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34 Pages

### Problemas Impares

Course: MATH 1301, Fall 2011
School: Al Akhawayn University
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Word Count: 5218

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Integration Section CHAPTER 4.1 Section 4.2 Section 4.3 Section 4.4 Section 4.5 Section 4.6 4 Antiderivatives and Indefinite Integration . . . . . . . . . 177 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Riemann Sums and Definite Integrals . . . . . . . . . . . 188 The Fundamental Theorem of Calculus . . . . . . . . . . 192 Integration by Substitution . . . . . . . . . . . . . . . . . 197...

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Integration Section CHAPTER 4.1 Section 4.2 Section 4.3 Section 4.4 Section 4.5 Section 4.6 4 Antiderivatives and Indefinite Integration . . . . . . . . . 177 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Riemann Sums and Definite Integrals . . . . . . . . . . . 188 The Fundamental Theorem of Calculus . . . . . . . . . . 192 Integration by Substitution . . . . . . . . . . . . . . . . . 197 Numerical Integration . . . . . . . . . . . . . . . . . . . 204 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 CHAPTER Integration Section 4.1 4 Antiderivatives and Indefinite Integration Solutions to Odd-Numbered Exercises 1. d3 dx x3 dy dt y C d 3x dx 3 C 9x 4 9 x4 3. d 13 x dx 3 4x C x2 4 x 2x 2 5. 3t2 t3 C C 3t2 7. dy dx y x3 2 d3 t Check: dt 25 x 5 2 C 2 Check: d 25 x dx 5 C x3 2 Given 9. 3 Rewrite x1 3 dx Integrate x4 3 43 x 12 Simplify 34 x 4 3 x dx C C 11. 1 xx dx x 32 dx 12 1x2 22 C 2 x 1 4x2 C 13. 1 dx 2x3 x2 2 3x 1 x 2 3 dx C C 15. x 3 dx d x2 dx 2 3x C C x 3 17. 2x Check: 3x 2 dx d2 x dx x2 x3 x3 C C 2x 3x 2 Check: 19. x3 Check: 2 dx d 14 x dx 4 14 x 4 2x 2x C C x3 2 21. x3 2 2x d 25 x dx 5 1 dx 2 25 x 5 x2 x 2 x2 C x x3 2 C 2x 1 Check: 23. 3 x2 dx x2 3 dx d 35 x dx 5 3 x5 3 53 C x2 3 C 3 35 x 5 x2 3 C 25. 1 dx x3 Check: d dx x 3 dx x 2 2 C C 1 x3 1 2x2 C Check: 1 2x2 177 178 Chapter 4 Integration 27. x2 x x 1 dx 2 x3 23 x 3 2 x1 2 x 2 12 dx x3 2 25 x 5 x1 2 23 x 3 x 2 2x1 x2 2 C x x 1 21 x 15 2 3x2 5x 15 C Check: d 25 x dx 5 2 2x1 C 2 12 29. x 1 3x 2 dx x3 3x2 12 x 2 C x 2 dx 2x 3x2 x C x 1 3x 2 2 31. y2 y dy Check: d 27 y dy 7 y5 2 dy 2 27 y 7 y5 2 C y2 y C 2 Check: d3 x dx 12 x 2 2x 33. dx Check: 1 dx d x dx x C C 1 35. 2 sin x Check: d dx 3 cos x dx 2 cos x 2 cos x 3 sin x C 3 sin x 2 sin x C 3 cos x 37. 1 Check: csc t cot t dt d t dt csc t t C csc t 1 C csc t cot t 39. sec2 Check: sin d tan d d cos tan C cos sec2 C sin 41. tan2 y 1 dy C sec2 y dy sec2 y tan y tan2 y C 1 43. f x y cos x d Check: tan y dy 3 2 C 2C 3 x 2 2 0 2 3 C 45. f x fx 2 2x y 5 4 47. f x C f )x) 2x 2 1 x x3 3 x2 x3 3 x y 4 49. C dy dx y 1 2x 2x 1 x2 2 1, 1, 1 1 dx 1 x 1 x2 CC x C 1 fx f )x) f )x) x3 3 x 3 y f 3 f )x) 2x 3 2 x 3 2 1 2 3 3 2 1 3 2 x Answers will vary. f Answers will vary. S ection 4.1 Antiderivatives and Indefinite Integration 179 51. dy dx y 4 y cos x, 0, 4 cos x dx sin 0 sin x sin x C 4 53. (a) Answers will vary. y 5 CC 4 3 x 5 3 (b) dy dx y 2 2 y 1 x 2 x2 4 42 4 C x2 4 8t3 8t3 4 2 5t 32 1, 4, 2 x 4 C 4 