(5) = 5, he area above he ais is abou (3)(5) = 5. The area below he ais is abou boes, giving an area of abou ()(5) = 55. We have 5 f()d 5 55 = (a) The area shaded beween and is abou he same as he area shaded beween and, so he area beween and is abou.5. Since his area lies below he -ais, we esimae ha f() d =.5. (b) Beween and, he area above he -ais approimaely equals he area below he -ais, and so (c) We esimae f()d = (.5) + (.5) =. Toal area shaded = = The region shaded beween = and = appears o have approimaely he same area as he region shaded beween = and =, bu i lies below he ais. Since f()d = 4, we have he following resuls: (a) f()d f()d = 4. (b) f()d 4 4 =. (c) The oal area shaded is approimaely = The area above he -ais appears o be abou half he area of he recangle wih area 3 = 3, so we esimae he area above he -ais o be approimaely 5. The area below he -ais appears o be abou half he area of he recangle wih area 5 =, so we esimae he area below he -ais o be approimaely 5. See Figure 5.7. The value of he inegral is he area above he -ais minus he area below he -ais, so we esimae The correc mach for his funcion is IV. 5 f()d 5 5 =. 5 5 Figure

15 5.3 SOLUTIONS 3 5. The area below he -ais is bigger han he area above he -ais, so he inegral is negaive. The area above he -ais appears o be abou half he area of he recangle wih area = 4, so we esimae he area above he -ais o be approimaely. The area below he -ais appears o be abou half he area of he recangle wih area 5 3 = 45, so we esimae he area below he -ais o be approimaely.5. See Figure 5.8. The value of he inegral is he area above he -ais minus he area below he -ais, so we esimae The correc mach for his funcion is II. 5 f()d.5 = Figure The area below he -ais is bigger han he area above he -ais, so he inegral is negaive. The area above he -ais appears o be abou half he area of he recangle wih area 5 =, so we esimae he area above he -ais o be approimaely 5. The area below he -ais appears o be abou half he area of he recangle wih area 3 = 3, so we esimae he area below he -ais o be approimaely 5. See Figure 5.9. The value of he inegral is he area above he -ais minus he area below he -ais, so we esimae The correc mach for his funcion is I. 5 f()d 5 5 = Figure On he inerval 5, he enire graph lies above he -ais, so he value of he inegral is posiive and equal o he area beween he graph of he funcion and he -ais. This area appears o be abou half he area of he recangle wih area 5 =, so we esimae he area o be approimaely 5. See Figure 5.3. Since f() is posiive on his inerval, he value of he inegral is equal o his area, so we have The correc mach for his funcion is III. 5 f()d = Area 5. 5 Figure 5.3

16 3 Chaper Five /SOLUTIONS 8. (a) The area beween he graph of f and he -ais beween = a and = b is 3, so b (b) Since he graph of f() is below he -ais for b < < c, a c b f()d = 3. f() d =. (c) Since he graph of f() is above he -ais for a < < b and below for b < < c, c a f() d = 3 =. (d) The graph of f() is he same as he graph of f() ecep ha he par below he -ais is refleced o be above i. See Figure 5.3. Thus c a f() d = 3 + = 5. f() Area = 3 Area = a b c Figure In Figure 5.3 he area A is larges, A is ne, and A 3 is smalles. We have I = IV = b a e b f() d = A, II = c f() d = A + A 3, V = a f() d = A A, III = c b f() d = A. e a f()d = A A + A 3, The relaive sizes of A, A, and A 3 mean ha I is posiive and larges, III is ne larges (since A + A 3 is negaive, bu less negaive han A ), II is ne larges, bu sill posiive (since A is larger han A ). The inegrals IV and V are boh negaive, bu V is more negaive. Thus V < IV < < II < III < I. f() A a b c e A A 3 Figure 5.3

17 5.3 SOLUTIONS 33. (a) y = f() y A A (b) A = (c). (a) A = f()d =.667. f() d =.47. So oal area = A + A Noe ha while A and A are accurae o 3 decimal places, he quoed value for A + A is accurae only o decimal places. f()d = A A =.5. f()d = (b) f()d = f() d + f()d + f()d = + A = A I appears ha f() is posiive on he whole inerval, so we have 4 Area = We esimae he value of he inegral using lef and righ sums: A beer esimae of he area is he average of he wo: f()d. Lef-hand sum = = 345. Righ-hand sum = = 33. Area = Area= 3.99 f() = 3 Area=.65 4 Figure 5.33 Using a calculaor or compuer, we see ha 4 3 d The graph of f() = 3 is shown in Figure The funcion is negaive o he lef of = 3.73 and posiive o he righ of =.73. We compue.73 3 d = Area below ais.65

