solnshw6 - Seclion 5.4 Area and the Fundamental Theorem of...

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Unformatted text preview: Seclion 5.4 Area and the Fundamental Theorem of Calculus 173 M 58- (a) P = J, ‘125"_’/:” df (C) We solve fortin the equation P z I: = (— I25)e-'/3”(—20) + c = 250()e"/3” -!/2|1 l “ e = 25009"/3" + C .500 — Since P = 2500 when r = 0. C = 0. 1 _[ Hence. P = 2500e"/:“. ln( ) : ——~ 2500 20 (b) P(15) = 1500641sz 1181 fish I: 201n 2500 x 156.5 days 60. (a) R = J(50 — 0.02.): + (IEOI)(lX (C) When .1' = 1500. R z $53.231.39. : 50x _0.01x2 + [comm + I) + C (d) WhenR 60. 0..) 1949 or x 3054un1ts. Since R = 0 when x = 0. we have C = 0 and R = 50x - 0.01.:2 + 100 ln(.x + l). (1‘)) 70.000 62. (a) S = [(1.0‘“ + 544'694) dz 1 = 0.52!2 + 544.694 lnt + C Fort = 9, 256.6 = 052(9)2 + 544.6941n 9 + C C~ —982.335 S = 0.52t2 + 544.694lnt — 982.335. 8 S t S 13 (b) For 2002. t = 12 and S x 446.1 million. Section 5.4 Area and the Fundamental Theorem of Calculus 4 3 2. I 2 dx 4. f (2x + lldx ( U l 13.71 Area = (base)(height) = (4)(2) = 8 Area = 1mm + 7) = 12 174 Chapter 5 Integration and Its Applications .1 6. f Ix - 2| (Ix ~| Area =fim" =fi'rr(2)2 = 77 5 s 5 10. (a) I 2g(x)dx = 2] g(x) dx = 2(3) = 6 (C) I f(x) dx = 0 U U 5 0 5 5 5 (b) J f(x)dx = ‘f f(x)tlx = —8 (d) f UL!) ‘f(x)] dx = I 0111‘ = 0 5 (1 U 0 l 4 7 4 = _ .4 . : ~ = .1/2 = _ 2 12. A [4(1 .x )dA 14. A J: fldx 4x ]I 8 4 4 | = 2f (1—x4)dx 0 8 . = 3 square umts l l 4x— 2 4 2 4 16.A=J 2e-“/2dx=4e-"/2] =4e‘”—4~2.595 18. 14:] dx= f <1—-)dx=[x-21nx] o 2 2 0 x x 2 =(4-21n4)—(2—21n2) =2—21n2'~'0.614 7 1 '9 '7 _3"'7_3_(Zt_l22:_§ _ _.1_32 20. J; 3vdv— 2]: — 2 2 -2(49 4)— 2 5 2 , 22. I (—3x + 4) dx = + 4x]; = (—2 + 20) — (—6 + 8) = 322 3 2 2 l _ 2 _-l£fixfl]'__l__ -1 24.J0(1 2x)dx— 2 3 0- 6(1 1)—3 : (x—3)5]2 1 1 _ 4 2 :__ -—_- 26.L(x 3)dx 5 2 S+5 0 4 4 28.f\/%dx=f fix-I/Iczx: J2(2x'/1)]4=2\/Z]:=2\/§—2fi=4fl~2 2=2J§ l l 36. 38. 40. 42. 46. 48. 50. J 3/2 .J I2x-3ldx=J‘ —(2x—3)tix+ 0 (l Section 5.4 (L U (4 — |.x‘l)dx =f (4 + x)dx +J ‘4 (J .z/z [—xl + 3x] 0 I + [x3 — 3x] = 3/2 e- (h _ (€_" I l)"/3( e”"')d.\' u 9“ + I n 71 = —2(e‘-‘ + Um] l) l = -2(e"+l)1/3 + 2(1 + I)”: = 2f: — 2 1+ ; _ iv _ ~ ' J; a 0.489 3 (2x — 3) dx <—2+%>+<9—9>~<2—3>= 3/ I 4 (4 — x)ch .24 +[4x—%] =0‘(—l6+8)+(16—8)—0=l6 J x — :ln 12 lln5 “40.438 2 x: () .—. <+— [41 2]—4 _ 0 1 2 I ’) 3€.\ 2 X v .r. 2 (2 +lnx)4]2 I 7 2— =~_+124—4x9.152 tl.\ [ 4 l 4( H) Area and the Fundamental Then/11m of Calculus 175 176 Chapter 5 Integration and Its Applications 52.] (.x-+4)dx=[£+4x]’=2+8=10 0 2 0 B = 3(e — 3—2) 3 7.75 2 ' 4 62. Average value = —l— (x — zfi) dx : 1(1 _ 313/2 1 l 54. f fx(1—x)dx=J (xl/Z-xmhu I) (l 0.75 -0.25 1.25 —0.25 4—0 0 4 2 3 4 3 3 To find the x-Values for which f(x) = —%, we let x — 2/— = -% and solve forxto obtain the following. x + g = 2/; 4 4 - + —- + — = x 3x 9 4x 8 4 2 — —- —— = x 3x + 9 0 9x3 ‘ 24x + 4 = 0 24 1t 4/432 4 i 2/? x z 18 = 1 Average value = (1165. 