6 8 2 C x 2 55. f x fx f0 fx 4x, f 0 4x dx 6 2x 2 20 6 2 6 2x 2 C 6 57. h t ht h1 ht 61. f x f4 f0 5, h 1 5 dt 5 11 2t4 4 5t C 11 CC CC 2t4 x 2 0 x 2 2 59. f x f2 f2 fx f2 fx fx f2 fx 2 5 10 2 dx 4 2x 2x 6 x 2 2x C1 1 fx f4 x 4 C2 fx fx f0 fx 0 32 dx C1 3 12 2x 12 C1 3 2 x C1 C1 1 5 C1 2 C1 1 dx x2 2 x 2x 0 4x1 2 C2 x 10 C2 4 3 dx 4x1 2 3x C2 C2 3x 0 C2 4x 0 3x 63. (a) h t h0 ht (b) h 6 0 1.5t 0 5 dt C 5t 2 0.75t 2 12 C 12 5t 12 C 0.75t2 0.75 6 56 12 69 cm 180 Chapter 4 Integration 65. f 0 (a) f 4 4. Graph of f is given. 1.0 (f) f is a minimum at x (g) 6 4 2 x 2 3. (b) No. The slopes of the tangent lines are greater than 2 on 0, 2 . Therefore, f must increase more than 4 units on 0, 4 . (c) No, f 5 < f 4 because f is decreasing on 4, 5 . (d) f is an maximum at x 3.5 because f 3.5 the first derivative test. 0 and y 2 4 6 8 (e) f is concave upward when f is increasing on ,1 and 5, . f is concave downward on 1, 5 . Points of inflection at x 1, 5. 67. a t vt v0 st s0 st 6 60 32 ft sec2 32 dt C1 32t C2 16t2 60t 6 Position function 60 dt 16t 2 60t C2 32t C1 6 69. From Exercise 68, we have: st 16t2 v0t 0 when t v0 32 v0 32 time to reach st 32t v0 maximum height. s v02 64 v0 32 v02 32 v02 v0 16 550 35,200 v0 32 2 v0 550 The ball reaches its maximim height when vt 32t 60 32t t s 15 18 16 9.8 9.8 dt v0 9.8t C1 9.8t 4.9t 2 0 60 15 seconds 8 2 187.617 ft sec 15 8 60 15 8 6 62.26 feet 73. From Exercise 71, f t vt 9.8t v0 t v0t v0t C2 s0 f 10 9.8 9.8t 10 10 9.8 7.1 m 10 4.9t2 10t 2. 0.) 71. a t vt v0 ft f0 s0 a vt st s 20 0 (Maximum height when v C1 v t 9.8t v0 dt C2 f t 1.6 1.6 dt 1.6t dt 0 1.6t 4.9t 2 75. v0 s0 s0 s0 1.6t, since the stone was dropped, v0 0. 0.8t2 0.8 20 2 0 320 Thus, the height of the cliff is 320 meters. vt v 20 1.6t 32 m sec S ection 4.1 Antiderivatives and Indefinite Integration 181 77. x t (a) v t t3 6t2 xt 3t 2 9t 3t 4t 6t 2 2 12t 3 12 0t5 9 3t 6t 1t 2 3 79. v t xt x1 xt at 1 t t 12 t>0 2 v t dt 4 2t1 2 vt 21 2t1 C 2 at vt CC (b) v t > 0 when 0 < t < 1 or 3 < t < 5. (c) a t v2 81. (a) v 0 v 13 at vt v0 v 13 6t 31 2 1 0 when t 3 25 80 1000 3600 1000 3600 250 m sec 36 800 m sec 36 2. 2 position function 1 t 2 32 1 acceleration 2t 3 2 25 km hr 80 km hr 83. Truck: v t st 30 30t Let s 0 6 6t Let v 0 3t2 Let s 0 0. 0. 0. Automobile: a t vt st a constant acceleration at C at 250 36 250 36 250 vt 36 800 36 550 36 a 13a 13a 550 468 250 t 36 2 At the point where the automobile overtakes the truck: 30t 0 0 (a) s 10 3t2 3t2 3t t 3 10 6 10 30t 10 when t 2 10 sec. 300 ft 60 ft sec 41 mph 275 234 s0 250 13 36 22 ft sec 15 (b) v 10 1.175 m sec2 0 189.58 m (b) st s 13 a t2 2 275 13 234 2 85. 1 mi hr 5280 ft mi 3600 sec hr (a) t V1 ft sec V2 ft sec (c) S1 t S2 t 0 0 0 5 10 10.27 55.73 15 23.47 74.8 0.0416 2 t 2 6.7991t2 2 20 42.53 88 25 66 93.87 30 95.33 95.33 (b) V1 t V2 t 0.1068t2 0.1208t2 0.0416t 6.7991t 0.3679 0.0707 3.67 30.8 V1 t dt V2 t dt 0.1068 3 t 3 0.1208t3 3 0.3679t 0.0707t S2 0 0 In both cases, the constant of integration is 0 because S1 0 S1 30 S2 30 953.5 feet 1970.3 feet The second car was going faster than the first until the end. 182 Chapter 4 Integration 87. a t vt st k kt k2 t since v 0 2 s0 0. 