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solnshw8 - C H A P T E R 6 Techniques of Integration...

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Unformatted text preview: C H A P T E R 6 Techniques of Integration Section 6.1 Integration by Substitution Solutions to Even-Numbered Exercises 2 4 2 . + 3/3 =-— + 5/2+ . = + 2f(x 5) dx 5(x 5) C 4_[(l—t)3dt (1-02 C 6. 2"} dv=11n(1+v4)+c 8. (3+x)5/3dx=3(3+x)7/2+C y4+l ' 2 " 7 6x1+2 _ 3x3+1 . 3 l -1 _2 10'jx3+.rdx_2fx3+xdx_ Zlnlx +xl+C 12.f(3x+l)2dr—3f(3x+l) (3)dx : 3 2 ._ ln(.r +x) +C 2—31-(3x+1)“+C — _] +C ‘3(3x+1) 14 __.' (1le (5x+])"/3(5)dx 16. 4€2de=21n(1+e2")+C . V5x+l 5_ ‘ l+e2" = %(2)(5x + l)‘/2 + C =§J5x+l+C JET __ 18. e dLr=2e~"-"+'+C 20. 2" dx=2 [1+ 4 ]dx x+l x~4 x—4 =2[x+4lnlx—4+C I __l _ 2_1/2_ ,1” =1 ,2” 21,241+ 22.]der— 2]“ t) (2!)dt 24.]te dt 2 e (2t)dt 2e C = -%(2)(1 - mm + c =—\/1——_12+C 26.] 6' de=ln(l+e"')+C l+e-‘ 28.Letu=x+l.thenx=u—landdx=du. x2 (u—l)2 (AZ-214+] LxHdeI—uz—WITW 2 l ————+c 2 l = *1.— -2 *3 = — _ —. + = + + fol 214 + u )du lnlul + C lnIx 1! x +1 2(x + ”2 14 2112 194 ”ﬂieu‘r‘w'jwm“? ‘ Section 6.] Integration by Sulmtitution 195 W 30. Let u = .r — 4. then .x‘ = n + 4 and 111' = (in. 5.‘ 5 ‘ ’ w ‘ 41x: JET—“(In = 5 (if- + 4n") (It! 1.\' — 4)‘ u‘ 7 7 ' v _ n 5[ l .1]+ C 5[n +-] + C _ 5(.\ -1) + C u n- _ u— (.x‘ - 4)‘ 32. Let u = l + V: then .\‘ 2 (u - l)2 and dx = 2(u — “till. I I . l -L _ a _ __ f l + ‘4. (1.x [(u)2(u ”(In _f[l u] (I'M =2[u—lnu]+C =21+¢I~~1n1+ﬁ +C =2¢I~—21n1+ﬁ +C I | 6x + J; 7 H ,. 34. ——zl.\' = (6 + x '/‘) (I.\' = 6L“ + ZV’X + C .x' 36. Let u = .\~ ~ 1.1/11 = dxhr = u + l. 7 7 x— (14+ 1)- , ,, _ , (IN = I ‘ ,_ (In = (”V- + 2n'"— + n ”3) (/11 VA‘ - l V/u 2 4 , W 2 _u5/l + _u3/2 + 2"]/_ + C 5 3 2 ‘ 7 4 M /1 = 2U — |)“/’ + :(x * I)?“ + 2H — I)“ + C 38. Let)‘ + l = u.d_\' = du._v = u — l. fylb‘ + HWII')‘ = [(11 - ”3141/3 (In = [“47” — 2H4” + u'"/3)du = .I‘iOHHVK _ gum/3 + 2114/: + C x : %(\ + l)lU/1 __ g(\ + ”7/3 + it“ + ”AM + C l 4 4 4 l / l - -:_ . 1/2, -:_ .+ 2/2 :~ 1/2_2 40.[ \ 41 + Id.x 4L (4.x + I) (4) (1.1 6(4.\ 1) ] 6(17 7) , 2 _,\ __ ‘1 2 _7\. _ ..,__\ 2 _l ,4 -1 ‘4 42.1)e "(Ix—T e "( 2)tl.\— 8 ]~ (v l)—7(_l~( ) " (t 44. Let u = x + 5. thenx = u — 5 and dx = (In. u = 5 when x = 0 and u = 6 when .\‘ = l. l 6 (\ I .1'(.