1436) 2e — 2 = 2M e - l = 6““ x = 4ln(e — I) 2 2.165 4—0 =Ze—2z3.437 4 I 28x/4dx = _‘(8)ex/4] () 1 4 4 0 Suction 5.4 Area and {/10 FIIIIt/Cllllt'llml Tln'm'vm of Calculus 177 1 3 1 1 3 1 1 1 1 . r12 '1 : ,x=- .-—3’3s=— =~*—=- 66 Amer 19:: va us 3 _ “II (J _ 3)_ (/1 2 I) (\ ) c/\ 2va _ ML 2 6 3 To find the .\'-Value(s) for which 11,1») = a we let I l 1.1- ~ 313 E and solve for .1‘ to obtain the following. 3 = (.1 — 3): i J; : .\‘ — 3 '31 3:\, =.\‘ x = 3 4 J? x 1,268. which is in the interval [0.2]. 1 ‘ 4; , 1 68. Average value = — , ‘ dx = 2 1h1x- + 1)] = 21h 2 2 L386 1 — 0 (I "a + l 11 4(l_4l)3,1_3h’7l 21h 2 = ‘4'” x- + 1 Solving for x. you obtain .1‘ 2 0.403. 70. g(.1') = ,1“ - 2x is odd. 72. f1!) = Sr4 + l is even. [gt-1') = ‘gt-xfl [fm =.f’(—r)] (J 3 I3 9 74. (a) I .\‘3dx = ‘f X31111: —4 76. AC = I ‘0300 dx —2 H II) ' 3 7 13 (b) I .de = 0 since f(.\') = X3 is an odd function. = — '0'300] -2 . 11) 1c) _ 3.134.1- = 3 huh- = 3141 = 12 = —20.00'0(i — i) a $461.54 11 1) I3 10 5113 78. AR = J 75(20 + 1121‘ 5111) X 503 = [1500.11‘ + 67,5001nx] 5011 2 $4903.79 128 25 P 80. AP =J 3‘(40 — 3mm» 125 2. 5 , '3“ 25 5 = —[40.\' — 21311] = 7[401 128) ~ 2(_128)»‘/3] — 2791011251 — 21125W1] z $234.69 2 1*; l) T T 82. Amount = W] (We—"(It 84. Amount = ("’J c1t)v""’dt 1) I5 4 : ,1).11:1151 QUOOrU-W' {1% . 37.88748 = 81109141.! 5006—1111wdf : $240739 ( Jo ( t $ 11 q 86. J 1001(11 = 5013] = $l250 U l I 178 Chapter 5 Integration and Its Applications M 5 '1 = 1 0 (t- + 2)2 r- + 2 5 88. Capital accumulation = 11000, — 6000] 1 1 = 6 0 —— + — z 00( 27 2) $2777.78 1) ‘1 3 2 90. v: I 100000 - 6) d; = 100006 — 61)] = 10.000(2 — 18) = —$135.000.00 0 a 2 U 92. (a) M = [13.439912 + 6603.7e"]dt = 1.8133t3 ~ 6603.762" + C Att = 3: 3119 = 1.8133(3)3 — 6603.7e‘] + C C 2 3398.8 M = 1.8133t3 ~ 6603.7e“’ + 3398.8 12_ if (1.8133!3 — 6603.7e" + 3398.8)(1’! ~ 3 (b) Average = 12 539623.89) m 4403 (billions of dollars) 94. (a) (1%, = k[650 - N(t)]' dN 650 - N dN f650-N_fkdt -1n1650 — NI = kt + _cl 1111650 ~ M = —k1 ~ C, 650 — N = (“-61 = Ce-kr N = 650 — 02"” MO) = 300:: 300 = 650 — C: C = 350 N0) = 500 =: 500 = 650 — 350w“ =kdt 1—50 = ‘2k 3 350 e l 3 I 7 z” ‘ ‘5‘"7 — 2mq ~ 0.42265 M!) = 650 — 350e—0.42365, 1 5 1 5 (b) — N(!)dt = — (650 - 3506’042365t) d, z 504 coyotes 5 _ 0 0 5 0 l r—- 17 ‘ 1921 96. +1 1- dx = -—J§ 93. f 3 .1 + 3 = 1/206 ) x 60 1, x (x 1) dx 1820 Section 5.5 The Area 0ft: Region Baum/ed by Two Graphs 179 Section 5.5 The Area of a Region Bounded by Two Graphs x3 (—x1 + 4mi- = [n— + 4x]- = A: 3 ~2 | I ‘3 ‘4 l 1 ' .A— :3 -.'—'- '— —— . =2 .'— 3* — A 4 Lb x)dx [3 4]“ ’2 6 A 0 [(t l) (X 1)](l.x : 2[(-16 — l)‘ _ (x —1)3]': 1 ,4 2 o 2 3 7 8. A = I [fix + 3) _ (1x 10. The region is bounded by the graphs of y = 1 '- x2 and I " y=x3— Lfromx= -1tox=1.asshowninthe 3 I t‘ me. = + 3x — 2 in .t] 1g .. I \ l t =(4—21n2)—<§) I q! 3 IV‘ = 5 — 2 ln 2 l 12. The region is bounded by 14. The region is bounded by A the graphs of y = .r — 6 - F the graphs of x = y + 6 ’ » and y = x3 + 5x - 6 from and x = y: from y = —2 x = —4 to x = 0, as shown to y = 3. as shown in the figure. . V .1 , t w in the figure. .