5.5 - 468 CHAPTER 5 INTEGRALS (5)At=b‘a:fl:2. n 3 Q(6) z...

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Unformatted text preview: 468 CHAPTER 5 INTEGRALS (5)At=b‘a:fl:2. n 3 Q(6) z M3 = 2[r(1) + r(3) + r(5)] 2 2(10 + 36 + 54) = 2(100) :200 tonnes. 61. From the Net Change Theorem. the increase in cost if the production level is raised from 2000 yards to 4000 yards is C(4000) — C(2000) = 4000 C'(x) day. 2000 4000 4000 / 0’00) dac : / (3 — 0.011: + 0000006922) data 2000 2000 4000 2000 : 60.000 — 2.000 : $58000 : [3x — 0.005302 + 0000002233] 62. By the Net Change Theorem. the amount of water after four days is 25000 + fo“ r(t) dt % 25000 + M4 = 25000 + 4&0 [r(0.5) + 7’(1.5) + «2.5) + r(3.5)] % 25.000 + [1500 + 1770 + 740 + (7690)] : 28.320 liters 63. (a) We can find the area between the Lorenz curve and the line 3; : :c by subtracting the area under 3; : L(w) from the area under y : 3c. Thus. area between Lorenz curve and line y = :0 fol [at i dx area under line 3/ = x _ f1 x dx :fo [miL(a:)ldw _M:2/Ol[xL(w)]dx coefficient of inequality 2 [962/215 1/2 (b) L(:c) : 155.52 + 11255 => L(50%) 2 LG) 2 438 + 214 = 31—: : 0.39583. so the bottom 50% of the households receive at most about 40% of the income. Using the result in part (a). coefficient ofinequality = 2f01 [an — dx 2 2f01(x — 1—52302 7 dsc amaze—5w2>dz=2155<x7x2>dw 2 3 1 =2[%w —%mlo:%(%—%):%(é)=% 64. (a) From Exercise 4.1 .72(a). v(t) : 0.00146t3 ~ 0.115532:2 + 249816925 — 21.26872. (b) h(125) i h(0) : [0125 o(t) dt : [0.000365t4 — (103851153 +12.490845t2 — 212687225]125 0 z 206.407 ft 5.5 The Substitution Rule 1. Let u : 336. Then du : 3dx. so dac : fidu. Thus. fcos 322612: : fcos du) : éfcosudu = ésinu + C’ : ésin3w + 0. Don’t forget that it is often very easy to check an indefinite integration by differentiating your answer. In this case. di sin 3m + C) = %(cos 3:3) - 3 : cos 32:, the desired result. :5 2. Letu = 4+:c2. Then du : 2:1:dx andmdw : édu. so fm<4+$2)10 d$ : fu10(%du) : é o in” +0: %(4+x2)11 +0. 3. Let u : m3 + 1. Then du : 35“ d3: and 1:2 (190 = gdu. so 3/2 fx2vw3+1d$:.lx/5(%du) :égfl +C=§-§u3/2+C: 3(x3+1)3/2+C. SECTION 5.5 THE SUBSTHUTION RULE 1 1 4.Letu:\/E.Thendu=mdscandfidw=2dmso fsmfidw: sinu(2du):2(—cosu)+C=~2cos\/§+C. fl 5.Letu:1+2m.Thendu=2dacanddm=édufio 4 _ #31 1 E __i C_1___1 C /(1+2m)3dx#4/u (2du)—2_2+Cg u2+ i (1+2$)2+. 10. 11. 12. 13. 14. 15. 16. 