\' + 5V [I.\' = J (u - Stu4 du = I (Hi ~ 5114) du I) i 5 u" . 6 115 , 6 3|25 3125 = ._L s _ _ 0_ __ [6 ll:L ()[u b]; 6 t 1} 6 196 Chapter 6 Techniques of Integration W 46. Letu= l —x,thenx= l —uanddx= —du. u= l whenx=0andu=0.5whenx=0.5. 0.5 0,5 0.5 f x3(l - .1‘)3 dx = J -(l — u)2u3 du = f (—u5 + 2144 - 143M114 0 1 1 uﬁ 2M5 “4]1LS “4': :lUﬁ 7 =—~+————- =——10-—2 +5 =— [ 6 5 4 1 60 “ 4“ I 1 640 48. Letu=2x+ Lthenx= u; l anddx=%du. Whenx=0,u= I,andwhen.x=4,u=9. 4 9 u-l 1 ”’1 foZX+ ldx=f ( )ﬁ(-)du=f ‘(u3/3—Lt'/3)du O 1 2 2 [4 _125/2_33/2]9:l[.2. 5 __2. 3 _l[_2._.2_]-39_8~ ~4[5u 31. 1 45(3) 3(3)l 45 3'13 19.867 50. Letu=x+2,thenx=u—2.anddu=dx. Whenx=0,u=2.3ndwhenx=7,u=9. 9 7 9 fszx+ 2dx=f (u—2)3\/;du=f (us/2—4u3/2+4u'/2)du 0 2 , 8 2 , 8 I , 9 = 71W- — 'S-u-h + gui- -~ 308.057 — 1.724 = 306.333 Ix: =3andwhenx=6 u=9. tel—- 52. Letu= x+3,.=thenx u—3anddx= du. Whenx= 6 9 I 2x 1 dx =f9 Z-(-u—3—)-——ldu =f (214V2 — 714"”) du 1/2 x + 3 7/2 J; 7/2 9 = [ﬂu-V3 - 1414“] ~ (—6) — (—17.46!) = 11.46] 3 7/2 54.Letu=x-2,thenx=u+2anddu=dx.Whenx=2,u=0.andwhenx=10,u=8. 8 J40 xmdx =J8 (u + 2)u‘/3 du =[ (144/3 + 2u1/3)du 2 0 0 3 7/3 3 4 :l8 = _ . + _ /3 1: . [714 214 0 78857 56. Letu= l—x,thenx=1—uanddx= —du.u= 1whenx=Oandu=Owhenx= 1. 11 1 0 0 xmdx=f —(l -u)§/;du=f (u4/3-ul/3)du 1 1 .33 2': 1.2 , 3 3 0 9 = _ 7/3 - _ 4/3 = _ ‘ [714 414 1 28 square umt 58. Letu= 1 + J},thenx=(u— 1)2anddx=2(u—1)du.u= 1 whenx=0andu=3whenx=4. 3 204;” du=2f <1-l>du , u =2[1~+1n|u|]:: 4 — 21n 3 z 1.803 square units 4 A: 0 St’t‘IIUII 6.] Integration /7_\' Su/M'H'mrimz ‘l | (.0. v = ‘ITJ [ft-ti — .\*l3]3¢/.\' = w] m w 04m 0 t) Let u = l '- .\'. then .\* l e u and (It ‘- —clu. u =1 when .\ =00nd 11:0 when .\ 5 —l “' U 11 ug " “ V = 7TJI ~ll — uliﬁzlu = 7f] (11‘ ~114ltlu= ”ii-[é — 3—] = Tbcubic units l 3 l ‘3 ~— ,— 1 3 , 3 . 62. q _ (ll [fm - gt.\')]d.\' = :J [xv-1.x- + l — 205"]45' = 31” .rV 4x + 1 dx — f\ V (1.x - o * (J ~ 0 H - . — l I For the first integral. let u = 4x + I. then .x‘ : u 4 and (Ix : 1d”. 11 = I when x = 0 and u = 9 when .\' = 2. l ”0-1) .—(I ) 3 V3 l ‘l .. . [2..3 _ L = 5/- _ I/‘_ _ _ .5’- 21 ( 4 /u ‘4 (Iu 0 (I\ 32 I (u u lc/u 52 U l 2 5. 2 V. " so? 149 80/2 — -—— — :——~‘ 2 7 .52l5“ 3:" V. 5 60 5 0'2‘1 ll | r‘ b .. 64. Let u = l — .\‘. then .x‘ = l - u and c/.\' It‘ll 5‘ All/311x : [(1 — ”Fug/3d“ = fl“; — 31/3 + 311 - llul'lzdu 2‘ flu”: — 3117/3 + 3115": 5- ”WM/u Fl It fl ) ° ° :4: -L3:‘/3+3(:72)_:52 11’ <)” l 7“ 5“ T C 2 ., = ”ﬂu" 3(105u‘— 385“2 + 495” — 2m + C ’3 = 11555“ — .\*l-""3[IO5(1 — 3x + 35-3 — x3) — 385(1 — 25 + x3) + 495(1 — x) — 251] + C 2 = 1155” x)“( 105x‘ 70.x2 40x to) l C 2 I = -m (I — r) ”(105‘ + 70x” + 40x + m) + C Therefore. we have the following. . l, — l’ H55 1 S 5 -\ “l \= A\ l 11.53.“ — -\')"/-c/~\‘ = [-itl — 05" -(105\-3 + 705- + 40.x + 16)] ’_“_‘— ", L ., 32 10 .. \ , ‘PSm \<IH (at u = 0 and b = 0.25 P(0 s x s 0.25; ~ 0.0252 2.52% /7\ (b) a = 0.50 and [2 = l P(0.50 s x: I) 207361 73.61% ‘ 66. (a) The average daily revenue is given by the following. 1 5(5) 1 “(‘5 l 105 :;’§. I’: 7’ 13 -_ 1/- _ 7; 305 _ 0 J0 [0.00r (.0. r) + 1-50M: 365 J“ [0.0m (.65 r) M T 305 1-. 0 (Ir [t = 36——5 0 0(1( 365 ~ u ) til-2n) )rlu + 1250— \$34 520. 95 x (—55 (b) Certain fruit and vegetable sales resemble this pattern. ‘l 68. A -: J .\'\/l 2- x: (I.\ '3 1.346 197 198 Chapter 6 Techniques of Integration W Section 6.2 Integration by Parts and Present Value 2. Let u = x and dv = e‘-"dx, then du = dx and v = —e"". fxe‘-‘dx = —xe‘-" - J' —e‘-"dx = -xe’-" — e" + C 4. Let u = x2 and dv = eZ-‘dx, then du = Zxdx and v = %el‘. fxzel‘dx = %x2e1" — fxehdx Let u = xand dv = el‘dx, then du = dx and v = 21!: _l 2 2.x..[l 2V... 1 Z\' ] fxe dx—zxe 2xe 2e dx 1 l l l ._ 2 1r _ _ 1t _ 2r + = .. 2x 2 _ + + 2x e 2xe + 4e C 4e (2x 2x 1) C 6. flnxzdx= 2flnxdx Letu = lnxanddv =dx,thendu = idxandv =x. e1". ul— flnxzdx= 2[xlnx—fdx] = 2xlnx— 2x+ C=2x(lnx-1)+ C 8. [8—3 = —%J‘e‘1‘(—2) dx = -ée-Zx + C 10. Let u = x and dv = e‘l" dx, then du = dx and v = -%e‘l". l l l 1 I —2x = __ -2x _ ~_ —2x = __ —Zr _ _ -?_r = __ «2x + fxe dx 2xe I 2e dx . 2xe 4e + C 4e (2x + l) C 12. fxze" dx = %fe"‘3 (3x2) dx = %e-"‘ + C 14. fgdx = fxe‘ﬁix Let u = x and dv = e‘x dx, then du = dx and v = —e‘~“. + gdx = -xe“‘ — f—e‘ﬁlx = —xe’-" - e“ + C = —e’-"(x +1)+ C = —xex1 + C 16. Letu = lnxanddv = x3dx.Thendu = i-dxandv = ' L 4 4 4 4 a :L _ £1) :1 _ [xlnxdx 4lnx [4(x dx 4lnx f Sm'riuu 6.2 Integration I7_\' Paris and Prawn! Value 199 W 18. 20 22. 26 28 30 32. . I ‘ x 1 (Inxrl I V : . v _, . 2 + : _ faint-I‘d" I ”n“ Id" ~2 C mun-I3 + C I Let u = In it and (Ir = c/.\'. then (In = T (I.\‘ and v = .\'. flan) (Ix = .\‘In(3.XI — J (Ix = x |n(3.\‘) - .\' + C = .\'[In(3.\') — I] + C I (Ix = —l- (1) dx = ln(In.\‘) + C .\‘ In .\' In .\‘ 1', Let u = .\' ' I.(/u = (lax = u + I. .‘ + I I 1 J, y/ \ (Ix = I“ _ (/11 = [(11'1’2 + u"'/-)zlu V’x * I V/u = gum + 2n”: + C = %(.\’ — III-"3 + 2(3‘ ~ II“2 + C' Let u = .\' and (Iv = (2 + MY”: (Ix. then clu = (Ix and r = %(2 + 3x)‘”. .\' 2 3 2 4 [mm = 3m + 3x)”: — KI” + 3x)‘/3dx = 3H2 + 3x)”: — 5‘2 + MW: + C a a = in + 3.x)”3I9x — 2(2 + 3n] + C = iﬁﬁm — 4) + C I Let u = .x'ze": and LIV = XII: + [I‘ll/.\‘. then (/11 = 2xe""(.\'3 + IIdX and v = — 130‘: V X x23": \ __ t z I. __ _—. ——.- + '.,,\» I”: + l)3 (1.x In e )[Ug + 1):]ch 2(.\‘3 + I) In d\ 2(.\‘3 + I)’ II x1 3x3 1 V _x3‘,.\'-' + ex: .\': + I EA: : -1—(_+_e,\‘+C:—__7._.(___)+C:_~T_+C 2(x- + I) 2 2H- + I) 2(x- + I) Let u = .\'Z and (Iv = 9“" (IX, then (In = 2x (Ix and v = —e“‘. 3 .\‘: ﬂ 2 3 , _ —‘.(/.\' 2 —x-e"] + 2 m“ (IX 0 " ‘I 0 Let u = x and (IV = P‘de. then du = (Ix and v = —e#‘. I 1:11.): : -4e‘2 + 2” .re“-‘]_ f ("‘dx} — 46—: 46‘2 2[e"‘]- = —10e‘3 + 2 2 0.647 0 L" 1) U [I Let u = In .\' and (Iv = 24'. Then du = l/x (1.x. and v = x3. A 2 J lvlnx (1x = [.\‘3 In .\' ;J - <93 6:) — (0 %) ~ :(I + C3) I .'. I ._ _ .. , a l , .\‘: [Zr 1n .\'(l.\' = x- In .\' — Jr —_dx = x- lnx — - + C 200 Chapter 6 Techniques of Integration W 2 I+2x I | I 2X I 1 1 1+7- ,-=,1 + — = 3_ _ L n( .0111 rn(l 20]“ J; l+2xdx 1n L[l 2x+ l]dx I I I 3 = 3— —— -+ = 3—— +— > =‘— _ — z . 1n [x 21n(2x ”1) In I 2111 3 2 ln 3 1 0648 34. Let u = In(I + Zr) and dv = dx. then du = dx and v = x. | I 36. Area = —I (x2 * l)e-‘dx = I (I — x2)e“’dx I Let u = I - x2. dv = e-‘dx. du = —2x dx, and v = e-". I I [(1 - x3)e-"dx =(1— x2)e‘ - f—Je-‘dt = (l — x2)e‘ + 2fxe‘dx = (l - x2)e~“ + 2(xe" — e“) (Example 1) Hence. I —I I Area = J (l - x3)e“'dx = [(1 - x2)e“' + 2xe-" - 29‘] —1 = (0 + 2e — 2e) - ("Ze‘l - 2e“) =3= [.472 38. Area=fx m—_de I l Letu = Inx,dv=—;dx,du = -dxandv = —-. x- x x 1nxdx= ~llnx — l-Zzldx— — ~llnx —1 x~ x2 x x Hence Area=fx11£dx=[—-lnx—l] = (—1—1) — (—1) = 1 —220.264 x: e e e 40. (a) La u = xand dv = \/4 + xdx, then du = dx and v = %(4 + x)”. Jmedx = -:'x(4 + x)3/2 - i—J‘M + x)3/2dx = %x(4 + x)3/2 — 74;“ + x)”2 + C = 12—44 + x)3/2[5x — 2(4 +x )1 + C— — —5(4 + xthx — 8) + C (b) Letu= \/4+x,thenx=u2-4anddx= 2udu. [Xv4+de=J(u2—4)u(2u)du=2JIu4-4u2)du=2l:E5-s-—4—:i]+ C : 1—25.” (3u2 — 20)+ C=1—25(\/4 + x) [3( 4 + x) — 20] + C = _(4 + X)3/2(3x _ 8) + C Section 6.2 Integration by Party and Present Value 201 ﬁll.) 42. (a) Let u = .\‘ and (Iv = \/4 — .t‘ (1x. then du = (It and v = — (4 — X)“. ’7 1 A, fivﬁdx ”114 x)” + H I (4 x)“ (Ix = *ixH- _ XW2 H i(4 — I)“ + C 3 3 15 2 2 = "FM — .\')"3[5X + 2(4 — N] + C = _l_5(4 — .\')V3(3,\* + 8) + C (1')) Let u = v4 - x. then .\' = 4 — n2 and (IX = — Zn (In. . , , , 7 4 411‘ u‘ .\‘/4 — xdx = (4 ‘- tr)u(*2u)¢lu = *2 (4M- ~14 )du = ~2 T ~ ? + C \ 2 2 x 2 , {f = ’ﬁtl3(20 - 3uz) + C = —ﬁ(\/4 H x) [20 — 3(4 — x)] + C = —l—§(4 - XWZUX + 8) + C l 44. Let u = x” and (Iv = e‘” (Ix. then (In = nx”‘ ' (1x and v = -e"‘. a .n ,u\ J'\~I1€tl.\' d.“ = A i _ KIA-11* lent d"- (l a 46. Using n = l and a = -3. we have the following. . —_1\ .: ".9va + l )‘h -: __1_ -,-l\ _ l ,—K.\ + 2 .1 ~31 }‘+ + fl? (1.x “3 3 ( (1.x 3.\( 9L C 9? (..\ l) C 48. Using n = % yields A.l/Z+l 4 3 .x-'/31nxdx = l ,[—1+<— + 1) lnx] + C = —.x--‘/3[—1 + —lnx] + C, (: + l)‘ 9 2 to— 50. Let u = x and dv = 6"” (Ix. then du = dx and v = — 36“”. 1 J 3 l .1 7 = -“- ire—‘43] + 3f e""""dx] = —{- 9e“ - 9[e‘~"/3] } = —Ze" + l = 1 — 5 z 0.264 9 0 l) 9 0 e 52. The areais J {3 lnx dx. 025 I . . l , Use Integrauon by parts. Let u = In x. (In = -jdx. dv = x“ (Ix. v = - x“- l .t 2 a ,._1‘.: . 1-—:'- Ix 3nx d.\ — —§" ln.x + I? (Em) ~°-25 _ —ln.x l f} — 2x3 + 2]} [Ix 1].”) 1e. 0) 202 Chapter 6 Techniques of Integration W ' 1 xe‘ dx = [xe‘ — e‘] = 1 () 54. (a) Area = f 0 1 (b) Volume = 71-fx3el'dx U 1 2,‘ 1 ,, 1 ,]l = .... - _‘_ _\+_ -V ”[er 2X8 46’ 0 1 2 1 2 l , I] 71' , _ ___. +_..__ :—- -_ #5 77[2e 22 4e 4 4(e 1) 018 11 I\. f 4 P ‘ , 112 1, 9 10 l . .‘ " =— *‘ R‘s. . - . . =— +—‘z ., 56 J; In 1(x +4)dx 3 ln4 19 32 755 58 J: r 1nxdx 100e 100 1982 "492 60. (3) 2.000000 3 0 5 0 dl Since the graph of I is increasing on [0, 5]. or equivalently. E > 0 on [0. 5], the gift income is expected to increase. 