160 and k 2 t2 0.7. Since k 2 t2 0.7, At the time of lift-off, kt t v 1.4 k 1.4k k 1.4 k 1.4 k 160 1602 1.4 1602 k 18,285.714 mi hr2 7.45 ft sec2. 89. True x3 3 91. True x2 2 x2 2 93. False. For example, x x dx x dx x dx because C C1 C2 95. f x 1, 0 x < 2 3x, 2 x 5 x C1, 0x<2 3x2 C2, 2 x 5 2 31 C1 3 C1 2 2: fx f1 f is continuous: Values must agree at x 4 fx 6 C2 C2 2 0x<2 x 2, 3x 2 2, 2 x 5 2 2 do not agree. Hence f is not differentiable at x 2. The left and right hand derivatives at x Section 4.2 5 5 Area 5 1. i 4 1 2i 1 0 9 1 2 i i 1 i 1 1 21 1 10 9. j 1 2 1 17 8 3 158 85 4 5 5 35 4 3. k k2 1 1 1 2 1 5 5. k 1 c 2n ni 1 20 c 2i n 1 2 3 c c 2i n 19 c 4c 3n 21 ni 1 3i n 2 7. i 1 3i 1 20 5 j 8 3 11. 13. 20 15. i 1 2i 2 i i 1 2 20 21 2 420 17. i 1 i i2 i 1 19 20 39 6 2470 S ection 4.2 Area 183 15 15 15 15 19. i 1 ii 1 i 1 2 i 1 i 1 20 > 2 i3 152 16 4 14,400 12,040 2 i2 i 15 16 2 21. sum seq x i2 i 1 2 3 3, x, 1, 20, 1 20 20 2930 (TI-82) 1 3 20 15 16 31 2 6 2,480 120 1 2 20 6 60 20 21 41 6 2930 23. S s 3 1 4 3 9 2 51 9 2 33 2 25 2 16.5 12.5 25. S s 31 44 1 4 1 1 2 8 1 5 1 6 1 6 1 7 81 4 2 8 3 3 2 3 3 2 2 51 31 11 7 4 1 27. S 4 s4 0 1 4 1 5 11 44 11 44 11 655 11 755 2 11 24 11 24 11 755 11 855 81 lim 4 n 18 lim 2 n n4 1 31 44 0.768 0.518 29. S 5 s5 1 11 855 11 955 2n3 n4 n n2 11 955 11 25 n2 81 1 4 1 7 1 8 1 8 1 9 1 9 1 10 0.746 0.646 11 655 31. lim n 81 n2 n 1 n4 4 18 n n 1 n2 2 1 n2 S 10 33. lim n n2 18 1 2 9 n 35. i 1 2i 1n 2i n2 i 1 12 10 1.02 1.002 1.0002 1.2 1 1 nn 1 2 n2 2 n n n 2 Sn S 100 S 1000 S 10,000 n 37. k 6k k 1 n3 1 6n 2 k n3 k 1 6 2n2 n2 k 3n 1 6 6 nn n3 3n 1 2n 6 3 1 nn 2 2 1 1 2n2 n2 Sn S 10 S 100 S 1000 S 10,000 n 1.98 1.9998 1.999998 1.99999998 16i n2 16 n i n2 i 1 16 n n 1 n2 2 n2 n2 n 1 n 39. lim n i 1 n lim n lim n lim 8 8 lim 1 n 8 184 Chapter 4 Integration n 41. lim n i 1 i n3 1 1 2 n lim 1n n3 i 1 i2 1 n lim 1n n3 n n 1 n 2n 6 lim 12 6 1 n n 1 3n 1 1 nn 1 n 2 1 n2 1 3 2 lim n 1 2n3 lim n 6 n 3n2 n3 n 43. lim n i 1 1 i n 2 n 2 lim n 1 n 1 i 1 1n i ni 1 2 lim n 1 n2 n 2n2 21 1 2 3 45. (a) 3 y (e) x sn Sn n 5 1.6 2.4 10 1.8 2.2 2 n 2 n 50 1.96 2.04 100 1.98 2.02 4n i n2i 1 2 1 x 1 3 (f) lim n i 1 i 1 n lim lim 1 n 4 n 2 (b) x 2 n 0 2 n n 4 nn 1 n2 2 2n n 4n i n2i 1 4 nn 1 n2 2 2n n 1 2 1 Endpoints: 0 <1 (c) Since y sn i n 1 n lim 2 2 <2 <...< n n n x is increasing, f mi n 1 2 2 <n n n 1 2 n n lim i 1 i 2 n 2 n n lim lim lim f xi on xi 1, xi . f xi f i 1 1 x 2 2 n n n 2i n 1, i i 1 2 1 n 2 n n (d) f Mi Sn i f xi on xi n xi n f xi 1 x i 1 f 2i 2 nn n i i 1 2 n 0 2 n 1 n 47. y 2x sn i 3 on 0, 1 . n Note: x n 1 n 3 2 y f 1 i n 2 n2i n 1 n i 1 2 i 1 i n 1n 1 n 1 n 2 3 Area 49. y x2 Sn i 1 n 3 2n 2n2 1 x lim s n 2 Note: x n i 1 1 2 3 2 on 0, 1 . n 1 n 2 y f 1 n3 i i n n 1 n i2 2 7 3 i n nn 2 1 2n 6n3 1 n 1 2 1 2 6 3 n 1 n2 2 3 1 1 x 1 2 3 Area n lim S n S ection 4.2 Area 185 51. y sn 16 n x 2 on 1, 3 . Note: x f1 i 1 2 n 1 2i n 2 y 18 2i n 2 n 4i2 n2 n 16 i 1 2 n 2n 15 ni 1 2 15n n 30 Area 4i n 1 2n 6 1 4 1 4 n n 70 3 4 n 1 9i n 27 n n n2 1 2n 27 2 1 n 1 2n 6 1 513 4 1 1 9nn 1 n 2 4nn 1 n 2 1 23 1 3 3 n 3i n 3 1 14 12 10 8 6 4 2 x 1 2 3 4 nn n2 1 2n 30 8 n 6n2 n lim s n 8 3 53. y sn 64 n x3 on 1, 4 . Note: x f1 i 1 1 y 