18 34 Chaper Five /SOLUTIONS and 4 See Figure Then Using a calculaor or compuer, we find.73 3 d = Area above ais d = Area above ais Area below ais = = ( 3ln )d.36. The funcion f() = 3 ln crosses he -ais a.86. See Figure f() = 3ln 3 4 A A 5 Figure 5.34 We find.86 4 Thus, A.347 and A.483, so 4 ( 3 ln) d = Area above ais.347 ( 3 ln) d = Area below ais ( 3ln )d = Area above ais Area below ais = cos d =.8 = Area A Area A. See Figure cos A ( π ) A 4 Figure The graph of y = is shown in Figure 5.36, and he relevan area is shaded. If you compue he inegral 3 ( )d, you find ha 3 ( )d = 3.. However, since par of he area lies below he -ais and par of i lies above he -ais, his compuaion does no help us a all. (In fac, i is clear from he graph ha he shaded area is more han 3.) We have o find he area above he -ais

19 5.3 SOLUTIONS 35 and he area below he -ais separaely. We find ha he graph crosses he -ais a =.44, and we compue he wo areas separaely:.44 ( )d =.886 and ( )d = As we epec, we see ha he inegral beween and.44 is negaive and he inegral beween.44 and 3 is posiive. The oal area shaded is he sum of he absolue values of he wo inegrals: 3 Area shaded = = 6.77 square unis y y = 3 Figure A graph of y = 6 3 shows ha his funcion is nonnegaive on he inerval = 5 o =. Thus, The inegral was evaluaed on a calculaor. Area = 5 (6 3 ) d = 4, A graph of y = cos(/) shows ha his funcion is nonnegaive on he inerval = o =. Thus, The inegral was evaluaed on a calculaor. Area = cos d = The graph of y = 5 ln() is above he line y = 3 for 3 5. See Figure Therefore The inegral was evaluaed on a calculaor. Area = 5 3 (5 ln() 3) d = y y = 5ln() y = Figure 5.37

20 36 Chaper Five /SOLUTIONS 3. The graph of y = sin + is above he line y =.5 for 6. See Figure Therefore The inegral was evaluaed on a calculaor. y Area = sin +.5 d = y = sin + y =.5 6 Figure The graph of y = cos + 7 is above y = ln( 3) for 5 7. See Figure Therefore The inegral was evaluaed on a calculaor. y 7 Area = cos + 7 ln( 3) d = y = cos + 7 y = ln( 3) 5 7 Figure The graph of y = 4 8 has inerceps = ± 4 8. See Figure 5.4. Since he region is below he -ais, he inegral is negaive, so The inegral was evaluaed on a calculaor. Area = 4 8 y ( 4 8) d = y = Figure 5.4

21 5.4 SOLUTIONS The graph of y = e + e ( ) has inerceps where e = e ( ), or where = ( ), so =. See Figure 5.4. Since he region is below he -ais, he inegral is negaive, so The inegral was evaluaed on a calculaor. Area = e + e ( ) d =.76. y y = e + e ( ) Figure The graph of y = cos is above he graph of y = sin for π/4 and y = cos is below y = sin for π/4 < < π. See Figure 5.4. Therefore, we find he area in wo pieces: Area = π/4 The inegral was evaluaed on a calculaor. y (cos sin )d + π (sin cos ) d =.88. π/4 y = sin π/4 π Figure 5.4 y = cos Soluions for Secion 5.4. (a) The inegral 3 f()d represens he oal emissions of nirogen oides, in millions of meric ons, during he period 97 o. (b) We esimae he inegral using lef- and righ-hand sums: Lef sum = (6.9)5 + (6.4)5 + (7.)5 + (5.8)5 + (5.5)5 + (5.)5 = Righ sum = (6.4)5 + (7.)5 + (5.8)5 + (5.5)5 + (5.)5 + (.6)5 = 76.. We average he lef- and righ-hand sums o find he bes esimae of he inegral: 3 f()d = 77.8 million meric ons. Beween 97 and, abou 77.8 million meric ons of nirogen oides were emied.

