solnshw9 - ,l m 5 m 64 V = f C(t)e”"dt = f...

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Unformatted text preview: ,l m 5 m 64. V = f C(t)e”"dt = f 450e‘”‘”4’dt = 4- 0 "”"M’] r) n ‘ — 0.04‘ 1 Section 6.3 Partial Fractions and Logistic Growth z $3708.90 II b 66. V = f c(l)e“"’dt = I (30.000 + 500t)e’“-”7’dt z $153.8l6.01 (l I) r. 10 68. V = J C(t)e“"’dt = f (5000 + 25f e’/“’)e_”-““’dt -v $39,238.17 (1 n 3 70. (a) J (300.000 + 125.0000 dt = 300,000t + 62.500t2] 1) 1 , (b) I (300000 + l25.000t)e“‘-”5'dt *3 134504350 1) g l (l = $1.462.500 t, . 72. Future value = e”: f f(t)e""dt = WW5] (3000e”-“5’)e“’-"dt *3 $21,881.75 1) l) m 74. Future value = e‘“"""’f (400t)e'“'"dt 0 = e"8[—4000(t + nah-01“]? = c"-"[8000(5e9/5 — l4)e‘9/5] ~ $129,986 Four years at Penn State costs (19,843) X 4 = $79372. Yes. the fund would be large enough. 4 4 10 1 1 76. V: dx= 100 Td‘ ”J: (fl?) W I xe~‘ 1 -MEWSwe)”(‘75)+f<e>+f<%)+r(%>+f<%> we) were)”ermine Section 6.3 Partial Fractions and Logistic Growth 3x+ 11 A B 2.~—~= + (x-3)(x+1) x—3 x+l Basic equation: 3x + 11 = A(x + l) + B(x — 3) When .r = 3: 20 = 4A, A = 5. Whenx= ~12 8: ~43. B: -2. 3.x+ll 5 2 1m+3_§+ B 4‘x(x+l)_x x+l Basic equation: 10x + 3 = A(.r + 1) + Bx When x = 0: 3 = A. Whenx=—l: —7=—B. B=7. 203 204 Chapter 6 Techniques of Integration W 7x+5 _1(A + B) '6(2x+1)(x+]) 62x+1 x+l Basic equation: 7.x- + 5 = A(x + 1) + B(2x + l) Whenx= -: Whenx= —1: —2 = —B, B: 2. 6 .4 A. A=3. m... ml ul— L:1( 3 + 2 )=__|__+ 1 6(2x2+3x+1) 62x+1 x+l 2(2x+1) 3(x+l) 3x3—x+1 A B C 8.——=—+ + x(x +1)2 x x+l (x +1)2 Basic equation: 3x2 - x + l = A(x + 1)2 + Bx(x + l) + Cx Whenx=02 I=A. Whenx=—I: 5=—C, C=—5. Whenx=1t3=4A+ZB+C,B=2. 3x2—x+1_1 2 5 + x(x +1)2 x x +1 (x + l)2 I 3x ~ 4 A 3 6x2 — 5x A B C 10. = + . —_ = + 7 + (x ~ 5)2 x - 5 (x — 5)2 12 (x + 2)‘ x + 2 (x + 2)- (x + 2)3 Basic equation: 3x — 4 = A(x — 5) + B Basic equation: 6x2 — 5x = A(x + 2F + B(x + 2) + C Whenx=5: 11:8. Whenx=~2z 34=C. Whenx=0: —4=-5A+11=>A=3. Whenx=02 O=4A+ZB+34. ”‘4: 3 + ” Whenx=lz 1=9A+3B+34. (x—S)2 x-S (x—5)2 Solving these equations yields A = 6 and B = * 29. 6x3-5x= 6 ‘ 29 + 34 (x+2)3 .x+2 (x+2)2 (x+2)3 A 14. 