Pre-Calc Homework Solutions 230

# Pre-Calc Homework Solutions 230 - 230 Chapter 5 Review dy...

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

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

Unformatted text preview: 230 Chapter 5 Review dy dx d dx 2x 0 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 42. 1 t 2 x 1 dt 0 1 t2 1 1 dt 1 (2x)2 1 2 4x 2 1 x 2 1 x2 1 x2 1 2(0.020) 0.035] 2(0.021) 4.2 L ... ... ... 43. c(x) 2(0.031) 25 2 t dt x 50 50 4t1/2 4 1.11) 0) ... 103.05 ft 87.15 ft c(2500) 2(1.11) 0] 4 4 x x 25 34. (a) Upper estimate: 3(5.30 5.25 5.04 Lower estimate: 3(5.25 5.04 4.71 (b) 3 [5.30 2 20 30 2500 50 30 230 2(5.25) 2(5.04) The total cost for printing 2500 newsletters is \$230. 44. av(I) 1 (600 600t) dt 14 0 14 1 [600t 300t 2 4800 14 0 14 95.1 ft 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 0 4 Rich's average daily inventory is 4800 cases. c(t) av(c) 0.04I(t) 1 14 14 36. 4 f (x) dx 0 f (x) dx 4 0 f (x) dx 4 24 24t 24t) dt 1 24t 14 (24 0 12t 2 14 192 0 (x 4 2) dx 0 x 2 dx 4 0 1 2 x 2 0 2x 4 1 3 x 3 Rich's average daily holding cost is \$192. We could also say (0.04)4800 192. x [0 37. Let f (x) max f min f 1 16] sin2 x 64 3 0 16 3 45. 0 (t 3 2t 3) dt 1 4 t 4 1 4 x 4 t2 x2 x 3t 0 3x 2 since max sin2 x 1 since min sin x 1 1 0 1 4 x 4 1 4 x 4 x2 x2 4x 2 3x 3x 12x 4 4 16 0 0 3.09131 2 x4 (max f)(1 0) (min f)(1 1 0) 0 1 sin2 x dx 4 sin x dx 2 1 2 3/2 x 4 3 2 Using a graphing calculator, x or x 1.63052. f (x). f (1) f (1) f (1) f (1) 0 1 0 1 1 4 0 4 0 38. (a) av(y) x dx 0 1 16 4 3 0 4 3 46. (a) True, because g (x) (c) True, because g (1) (b) True, because g is differentiable. 0. 0. 0 and g (1) 0. f (1) 0. (d) False, because g (1) (e) True, because g (1) (f) False, because g (1) (b) av(y) 39. 40. dy dx dy dx d dx 1 a 0 a a 0 x dx 1 2 3/2 ax a 3 a 0 2 3/2 a 3 2 2 cos3 x cos3 (7x 2) x 1 (g) True, because g (x) f (x), and f is an increasing function which includes the point (1, 0). d (7x 2) dx 6 3 x4 14x 2 cos3 (7x 2) 1 47. 0 1 x 4 dx x F(1) 3 F(0) dy 41. dx 6 3 t4 dt 48. y(x) 5 sin t dt t ...
View Full Document

## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online