22 38 Chaper Five /SOLUTIONS. We use lef- and righ-hand sums o esimae he oal amoun of coal produced during his period: Lef sum = (.8)5 + (3.6)5 + (4.6)5 + (4.99)5 + (8.6)5 + (9.33)5 = Righ sum = (3.6)5 + (4.6)5 + (4.99)5 + (8.6)5 + (9.33)5 + (.46)5 = We see ha Toal amoun of coal produced = quadrillion BTU. The oal amoun of coal produced is he definie inegral of he rae of coal producion r = f() given in he able. Since is in years since 96, he limis of inegraion are = and = 3. We have Toal amoun of coal produced = 3 f() d quadrillion BTU. 3. The inegral 3 v()d represens he change in posiion beween ime = and = 3 seconds; i is measured in meers. 4. The inegral 6 a()d represens he change in velociy beween imes = and = 6 seconds; i is measured in km/hr. 5. The inegral 4 f()d represens he change in he world s populaion beween he years and 4. I is measured in billions of people. 6. The inegral 5 s()d represens he change in saliniy (sal concenraion) in he firs 5 cm of waer; i is measured in gm/lier. 7. For any, consider he inerval [, + ]. During his inerval, oil is leaking ou a an approimaely consan rae of f() gallons/minue. Thus, he amoun of oil which has leaked ou during his inerval can be epressed as Amoun of oil leaked = Rae Time = f() and he unis of f() are gallons/minue minues = gallons. The oal amoun of oil leaked is obained by adding all hese amouns beween = and = 6. (An hour is 6 minues.) The sum of all hese infiniesimal amouns is he inegral Toal amoun of oil leaked, in gallons = 6 f() d. 8. If H() is he emperaure of he coffee a ime, by he Fundamenal Theorem of Calculus Therefore, Change in emperaure = H() H() = H ()d = 7(.9 ) d. H() = H() + 7(.9 )d = 45.8 C. 9. The oal amoun of anibodies produced is Toal anibodies = 4 r()d.47 housand anibodies. (a) In 99, when =, gas consumpion was 77 millions of meric ons of oil equivalen. In, when =, gas consumpion was N = () = 83 million meric ons of oil equivalen. (b) We use an inegral o approimae he sum giving he oal amoun consumed over he -year period: Toal amoun of gas consumed = ( ) d = 46, million meric ons of oil equivalen.. The quaniy S is he rae of producion, in megawas per year. Use an inegral o approimae he sum giving oal producion: Toal PV producion = 77e.368 d = 9,89.83 megawas.

23 5.4 SOLUTIONS 39. (a) The disance raveled in he firs 3 hours (from = o = 3) is given by 3 (4 )d. (b) The shaded area in Figure 5.43 represens he disance raveled. v() 4 3 Figure 5.43 (c) Using a calculaor, we ge So he oal disance raveled is 9 miles. 3 (4 )d = Since v() for 3, we can find he oal disance raveled by inegraing he velociy from = o = 3: 3 Disance = ln( + ) d = 3.4 f, evaluaing his inegral by calculaor. 4. Suppose F() represens he oal quaniy of waer in he waer ower a ime, where is in days since April. Then he graph shown in he problem is a graph of F (). By he Fundamenal Theorem, F(3) F() = 3 F ()d. We can calculae he change in he quaniy of waer by calculaing he area under he curve. If each bo represens abou 3 liers, here is abou one bo, or 3 liers, from = o =, and 6 boes, or abou +8 liers, from = o = 3. Thus 3 F ()d = 8 3 = 5, so he final amoun of waer is given by F(3) = F() + 3 F ()d =, + 5 = 3,5 liers. 5. (a) The weigh growh rae is he derivaive of he weigh funcion. The fac ha he weigh growh rae is increasing means ha he weigh funcion has an increasing derivaive. Thus he graph of he weigh funcion is concave up. (b) The oal weigh growh during he fory week gesaion equals he inegral of he growh rae over he ime period. Thus he birh weigh equals he area under he graph. The region is riangular, wih base 8 weeks and heigh. kg/week. Thus, Toal weigh = 8(.) = 3.8 kg 3. kg.

