solnshw9

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

This preview shows pages 1–10. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern