ch 5 review - 478 CHAPTER 5 INTEGRALS 10 Let Ioga:c r and...

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Unformatted text preview: 478 CHAPTER 5 INTEGRALS 10. Let Ioga :c : r and loga y = 3. Then aT : m and as : y. (a) my : aras : ah” :> loga(xy) = 7" + s = loga z + loga y :3 ar . as b — 2 — 2 rib :> 1 — : i : — ( ) y as a oga y r s logaw logay (c) my : (ar)y : ary => loga(a:y) : Ty = yloga a: 5 Review CONCEPT CHECK 1. (a) 2le f (an?) Act is an expression for a Riemann sum of a function f. at: is a point in the ith subinterval [3-1.561] and A2: is the length of the subintervals. (b) See Figure 1 in Section 5.2. (c) In Section 5.2. see Figure 3 and the paragraph beside it. 2. (a) See Deﬁnition 5.2.2. (b) See Figure 2 in Section 5.2. (c) In Section 5.2. see Figure 4 and the paragraph above it. 3. See the Fundamental Theorem of Calculus after Example 9 in Section 5.3. 4. (a) See the Net Change Theorem after Example 5 in Section 5.4. (b) fit: 7’(t) (it represents the change in the amount of water in the reservoir between time t1 and time t2. 5. (a) [61020 v(t) dt represents the change in position of the particle from t = 60 to t = 120 seconds. (b) £31020 |v(t)| dt represents the total distance traveled by the particle from t = 60 to 120 seconds. (C) [61020 a(t) dt represents the change in the velocity of the particle from t = 60 to t = 120 seconds. 6. (a) f f(;z) dz is the family of functions {F | F’ = f}. Any two such functions differ by a constant. (b) The connection is given by the Evaluation Theorem: f (3:) dsc : [f f (x) dx]: if f is continuous. 7. The precise version of this statement is given by the Fundamental Theorem 0f Calculus. See the statement of this theorem and the paragraph that follows it at the end of Section 5.3. 8. See the Substitution Rule (5.5.4). This says that it is permissible to operate with the dm after an integral sign as if it were a differential. TRUE—FALSE OUIZ 1. True by Property 2 of the Integral in Section 5.2. 2. False. Try a : 0. b : 2. f(:c) 2 9(93) : 1 as a counterexample. 3. True by Property 3 of the Integral in Section 5.2. 10. 11. 12. 13. 14. 1. CHAPTERS REVIEW B 479 False. You can’t take a variable outside the integral sign. For example. using f : 1 on [0, 1]. fol :cf(m)d\$ : fol wdrc : E312]; : % (a constant) while xfol 1d:c : a: : :1: - 1 = x (a variable). . False. For example. let ﬂag) 2 :32. Then fol V552 drc 2 folxdm = %.but t/fol x2 dar = ﬂ : True by the Net Change Theorem. . True by Comparison Property 7 of the Integral in Section 5.2. False. For example. leta = 0. b : 1, f(z) : 3. g(m) : 9:. f(m) > g(;1c) for each x in (0. 1). but f’(ac) = O < 1 : g’(:c) form 6 (0.1). . True. The integrand is an odd function that is continuous on [—1. 