24 3 Chaper Five /SOLUTIONS 6. (a) (i) The income curve shows he rae of change of he value of he fund due o inflow of money. The area under he curve, 5 I() d, represens he oal change in he value of he fund ha is due o income. I is he quaniy of money, in billions of dollars, ha is projeced o flow ino he fund beween and 5. (ii) The ependiure curve shows he rae of change of he value of he fund due o ouflow of money. The area under he curve, 5 E() d, represens he magniude of he change in he value of he fund ha is due o epenses. I is he quaniy of money, in billions of dollars, ha is projeced o flow ou of he fund beween and 5. (iii) The area beween he income and ependiure curves, 5 I() E()d, represens he difference beween oal income and oal epenses beween and 5. I is he projeced change in value of he fund beween and 5. (b) In he figure, we see ha he value of he fund was abou billion dollars in and is projeced o be abou 35 billion dollars in 5. The fund is projeced o increase in value by abou 5 billion dollars, and ha is he area beween he income and ependiure curves on he graph. 7. (a) When income is greaer han epenses, before 3, he value of he fund increases. When he epenses are greaer han income, afer 3, he value of he fund decreases. The value increases from hrough 3, and hen declines. The value is a maimum in 3, where he income and ependiure curves cross. (b) The value of he fund is represened by he heigh of he curve. Is highes poin is in 3, when he value is a maimum. 8. The rae of change of he value of he fund is he difference I() E() beween he rae of increase due o income and he rae of decrease due o epenses. To find he oal change in he value, inegrae he rae of change: Change in value of fund = 3 I() E()d. 9. The change in he number of acres is he inegral of he rae of change. We have Change in number of acres = 4 (8 )d = 67. The number of acres he fire covers afer 4 hours is he original number of acres plus he change, so we have Acres covered afer 4 hours = + 67 = 67 acres.. The change in he amoun of waer is he inegral of rae of change, so we have 6 Number of liers pumped ou = (5 5e. )d = 58.4 liers. Since he ank conained liers of waer iniially, we see ha Amoun in ank afer one hour = 58.4 = 74.6 liers.. Looking a he area under he graph we see ha afer 5 years Tree B is aller while Tree A is aller afer years.

25 5.4 SOLUTIONS 3. The area under A s curve is greaer han he area under B s curve on he inerval from o 6, so A had he mos oal sales in he firs 6 monhs. On he inerval from o, he area under B s curve is greaer han he area under A, so B had he mos oal sales in he firs year. A approimaely nine monhs, A and B appear o have sold equal amouns. Couning he squares yields a oal of abou 5 sales in he firs year for B and 7 sales in firs year for A. 3. (a) The black curve is for boys, he colored one for girls. The area under each curve represens he change in growh in cenimeers. Since men are generally aller han women, he curve wih he larger area under i is he heigh velociy of he boys. (b) Each square below he heigh velociy curve has area cm/yr yr = cm. Couning squares lying below he black curve gives abou 43 cm. Thus, on average, boys grow abou 43 cm beween ages 3 and. (c) Couning squares lying below he black curve gives abou 3 cm growh for boys during heir growh spur. Couning squares lying below he colored curve gives abou 8 cm for girls during heir growh spur. (d) We can measure he difference in growh by couning squares ha lie beween he wo curves. Beween ages and.5, he average girl grows faser han he average boy. Couning squares yields abou 5 cm beween he colored and black curves for.5. Couning squares beween he curves for.5 8 gives abou 8 squares. Thus, here is a ne increase of boys over girls by abou 8 5 = 3 cm. 4. (a) In he beginning, boh birh and deah raes are small; his is consisen wih a very small populaion. Boh raes begin climbing, he birh rae faser han he deah rae, which is consisen wih a growing populaion. The birh rae is hen high, bu i begins o decrease as he populaion increases. (b) baceria/hour baceria/hour B D B D 6 5 ime (hours) 6 5 ime (hours) Figure 5.44: Difference beween B and D is greaes a 6 The baceria populaion is growing mos quickly when B D, he rae of change of populaion, is maimal; ha happens when B is farhes above D, which is a a poin where he slopes of boh graphs are equal. Tha poin is 6 hours. (c) Toal number born by ime is he area under he B graph from = up o ime. See Figure Toal number alive a ime is he number born minus he number ha have died, which is he area under he B graph minus he area under he D graph, up o ime. See Figure baceria baceria B D 5 5 ime (hours) Figure 5.45: Number born by ime is B()d N B D ime (hours) 5 5 Figure 5.46: Number alive a ime is (B() D())d From Figure 5.46, we see ha he populaion is a a maimum when B = D, ha is, afer abou hours. This sands o reason, because B D is he rae of change of populaion, so populaion is maimized when B D =, ha is, when B = D. 5. The quaniies consumed equal he areas under he graphs of he raes of consumpion. There is more area under he graph of fa consumpion rae, so more fa is consumed han proein. 6. (a) Figure 5.47 is he graph of a rae of blood flow versus ime. The oal quaniy of blood pumped during he hree hours is given by he area under he rae graph for he hree-hour ime inerval. The area can be esimaed by couning grid boes under he graph.