\ AH-Ilt 6 In [4 16. The point of intersection of the two graphs is found by 18. The points of intersection of f and g are found by setting equating )r-values and solving for x. f(.x') = g(x) and solving for x. x3—2x+l=-2x \/3x+1=x+l x3+l=0 \/3x=x 3x = r- t: - 3x = 0 x(r — 3) = x = 0, 3 180 Chapter 5 Integration and Its Applicatimzs 20. 4 — x3 = x3 x 6"“ m]; o e ‘ k) II p . 5: 1 (1.2.718) l (SJHSSt l a ‘ ' 1 26- A : [‘ * VFW] dx 28. 2y ~ )‘3 = —_v H: X _l I r — = Inx + e‘] _ 1 1 1/2 y — 0. 3 A =f (- — x3) dx 1 3 m x = e — (Ini + e'”) A =f [(2)’ - yz) - (-yfldy x4 ‘ o = [lnx _ —] _. _ 1/: z 3 4 1/2 — e + 1n 2 e 1.763 2 I (3y _ yady _ -1 -1 1 + L \ o h 4 n2 64 ‘ G2) 15 _ 2 3 o =1n2—E=0.459 N H 2 - (%.—1.M9) 2 (L—Z‘7IX) / (0,1)) Sm‘limz 5.5 T/w Area Ufa Region Baum/ed by Two Graphs 181 30. ,\'3+ l =4—2y ‘ )3+2y‘3=0 (y + 3)(,\' ~ I) = 0 "yahv‘ K [H II Intersections: (2. l). (10, —3) I . A = f [(4 — 2y) ~ (V3 + I]¢IV\' ‘x {V —} 7 )3] l [3y y- 3]_3 [3 I 3] ( 9 9+9)— 3 1 l, 32. XX — 3x3 + 3x = x— .x" —- 4x3 + 3x = 0 (3.9) x(x — l)(x - 3) = 0 x = 0,1,3 I 1 A = f [(x3 — 3x3 + 3x) — 1‘3] (be + I [x2 — (I? — 3x3 + 3.0] (Ix 4 U I l x = f (1" ~ 4x2 + 3x) (Ix + f (*x‘ + 4x3 - 3.x) zlx U l _ [g _ 4-_v* + + + _ 3—HT 4 2 1) 4 3 2 ! -i+§ ‘u 3 _E ‘n 34. x3—4x3+ l =x—3 x3~4x3-x+4=0 (x — l)(x — 4)(x +1): 0 Intersections: (l, —2), (4. I). (— 1. —4) I A = f [U3 — 4x2 + I) - (x — 3)] dx ‘1 4 + I [(x — 3) — (x3 — 4x3 + H] dx 1 _E 91-253 + 3 4 12 182 Chapter 5 Integration and Its Applications 36. The points of intersection of f and g are found by setting 38. The points of intersection of f and g are found by setting f(.r) = g(x) and solving for x. f(x) = g(x) and solving for x. 3-2x—x3=0 “x1+4.v+2=x+2 (3 + x)(l — x) = 0 0 = x: — 3x : x(.r — 3) x = -3, l x = O, 3 I ‘3 1 3 A=f (3*2x~x2)dx=<3x-x3—'?)] A=f[(—x3+4x+2)—(x+2)]dx —3 ~ ‘3 u 1 32 3 l .6 3x13 9 = - _———— _9+9=—_ = —2+3 '=__+—__ :— (313)(9 )3 If" mi" 3202 5 I 7 40. The line through (6, 4) and (O, 0) is y = %x and the line through (6, 4) and (4. 0) is y = 2x — 8. Therefore, the area is given by the following. 4 4 v+8 3 v24 A: ' -——y I: 41,—» r:4t—~' :8 H 2 2"]"V [H “‘1’ [~‘ 2L (6.4) 42. The point of equilibrium is found by equating 300 — x = 100 + x to obtain x = 100 and p = 200. mo vi 100 Cs=f [(300—x)-200]dx=00x_'3_] =5000 O l) 100 x2 100 133:] [200-(100+x)]dx=[100 —3] =5000 0 0 44. The point of equilibrium is found by equating lOOO — 0.4):2 = 42x to obtain x = 20 and p = 840. 0.4x3]2° 3 20 CS =f [(1000 - 0.4x2) — 840] dx = [l60x — z 2l33.33 r) (l 3!) 20 PS 2 f [840 — 42x] dx = [840x — 21x2] = 8400 o 0 ...
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solnshw6 - Seclion 5.4 Area and the Fundamental Theorem of...

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