Business Calc Homework w answers_Part_46

Business Calc Homework w answers_Part_46 - 226 Section 5.5...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 226 Section 5.5 5. (a) f (x) 4 x, h 0 1 4 x 0 1 f (x) 0 1 1 (0 2 T 2 3 2( 4 x 3 2 f (x) 2( 2 2(1) 3) 2) 2) 1 3/2 x 4 1 4 0 6. (a) f (x) 0 f (x) 0 8 (b) f (x) 0 16 3 0 4(0) 4 (x 3 2x) dx 2 2 2 14 x 4 2 0 sin x dx cos x 7. 5 (6.0 2 1.896 1 (126 2 2 65 1 12 … 2 66 841 3.13791, S100 2(800) 1.08943, S100 15. S50 1.37066, S100 limit S50 1.37076, S100 lower limit 26,360,000 ft3 2(860) 0.0001 as lower 1.37076 using a 0.000000001 as 0.82812 17. (a) T10 1.983523538 T100 1.999835504 T1000 1.999998355 (b) ET n 10 988.5, the town can sell at most 988 licenses. 2 0.016476462 1.6476462 100 1.64496 1000 1 to change seconds 3600 Tn 1.645 (3.2 2.2)(30 (4.5 3.2)(40 50) (5.9 (7.8 5.9)(60 70) (10.2 (12.7 10.2)(80 90) 16.0)(100 20.6)(110 26.2)(120 130)) (d) b 80) 12.7)(90 a 100) ET n ET 10n 1 3600 2 10 10 ET n , h2 2 120) (37.1 7.8)(70 60) 110) (26.2 10n 4.5)(50 (16.0 (20.6 (c) ET 40) 0.9785 miles 12 n 2 3 12(10n)2 2 ,M n2 3 1 12n 2 10 2 4 10 to hours 30) 841 1.37066 using a 0.82812, S100 16. S50 19,770 fish to be caught. Since 9. Sum the trapezoids and multiply by 110) 1.08943 … 2(1000) 2 58 3.14029 14. S50 (b) You plan to start with 26,360 fish. You intend to have 1 (2.2(0 2 0. 70.08 13. S50 0)(20) 19,770 20 1 0 since f (4) 2x, Mf (4) 13.0)(30) 2(12.7) 2(520) (0.75)(26,360) 1 4 9 (b) We are approximating the area under the temperature graph. By doubling the endpoints, the error in the first and last trapezoids increases. 15,990 ft3 200 8. (a) (0 2 12. (a) 0 on [0, ] 2 … 2(9.1) x3 av( f ) 0 2 0 2(8.2) 3 11. The average of the 13 discrete temperatures gives equal weight to the low values at the end. 2 2 sin x 0 12 (d) Simpson’s Rule for cubic polynomials will always give exact values since f (4) 0 for all cubic polynomials. The approximation is an underestimate. (c) x2 21) 0 (c) For f (x) 2 cos x, f (x) 4(4) 12 4 2(1) 2 21 1 3 4 1 2 4 81 4 Es 2 3 2( 1) 5.333 4 2 1 1 0 sin x, h x T 4 2 3/2 x 3 x dx 1 0 The approximation is an underestimate. (c) ( 1) 4 1 0 1 (1 3 (b) 0 on [0, 4] 3 2x, h 1 S 5.146 3 1 1/2 x , f (x) 2 (b) f (x) x3 10. (a) f (x) ET n 6 10 2 Chapter 5 Review 22. Note that the tank cross-section is represented by the shaded area, not the entire wing cross-section. Using Simpson’s Rule, estimate the cross-section area to be 18. (a) S10 2.000109517 S100 2.000000011 S1000 2.000000000 (b) ES n 10 2 1.09517 100 10 1000 (c) ES 10 a ,h 4 ES 4 n4 ,M n ES 10n 1 h[ y0 yn 1 180(10n)4 y6] 2(2.0) 2yn 1 10.63 ft yn] h[ y1 y2 … 4y3 … 2y2n RRAMn LRAMn 4 10 ES 2 n 2x cos (x 2) 2x 2x sin (x 2) 2 cos (x 2) 4x 2 sin (x 2) 2 cos (x 2) 24. S2n h (y 30 4y1 2y2 4y2n 1 [h(y0 3 (b) y2n ) 2y2 2y1 1 … (2h)(y1 2T2n MRAMn 3 y3 2y2n … y5 b , where h 1 a 2n y2n) y2n 1)] . [ 1, 1] by [ 3, 3] (c) The graph shows that for 1 x 1. 