70 3i n 3 n 27i3 n3 n 64 i 1 3 n 60 50 40 30 20 10 1 x 1 2 3 5 6 3n 63 ni 1 3 63n n 189 Area 27i2 n2 2 27 n2 n 1 n3 4 81 n 4n2 1 189 2 81 n 6n2 81 4 27 27 n 1 2 n 128.25 n lim s n 55. y x2 x3 on 1, 1 . Note: x 2 n 2 y Again, T n is neither an upper nor a lower sum. n Tn i n 1 f 1 i n 1 1 4i n 10i n 20 n2 10 1 4 2i n 2 n 4i2 n2 16i 2 n2 n 1 i 1 2i n 2 1 8i3 n3 20 n i n2 i 1 1 2 n 2i n 2 n 3 2 n 1 1 1 8i3 n3 32 n3 3 n 4 2 n 6i n 12i 2 n2 4n 1 ni 1 1 2n 6 41 x 1 2 i 1 32 n 2 i n3 i 1 16 n4 1 n2 n2 n 4 16 n 3 i n4 i 1 1 2 4 n n 4 Area nn 2 1 n 10 1 16 2 3 32 3 nn 1 n2 2 3 n lim T n 186 Chapter 4 Integration 57. f y Sn 3y, 0 y 2 n Note: y n 2 n 2 n 0 n 2 n 2i n 1 n 2 n 6 6 n 4 y f mi i 1 y i 1 f 12 n2 2i n 2 6 n 3 i 1 2 12 n i n2 i 1 Area n nn 1 6n x 2 4 6 lim S n n lim 6 6 2 59. f y Sn y 2, 0 y 3 n Note: y 3 n n i 1 3 n 2 0 3 n 6 4 2 y f i 1 3i n nn 3i n 1 27 2n 3 n 27 n 2 i n3 i 1 3n 2 9 3 n 1 2 n 10 8 6 27 n3 Area 61. g y Sn i n 1 n 1 2n 6 n 9 2n2 n2 9 2n2 1 9 27 2n 9 2n2 x 2 2 4 4 6 8 10 lim S n lim 9 4y2 n y3, 1 y 3. g1 41 2i n 2i n 4i n 10i n 6 2 n 2 Note: y y 1 4i2 n2 4i2 n2 10 i 1 2i n 1 8i3 n3 8 3 4 3 2 n 6i n 2 3n n 4 44 3 65. f x Let ci 12i2 n2 8i3 n3 4 nn n2 2 4 2 2 4 x 2n 41 ni 1 2n 3 ni 1 Area 63. f x Let ci x n n 10 n n 1 n 2 1 2n 6 1 8 n2 n 1 n2 4 2 lim S n x2 xi 3, 0 x 2, n xi 2 1 tan x, 0 x xi xi 2 , c1 x i 1 4 ,n 4 . 1 ,c 42 4 . , c2 4 1 ,c 21 f ci i 1 3 ,c 43 ci2 3 3 5 ,c 44 1 2 25 16 7 4 Area x n 16 32 3 ,c 32 3 tan ci 16 tan 5 32 5 ,c 32 4 7 32 Area 1 2 69 8 67. f x n x i 1 f ci i 1 1 1 16 3 9 16 3 49 16 3 16 tan 32 tan 3 32 tan 7 32 0.345 x on 0, 4 . 4 5.3838 8 5.3523 12 5.3439 16 5.3403 20 5.3384 Approximate area Exact value is 16 3 S ection 4.2 Area 187 69. f x n tan x on 1, 3 . 8 4 2.2223 8 2.2387 12 2.2418 16 2.2430 20 2.2435 Approximate area 71. We can use the line y x bounded by x a and x b. The sum of the areas of these inscribed rectangles is the lower sum. y The sum of the areas of these circumscribed rectangles is the upper sum. y x a x a b b We can see that the rectangles do not contain all of the area in the first graph and the rectangles in the second graph cover more than the area of the region. The exact value of the area lies between these two sums. 73. (a) 8 6 4 2 x 1 2 3 4 1 2 3 4 y (b) 8 6 4 2 y x Lower sum: s4 0 (c) 8 6 4 2 4 51 3 6 151 3 46 3 15.333 Upper sum: 4 S4 51 3 6 62 5 2111 15 326 15 21.733 y (d) In each case, x 4 n. The lower sum uses left endpoints, i 1 4 n . The upper sum uses right endpoints, i 4 n . The Midpoint Rule uses midpoints, i 1 4 n . 2 x 1 2 3 4 Midpoint Rule: M4 22 3 (e) n sn Sn Mn 4 15.333 21.733 19.403 44 5 55 7 8 17.368 20.568 19.201 62 9 6112 315 19.403 100 18.995 19.251 19.125 200 19.06 19.188 19.125 20 18.459 19.739 19.137 (f) s n increases because the lower sum approaches the exact value as n increases. S n decreases because the upper sum approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the exact value, whereas the upper sum is always larger. 