26 3 Chaper Five /SOLUTIONS Each grid recangle has area 3 minues / lier/minue = 5 liers, represening 5 liers of blood pumped. The grid boes in he graph are sacked in si columns. Esimaing he number of boes in each column under he graph gives Number of boes = = 8.75 boes. Approimaely Amoun of blood pumped = (8.75)(5) = 43.5 liers. Thus, abou 43 liers of blood are pumped during he hree hours leading o deah. (b) Since f() is he pumping rae in liers/minue a ime hours, 6f() is he pumping rae in liers/hour. Thus 3 6f() d gives he oal quaniy of blood pumped in liers during he hree hours. (c) During hree hours wih no bleeding, he hear pumps 5 liers/minue for 3 6 = 8 minues. Thus Toal blood pumped = 5 8 = 9 liers. This is = liers more han pumped wih lier bleeding. Thus, abou 47 liers are pumped. This corresponds o he area on he graph beween he 5 liers/minue line and he pumping rae for he lier bleed. See Figure pumping rae of hear (liers per minue) Bled lier Bled liers Reurn o normal 3 Figure 5.47 hours Deah 7. (a) Figure 5.48 is he graph of a rae of blood flow versus ime. The oal quaniy of blood pumped during he hree hours is given by he area under he rae graph for he hree-hour ime inerval. The area can be esimaed by couning grid boes under he graph. Each grid recangle has area 3 minues / lier/minue = 5 liers, corresponding o 5 liers of blood pumped. The grid boes in he graph are sacked in si columns. Esimaing he number of boes in each column under he graph gives Approimaely Number of boes = = 5 boes. Amoun of blood pumped = 5 5 = 75 liers. Thus, abou 75 liers of blood are pumped during he hree hours leading o full recovery. (b) Since g() is he pumping rae in liers/minue a ime hours, 6g() is he pumping rae in liers/hour. Thus 3 6g()d gives he oal quaniy of blood pumped in liers during he hree hours. (c) During hree hours wih no bleeding, he hear pumps 5 liers/minue for 3 6 = 8 minues. Thus, Toal blood pumped = 5 8 = 9 liers. This is 9 75 = 5 liers more han pumped wih lier bleeding. This corresponds o he area on he graph beween he 5 liers/minue line and he pumping rae for he lier bleed. See Figure 5.47.

27 5.4 SOLUTIONS 33 pumping rae of hear (liers per minue) Bled lier Bled liers Reurn o normal 3 Figure 5.48 hours Deah 8. Since W is in ons per week and is in weeks since January, 5, he inegral 5 W d gives he amoun of wase, in ons, produced during he year 5. Toal wase during he year = e.8 d = ons. Since wase removal coss $5/on, he cos of wase removal for he company is = $ The velociy is consan and negaive, so he change in posiion is 3 5 cm, ha is 5 cm o he lef. 3. From = o = 3 he velociy is consan and posiive, so he change in posiion is 3 cm, ha is 6 cm o he righ. From = 3 o = 5, he velociy is negaive and consan, so he change in posiion is 3 cm, ha is 6 cm o he lef. Thus he oal change in posiion is. The paricle moves 6 cm o he righ, followed by 6 cm o he lef, and reurns o where i sared. 3. From = o = 5 he velociy is posiive so he change in posiion is o he righ. The area under he velociy graph gives he disance raveled. The region is a riangle, and so has area (/)bh = (/)5 = 5. Thus he change in posiion is 5 cm o he righ. 3. From = o = 4 he velociy is posiive so he change in posiion is o he righ. The area under he velociy graph gives he disance raveled. The region is a riangle, and so has area (/)bh = (/)4 8 = 6. Thus he change in posiion is 6 cm o he righ for = o = 4. From = 4 o = 5, he velociy is negaive so he change in posiion is o he lef. The disance raveled o he lef is given by he area of he riangle, (/)bh = (/) =. Thus he oal change in posiion is 6 = 5 cm o he righ. 33. From = o = 3, you are moving away from home (v > ); hereafer you move back oward home. So you are he farhes from home a = 3. To find how far you are hen, we can measure he area under he v curve as abou 9 squares, or 9 km/hr hr = 9 km. To find how far away from home you are a = 5, we measure he area from = 3 o = 5 as abou 5 km, ecep ha his disance is direced oward home, giving a oal disance from home during he rip of 9 5 = 65 km. 34. (a) A = minues, she sops moving oward he lake (wih v > ) and sars o move away from he lake (wih v < ). So a = minues he cyclis urns around. (b) The cyclis is going he fases when v has he greaes magniude, eiher posiive or negaive. Looking a he graph, we can see ha his occurs a = 4 minues, when v = 5 and he cyclis is pedaling a 5 km/hr away from he lake. (c) From = o = minues, he cyclis comes closer o he lake, since v > ; hereafer, v < so he cyclis moves away from he lake. So a = minues, he cyclis comes he closes o he lake. To find ou how close she is, noe ha beween = and = minues he disance she has come closer is equal o he area under he graph of v. Each bo represens 5/6 of a kilomeer, and here are abou.5 boes under he graph, giving a disance of abou km. Since she was originally 5 km away, she hen is abou 5 = 3 km from he lake. (d) A = minues she urns around, since v changes sign hen. Since he area below he -ais is greaer han he area above, he farhes she is from he lake is a = 6 minues. Beween = and = 6 minues, he area under he graph is abou.8 km. (Since 3 boes 5/6 =.8.) So a = 6 she will be abou 3+.8 = 3.8 km from he lake. 35. Looking a he figure in he problem, we noe ha Produc B has a greaer peak concenraion han Produc A; Produc A peaks sooner han Produc B; Produc B has a greaer overall bioavailabiliy han Produc A. Since we are looking for he produc providing he faser response, Produc A should be used as i peaks sooner.

28 34 Chaper Five /SOLUTIONS 36. Looking a he figure in he problem, we noe ha Produc A has a greaer peak concenraion han Produc B; Produc A peaks much faser ha Produc B; Produc B has greaer overall bioavailabiliy han Produc A. Since Produc B has greaer bioavailabiliy, i provides a greaer amoun of drug reaching he bloodsream. Therefore Produc B provides relief for a longer ime han Produc A. On he oher hand, Produc A provides faser relief. 37. C Drug A Drug B Figure The oal number of worker-hours is equal o he area under he curve. The oal area is abou 4.5 boes. Since each bo represens ( workers)(8 hours) = 8 worker-hours, he oal area is 6 worker-hours. A $ per hour, he oal cos is $, The ime period 9am o 5pm is represened by he ime = o = 8 and = 4 o = 3. The area under he curve, or oal number of worker-hours for hese imes, is abou 9 boes or 9(8) = 7 worker-hours. The oal cos for 9am o 5pm is (7)() = $7. The area under he res of he curve is abou 5.5 boes, or 5.5(8) = 44 worker-hours. The oal cos for his ime period is (44)(5) = $66. The oal cos is abou = $3,8. 4. (a) The amoun when = is four imes he maimum accepable limi, or 4.6 =.4. Therefore he level of radiaion is given by R() =.4(.996). We wan o find he value of making R() =.6. The graph of he funcion in Figure 5.5 shows ha R() =.6 a abou = 346, so he level of radiaion reaches an accepable level afer abou 346 hours (or abou weeks). Alernaively, using logarihms gives (ime).6.4 = (.996) ln(.5) = ln(.996) = ln(.996) = ln(.5) ln(.996) = R() =.4(.996) Figure 5.5 (b) Toal radiaion emied beween = and = 346 is 346.4(.996) d 449 millirems.

29 5.5 SOLUTIONS (a) The disance raveled is he inegral of he velociy, so in T seconds you fall (b) We wan he number T for which T T 49(.887 ) d. 49(.887 ) d = 5. We can use a calculaor or compuer o eperimen wih differen values for T, and we find T 7 seconds. 4. (a) The acceleraion is posiive for < 4 and for a iny period before = 6, since he slope is posiive over hese inervals. Jus o he lef of = 4, i looks like he acceleraion is approaching. Beween = 4 and a momen jus before = 6, he acceleraion is negaive. (b) The maimum aliude was abou 5 fee, when was a lile greaer han 4 (here we are esimaing he area under he graph for 4). (c) The oal change in aliude for he Mongolfiers and heir balloon is he definie inegral of heir velociy, or he oal area under he given graph (couning he par afer = 4 as negaive, of course). As menioned before, he oal area of he graph for 4 is abou 5. The area for > 4 is abou. So subracing, we see ha he balloon finished 8 fee or so higher han where i began. Soluions for Secion 5.5. The unis for he inegral ( ) C (q)dq are dollars ons (ons) = dollars. C (q)dq represens he cos of increasing producion from 8 ons o 9 ons.. We are old ha C (q) = q 5q + 7 for q 5. We also know ha 5 Since we are old ha fied coss are $5 we know ha C (q)dq = C(5) C(). C() = $5. Thus 3. (a) 5 C(5) = C (q)dq + $5 $4,667. C (q) q 5 Figure 5.5 The oal variable cos of producing 5 unis is represened by he area under he graph of C (q) beween and 5, or 5 (.5q q + 56)dq.