5 5 . _ 5 (b) Incoming during5 years = I 2000(375 + 681e“’-3’)dt = 2000[375!] + [36,000] te‘O-z'dt 0 0 0 . te—(m 5 ] 5 0 = 3 + 3 ’ + __ — .21 .,750~000 l 6000 [—0.2]0 0-2 o e d, 5 3,750,000 + 136,000[-9.196986 - [255,413] } 0 ' = 3,750,000 + 136,000[—9.196986 — 25(e“ ~ 1)] ‘6 \$4,648,419.80 (c) Average = w = \$929,683.96 [(3) = \$973,915.15, which is close to the average. 62. Using Exercise 44 twice with n = 2 and a = 3515 and with n = 1 and a = -%, we have the following. [(410.512e‘730 + 25,000) dt = 410.5[-30t e“’/3" + 60f “3‘”30 dt] + 25,000! = 25.000t + 410.5[—30!2e“'/3‘0 — 1800te"/3‘) + 1800]?”30 dz] = 25.0001+ 410.5[—3011e-r/3O — 1800te"/3" — 54,000e"/3”] + C = 25.0001— 12315510002 + 60; + 1800) + C Therefore, we have the following. 90 90 (a) 1 (410.5t2e_’/3" + 25.000)d1 = i[25.0001 — 12,315e”/30(t3 + 601+ 1800)] m \$167,068.28 90 — 0 0 90 0 1 365 1 365 _.__. “ 2 -1/30 + = __ _ _ 4/30 2 z (b) 365 _ 274 In (410 5: e 25,000)d1 91[25’000’ 12,215e (1 + 601+ 1800)]274 \$26,253.43, 365 (c) I (410.5!2e““/30 + 25,000) dt z 31.281.948.97 () Section 6.3 Partial Fractions and Logistic Growth 203 10 450 NW] 233708.90 ( 1l 1!) . V : . —n = 5 ,—lm4t = 64 I) ((116 (It I) 4 0c (It _ 004i ’ I, a 66. V = f C(t)e""dt = I (30.000 + 500t)e""”7’clt z \$153,816.01 (1 ‘ l) r, m 68. V = f C(t)e""dt = f (5000 + 25f P’/”’)€“0“"’dt z \$39,238.17 () n ‘ K R / 70. (a) f (300,000 + 125,000!) dt = 300.000t + 62,500t3] = \$1,462,500 0 l) 3 . (b) I (300.000 + 125.0001‘)e‘“‘”5’dt2 1,345,043.50 (1 g "I 72. Future value = e”: I f(t)e""dt = NW5] (300061)“5’1e‘0'l’tlt z \$2l.881.75 0 0 18 74. Future value = e“’~"““1f (400t)e“’-“dt l) = e‘~“[—4000(t + lO)e""-"]': = e”“[8000(,’ie"/S — l4)e“’/5] z \$129,986 Four years at Penn State costs (19,843) x 4 = \$79,372. Yes, the fund would be large enough. 4 10 2 4 1 76.V= . d.-=100 7d- 7TJ; <ﬁé") 1 77 I xe‘.‘ .X ~ 10014242142) 124) +1?) +11% Howe) ”(25> we»+f<2§>+f<21>+fe>+fe>l~m Section 6.3 Partial Fractions and Logistic Growth 2.ﬁrl—1‘T)=Xf3+xfl 4'.1(,):++13):f‘7+xfl Basic equation: 3x + 11 = A(x + l) + B(x — 3) Basic equation: 10x + 3 = AU + 1) + BX Whenx=32 20=4A. A =5. Whenx=0z 3=A. Whenx=—1: 8=—4B. B=—2. Whenx=-l: —7=-B, B=7. 3x+11 5 2 10x+323 7 __ —+ xz-Zx—3~ x—3 x+l x3+x .x' x+l ...
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solnshw8 - C H A P T E R 6 Techniques of Integration...

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