1]. so the result follows from Theorem 5.5.7(b). True. f55(am2 +bsc+c) date = f35(ax2 +0) dm+f_55 bmdm : 2 [05(mc2 + 0) due [by 5.5.7(a)] + 0 [by 5.5.7(b)] False. The function f(m) = 1/11:4 is not bounded on the interval [—2, 1]. It has an inﬁnite discontinuity at 9: = 0. so it is not integrable on the interval. (If the integral were to exist. a positive value would be expected, by Comparison Property 6 of Integrals.) False. See the remarks and Figure 4 before Example 1 in Section 5.2. and notice that y = .7: — x3 < 0 for 1 < m S 2. False. For example. the function y : is continuous on R. but has no derivative at x : 0. True by F'I‘Cl. EXERCISES (3) L6 6 I; f(acz-_1)A:c [Ax : % 1] f(\$0)‘1+f(931)‘1+f(\$2)'1 +f(x3)-1+f(\$4)'1+f(935)'1 m2+3.5+4+2+(~1)+(v2.5):8 The Riemann sum represents the sum of the areas of the four rectangles above the ac—axis minus the sum of the areas of the two rectangles below the m—axis. X6: f(f;)Aa: [Am : 1&9 1:1 f(E1)‘1+f(Ez)‘1 +f<i3) '1 +f(E4) ‘ 1+ f(is)~1 was) ~ 1 = f(0.5) + f(1.5) + f(2.5) + f(3.5) + f(4.5) + f(5.5) m 3 + 3.9 + 3.4 + 0.3 + (72) + (~29) : 5.7 The Riemann sum represents the sum of the areas of the four 0)) M6 1] rectangles above the x—axis minus the sum of the areas of the two rectangles below the m—axis. 480 CHAPTER 5 INTEGRALS Z. (a) f(sc):x2—mandAm:¥:0.5 => R4 : 0.5f(0.5) + 0.5f(1) + 0.5f(1.5) + 0.5f(2) = 0.5(7025 + 0 + 0.75 + 2) = 1.25 The Riemann sum represents the sum of the areas of the two rectangles above the x-axis minus the area of the rectangle below the m-axis. (The second rectangle vanishes.) (b) [0208 — x) dx ‘ ; [Ax = 2/n and z, = 2i/n] n2 2 _ 4 n+1 2n+1 n+1 =11m —-—-—72~ n—boo 3 n n n [02 (x2 — m) dac = A1 — A2. where A1 and A2 are the areas shown in the diagram. 11 can be interpreted as the area of the triangle shown in the ﬁgure and 12 can be interpreted as the area of the quarter-circle. Area = + i(7r)(1)2 : % + g. n 4. On [0,71']. lim 2 sinsc, Ax 2 f0" sinscdx : [7 60336];r : —(—1)—(—1): 2. naoo 7;_1 5. f06f(w)dm:f04f(m)dx+fff(a:)dm :> 10=7+f46f(:c)d;c 3 fff(m)d:c:1077:3 6. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. CHAPTER 5 REVIEW (a) ff (m+2x5)d93: lLrgto(mi)Aa: [A:c=——— = —.\$i:1+— i=1 " 41' 415 4 =‘ 1— 21— .— .1L%;(+n>+(+n>i n 1. 1305n4 + 3126723 + 2080112 — 256 4 : 1m ————— > — n—voo 77,3 TL a» If (me) am =11x2+§x613:<a +15~§25>—e+%>=12+5208 : 5220 = 5220 First note that either a or I) must be the graph of fox f(t) dt. since foo f(t) dt = O, and 0(0) 7é 0, Now notice that 3 481 b > 0 when c is increasing, and that c > 0 when a is increasing. It follows that c is the graph of b is the graph of f'(x), and a is the graph of fox f(t) dt. ,_. >4 \ A | H I: arctan :0] 1 d 311' an m (a) By the Evaluation Theorem (FTCZ),/ E (e