188 Chapter 4 Differentiation 75. y 77. True. (Theorem 4.2 (2)) 4 3 2 1 1 2 3 4 x b. A 79. f x 6 square units sin x, 0, y 2 1.0 0.75 0.5 0.25 sin x, the x-axis, x 0 and x 2. Let A2 area of the rectLet A1 area bounded by f x angle bounded by y 1, y 0, x 0, and x 2. Thus, A2 21 1.570796. In this program, the computer is generating N2 pairs of random points in the rectangle whose area is represented by A2. It is keeping track of how many of these points, N1, lie in the region whose area is represented by A1. Since the points are randomly generated, we assume that A1 A2 N1 A1 N2 N1 A. N2 2 f (x) = sin(x) ( , 1) 2 4 2 x The larger N2 is the better the approximation to A1. 81. Suppose there are n rows in the figure. The stars on the left total 1 n n 1 stars in total, hence 21 1 2 2 ... ... n n nn 1 2 2 ... n, as do the stars on the right. There are 1 1. nn 83. (a) y 4.09 10 5 x3 0.016x2 2.67x 452.9 (b) 500 (c) Using the integration capability of a graphing utility, you obtain A 76,897.5 ft2. 0 0 350 Section 4.3 1. f x xi lim i 1 Riemann Sums and Definite Integrals 0, x 1 n2 n 2 x, y 3i2 n2 n 0, x 3 2i n2 n 3, ci 1 3i2 n2 3 2 y 3i y= x n f ci xi lim i 1 3i2 3 2i n2 n2 i 1 1 ... 3 n2 3(2)2 . . . 3(n 1)2 3 n2 n2 x 3 3n lim 2i2 n n3 i 1 n lim 3 3 nn 2 n3 n 1 2n 6 1 2n 3n2 23 1 1 n nn 2 1 2n2 1 n lim 3 3 2 3 0 33 3.464 S ection 4.3 Riemann Sums and Definite Integrals 189 3. y 6 on 4, 10 . n n Note: x f4 i 1 10 n 6 n 4 n 6 , n 6 6 n 0 as n n i 1 f ci i 1 10 xi 6 dx 6i n 36 i 1 36 n 36 4 n lim 36 5. y x3 on n 1, 1 . n Note: x f i 1 1 n 2 n 24 n3 i 1 n n 1 n 2 , n 1 0 as n 2i n n 3 f ci i 1 xi 2 2 1 12 n2 i n 2i n i 1 i 1 2 n n 1 i 1 6i n 12i 2 n2 8i 3 n3 2 n i2 1 16 n4 i 3 n i3 1 61 2 n 0 42 1 n2 41 2 n 1 n2 2 n 1 x3 dx 1 n lim 7. y x2 n 1 on 1, 2 . n Note: x i n 1 n 2 n 1 1 , n 1 i n 0 as n 2 n i 1 f ci i 1 xi i 1 f1 1 1 n 1 n 1 2 6 n 1 i 1 2i n 1 n2 i2 n2 10 3 1 3 2n 1 n 1 6n2 2 2 2n i n2 i 1 10 3 5 1n2 i n3 i 1 3 2n 1 6n2 2 10 3 1 3 n x2 1 1 dx n lim n n 3 9. lim 0 i 3ci 1 10 xi 1 3x 10 dx 11. lim 0 i ci2 1 4 xi 0 x2 4 dx on the interval 5 1, 5 . 4 on the interval 0, 3 . 2 13. 0 3 dx 15. 4 4 x dx 17. 2 4 x2 dx 2 19. 0 sin x dx 21. 0 y 3 dy 23. Rectangle A A 0 y bh 3 34 4 dx 12 5 3 2 1 Rectangle x 1 2 3 4 5 190 Chapter 4 Integration 25. Triangle A A 0 y 1 bh 2 4 1 44 2 8 4 27. Trapezoid b1 b2 h A 2 Triangle 2 y 5 2 5 dx 9 2 9 6 x dx 2 A 0 x 2 4 2x 14 3 Trapezoid x 1 2 3 29. Triangle A A 1 y 31. Semicircle Triangle y 1 bh 2 1 1 21 2 1 x dx 1 1 1 A x 12 r 2 3 1 2 9 4 Semicircle 3 2 1 A 3 x2 dx 9 2 x 4 2 2 4 4 4 4 In Exercises 3339, 2 x3 dx 60, 2 x dx 6, 2 dx 2 2 4 4 4 33. 4 x dx 2 x dx 6 35. 2 4 4x dx 4 2 x dx 46 4 24 4 4 4 4 4 37. 2 x 8 dx 2 x dx 8 2 dx 6 82 10 39. 2 13 x 2 3x 2 dx 1 2 x3 dx 2 3 2 x dx 22 6 2 2 dx 1 60 2 7 5 7 6 6 36 f x dx 16 41. (a) 0 f x dx 0 0 f x dx 5 5 f x dx 10 10 3 13 43. (a) 2 fx g x dx 2 g x dx 2 (b) 5 5 f x dx 0 f x dx 0 5 10 6 6 2 g x dx 8 6 (b) 2 gx f x dx 2 f x dx 2 (c) 5 f x dx 5 2 3f x dx 3 0 10 2 2 12 4 30 (d) 0 f x dx 3 10 30 6 6 (c) 2 2g x dx 6 2 2 g x dx 6 (d) 2 3f x dx 3 2 f x dx 3 10 45. (a) Quarter circle below x-axis: (b) Triangle: 1 bh 2 (c) Triangle 1 2 1 4 r2 1 4 2 2 42 4 1 2 Semicircle below x-axis: 1 2 21 3 1 20 2 1 2 2 2 1 2 (d) Sum of parts (b) and (c): 4 (e) Sum of absolute values of (b) and (c): 4 (f) Answer to (d) plus 2 10 20: 3 2 2 23 5 2 2 47. The left endpoint approximation will be greater than the actual area: > 49. Because the curve is concave upward, the midpoint approximation will be less than the actual area: < S ection 4.3 Riemann Sums and Definite Integrals 191 51. f x 1 x 4 53. 4 3 2 1 y is not integrable on the interval 3, 5 and f has a discontinuity at x 4. 1 2 3 4 x a. A 5 square units 55. 2 3 2 y 1 1 2 x 1 2 1 3 2 2 1 d. 0 3 2 sin x dx 1 12 2 1 57. 0 x3 n Ln Mn Rn 2 x dx 4 3.6830 4.3082 3.6830 8 3.9956 4.2076 3.9956 12 4.0707 4.1838 4.0707 16 4.1016 4.1740 4.1016 20 4.1177 4.1690 4.1177 59. 0 sin2 x dx n Ln Mn Rn 4 0.5890 0.7854 0.9817 8 0.6872 0.7854 0.8836 12 0.7199 0.7854 0.8508 16 0.7363 0.7854 0.8345 20 0.7461 0.7854 0.8247 61. True 63. True 65. False 2 x dx 0 2 67. f x x0 x1 c1 x2 0, x1 3x, 0, 8 1, x2 3, x3 7, x4 4, x4 8 f2 10 2 x2 f5 40 4 x3 f8 x4 272 1 8 1, x2 1, c2 4 2, x3 2, c3 x 5, c4 f1 41 f ci i 1 x1 88 1 192 Chapter 4 Differentiation 69. f x 1, 0, x is rational is x irrational 0, f ci 1 or f ci 0 in each subinterval since there is not integrable on the interval 0, 1 . As are an infinite number of both rational and irrational numbers in any interval, no matter how small. 71. Let f x n x2, 0 x 1, and xi n 1 n. The appropriate Riemann Sum is f ci i 1 xi i 1 i n 32 2 1 n 1n2 i. n3i 1 ... n2 n 1 lim 3 12 n n 22 lim lim 1 n3 2n2 n 2n 3n 6n2 1n 6 1 1 1 3 1 2n 1 6n2 1 3 n n lim Section 4.4 1. f x 4 0 The Fundamental Theorem of Calculus 5 4 x2 x2 1 1 dx is positive. 5 3. f x 2 x x2 x x2 1 0 5 5 1 dx 5 5 2 5 2 1 1 0 5. 0 2x dx x2 0 1 0 1 7. 1 x 2 dx x2 2 0 2x 1 0 1 2 2 5 2 1 9. 1 t2 2 dt t3 3 1 1 2t 1 1 3 2 1 3 43 t 3 2 10 3 1 1 11. 0 2 2t 1 2 dt 0 4t2 4t 1 dt 2t2 t 0 4 3 1 2 2 3 2 1 1 3 13. 1 3 x2 u 1 dx 3 x 4 2 x 1 3 2 2 3 1 4 15. 1 1 2 du u 3 u1 1 2 2u 12 du 23 u 3 3 4 1 x2 32 4 2 4u1 2 1 4 3 44 2 3 4 2 3 17. 1 1 t 2 dt 34 t 4 1 3 1 1 3 2t 1 2 3 4 23 x 3 3 4 1 2 0 2 4 19. 0 0 x 3 x dx x 0 x1 2 dx 11 32 3 5 27 20 2 3 1 18 21. 1 t1 3 t2 3 dt 34 t 4 32 3 35 t 5 0 3 1 0 3 3 23. 0 2x 3 dx 0 3 3x x2 2x dx 32 32 2x 3 3 dx 9 2 split up the integral at the zero x 9 4 0 9 9 9 4 9 2 3 2 2 9 2 9 4 9 2 x2 0 3x 32 S ection 4.4 The Fundamental Theorem of Calculus 193 3 2 3 25. 0 x2 4 dx 0 4 4x 8 23 3 x 2 dx 2 x2 x3 3 4x 4 dx 3 2 x3 3 8 3 2 0 9 12 8 3 8 27. 0 1 sin x dx x cos x 0 1 0 1 2 6 6 29. 6 sec2 x dx tan x 6 3 3 3 3 3 23 3 3 31. 3 4 sec tan d 4 sec 3 42 42 0 3 33. 0 10,000 t 3 6 dt 10,000 3 t2 2 3 1 6t 0 \$135,000 35. A 0 x x2 dx x2 2 12 3 5 x3 3 1 0 1 6 37. A 0 3 x x dx 0 3x1 2 x3 2 dx 2x3 2 25 x 5 3 2 0 xx 10 5 3 2x 0 2 2 39. A 0 cos x dx sin x 0 1 41. Since y 0 on 0, 2 , 2 2 A 0 3x2 1 dx x3 x 0 8 2 10. 43. Since y 0 on 0, 2 , 2 2 45. 0 x 2 x dx fc 2 c 0 2c 1 1 c 2 x2 2 6 3 3 6 4x3 3 82 3 42 3 42 3 42 3 6 22 2 0 82 3 A 0 x3 x dx x4 4 x2 2 2 4 0 2 6. c 2c c 1 1 c c 42 3 6 42 3 2 1 0.4380 or c 1.7908 194 Chapter 4 Integration 4 4 47. 4 2 sec2 x dx fc 2 tan x 4 21 2 1 4 4 4 2 sec2 c sec2 c sec c c 4 8 4 2 2 arcsec arccos 2 13 x 3 0.4817 2 2 49. 1 2 2 2 4 2 x2 dx 8 3 1 4x 4 1 4 8 8 3 8 8 3 8 3 ( 2 3 3 ,8 3 ( 5 ( 2 3 3 ,8 3 ( Average value 4 x2 3 3 1 8 when x2 3 4 8 or x 3 23 3 1.155. 51. 1 0 0 sin x dx 2 2 1 cos x 0 2 2 2 (0.690, ( 2 (2.451, ( 53. If f is continuous on a, b and F x b f x on a, b , Average value sin x x 2 then a f x dx Fb Fa. 2 1 3 2 0.690, 2.451 6 2 6 55. 0 6 f x dx area of region A 6 6 1.5 57. 0 f x dx 0 f x dx 2 f x dx 1.5 5.0 6.5 59. 0 2 f x dx 0 2 dx 0 f x dx 15.5 (b) 1 3 3 12 61. (a) F x F0 Fx k sec2 x k 500 3.5 0 500 sec2 x dx 0 1500 1500 3 tan x 0 500 sec2 x 3 0 826.99 newtons 827 newtons 1 5 0 5 63. 0.1729t 0 0.1552t2 0.0374t3 dt 1 0.08645t2 5 5 0.05073t3 0.00935t4 0 0.5318 liter S ection 4.4 The Fundamental Theorem of Calculus 195 65. (a) 0 1 (b) 24 10 1 0 0 24 The area above the x-axis equals the area below the x-axis. Thus, the average value is zero. 67. (a) v (b) 8.61 90 The average value of S appears to be g. 10 4t3 0.0782t2 0.208t 0.0952 10 10 70 60 (c) 0 v t dt 8.61 4 t2 2 8 25 2 8 10 4t 4 0.0782t3 3 x2 2 0.208t 2 2 60 0.0952t 0 2476 meters x x 69. F x 0 t 4 2 25 2 64 2 5 dt 52 55 58 5t 0 5x 71. F x 10 2 dv 1v 10 x 10 5 8 35 4 x x 10v 1 2 dv 10 v x 1 F2 F5 F8 10 1 1 x F2 F5 F8 10 10 10 1 2 4 5 7 8 x x x 73. F x 1 cos d sin 2 sin 5 sin 8 x 3 8 sin 1 sin x sin 1 75. (a) 0 t d 12 x dx 2 2 dt 2x t2 2 x x 2t 0 12 x 2 2x F2 F5 F8 sin 1 sin 1 sin 1 34 t 4 x 3 8 0.0678 1.8004 0.1479 34 x 4 x1 3 3 (b) 2 77. (a) (b) t dt 3 3 16 34 x 4 x 3 x 12 79. (a) x4 sec2 t dt d tan x dx 1 tan t x4 tan x 1 d 34 x dx 4 x 12 x x (b) sec2 x x 81. F x 2 t2 x2 2x 2t dt 83. F x 1 t4 x4 1 1 dt 85. F x 0 t cos t dt x cos x Fx Fx Fx 196 Chapter 4 Integration x 2 87. F x x 4t x 1 dt 2 Alternate solution: x 2 2t 2 t x Fx x 0 4t 4t x x 1 dt x 2 2x 8x Fx 8 2 10 2 x 2 2x2 x 1 dt 0 x 4t 2 1 dt 4t 1 dt 8 4t 0 1 dt 0 Fx sin x 4x x3 1 4x 2 1 89. F x 0 t dt sin x 1 2 cos 23 t 3 x 2 sin x 0 2 sin x 3 32 91. F x 0 sin t 2 dt sin x3 2 Fx cos x sin x Fx 3x2 3x2 sin x 6 Alternate solution sin x Fx 0 t dt sin x x Fx 93. g x 0 d sin x dx sin x cos x x f t dt 0, g 1 1 ,g 2 2 1, g 3 1 ,g 4 2 0 95. (a) C x 5000 25 5000 25 5000 25 3 0 t1 4 dt 45 t 5 4 x 4 0 g0 y 2 1 3 12 5 x 5 1000 125 \$137,000 54 54 12x5 4 f (b) g x 1 2 3 4 C1 C5 1000 125 1000 125 1000 125 12 1 12 5 \$214,721 \$338,394 1 2 C 10 12 10 g has a relative maximum at x 97. True 2. 1 0 1 99. False; 1 x 2 dx 1 x 2 dx 0 x 2 dx 2 Each of these integrals is infinite. f x has a nonremovable discontinuity at x 1x x 0. 101. f x 0 1 t 2 x 1 dt 1 1 2 0t dt By the Second Fundamental Theorem of Calculus, we have fx 1 1x 1 Since f x 2 1 x2 1 x2 1 x2 1 1 x2 0. 1 1 0, f x must be constant. S ection 4.5 103. x t xt t3 3t 2 3 t2 6t 2 12t 4t 3t 5 Integration by Substitution 197 9t 9 3 1 2 3t Total distance x t dt 0 5 3t 0 1 3t 4t 20 1 dt 3 5 3 0 t2 4 3 dt 3 1 t2 4t 3 dt 3 3 t2 4t 3 dt 4 28 units 4 105. Total distance 1 4 x t dt v t dt 1 4 1 1 t 4 2 1 dt 2t1 22 1 2 units Section 4.5 f g x g x dx Integration by Substitution u gx du g x dx 1. 3. 5. 5x2 x x 2 1 2 10x dx 5x2 x2 tan x 5 1 1 10x dx 2x dx sec2 x dx 1 dx tan2 x sec2 x dx 7. 1 Check: 2x 4 2 dx d 1 2x dx 5 x2 12 1 5 5 2x C 21 x2 32 C 32 C 9 2x 4 9. 9 Check: 2x dx x2 32 C 2 3 3 9 2 2 9 3 x2 12 x2 32 C 9 x2 2x d2 9 dx 3 2x 198 Chapter 4 Integration 11. x3 x4 Check: 3 2 dx 1 4 3 x4 C 3 2 4x3 dx 2 1 x4 3 4 3 4x3 3 C 3 2 x4 12 x3 3 3 C d x4 3 dx 12 1 3 3 x4 3 12 x4 13. x2 x3 Check: 1 4 dx x3 5 1 4 3x2 dx 1 x3 1 3 5 5 C 1 4 x3 15 1 5 C d x3 1 dx 15 C 5 x3 1 4 3x2 15 x2 x3 15. t t2 Check: 2 dt d dt t2 1 2 2 3 t2 32 2 C 12 2t dt 3 2 t2 1 t2 2 2 32 2 3 12 32 C t2 2 t2 3 1 2t 2 32 C 2t 17. 