1 (d) ET (e) For 0 1 (f) n 3 h2 2 0.1, ET ( 1) h 2 0.1 f (x) 2 so f (x) 3 s Chapter 5 Review (pp. 298–301) 1. h2 2 ( 1) 2 (h )(3) 12 h f (4)(x) y 4 0.12 2 0.005 0.01 20 2 4x2 2x cos (x 2) 8x sin (x 2) 4x sin (x 2) 8x3 cos (x 2) 12x sin (x 2) 8x3 2x sin (x 2) 24x2 cos (x 2) 12x 2x cos (x 2) 12 sin (x 2) (16x 4 12)sin (x 2) 48x 2 cos (x 2) 20. (a) f (x) 1 2. 2 1 (b) 2 x y 4 [ 1, 1] by [ 30, 10] (c) The graph shows that 30 f (4)(x) so f (4)(x) 30 for 1 x 1. 1 (d) Es (e) For 0 (f) n 21. h ( 1) 4 (h )(30) 180 h 1 24 in. 6 2 0.4 4 h 3 0.4, ES ( 1) h h4 3 2 10 4 0.4 3 0.00853 0.01 LRAM4: 5 4 in. Estimate the area to be 4 [0 3 0] 4(18.75) 466.67 in2 yn] 2 180n 4 5 19. (a) f (x) f (x) … 5 180 n 4 4(1.9) 1 1 42 lb/ft3 11.2 ft2 2y2 … 4y5 11.2 ft2 2.1] 2y1 y1 2y4 2(1.8) (5000 lb) h [y 20 23. Tn n 4 4y3 4(2.1) Length 4 ES 8 0 10n (d) b 4 10 1.1 1 [y 4y1 2y2 30 1 [1.5 4(1.6) 3 Sn 2(24) 4(26) 2(24) 4(18.75) 1 0 2 15 8 3 x 21 8 15 4 3.75 2 227 228 Chapter 5 Review 3. 2 y 5 f (x) dx 9. (a) f (x) dx 5 4 3 2 The statement is true. 5 (b) 2 [ f (x) 2 g(x)] dx 5 5 f (x) dx g(x) dx 2 2 1 MRAM4: 4. 1 63 2 64 165 64 f (x) dx x 2 2 5 195 64 4 105 64 5 f (x) dx 2 g(x) dx 2 3 2 2 9 4.125 The statement is true. y 5 (c) If f (x) 4 5 g(x) on [ 2, 5], then f (x) dx g(x) dx, 2 2 but this is not true since 5 2 5 f (x) dx 2 f (x) 2 5 2 g(x) dx f (x) 4 3 7 and 2 2. The statement is false. 2 1 1 15 RRAM4: 28 21 8 3 10. (a) Volume of one cylinder: r 2h x 2 15 4 0 Total volume: V lim n→ 3.75 y ∑ sin2(mi ) x i1 sin2 x on [0, ]. (b) Use 5. sin2 (mi ) x n NINT( sin2 x, x, 0, ) 4.9348 4 11. (a) Approximations may vary. Using Simpson’s Rule, the area under the curve is approximately 2 1 [0 3 4(0.5) 2(4.75) 1 (LRAM4 2 T4 2 6. 2 (4x x 3) dx 2x 2 7. 14 x 4 1 15 24 15 4 3.75 0] 26.5 s 4 4 0 LRAMn MRAMn RRAMn 1.78204 1.60321 1.46204 20 1.69262 1.60785 1.53262 30 1.66419 1.60873 1.55752 50 10 1.62557 1.60937 1.59357 The curve is always increasing because the velocity is always positive, and the graph is steepest when the velocity is highest, at t 6. 1000 1.61104 1.60944 1.60784 1 dx x 5 ln x ln 5 1 ln 1 ln 5 t Time (sec) 1.64195 1.60918 1.57795 100 1 4(2) 4(4.5) 30 8 10 5 (b) 2 n 8. 2(3.5) 2(3.5) The body traveled about 26.5 m. RRAM4) 0 4(4.5) 4(2) Position (m) 1 x 2(1) 1.60944 229 Chapter 5 Review 10 e x 3 dx 12. (a) 1 10 (b) 1 x sin x dx 25. 0 10 ln x 1 36(2x dx 1)3 1) 9(2x 1 2 10 (e) 1) ( 9) 0 1 13. The graph is above the x-axis for 0 x-axis for 4 x x 2 1 dx x2 x 26. x sin dx 10 2 9 8 x 2) dx (x 1 12 x 2 4 and below the 1 x 1 2 2 0 0 Total area 6 (4 x) dx (4 0 27. x) dx 2 1 3 2 6 4 sec x tan x dx sec x 1 /3 4 12 x 2 [8 0] 0 [6 8] 1 4 28. 10 x 2) cos (1 1 1 x 2 and below the 29. 1 0 2 2 y 0 1 1 1 dy 2 ln y 1 2 ln 3 0 2 ln 3 0 30. Graph y x 2 x 2) dx 2x sin (1 2 14. The graph is above the x-axis for 0 1 6 12 x 2 4x 2 /3 4 4x x-axis for dx 1 dx x2 1 2 0 1 10 3 0 2)2 dx x(3x 0 1 1 36 0 (d) 0 1 (2x 0 (c) e 1 dx x 24. 