17. Letu : sine. Then du : c086d6, so fesmgcosfidfi : feudu = 6" +0 2 65mg +C’. . Let u : x3 + 5. Then du : 3x2dm and mgda: : édu. so lion” + C = $(w3 + 5)10 + C, f$2($3+5)9dw=fu9(%du) = %- Letu = 3x—2. Then du : 3dzc and dx = fidu. so f(3m—2)2°dm:fu2°(§du) : é. Let u : 2 — ac. Then du = —dx and dsc : —du. so f(2—m)6dac: fu6(—du) = —%u7+C : —%(2—x)7+C. Letu = 1 +m+2cc2. Then du = (1 +4x)dm, so 1+4a: du —-——-dac= —: V1+m+2x2 x/T—L Letu : x2 + 1. Then du = 2mda: and mdm : édu. so 1/2 1/2 NI 2—1111” +0 : $(3x ~ 2)21 +0. —1 :2: _ —2 l ini _—_l _ /—($2+1)2dw—/u (203107 u+Ci2u+C—2(x2+1)+0. Letu = 5 — 3x. Then du = —3d:1: and dm 2 —§du. so M2. :/%<—%du>= Letu : m2 + 1. Then du : 2xdac and soda: : %du. so 1 -d f m dac—/2 u:éln|u|+C=élnlz2+1|+C=%1n(ac2—+—1)+C [sinceac2+1>0] 3112-1-1 _ or ln(m2+ 1)1/2+C=1n\/;1:2+ 1+C’. u Letu = 2y+ 1. Then du : Zdy anddy : §du, so 31 —3 _3__ _ —s; ,_._—4 1.—(2y+1)5dyT/3u (“m—2 i4“ +C_8(2y+1) Letu=5t+4.Thenduz5dtanddt=édu.so _1___ _ -2~7 1 _l. 1 /(5t+4)2.7dt*/u (sdu)—5 —1.7“ Let u : 4 — t. Then du : idt and dt : —du. so fx/4 — tdt : fu1/2 (—du) 2 gas/2 +0 : —§(4 — t)3/2 +0. —1.7+C: __u —%ln|u}+C=—%ln|573$|+0. 4 /u_1/2du:L+C:2\/1+x+2x2+0. .Letuzsc2 +3.Thendu=2xd$.sof2$(ac2+3)4d$ = fu4du= éus +C : %(m2+3)5 +C. 469 470 U CHAPTER5 INTEGRALS 18. Let u 2 23/4 — 1. Then du 2 8313 dy and y3 dy 2 édu. so 19. 20. 21. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. . Let u 2 tan—1 1:. Then du 2 fysx/Wdy =fu1/2(§du):§ 3 . 2u3/2+C: Let u 2 mt. Then du 2 7rdt and dt 2 fidu. so fsinfitdt 2 fsinu(%du) 2 1 7r Let u 2 29. Then du 2 2d0 and d0 2 ) du, so 1—12(2y4 21)?”2 + C. (—cosu)+C’2 ficosm‘A—C. fsec26 tan20d0 2 fsecutanugdu 2 ésecu—FC’: ésec20+0 Letu 2 111$. Then du 2 so I / dcc 1+x2’ T Let u 2 Then du 2 dt /cos\/i \fl 2x/ix/i (1n 9:)2 so/ — and —1—dt 2 2du. so dm2fu2du2 §u3+C2 %(Inw)3+C’. 21 2 tf1$dx2fudu2u§+0 $2 dt 2 fcosu(2du) 22sinu+C2 28inx/i-i—C". Letu 2 1 +933”. Then du 2 g331/2dsc and fidw 2 gdu, so 2 ffisin(1+m3/2)d$ 2 fsinugdu) 2 2 3 ~(—cos’u) + C 2 2% cos(1 +w3/2) + C. Letu 2 sine. Then du 2 cosOd6,sofc050 sin69d6 2 fuedu 2 %u7 +0 2 §sin76+0 Let u 2 1 + tan 6. Then du 2 seCZOdB. s0 f(1+tan6)‘r’seC20d€ 2 fu5du 2 éus +0 2 %(1 +tan6)6 +0. Letu21+ez.Thendu2emdw,sofeI\/1+exdx2f\/1Idu2gu3/2+C2 (1 +633” +0. ODIN 0r: Letu 2 V1 +61. Then uz 21+ ex and 2udu 2 6”” dm. so femx/1+emdm2fu-2udu2 %u3+02 §(1+e$)3/2+C. Let u 2 cost. Then du 2 — sintdt and sintdt 2 idu. so feCOStsintdt 2 fe" (—du) 2 ie” + C 2 —eC°St + C'. Letu 21+ 23. Then du 2 322 dz and 22 dz 2 l du, so 3 22 —1/3 1 /?—/1+—Z3d2:/u (édu):§‘ Let u 2 (1.752 + 21m + 6. Then du 2 2(aa: + b) day and (a2: + 1)) da: 1 guZ/s—i-C: %(1+23)2/3+C. — 5 du. so 2éfuil/zdu2ul/2—1—C'2x/asc2+2bm+c+C. : :1niu|+021nllnml+0. ‘ ’LL / (ax+b)dm _ édu x/ax2+2bm+c fl Letu 2 lnx. Then du. 2 $.50 day so xlnm Letu2em+1.Thendu2eIdx.so/ e em+1 dsc2 @ :1n|u| +0: ln(e1 +1) +0. u SECTION 5.5 THE SUBSTiTUTlON RULE :1 471 33. Letu : cotx. Then du : — csc2 :cdac and CSC2 scdw = Adu. so 3/2 /\/cotmcsc2md9:—/\/1T( du): 1:3/2 +C—A§(cotav)3/2+C. 7r 7r 1 1 34. Letu : —. Then du = ———dx and —d$ : ——du. so an m2 2:2 7r ' 1 1 /is(7;—/m—)dz:/C05u(ildu> = A—sinu+C= ——sin£ +C'. . :c 7r 7r 77 w 35. /cotxdac 2 / Césw dw. Let u : since. Then du = cosxdx. so smzc footscdxzfidu :lnlul +C= lnlsinml +0. 35. Let u : cos 3:. Then du = — sinwdx and sinzcdx : —du. so sinm ‘ _du _1 71 /———1+COS2xd:c:/1+u2:_tan u+C:Atan (cosm)+C. 37. Let u 2 sec ac. Then du 2 seem tang: dm. so fsec3$ tanmdw: fsec2x(secac tanx)dx: fu2du= §u3+C : gsecscc—kC. 38. Letu = $3 + 1. Thenm3 zue 1 and du : 3x2dac. so 1W +1m5dm=f W +1-m3 ~m2dw :fu1/3<u—1>(%du) = Hue/3 -u1/3)du 2 l (gum L amt/3) +0 2 3 7 4 (m3+1)7/3 — i(;1;3+1)4/3 +0 \IIH 39. Let u : b + of“. Then du : (a + 1)ca:“ dw. so /xa\/b+cata+1 d2: = /u1/2 1 du 1 Gus/2) + C = ———2 (b+C£lZa+l)3/2 +0. ((1 +1)c : (a + 1)c 3c(a + 1) 40. Let u : cost. Then du : * sintdt and sint dt : —du. so fsint sec2(cos t)dt = fsec2 u . (—du) : A tanu + C = — tan(cos t) + C. 41. Letu : 1+ :52. Then du = 2x dw. so 1+:c 1 :c ,1 ldu V d,: = 2—: 1 l /1+z2 5r /1+$2d$+/1+$2dm tan x+ u tan x+2ln|u|+C =tan_1w+%ln11+a:2| +C:tan_1x+%ln(1+a:2) +C’ [since1+.7:2 > 0]. 1 _ 2 _ m 7 Ed“ ,1 —1 _1, 71 2 42.Letu—x.Thendu—Zxdsc.so/1+x4dcc—/1+u2 fitan u+C—§tdn ($)+C. 43. Letu : III+ 2. Then du : dm. so —2 fé/ELZdez/uw du2/(u3/4A2uTl/4)du: : gm +2)”4 7 593+ 2):“4 +0 u7/4—2-gu3/4+C' «ms 44. Letu:le.Thena::l—uandd$:~du.so 2:2 (17102 1—2u+u2 d d : i _ —1/2 _ 1/2 3/2 /‘/i__x 567/ \fl—L ( U) W du 2u +u )du :—(2u1/2—2~gu3/2+§u5/2>+0:—2\/1—x+%(1im)3/2—%(1~x)5/2+C 472 1: CHAPTERS INTEGRALS In Exercises 45—48, let f(:c) denote the integrand and F(m) its antiderivative (with C : 0), 3x — 1 45. fix) : F2 _ 2m + D4. uz3m2—2m—l—1 => du:(6x—2)dm:2(3m~1)dw.so *0.75 3w—1 1 1 1 _4 “d : — —d :— d /(3L‘62—2$+1)4 m /u4<2 u 2/u u Notice that at :c : f changes from negative to positive. and F has a local minimum. 45. fix) I x .u:w2+1 => du=2xdcc,so 2 9: 1 1 1 /de= W(Eda) :E/u‘l/zdu —2 2 =u1/2-i-C: i/ac2+1+C. Note that at an : 0. f changes from negative to positive and F has a local —‘ minimum. 47. flats) : sin3 ac cos x. u : sinac 2) du : cosmdx, so 035 fsin3sccosasdac= fu3du= iu4+C: isin4zc+C Note that at x : 5. Chan es from ositive to ne ative and F has a local 2 g P g maximum. Also. both f and F are periodic with period 77. so at w = 0 and at x : 7r. f changes from negative to positive and F has local minima. 48. f(9):tan21‘)sec26.u:tan6 => du=sec20d6, so ftanzfi sec29dl9 = fuzdu: %u3+C’= §tan30+C Note that f is positive and F is increasing. At a: = 0. f : 0 and F has a horizontal tangent. 49. Letu : :c — 1. so du : dw. When 1: = 0. u : —1; when :0 = 2, u : 1. Thus, f02(:c i 1)?5 day : L11 u25 du : 0 by Theorem 7(b), since f (u) : u25 is an odd function. 