9 B mzx—3+x+3 Basic equation: 9=A(x+3)+B(x-3) Whenx=3z 9=6A, A=§. Whenx=—3: 9=—6B. B=—‘§. x—3 x+3 9 3 1 3 1 3 3 3 z- __ =_ _ _— +3 + :— fx2_9dx 2fx_3dx 2] dx 2lnlx 3| 2lnlx I C 21n [+c __;4___ A + B (x+2)(x—2) x+2 x—2 Basic equation: —4 = A(x ~ 2) + B(x + 2) Whenx = —2: —4 = —4A, A =1. Whenx=2: —4=4B, B=—l. —4 l l ‘2 — '= — '— + —-‘ [xi 4_ 4th Ix + 2ch Ix _ 251,1 lnlx + 2] [nix 2] C 1n l6. x+2 x—2 be Suction 6.} Partial Frat'lions and Logistic Growl/1 205 3 _ A + B .\'(_\ — 3) .\‘ .\‘ ‘ 3 18. Basic equation: 3 = AU — 3) + Bx When x = (1: 3 = ~3A. A = -11 When .\' = 3: 3 = 38. B = 1. ? 1 ‘ I - 3 I a , d.\'= -I—¢lx+I dx=—1nIxI +1nIx—3I +C= In\ I +C .\'~ — 3x .\' .r — 3 .x 5 A B . ———————— = + 20 I." - 2)(.\' + 3) x — 2 .\‘ + 3 Basic equation: 5 = A(.\' + 3) + B(_x — 2) When.\'=2: 5=5A. A: I. When .x‘ = —3: 5 = —SB. 8 = -l. ——-§—I\'= 1 dr» 1 Ix~=in|\-—2|—1nI\-+3[+C—1n"_3+C .\'3+.\‘-6(' .\'—2' .\'+3(' ' .\'+3 1 A B 22‘ (2x — 31m + 3) _ 2.1- ~ 3 + 2.1- + 3 Basic equation: 1 = AIZY + 3) + BIlt~ - 3) Whenx=%z 1: 6A, A =):. When .\' : —%: 1: —6B. 3 = ~%. 1 1 1 1 1 l | 2x — 3 _: _ ._ _ . z 7 _ _ o 3 _ __ I4xl-9d" 6I2‘-_3d.\ 6I2x+3dA I—zlnlmx3I—211nI_x+ I+C—121n2x+3I C (+1 1 24. I(—:-— + 11H + 3)tlx {1+ 3th 1nI.\ + 3| C 3x3 - 7x - 2 A B C . ——-— = — + + 26 .\‘(.\' + l)(x - l) x x + l x - 1 Basic equation: 3x3 — 7x — 2 = A(x + |)(.\' — 1) +IBx(x — I) + C.v(x + When x = O: -2 = ~A. A = 2 Whenx=-l: 8:28, 8:4. Whenx=]: —6=2C. C=—3. 3r — 7.x — 3 1 1 1 ;-*—V—_d.\‘=2 —dx+4 (Ix-3 (Ix =21nIxI+41nIx+1I—31nIx— lI+C x3 - x x x + l x — 1 4x3 4- 2v - | A B C 28. —,-—— = — + —, + xiv + 1) X .t- x + 1 Basic equation: 413 + 2x — 1 = A.x'(.x +1) + B(\ + )+ C1- Whenx=0: -1=B. B=—l. When_\'=*1: 1=C. C=l. Whenx=1z .:2A+23+C. A=3. ,~Z+2_._ 4;-—“—ld.t=3 “(IA— :(IX‘I‘ (1.\' .\“+.\'- +1 =3lnIAI+~+lnI\+ II+C=1+1nIUI+1nI.\'+1I+C=l+1nI_.\‘4+.\“‘I+C t\ .\ .\' 206 Chapter 6 Techniques of Integration x4 6x2—8x+4 30 r‘—hz+3x-l x (x—l)‘ 6x3—8x+3= A B C (dc—l)3 x-l+(x—l)2+(x-l)3 Basic equation: 6;:2 — 8x + 3 = A(x — I): + B(x — l) + C = Ax2 + (B — 2A)x + (A — B + C) Therefore,A = 6,3 — 2A = —8.andA — B + C: 3. Solving these equations yields A = 6, B = 4. and C = 1. x4 _ l . l l J( _l)3dx~f(x+3)dx+6fx_ldx+4j(x_[)2dx+[(x._U3dx x2 4 1 =—+ + — — ————-—+ 2. 