0 4 x 2 on [0, 2]. /2 Total area cos x dx cos x dx 0 /2 /2 sin x sin x 0 (1 /2 0) (0 1) 2 [ 1.35, 3.35] by [ 0.5, 2.6] The region under the curve is a quarter of a circle of radius 2. 2 2 15. 5 dx 5x 10 ( 10) 50 2 8 20 2 2 5 4x dx 16. 2x 2 2 42 0 2 31. Graph y /4 /4 cos x dx 17. 2 sin x 0 1 (3x 2 4x x3 7) dx 1 2x 2 7x 1 6 1 (8s 3 12s 2 ( 10) 2s4 5) ds 16 20. 1 5s 4 x 4 dx x2 y 4 1 4/3 8 x dx 2 ( 4) 2 32. Graph y 1 dy 3y 1/3 1 (4)(4) 2 64 1 (8)(8) 2 40 x 2 on [ 8, 8]. 27 1 ( 3) 2 1 ( 2) 1 dt tt sec2 0 3 4 4 t 3/2 dt 2t 1 1/2 4 1 d tan 3 0 1 [ 9.4, 9.4] by [ 3.2, 9.2] The region under the curve y of radius 8. /3 /3 23. 0 2 1 22. 3 0 27 21. [ 4, 8] by [0, 8] The region under the curve consists of two triangles. 1 4s3 0 2 x dx on [ 4, 8]. 2 1 19. 1 (2)2 4 2 0 2 0 18. x 2 dx 4 5 0 3 8 2 8 64 x 2 dx 8 2 64 8 64 x 2 dx x 2 is half a circle 2 1 (8)2 2 64 230 Chapter 5 Review 33. (a) Note that each interval is 1 day 24 hours Upper estimate: 24(0.020 0.021 0.023 0.025 0.028 0.031 0.035) 4.392 L Lower estimate: 24(0.019 0.020 0.021 0.023 0.025 0.028 0.031) 4.008 L 24 (b) [0.019 2 2(0.020) 0.035] 3 [5.30 2 x 2 25 2(0.031) dt 1 0 1 t2 1 dt 1 2 x2 1 1 x2 dt t 4t1/2 1 50 x 50 25 4 1.11) … 0) 2(1.11) 20 50 4 x 30 4 2500 87.15 ft … x c(2500) … 2(5.04) 2(5.25) t 0 x 1 2 1 (2x)2 1 2 4x 2 1 4.2 L 34. (a) Upper estimate: 3(5.30 5.25 5.04 Lower estimate: 3(5.25 5.04 4.71 (b) 2x d dx 43. c(x) … 2(0.021) dy dx 42. 103.05 ft 30 230 0] The total cost for printing 2500 newsletters is $230. 95.1 ft 14 44. av(I) 35. One possible answer: The dx is important because it corresponds to the actual physical quantity x in a Riemann sum. Without the x, our integral approximations would be way off. 4 36. 0 f (x) dx 4 0 0 x 2 dx 2) dx 4 13 x 3 2x 4 [0 64 3 16] 2 1 since min sin x 0) 16 3 1 1 1 x4 sin x dx sin2 x dx 1 (max f )(1 0) 2 4 1 4 0 0 2t (24 0 14 t 4 14 x 4 3) dt 4 1 16 43 0 0 3x a 1 a 0 a 1 2 3/2 ax a3 x dx 0 39. dy dx 2 4 3 x2 3x 4 dy dx 2 cos3 (7x 2) 4x 2 12x 16 dy 41. dx d dx 1 6 3 t4 x2 3x 0 0 3.09131 1.63052. f (x). f (1) (d) False, because g (1) f (1) (e) True, because g (1) f (1) f (1) 0. 0. 0 and g (1) f (1) 0. (g) True, because g (x) f (x), and f is an increasing function which includes the point (1, 0). d (7x 2) dx 14x 2 cos3 (7x 2) 1 x 4 dx 1 47. F(1) 0 x 3t (b) True, because g is differentiable. 2 3/2 a 3 0 192 0 x t2 (f) False, because g (1) a 14 0 46. (a) True, because g (x) cos3 x 40. 12t 2 4 (c) True, because g (1) (b) av( y) 1 24t 14 24t) dt x2 or x 1 2 3/2 x 43 x dx 24t Using a graphing calculator, x 0 38. (a) av( y) (t 3 14 x 4 14 x 4 2 24 14 0 0 1 0 45. 0 1 (min f )(1 x 0 0 2 since max sin2 x min f 4 sin2 x 1 1 14 Rich’s average daily holding cost is $192. We could also say (0.04)4800 192. 0 0 12 x 2 0.04 I(t) av(c) 4 (x max f c(t) f (x) dx 4 37. Let f (x) Rich’s average daily inventory is 4800 cases. 4 f (x) dx 1 (600 600t) dt 14 0 14 1 [600t 300t 2 4800 14 0 dt 6 3 x4 x 48. y (x) 5 sin t dt t 3 F(0) 0. ...
View Full Document

Ask a homework question - tutors are online