50. Letu :4+3x.sodu :3daci Whensc :0.u:4;when$ : 7.u:25. Thus, u3/2 7 2-5 t 1 25 2 32 3/2 2 234 /\/4+3;L'd:c:/ fi(§du):§[fi] :§(25/ i4 )=§(125—8):?:26 0 4 / 4 51. Let u =1+ 2953~ so du = 69:2 dm. When a: = 0, u : 1; when :1: = 1, u = 3. Thus, f01w2(1+2m3)5 dw=ffustédu) : éléusli = 3%(36 —16>: 3—2<729~ 1) = : 52. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. SECTION 5.5 THE SUBSTITUTION RULE Letu 2 $2.30 (in 2 2wd93. Whencc 2 0.11 2 Ozwhenm 2 2 7r. Thus. ffimcos(zc2) dsc 2 f; cosu du) 0 2%[sinu]g2%(sin7r—sin0)2%(0—0)20. Letu 2 t/4, so du 2 idt. Whent 2 0.11 2 0: whent 2 7r. u 2 77/4. Thus. for sec2(t/4)dt 2 fofl/Al sec2 11 (4 du) 2 4[tanu]:/4 2 4(tang — tan 0) 2 4(1— 0) 2 4. Letu 2 7rt.sodu 2 7rdt. Whent 2 %.u 2 %;whent 2 $.11 2 Thus. 7r/2 _ 77/6 — f11//62CSC7TtCOt7Ttdt: ffl/Z cscu cotufidu) 2 %[~cscu] —%(1 — 2) 2 7r/6 flick tan3 9 d9 2 0 by Theorem 7(b). since f(0) 2 tan3 6 is an odd function. MK» 1 . . . . . 2 has an 1nfin1te dlscontmuity at w 2 2 f0 Easel—f3): does not exist since f(:1:) 2 m Letu 2 l/cc. so (111 2 21/m2d93. Whenx 21.11 2 1; whenzc 2 2. u 2 Thus. 2 l/m 1/2 / e dsc 2 f e" (—du) 2 — [tau]?2 2 1 1 —(el/2—e)2e—\/e. :62 Let u 2 2:32. so du 2 —2xd:c. When cc 2 0. u 2 0; when x 2 1.11 2 —1. Thus. flaw—m2 da: 2 f0_le"(—%du) 2 —l[eu]T1 2 —%(e_1 —e°) 2 %(1—1/e). 0 2 O Letu 2 c086. sodu 2 —sin6d6. When6 2 0.11 21:when6 2 2 Thus. 7r/3 ‘ 1/2 _ 1 1 / sm20 dQZ/ (in :/ u22du: Pl] : —1—(—2):1. 0 cos (9 1 U 1/2 “ 1/2 2 . x smm . . 6 IS an odd function. 1 + :c "/2 2:2 sin 1: . /W/2 1 + $6 dzt 2 0 by Theorem 7(b). smce f(.27) 2 Letu: 1+2m.sodu22dm.Whenx20.u21;whena: 2 13.21 227. Thus. /13 dm —/27u 0 ,3/(1+2m)2 1 Letu 2 sinx. so du 2 cosrdm. When as 2 0.1L 2 0; when as 2 u 2 1. Thus. _2/3(%du) : '3u1/3]27 : $6371) :3. 1 fen/2 cossc sin(sinx) d1: 2 fol sinudu 2 [— cosu]1 2 0 —(c03121)21—cosl. Letu2x21.sou+12manddu2dm. WhenzL’21.u20:whenx22.u21. Thus, 1 [fan/2r — 1dsc 2 +1)\/27du 2 f01(u3/2 + u1/2)du 2 [2115/2 + §u3/2]0 2 CHIN + COIN Letu21+2m,sow2%(u—1)anddu22dx.When$20.u21;Whenx24.u29.Thus. gun—1) /4 zda: i o V1+2I 1 fl 9 d 9 _ 3U : 411/ (u1/2 _u 1/2)du : % Bug/2 _2u1/2] 1 1 [113/2 — 3111/19 2 §[(27 9) (1 1 WIN 1 4 :1; :1 473 474 C CHAPTERS INTEGRALS 65. Letu : lnm,sodu : Whenw : 6.1L :1;whenm : e4;u = 4. Thus. :0 4 5 d 4 4 f m 2/ u_1/2du:2[u1/2] :2(271):2. e, m Inst 1 1 66Let—"1 d—d—mWh —0 —0'h —1 ~7‘ . uism 13,50 u—W. enm— .u— .wenxiiflriaThus. 1/2 - —1 7r/6 2 "/6 2 / Mm: udu:[u_] 2L 0 V1—502 0 4 d . , 1 . . . . 67. /0 fl does not extst Since f(a:) = W has an infinite discontinuity at x : 2. 