3x 6ln|x l] x _1 2(X _1)2 C 3x A+ B 32‘(x—3)2=x—3 (x—3)2 Basic equation: 3x = A(x — 3) + B = Ax + (B — 3A) Therefore, A = 3 and B — 3A = 0. Solving these equations yields A = 3 and B = 9. 3x 1 l 9 —— = + -=3 —3— + fxl—6x+9dx 3fx_3dx 9f(x_3)2dx lnlx [ x—3 C 3 A B .—+—= + 34 (2x+l)(x+2) 24+] x+2 Basic equation: 3 = A(x + 2) + B(Zx + l) Whenx=-%z 3=%A, A=2. Whenx=—2: 3=—3B, B=—l. l I I 3 1 1 I ]1 L————2x2+5x+2¢r 2L2x+ldx fox+2dx lnllx ill) 1an 2:0 =ln3—(ln3—ln2)=ln2~0.693 l 2 l 1 x —x 2x+1 . —— — _____ _ __ + + = -1 z... 36 fox2+x ldx—J; [l x2 x de—[x ln(12 x 1)] 1 n3 0099 l) 1 l l x3-l 4x—1 . = + 38 x1*4 x x2—4 4x-l_ A B x2-4flx—2+x+2 Basic equation: 4x - l = A(x + 2) + B(x — 2) Whenx=2: 7=4A=A=g Whenx=—2 —9=-4B=>B=§ '1 1 a 1 x'—l 7 1 9 l x- 7 9 = +—. +_. =_+_ _. +_ + ian—‘ldx i0(x 4 x—2 4 x+2>dx 2 4"‘lx 2’ 4'an zll) 1 7 9 9 .mm—Mmmva—‘fi Set/inn 6.3 A" - 4 w —3 40. , = .\'- + 1 + a x- - l x- - l —3 _ A B x: - l .\' * l .\' + I Basic equation: ‘3 = AM + I) + BU — I) ,1 Whenx: —l: —3 = —ZB::B= When.\:= I: —. = 2A2A 2—; 3 4 ,4 4 w .x —4 (7 3 I n I ) “2 ”1+ __. +_. >- J: x: - 1(l\ f3 ‘\ l 2 .v- I 2 .r + I, d‘ 62 3 :——3ln3+—ln5~19.785 3 2 3 ~4 3 "4 l 4 1 42“A‘JA,t,x+2)u-—3)‘l"‘L(5 .\‘+3 5 17—3)“ “ £[ln|x J 2| ln|.\‘ 3|]- — itln4+ ln4) 5 ~| J 8 =_ :16 51:14 _._ISI (1 N U X-+2x—l 7 I l l ) /. = = —— +— 44A [4 \3~4 (l\ J‘_I(l+4 x-Z 4 \+2d\ 7 H =[x+—ln|x—2|+-ln|x+2] 4 - 7 l 7 > = — 2+— 2 ~ — +— x (4m 4m) < 1 4In 7 =l+2ln2vlln3204637 46 l 5+ B 48. ' _x‘(.\' + a) _ x X + a Basic equation: I : AM + u) + Bx When x = O: 1: (1A. A = J]. ( Whenx = -—a l = —aB. B = “1!. ( __1_ _ l<i ‘ ' ) .x'tx + u) a .r .x‘ + a Purlia/ Frau/um and Lugislic Growl/1 l A+B (x + [Na — .v) : x + l u - _\‘ Basic equation: I Am - x) + B(_.\‘ + l) Whenx=a1 l=B(u+l)=>B= l 0+1 Whenx=~lz l=A(u+l)=>A= I Ll+l l _ I/m+l) , X/(u+l) (.\'+l)1u-‘\') .\'+l u—x 207 208 Chapter 6 Techniques of Imegration MM 1 X * 16 1677' 2 2 s s so.v= —,———dx=— ————— ‘ + ‘ 7TI,(JL‘+l)-()c—4)2 125 U[x+i x-4 (,x+l)2 0—4)de _m x+1_ 5(2x-3) ]~‘_m[( L5) ( 1 15) 125i:2]nx—4i (x+1)(x—4) (F125 21"4+4 21"4 4 1671' 15 =—- 2 +— ~-. 125(1n16 2) 5246 0 2 52.V=7Tf (#)dx -3 x-+x-6 (25x)~ A B C + D m=x+3+(x+3)3+x-2 (It—2): Basic equation: 625x2 = Aix + 3)(x — 2)2 + B(x — 2)3 + C(x - 2)(x + 3F + D(x + 3)2 When x = 2: 2500 = 25D=:D = 100 When x = —3: 5625 = 258=> B = 225 Whenx=02 0212A+900~18C+900~900=6A—9B Whenx=lz 625=4A+225—16C+l600~300=A—4C Hence,A = —60 and C = 60. 0 —60 225 60 100 = + + V "Lix+3 6+3)2 x—2+(x—2)2idx 225 100 ]" = w[—601nix+3| - x+ 1 +601nix—2I — r-2 ll 7T[(-601n3 — % + 601n2 + 50) — (—225 + 601n4 + 25)] z 212.04 x+1)(500 wad" I0 10 _f(x+1-x—5m)dx t=101n|x+1|—101n[500—x|+ C 54.I=5010f( l Since x = 1 when! = 0: 0: 101n2 — lOln499+ C=>C~55.l946. Thus: t= [Olnix + 1) — 101n(500 — x) + 55.1946. (3) lfx = (0.75)(500) = 375, then I m 66.2 hours. (b) When I = 100, you can use a graphing utility to solve for x: 100 = 101n(x + I) - 101n(500 — x) + 55.1946 2x z 494 individuals. Sci-Iii)” 6.4 Ilirvgmliun Tables and Comp/«ling r/n’ Squun' 209 I I ‘ l i ‘ 1 I , V 56.;lkf‘ ~~ (; 7 -__ ._ -. = ( ” film — _\‘ld‘\ mi r\' In ~~ \l‘l‘ 10”“ mi“) -‘l) + C' mmio — y) f C' , v Wk; » C. : in( > ) - . lo y. )I _ ‘ ,IHAI It) e '\' F (t . l \‘ l Stn‘e ':lwhent=0. -=C'nd ' :—)"“’. L 1‘ L) " to ~ _\' 9‘ Solving for y. 9.“ : emu ll) - ‘ . '\ (b) Since _\‘ = 2 whent = 2. L). : ,Iltki _ .,I(JAI _\ 101 _\£ 9(2) : 9mm, )‘(9 + PM") = 100mm ll) - 2 10W“ l0 9 , \‘ = W I m.1 2 ("HA => k 2 0.04055. _ c/N l()()t’““"’ . , . . 38. E = m MU) = 50 60. Both spec1es have posmve growth rates. The rate ol growth is increasing on [0. 3] and decreasing on [3. w) | for P. aurelia. The rate of growth is increasing on [0. 2] (a) N = 25 + C and decreasing on [2. 00] for P. caudatum. Hence. the former species has a greater limiting population. all + 40770,“) 350 N(()) = 50: 7 + USC: 0 _ 250 _ 11 + 4H“) (h) The virus will never infect half the students. N S 250 for all t. Section 6.4 Integration Tables and Completing the Square 2. Formula ll: L! = .X‘. (/11 = c/.\'. u = 2. [7 = 3 l l l l .r ixtz + 3.r)1d'\_ 5i: + 3x + 5'" 2 + 3.x] + C 4. Formula 29: u : x. du = (Ix. a = 3 l 4 x — 3 3 ,t' — 3 4Jx2—9‘IJA3(3lln.\'+3l r C_EIH,\'+31+C 6. Formula 22: u = .\‘. (In = dx. u z 3 . r—. l . —.— i f— 3" \/.\" + 9dr = §[.\'i2.\" + ‘Hx/X‘ + 9 _ 8l lnlx + v.\" * 9H + C x 8. Formula 37: u = in (lu = 2.\'(l.\' . . 7‘ J " Jil.\’=%Jr A Vt].\':é[.\‘:“lllil*("Vfli‘c l + t»- l + t» 10. Formula ll: u : .r. (In 2 (Ix. u z I: I l 12. Formula 25: u 1“ .\‘. (Ill : (/.\'. u : l ' 1 l l .- l - . “V ‘ _T(/.\‘ '—‘ T lllI \ i ( %(/X :: lll!.\ T V”.\>— “ i: ‘ C ,\ll ~t .\) ' I ', V "l . ' + ,t l + .\ {Tl 216 Chapter 6 Techniques of Integration W 4 5 40. f xK/x + 4dx 2 55.6246 (71 = 100); 42. f 10x3e“‘ (Ix z 11.0405 (11 = 100) 1 u 4 44. V = 77] x2(x + 4)2/3 (iv;- 0 ’~* n(4T(j‘:)9)[0 + 4(5)”3 + 2(4)(6)2/3 + 4(9)(7)3/-‘ + |6(4)] *~* 244.883 cubic units 46. y = %(x3/2 — 3x”2 + 2) ,r_l 9.1/2_§ —1/2) = (l)<x— 1) " :3<2x 2" 2 J; +(X_4_-xl)-_ (26:1)2 2 + I 1 5:21,)? /(x1+) dx = JON; x 1d x=éJ (XI/2 + x~1/2) dx = l[2x3/2 + le/z] = flunits 0 2 3 0 3 Simpson 5 Rule 15 inappropriate for this problem since the integral is undefined when x = 0. 6 48. % [l2 — 41n(t2 — 4t + 6)] dt z 5.78] grams per liter (1 (Simpson’s Rule with n = 100) Section 6.6 Improper Integrals 0 a 1 2. This integral converges since ez‘dx = lim [.1121] = - — 0 = _00 b—>—oc 2 2 :x: b 4. This integral diverges since I —1\/-_-dx = blim [2‘51 = 00 — 2 = 00. x ace 1 DC . . . 5 6. Th1s integral converges smce f _-21 dx = — 0 e IQILII °° h s s e-l' 2dx=——l =—‘— — =‘—. [0 1 1 im[e ] 2(0 1) 2 2 b—mc 0 8. This integral diverges since fx_1__dx=f _;_dx+f°_1_dx l/ZVZX—l l/ZVZX—l .VZx—l = lim[2x—1]l+lim[‘/2x—l]:=(1—0)+(oo—I)=oo I)-—>(1/2)* b c—DOC The second integral diverges. _, _| 10. This integral converges since I ~l:dx = lir_n Pi] = l + 0 = l. ‘9: Soc/ion O. 6 Improper Integrals 217 W 12. 14. 16. 18. 20. 22. 24. 26. 28. 36. (i This integral diverges since I , .\-— + “I This integral diverges since .\‘ .1" + l 1 [7"9‘1 7. ‘1) X f .\'3¢"“cl.\‘ = J t\‘:e“‘(l.t + I .\'3c"‘(l.\' 73C “1 (i ll .2 V‘ X This integral diverges since w 3 /_,1‘ I 1 I | 1 I ’ UM n2” H (x — 1):“ ‘ (l : lim [w l (1‘41 X — 1 27 This integral converges since J /. alx = if .r' [/1 (IX = 5 lim n V-V l) 3 l i, (.1‘ ~ 1rd" Ii __1_ val,” + 1' _l ,\.= L .— (__l_ . harp): 3e *3: ll)“; 38 l) — 3 4 . . . l . 9- J This integral converges smcef +CIX = lim 2./.\' — 3 = w . 0 (IX: lim J .1 (Ix: lim [élnlxli-llj : lim [Mllnl-b+ HM]— Iv—p A x /, ly—a — y_ I) l+ lim [*—l-] =(oo-l)+(¥l +oo)=oc. n .—>I‘ .\'—l. hall Th' ‘ t or I eroes s' e -———'\- I I'm [ 4 \ ] IS in eg a conv : .inc 1 1x — l V’ D H /4 _ .r- lr—>2 l . . . . I This integral diverges smce I —d.\= .\ 0 This integral diverges since 3_1_,.: ‘_1__ i, (.x‘ M 1W” ., (x — IW 1- l M 1w l ——~* 1* _ 1m 3/ _ l u .1111 Q/X’l /i—>l 11m[1nl\|:‘1=0—(‘001 b—fl)’ I 3 I + __ _. .c/x J: (x _ “4/3 dx 5 /«—-— 1 \/.\" “ 9 This inteoral converges smce 2 #— ¢I\ = lim —--. 1 \ va — C) lv—fl 9.x ‘1. I» H — lie“ (11‘ = lim [’Xt’i‘il l) /7‘—>.Z l) = l (b) V= 7T1 (6“): u 3 37 27 135 Mil, ‘5l7‘°)=7~ I) 77 (Ix = 7T11m ~—e‘3‘ = : /)—)T£ _- 36.393] 0.0043 .\‘e‘”-" 06065 673. 7947 ltfl'f. I) 7: I2 = 0 38. { Ne“ zl.\‘ = lim [—(X + He’l] = 0 + l = l .1) 218 Chapter 6 Techniques of Integration w 2" 75 000 3° 40. (a) Present value = I 75.000e“’""'dt = [— 0.08 e“"‘”"]0 z $748,222.01 0 ' . 3° 5‘ I) (b) Present value = 75.000e“’-“8’dt = lim [—7 OOOe‘Mg’] = $937,500.00 0 b—mc 0.08 0 ’ n 4'2. C 2 650,000 + I 0 25,000(l + 0.08t)e“’~‘2’dt = 650,000 + 25.000[*-;-(125 + 6t)e‘0-‘3’] 0 (a) Forn = 5, we have c = 650,000 + 25,000(—-1%e‘0'6 + 1?”)2 $760,928.32. 185 _ 7 125 ‘ (b) Forn = 10, we have C = 650.000 + 25,000 “76 l" + T x $842,441.86. II 125 (c) Forn = 00. we have C = 650,000 + 25,000 lim [-5125 + 6t)e‘0"2’] = 650,000 + 25,000(0 + T) 3 $997,222.22. n—nz: 0 h h e“("'“)J/2""dx=f h(x)dx (I l 44.Pssb= (“ x ’ Lam p. = 36. a = 0.2 Use numerical integration. :1: (a) P(35.5 S x < oo) = f h(x) dx —'= 0.9938 35.5 x (b) P(35.9 S x < oo) = I h(x)dx = 0.6915 35.9 Review Exercises for Chapter 6 3 2.f(x2+2x—l)dx=%+x2—x+C ' _ —l _ 4.] 2 fi=fZ(x-l)‘2dx=2£r:JL+C= 2 +C (x—l)2 l x-l _ 3- 6 3xe-‘dx=-§e‘+C 2r3-x _] 4x3~2x __l 4 2 8. x4_xl+ldx~2fmdx—zln|x x+1I+C 10. ~L~dx=lf(2x—9)“'/2(2)dx= ./2x—9 +C 12. [(x2 — l)e"j_3"’dx = %e“3_3"' + C 14. Use substitution and let u = l — x, x = 1 ~ u, du = -dx. (1‘20“ (1 ~XJ" 4 3 ll [x0 -x)2dx= f0 — u)u2(-du) = [(143 - u3)du =§~u§3+ C + C ...
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