t ) dsc : e o d 1 (b) d— / earcta“ 1 dx = 0 since this is the derivative of a constant. a” 0 d I arctan t arctan .7: B C1. — dt : . (c) y F I do: /0 e e .f12(89:3+3x2)dx:[8~ﬁx4+3-%ac3]f=[2w4+w3]::(2~24+23)—(2+1):4073:37 T /(w4—8m+7)dz: [gx5—4m2+7z]:=(gT5v4T2+7T)—0=§T5—4T2+7T 0 1 1 f0(1~z9)dw:[m—fam10]0:(1—ﬁ)_0:% Letu:1—ac.s0du:idxanddwz—du.Whenm=0.u=1;when9§:1.u=0.Thus. 1 1010 —m)9dz : ffu9<—du> : folugdu = e. [um]. : an — o) = a. 9 i 2 9 9 /'QE-CWZ/(u‘1/2—2u)du:[2u1/2—u2] :(6—81)—(2_1):—75 1 1 1 1 101%“)? du:/;<u1/2+2u1/4+1>du: [wage/wag = (§+§+1) -0: Letu : y2 +1,sodu : Zydy and ydy gdu. Wheny : 0.11 = 1; wheny = 1,11 : 2. Thus. 2 712(64—1):Q= foly(y2+1)5dy: 12u5(%du)=%l%u6]1: 12 %- Letu = 1 +y3.sodu =3y2dyandy2dy : édu. Wheny = 0.11 = 1; wheny = 2.11 = 9. Thus, 9 o2yzx/‘1+ysdy=ffu”2(%du) :% éuml. : 3(27— 1) 2 5 dt 1 /1 (t _ 4>2 does not exist because the function f(t) : (t _ 4)2 has an inﬁnite discontinuity at t : 4; that is. f is discontinuous on the interval [1. 5]. Let u = 3711,50 du = 37rdt. Whent = 0.11 : 1; whent : 1. u : 37r. Thus. /lsin(37rt)dt—/3ﬁsinu idu —i[—cosu]3”*i—1-(—1—1)—£ 0 i 0 371* _37r 0 _ 37r _37r- Let u : 113, so du : 3212 dv. When 12 = 0.21 2 0; when 11 21,11 = 1. Thus, fol 112 cos(v3)dv : fol cosudu) : %[sinu](l) = %(sin17 0) : gsin 1. 482 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 3 CHAPTER?) INTEGRALS 1 . . 5mg: smx d 2 Ob Th . . ' 2 ' ' /_1 1 + :62 :c y eorem 5 5 7(b), Since f(.7:) 1 + x2 is an odd function. f01 emdt 2 [ieﬂt]; 2 £02" — 1) Letu 2 2 — 39:. so du 2 —3dw. When at 2 Lu 2 —1;when x 2 2, u 2 74. Thus. —%(ln4—ln1) 2 iéln4. _ 74 1 1 1 —4 dsc — _ —(—§du) 2 i§[ln]u|]71 1 u f2 1 12—3£U 1+ — 2 4 1 2 4 2 1 1 4 —\$—2—ﬂ-U—dm2/ <7+£2—\$—2>dm2/ (x2+——1>d\$2[——+ln|:t|—x] x 2 :1; 3c :6 2 so an 2 ac +1n424)2(_§+1n2—2) =1n2—g 10 A \$274 that is. f is discontinuous on the interval [1,10]. dac does not exist because the function f 2 2 has an inﬁnite discontinuity at ac 2 2; as i 4 Letu 2 m2 + 4x. Then du 2 (25c + 4) dsc 2 2(x + 2)dm. so ———%xdx2/uil/2(%du) 2§~2u1/2+C2\/H+C2\/332+4x+0. Letu 2 3t. Then du 2 3dt. so fc5023tdt 2 fcscgugdu) 2 %(— cotu) +C2 —§cot3t+C. Letu 2 sinmﬁ. Then du 2 7t cosvrtdt. so fsin 7rt cosntdt 2 du) 2 % ~ 1u2 + C 2 %(sin7tt)2 + C. Let u 2 coscc. Then du 2 *sinmdm, so fsinx cos(cos;c)d;1: 2 — fcosudu 2 —sinu+ C 2 —sin(cosm) + C, d J5 Letu2ﬁ.Thendu225E.so i/Edac22/eudu22e“+C22e‘/E+C. d9: cos(1na:) . . Letu 2 lncc. Then du 2 —, so —— dac 2 cosudu 2 s1nu+ C 2 s1n(ln:c) + C. :c 36 Let u 2 ln(cosac). Then du 2 _sm\$ dm 2 itanxdz‘, so coscc ftanscln(cos;c) dZE 2 — fudu 2 —%u2 + C : '%iln(C°S-73)i2 + 0' 1-71 _i 53m u-i—C22 Let u 2 :02. Then du 2 236 due. 50/ sin’1(\$2) + C. _\$_dm_l/__d"__ x/l—w4 2 x/l—u2 \$3 1 d 2— 1+3:433 4 1 Leta:1+m4.Thendu24x3d:c,so/ /—du2ilnlul+C=iln<1+\$4)+C. u Letu 2 1+4x. Then du 2 4dac.sofsinh(1+4m)dm 2 ﬁfsinhudu 2 icoshu+C 2 icosh(1+4w)+C. Let u 2 1 + see 0. Then du 2 secB tanOdQ, so secOtanG 1 1 = : 2 :1 0:1 1 9 ——1+sec6 f1+sec0 (sec6 taanO) /udu n|u|+ n| +sec |+ CHAPTERS REVIEW E 483 36. Letu : 1 +tant, so du : sec2tdt. When t : 0.u : 1; whent = g. u = 2. Thus. few/4Q +tant)asec2tdt : ffuadu : [if]: : 411(24 —14):ﬁ(16 — 1) = a? 37. Since\$2—4<0for0§m<2andx2—4>0f0r2<ac£3,wehave|z2~4| =43: —4)=4#ar2for 03m<2and‘x2~4|=xg—4for2<\$§3.Thus. 3 2 3 m3 2 C53 3 /|m2—4|d:c:/ (4—w2)d:c+/ (1:2—4)d:c= [4xA—] + [——4m] 0 o 2 3 o 3 2 —(8§) 0+(912) ( 8)“%-3+1—f:3—§-§ 2: (.0 (ﬁlm 38.Sinceﬁ—l<0for0§at<landﬁ—l>0for1<m£4,wehave|\/E~1|=—(\/E—1):1—\/§ for0§w<1and|ﬁ71|=ﬁ-1for1<m§4.Thus. [WM dw=f01 <1—ﬁ>dm+f14<ﬁ—1>dx 0 —<1 a 0+0: 4) e 1)—%+16—4+§=6—4=2 H r——. H | ODIN: 8 w \ w .__.v ,_. + r_\ calm H on \ to I H u__4 ,t; In Exercises 39 and 40, let f (ac) denote the integrand and F(m) its antiderivative (with C : 0)4 39. Letu=1+sinx. Then du=cosmdm. so 3 cosmda: _1/2 1/2 . ' —: u du:2u +C:2\/1+smx+C. s/1+sin:c f r 17 311 7 *2 40. Letu:x2+1.Thenac2=u—1andmdm:%du,so 4 fvxmz—dem:/( ﬂ I (ul/ziu‘l/Z) du —25 l 41. From the graph, it appears that the area under the curve y : m \/a_: between at = 0 and w = 4 is somewhat less than half the area of an 8 X 4 rectangle. so perhaps about 13 or 14. To ﬁnd the exact value, we evaluate 4 f04\$ﬁd\$ : 04 z3/2 do: : [gx5/2]0 : §(4)5/2 : % = 12.8. 484 CHAPTER 5 INTEGRALS 42. From the graph. it seems as though f5" C082 :3 sin3 x dm is equal to 0. 0-2 To evaluate the integral. we write the integral as ‘ A I = [02” cos2 m (1 — cos2 ac) sinwdm and let u = cosm :> 0 277 du: —sin:cdac.Thus.I:f11u2(1—u2)(—du) :0. -0.2 43. By FTC]. : f: \/1 +t4dt => F’(ac) : \/1+a:4. 44. : tan(s2) d3 => F'(m) = tan(x2) 3 m tdt 45. a: 2 .Let = u andu::c3. 9( ) O m y g( ) dy dy du u m3 4 3m5 Then’w*————— 3x2———3x2:—. g() dsc dudx 1/1—1—113 \/1+x9 V1+a39 46. Letu = c0523. Then 3—: : isinw. Also. % = %%.so d COSI 3 d u 3 3 . —— 1—t2dt:— Vl—t2dt-—:\/1—u2(ismw) dx 1 du 1 d3: : \3/1— cos21:(—sin ac) : isinx \3/ sinzac = *(sin m)5/3 a: a 1 t x t J; t a: t 47-y=/ e—dt:/ e—dt+/ e—dtz—f e—dt+/ e—dt : ﬁt ﬁt 1 t 1 t 1 t dy (1 ﬂat d met —:—-—— — — —dt .Lt : .Th d9: da:(/1 tdt +dx/1t 6“ ﬁ 6“ d ﬁe‘dtid "adhi /“e_tdtd_u_§ 1 eﬁ. 1 _eﬁ E/l t _dm 1t _du 1t dxiu 2\/E_\/E 2ﬂi2x‘ 0 @ —eﬁ + i S d9: _ 2x a: 48. y : 23:“ sin(t4)dt : f2: sin(t4) dt + f03m+1 sin(t4) dt : 03\$“ sin(t4) dt 7 02m sin(t4) dt => . d , d , . y’ : Sln[(3.’E + U4] - a (3x +1) 7 sm[(2m)4] ~ E (22:) : 3sm[(3m + 1)4] v 231n[(2\$)4] 49.lf1§m§3.then\/12+33\/:c2+3§ 32+3 :> 2Sx/x2+332\/§.s0 2(3 7 1) g ff 1/352 + 3dx g 2\/§(3 — 1); that is. 4 : ff \/;132 + 3111: g 4J3. < 1 1 1 . < < . < < — —.. 