30 36 Chaper Five /SOLUTIONS (b) An esimae of he oal cos of producing 5 unis is given by, + 5 (.5q q + 56)dq. This represens he fied cos ($,) plus he variable cos of producing 5 unis, which is represened by he inegral. Using a calculaor, we see So he oal cos is approimaely 5 (.5q q + 56)dq,775. $, + $,775 = $,775. (c) C (5) =.5(5) = 8.5. This means ha he marginal cos of he 5h iem is 8.5. In oher words, he 5s iem will cos approimaely $8.5. (d) C(5) is he oal cos of producing 5 iems. This can be found by adding he oal cos of producing 5 iems (found in par (b)) and he addiional cos of producing he 5s iem (C (5), found in (c)). So we have C(5), = $, (a) There are approimaely 5.5 squares under he curve of C (q) from o 3. Each square represens $, so he oal variable cos o produce 3 unis is around $55. To find he oal cos, we had he fied cos Toal cos = fied cos + oal variable cos =, + 55 = $, 55. (b) There are approimaely.5 squares under he curve of C (q) from 3 o 4. Each square represens $, so he addiional cos of producing iems 3 hrough 4 is around $5. (c) Eaminaion of he graph ells us ha C (5) =. This means ha he cos of producing he 6h iem is approimaely $. 5. The area beween 97 and 99 is abou 5.3 grid squares, each of which has area.(5) =.5 million people. So Change in populaion = P () d 5.3(.5) 7.65 million people. The populaion of Tokyo increased by abou 8 million people beween 97 and Since C() = 5, he fied cos mus be $5. The oal variable cos o produce unis is C (q)dq = (q 6q + 7) dq = $ (using a calculaor). The oal cos o produce unis is he fied cos plus he variable cos of producing unis. Thus, Toal cos = $5 + $ = $, (a) Toal variable cos in producing 4 unis is We esimae his inegral: and so 4 C (q)dq 4 C (q) dq. Lef hand sum = 5() + () + 8() + () = 85; Righ hand sum = () + 8() + () + 8() = 88; = $865. Toal cos = Fied cos + Variable cos = $, + $865 = $8,65. (b) C (4) = 8, so we would epec ha he 4s uni would cos an era $8.

31 5.5 SOLUTIONS (a) Using he Fundamenal Theorem we ge ha he cos of producing 3 bicycles is or 3 C(3) = 3 C (q)dq + C() 6 dq + $ $ q + 5 (b) If he bikes are sold for $ each he oal revenue from 3 bicycles is and so he oal profi is 3 = $6, $6 $459.4 = $ (c) The marginal profi on he 3 s bicycle is he difference beween he marginal cos of producing he 3 s bicycle and he marginal revenue, which is he price. Thus he marginal profi is C (3) = 6 4 = $ (a) 5 R (q) q Figure 5.5 (b) By he Fundamenal Theorem, R (q)dq = R() R(). R() = because no revenue is produced if no unis are sold. Thus we ge R() = R (q)dq $,. (c) The marginal revenue in selling he s uni is given by R () = $8/uni. The oal revenue in selling unis is: R() + R () = $,8.. (a) The area under he curve of P () from o gives he change in he value of he sock. Eaminaion of he graph suggess ha his area is greaes a = 5, so we conclude ha he sock is a is highes value a he end of he 5h week. (Some may also conclude ha he area is greaes a =.5, making he sock mos valuable in he middle of he second week. Boh are valid answers.) Since P () < from =.5 o abou = 3.8, we know ha he value of he sock decreases in his inerval. This is he only inerval in which he sock s value is decreasing, so he sock will reach is lowes value a he end of his inerval, which is near he end of he fourh week. (b) We know ha P() P() is he area under he curve of P () from o, so eaminaion of he graph leads us o conclude ha P(4) < P(3) P() < P() P() < P(5).