68. Assume a > 0. Let u = a2 ~ $2. so du = —2xdz. When an : 0, u = a2; when m : at u : 0. Thus. 2 (1 (12 a to — was : Li: uW—é cm) = $10 New = % - Bus/2] : 0 69. Let u = m2 + (12, so du = 2m dm and Judy; : édu. When an : 0. u : (12; when as : a, u : 2a2. Thus. a 2‘12 20.2 20.2 f0 xx/m2+a2dm 2/ u1/2(%du) : égusfl] : Etta/2] a a 2 a2 : glam/2 — mas/2] = am —1>a3 70. fig x \/$2 + a2 due 2 0 by Theorem 7(b). since f($) = as V532 + a2 is an odd function. 71. From the graph. it appears that the area under the curve is about 1 + (a little more than % - 1 ~07), or about 1.4. The exact area is given by A : fol «23: + 1 (156. Let u : 2:): + 1. so du : 2dsc. The limits change to 2-0+1:1and2‘1+1:3.and 3 A:ffx/fi(%du)=%[§u3/2] :awfiinm/éi 1 m 1.399. wh— 72. From the graph. it appears that the area under the curve is almost % ~ 77 ~ 2.6. or about 4. The exact area is given by A : fOW(2 sinx — sin 2x)d9: : 72 [cos fig — f07r sin 2m dsc :—2(—171)—0:4 Note: f0" sin 22: day : 0 since it is clear from the graph of y : sin 2x 7r/2 0 sin 2£E dcc. that £32 sin 217d1c : ~ 73. First write the integral as a sum of two integrals: I = + 3)\/4 — 2:2 dcc : 11 + 12 : cc V4 — :32 dw + f: 3 V4 — x2 dx. [1 : 0 by Theorem 7(b). since f(:c) 2 an V4 — $2 is an odd function and we are integrating from so = —2 to ac : 2. We interpret 12 as three times the area of a semicircle with radius 2. so I : 0 + 3 - an - 22) : 677. SECTION 5.5 THE SUBSTITUTION RULE 3 475 74. Let u : 332. Then du : 2m dsc and the limits are unchanged (02 = 0 and 12 z 1), so 75. 76. 77. 78. 79. 80. 81. 82. I : fol act/1 — 2:4 dz = % fol v 1 — u2 du. But this integral can be interpreted as the area of a quarter—circle with radiusl.SoI:%-i(7r~12)= gm Letu : sow = u2 and dz = 2udu. When :2: : 0, u : 0; when ac : 1, u : 1. Thus. A1 =f01 efi dsc = fol e"(2udu) = 2f01ueudu. A2 : fol 2:106“c dzn = 2f01 ueu du. First Figure Second Figure Third Figure Let u = sin ac. so du = cosmdm. When m = 0. u = 0; when w = g, u 2 1. Thus, "/2 as“ sin 2x dm : W2 eSi”(2 sinzr cosx) dm : fol e”(2u du) : 2 fol ue” du. A3:0 0 Since A1 : A2 : A3. all three areas are equal. Let r(t) : me” with a : 450.268 and b : 1.12567, and n(t) 2 population after t hours. Since r(t) = n'(t), f: r(t) dt : n(3) — 71(0) is the total change in the population after three hours. Since we start with 400 bacteria, the population will be n(3) = 400 + f5 r(t) dt : 400 + [03 ma“ dt = 400 + 9 b 2(83b_1) 3 [ebt]0 : 400 + b x 400 +11.313 = 11,713 bacteria The volume of inhaled air in the lungs at time t is 2 02"”5 %sin 14% do) [substitute 2) : V(t) :fot f(u)du : jot ésin(%"u)du 27r i2_7r 5u.dv— 5 du] 27rt/5 _ zfikcosvh) — i i [~cos(2?"t) +1] 4” [1— cos(2?”t)] liters Number ofcalculators : 22(4) — w(2) : f; 5000 [1 ~ 100(t + 10W] dt 2 5000 [t + 100(t + 10W]: : 5000 [(4 + %) — (2 + %)] z 4048 Letu : 23:. Then du = 2dm, so f02 f(2:v)dz : f: du) 2 %f04 f(u)du 2 5(10) : 5. Letu : 2:2. Then du : 25cdx. so f03 xf(rc2) dm : fog du) 2 %f09 f(u) du = = 2. Let u : —x. Then du = —dx, so Y y = f(x) 1: f(—r) dx = 1:: f(u)(—du) : 1:: mo du = 1:: mo) dx From the diagram. we see that the equality follows from the fact that we are reflecting the graph of f , and the limits of integration, about the y—axis. Letu:;c+c. Then duzdm. so Y [5 f($ + 0) d1: : du : dar From the diagram. we see that the equality follows from the fact that y :f(x + c) y =f(x) we are translating the graph of f , and the limits of integration, by a distance c. b + c X k—c—H 476 C CHAPTERS INTEGRALS 83. Letu = 1 7:3. Then]: =1—uanddw : —du,s0 fol 950(1 7 m)” dx : f1" (1 — u)” ub(~du) = f0] ub(1 — u)“ du : f01 xb(17 my dm. 84. Letu: 7r—w. Thendu: Adm. Whena: :7r.u =0andwhenm:0,u:7r. So f0" scf(sin 11:) dsc = — ff(7r A u)f(sin(7r — 11)) du = f0"(7r — u)f(sin u) du :7rf07r f(sin u)du — f0” uf(sinu)du : fife” f(sinm)d$ — f0" xf(sin ac)d:I: :> 2ft)7r xf(sin:c)d:c = 7rf07r f(sin:c)dac => f0" mf(sinm)d:c = g f0" f(sinac)d:c. x sin w sin :c t 85. _— Z ._ Z . : , 1+cos2x x 2_sin2:c $f(51n$)~Whefe f(t) 2_t2. By Exerc1se 84. 7r xsinm " 7r 7‘ 7r 7' sins: — d : . d : _ I : — f0 1+Coszm :6 f0 If(sm:c) as 2/0 f(smm)dx 2/0 —1+m82mdflc Letu : COSm. Then du = —sinxd;z;_ Whenx : 7r‘u : ,1 and whsnx : O‘u :1. SO 71' /" sins: dw 71' f4 du 71' 1 du aha _1 ]1 _ —_——— :—— : — I _ n 2 0 1+cos2m 2 1 1+u2 2 _11+u2 2 “‘1 =g[tan7117 tan—1(—1)] : — (—EH : 7r— 5.6 The Logarithm Defined as an Integral We interpret In 1.5 as the area under the curve y : 1 /:c from x : 1 to w : 1.5. The area of the rectangle BCDE is i - g : The area of the 1. (a) trapezoid ABCD is % ~ %(1 + = Thus. by comparing areas. we 1 5 observe that g < In 1.5 < E, (b) With f(t) : 1/25, n : 10. and A3: : 0105, we have In 1.5 = [1"5(1/t)dt z (0.05)[f(1.025) + f(1.075) + - - ~ + f(1.475)] : (0'05)[14325 + 11575 + ’ ' ' + 1.4175] x 04054 2 — . i 1 . 2. (a) : ’ = —l. The 510 e of AD is 1/ 1 : ~l. Let cbe the t—coordmate of the mm on y : — w1th y t y t2 p 2 i 1 2 p t 1 1 1 2 . . . . slope 2. Then CZ — i ~> c — 2 => 0 : x/i smce c > 0. Therefore. the tangent line IS given by y $5: %(t —> y- étlx/i Since the graph of y : l/t is concave upward. the graph lies above the tangent line. that is. above the line segment BC. Now |AB| = *% + x/i and ICD| : —1 + So the area of the trapezoid ABCD is Hp; +\/§) + (—1+\/§)1] : _g +\/§z0.6642. So ln2 > area of trapezoid ABC'D > 0.66. ...
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This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.

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5.5 - 468 CHAPTER 5 INTEGRALS (5)At=b‘a:fl:2. n 3 Q(6) z...

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