50 If3_:c_5,then4ﬁ:c+1_6and67\$+1 4, 5 1</ 1dac<l. 3_3x+1 ‘2 .0 32:31: 030083331 => Kwangju2 :> fol :02 cosmdsc g fol \$2 (19: : §[m3]; 2 % [Property 7]. (5 — 3); that is. ’53 | 8 l/\ ex 0! ,_i A. H |/\ Jan—- 51 CHAPTERS REVIEW E 485 52. On the interval [g , m is increasing and sinx is decreasing. so 81:33 is decreasing. Therefore. the largest value sinzc 7r 7r . sin(7r/4) 2/2 zﬁ . 2¢§ —— ——=———=—.B Proert 8w1thM=——we et 0f 2: on [472]IS 7r/4 7r/4 7r y p y 71' g "/2 sinsc 2 \/2 7r 7r x/2 d.g_(___):_. 7|./4 IL' 7? 2 4 2 53. cosxgl => ewcosacgec : folexcosxdacSfolexd\$:[ez]é=e<1 54. ForO < as S 1.0 g sin’lm g §.so folccsin’lmdac £f01m(§)dx : [gig]; : g. 55. Let f(:£) : v1 + x3 on [0, 1]. The Midpoint Rule with n : 5 gives fol t/1 + :63 d1: z%[f(0.1)+ f(0.3) + f(0.5) + f(0.7) + f(0.9)] : g [J1 + (0.1)3 + 1+ (0.3)3 + ~ ~ ‘ + 1+ (0.9)3] m 1.110 56. (a) displacement = f05(t2 — t) dt : [ t3 — l152]5 : 125 — 25 = 1—355 : 29.16 meters 1 _._ 5 0 T 2 (b) distance traveled = [05 It2 it|dt = f05|t(t71)|dt= [01 (t 42) dt+f15(t2 —t)dt =1lt2~§t313+1§twt2ii —%7§—0+(1-2-5725)—(§—%)2%2295meters 57. Note that r(t) : b’(t). where b(t) : the number of barrels of oil consumed up to time t. So. by the Net Change Theorem. fog r(t) dt = b(3) — b(0) represents the number of barrels of oil consumed from Jan. 1. 2000. through Jan. 1, 2003. 58. Distance covered : f05'0 dt m M5 : 505— 0[11(05) + 11(15) + 12(25) + v(3.5) + v(4.5)] : 1(4.67 -l- 8.86 + 10.22 + 10.67 -l- 10.81) = 45.23 m 59. We use the Midpoint Rule with n 2 6 and At = 246’ 0 = 4. The increase in the bee population was f0“ ms) dt m M6 = 4M2) + r(6) + «10) + r(14)+ r(18) + 7122)] m 4[50 + 1000 + 7000 + 8550 + 1350 + 150] = 4(18.100) = 72.400 60. A1 = ébh : : 2.142 = §bh = = and since y=~x—1 y y = ix/l — m2 for 0 S at g 1 represents a quarter—circle with radius 1, A3 : ‘117r7‘2 :iw(1)2 : %. SO fig f(m)dx : A1 A2 A3 : 2 A”; 244 61. By the Fundamental Theorem of Calculus. we know that F(:v) : f: t2 sin(t2) dt is an antiderivative of ﬂat) : x2 sin(a:2). This integral cannot be expressed in any simpler form. Since f: f dt : 0 for any a. we can take a : 1, and then F(1) : O, as required. So F(x) 2 ff 2&2 sin(t2) dt is the desired function. 486 CHAPTER 5 INTEGRALS 62. (a) C’ is increasing on those intervals where C/ is positive. By the Fundamental Theorem of Calculus. d m C’(a:) = E No cos(§t2)dt] = cos(gx2). This is positive when \$352 is in the interval ((272 7 %)7r, (271+ 71 any integer. This implies that (271 — %)7r < €362 < (2n + %)7r <:> 0 g S 1 or x/4n — 1 < < x/4n + 1, 71 any positive integer. So 0 is increasing on the intervals [—1, 1], [x/E. x/S]. [—\/57—\/§], [We], [fa—ﬂ]. . (b) C’ is concave upward on those intervals where C” > 0. We differentiate C" to ﬁnd C": C’(m) = cos(§x2) => C'”(w) : * sin(§x2) ~2cc) : —7rw sin(§x2). For a: > 0, this is positive where (271 — 1)7r < €322 < 21m, 71 any positive integer <=> 2(2n — 1) < a: < 2 n any positive integer. Since there is a factor of —:c in C", the intervals of upward concavity for cc < 0 are (— —2 ﬂ n any nonnegative integers That is, C is concave upward on (—\/§. 