32 38 Chaper Five /SOLUTIONS. (a) The rae of ice formaion is dy d = inches/hour. So he amoun of ice formed in 8 hours equals 8 y(8) y() = d. Now, y() = because he ice sars forming a =. Thus here are 8 d 7.54 inches of ice in 8 hours. (b) A 8 hours, dy 8 d = = inches/hour. The ice is increasing a a rae of =.4 inches/hour.. The oal change in he ne worh of he company from 5 ( = ) o 5 ( = ) is found using he Fundamenal Theorem: Change in ne worh = f() f() = f ()d = ( ) d = 6, dollars. The worh of he company in 5 is he worh of he company in 5 plus he change in worh beween 5 and 5. Thus, in 5, Ne worh = f() = f() + Change in worh = Worh in 5 + Change in worh beween 5 and 5 = 4, + 6, = $56,. 3. We find he changes in f() beween any wo values of by couning he area beween he curve of f () and he -ais. Since f () is linear hroughou, his is quie easy o do. From = o =, we see ha f () oulines a riangle of area / below he -ais (he base is and he heigh is ). By he Fundamenal Theorem, so f ()d = f() f(), f() + f ()d = f() f() = = 3 Similarly, beween = and = 3 we can see ha f () oulines a recangle below he -ais wih area, so f() = 3/ = /. Coninuing wih his procedure (noe ha a = 4, f () becomes posiive), we ge he able below f() 3/ / / / / Soluions for Chaper 5 Review. (a) Since he velociy is decreasing, for an upper esimae, we use a lef sum. Wih n = 5,we have =. Then Upper esimae = (44)() + (4)() + (4)() + (4)() + (37)() = 48. (b) For a lower esimae, we use a righ sum, so Lower esimae = (4)() + (4)() + (4)() + (37)() + (35)() = 39.

33 SOLUTIONS o Review Problems For Chaper Five 39. (a) (i) Since he velociy is increasing, for an upper esimae we use a righ sum. Using n = 4, we have = 3, so (ii) Using n =, we have = 6, so Upper esimae = (37)(3) + (38)(3) + (4)(3) + (45)(3) = 48. Upper esimae = (38)(6) + (45)(6) = 498. (b) The answer using n = 4 is more accurae as i uses he values of v() when = 3 and = 9. (c) Since he velociy is increasing, for a lower esimae we use a lef sum. Using n = 4, we have = 3, so Lower esimae = (34)(3) + (37)(3) + (38)(3) + (4)(3) = We esimae he inegral by finding lef- and righ-hand sums and averaging hem: We have 6 f()d Lef-hand sum = = 4 Righ-hand sum = = = b a f()d is measured in ( ) miles (hours) = miles. hours 5. The unis of measuremen are meers per second (which are unis of velociy). 6. The unis of measuremen are dollars. 7. The unis of measuremen are foo-pounds (which are unis of work). 8. (a) Using recangles under he curve, we ge (b) Using recangles above he curve, we ge Acres defaced ()( ) = 3.6 acres. Acres defaced ()( ) = 6.9 acres. (c) The number of acres defaced is beween 3.6 and 6.9, so we esimae he average, 5.5 acres. 5 d =.44 ( + ) d = d = d = (r+) dr =.83 z + z dz = Since f() is posiive along he inerval from o 6 he area is simply 6. Since 3 for, we have The inegral was evaluaed on a calculaor. Area = 6 ( 3 )d =.83. ( + )d = 84.

34 3 Chaper Five /SOLUTIONS 7. Since / /3 for, we have 8. The inegral was evaluaed on a calculaor. y = Area = ( /3 / )d =.833. y y = 3 9 Figure 5.53 Inspecion of he graph ells us ha he curves inersec a (,) and (3,9), wih 3 for 3, so we can find he area by evaluaing he inegral Using echnology o evaluae he inegral, we see 3 (3 )d. So he area beween he graphs is See Figure (3 )d = 4.5. y y = y = Figure 5.54 Inspecion of he graph ell us ha he curves inersec a (,) and (,), wih for, so we can find he area by evaluaing he inegral Using echnology o evaluae he inegral, we see So he area beween he graphs is abou.667. ( )d. ( )d (a) An overesimae is 7 ons. An underesimae is 5 ons. (b) An overesimae is = 74 ons. An underesimae is = 59 ons.