0). (x/i, 2), (ix/6, —2), (x/é, . . .. (c) 0.8 0.8 -_ 0.7 0 2 0.6 From the graphs, we can determine that fox cos(%t2) dt = 0.7 at as m 0.76 and a: m 1.22 (d) The graphs of S and C have similar shapes. except that S’s ﬂattens out near the origin, while C ’s does not. Note that for a: > 0. C is increasing where S is concave up, and C is decreasing where S is concave down. Similarly, S is increasing where C is concave down, and S is decreasing where C is concave up. For x < 0. these relationships are reversed; that is. C is increasing where S is concave down. and S is increasing where C is concave up. See Example 5.3.3 and Exercise 5.3.57 for a discussion of S 63. Area under the curve y : sinh cw between cc : 0 and a: : 1 is 4 equal to 1 :> fol sinhcac dm : 1 => %[COShC\$]é : 1 :> I %(coshc i 1): 1 => coshc i 1 : 6 => coshc : 0+ 1. From the graph, we get 0 = 0 and c x 1.6161, but c = 0 isn’t a / solution for this problem since the curve y = sinh cm becomes y = 0 A and the area under it is 0. Thus, c x 1.6161. _2 0 2 64. Both numerator and denominator approach 0 as a a 0, so we use l’Hospital’s Rule. (Note that we are differentiating with respect to a. since that is the quantity which is changing.) We also use ETC 1: C I: e—(m—u)2/(4kt)du H 087(zvaﬁ/(4kt) Ce—z2/(4kt) lim T(a:,t) : lim : lim (1—0 “*0 a V 47rkt a—O v 41rkt \/ 47rkt 65. 66. 67. 68. 69. 70. CHAPTERS REVIEW E 487 Using FTCL we differentiate both sides of the given equation. f0m f (t) dt : \$62” + fox 6" f (t) dt, and get f(:c) : 62\$ + 268290 + e-wf(g;) :> f(\$)(1_ 67x) 2 622 + 2%“ : ﬂag) : A emm The second derivative is the derivative of the ﬁrst derivative. so we’ll apply the Net Change Theorem with F = h’. flzh" u) du = ff(h’)l(u) du = h'(2) — h'(1) = 5 — 2 : 3. The other information is unnecessary. Letu : f(a:) and du = f’(m)d\$. Sto: f(:c)f’(x)da: : 2ff(b)udu = [uzlﬂm ﬂy.) “(1) : _ F(2 + h) — F(2) _ hm 1 _ h—»0 h 2,133?) h m 2+h Let 2/ \/1+t3 dt. Then F'(2) / \/1+t3 dt, and 2 2 F'(x) : \/1 + 2:3, so lim h—vO 1 2+h / \/1+t3dt=F’(2):x/1+23:\/§:3. 2 E Letu =17 2:. Then du : #dx, so fol f(1— at)dm : flo f(u)(idu) = fol f(u) du = fol f(:c) dz. . 1 1 9 2 9 3 9 9 1 10 1 11m — (—> +(—> +<—> +-”+(E) / ccgdzc: x— :i naoon n n n n 0 10 0 10 The limit is based on Riemann sums using right endpoints and subintervals of equal length. ...
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