5x 1 Check: x2 d dx 13 dx 5 2 x2 1 43 x2 C 13 2x dx 15 8 4 1 3 5 2 x2 1 x2 43 2x 43 C 5x 1 x2 15 1 8 13 x2 43 C x2 15 1 8 1 2 13 5x 3 1 19. x 1 Check: x2 3 dx 1 C x2 3 2x dx 21 x2 11 2 3 x2 2 2x 2 C x 1 x2 3 1 41 x2 2 C 1 d dx 4 1 x2 1 3 1 31 2 1 4 21. 1 x2 dx x3 2 d dx 1 x3 C 2 3x2 dx 1 3 11 3 11 x3 x3 1 2 1 C 3x2 1 31 2 x3 C Check: x3 x2 1 x3 12 23. x 1 Check: x2 d dx 1 t 3 dx 1 1 2 x2 1 12 x2 C 12 2x dx 1 1 2 x2 1 1 x2 2 12 12 C x 1 4 1 x2 C 2x x2 25. 1 Check: 1 dt t2 1 1t 4 4 1 C 1 t 3 1 dt t2 1 41 4 1 t 1 3 1t 4 1 t2 C 1 1 t2 1 t 3 d dt 27. 1 dx 2x Check: d dx 1 2 2x 2x C 12 2 dx 1 2x 2 1 2x 1 2 2 12 12 C 1 2x 2x C 2 S ection 4.5 Integration by Substitution 199 29. x2 3x x 7 dx 2 x3 2x3 2 2 3x1 14x1 2 7x C 12 dx 25 x 5 3x x 2 2x3 2 14x1 2 C 2 5 x x2 5x 35 C Check: d 25 x dx 5 2 dt t d 14 t dt 4 2 x2 7 31. t2 t Check: t3 t2 C 2t dt t3 14 t 4 2t t2 t2 t C 2 t 33. 9 Check: y y dy d 23 y dy 5 2 9y1 15 y 2 y3 C 2 dy 9 23 y 3 2 2 25 y 5 2 2 C 9y1 23 y 5 2 2 15 2 y 9 C y y d 6y3 dy 25 y 5 C y3 35. y 4x 4 x dx 4 x2 2 2 4x dx 16 x2 2 16 16 x2 12 12 37. y 2x dx C 1 2 x x2 x2 2x 1 3 2 dx 2 2x 2x 1 1 2x 3 3 2x C C 2 dx x2 12 x2 1 x2 2 2 x2 dy dx y 1 2x2 4 16 C 3 39. (a) 3 y (b) x4 x4 1 2 2 4 3 x2, 2, 2 x2 dx x2 1 4 3 1 4 3 2 1 2 32 4 x2 12 2x dx x2 32 x 2 1 2 C 32 1 4 3 CC 2 C 2, 2 : 2 y 22 x2 2 32 2 2 1 41. sin x dx cos x C 43. sin 2x dx 1 2 sin 2x 2x dx 1 cos 2x 2 C 45. 1 2 cos 1 d cos 1 1 2 d sin 1 C 200 Chapter 4 Integration 47. sin 2x cos 2x dx sin 2x cos 2x dx sin 2x cos 2x dx 1 2 1 2 1 2 sin 2x 2 cos 2x dx cos 2x 1 sin 2x 2 2 2 C 2 12 sin 2x 4 C1 C2 C OR C1 OR 2 sin 2x dx 1 2 1 cos 2x 2 2 sin 4x dx 1 cos2 2x 4 2 sin 2x cos 2x dx 1 cos 4x 8 49. tan4 x sec2 x dx tan5 x 5 C 15 tan x 5 C 51. csc2 x dx cot3 x cot x cot x 2 2 3 csc2 x dx C 1 2 cot2 x C 12 tan x 2 C 1 sec2 x 2 1 C 1 sec2 x 2 C1 53. cot2 x dx csc2 x 1 dx cot x x C 55. f x Since f 0 fx cos x dx 2 3 2 sin 2 sin 0 x 2 C C, C 3. Thus, 2 sin x 2 3. 57. u x 2, x xx u 2 dx 2, dx u u3 25 u 5 2 2 du 2 u du 2u1 43 u 3 2 2 du C C 2 4 10 C C 2u3 2 3u 15 2 x 15 2 x 15 59. u 1 x, x 1 x dx u, dx 1 u1 23 u 3 2 10 32 2 2 3x 3x 32 du u 2 x2 1 u du 2 2u3 45 u 5 u5 2 du 2 2 2 27 u 7 C C x 8 15 1 C x 2 2u3 2 35 105 2 1 105 2 1 105 x x 42u 32 15u2 42 1 12x 35 15x2 C 32 S ection 4.5 Integration by Substitution 201 61. u 2x 1, x 1 u 2 1 , dx 12 u 1 u 8 1 8 1 du 2 1 u u2 2u1 43 u 3 10u 2 x2 1 dx 2x 1 11 du 2 1 3u 6u1 45 1 2 12 12 2u 2 4 du du C C 10 2x 52 13 1 C C 45 C u3 2 125 u 85 2 2 2 u1 2 2 3u 60 2x 1 3 2x 60 1 60 1 15 63. u x x 1, x 1 u x x 1 1, dx dx 2x 2x 1 12x2 1 3x2 8x 2x du u u u u 1 u u x x x where C1 u 2u1 2u 1 2x 2x 1 1 du u 1 u 12 u 1 du C C 1 du 2 2x 1 1 C. 1 1 C C C1 65. Let u 1 x2 x x2 1 1, du 1 3 dx 2x dx. 1 2 1 x2 1 1 3 2x dx 12 x 8 1 1 4 1 0 67. Let u 2 x3 1, du 1 dx 3x2 dx 2 1 3 2 2x2 x3 1 x3 1 1 12 3x2 dx 2 x3 1 3 32 43 x 9 4 27 9 1 32 2 1 2 32 1 22 12 8 9 2 202 Chapter 4 Integration 69. Let u 4 0 2x 1 2x 1, du 1 dx 2 dx. 1 2 4 4 2x 0 1 12 2 dx 2x 1 0 9 1 2 71. Let u 9 1 1 1 x1 x, x 1, u x 1 x, du 1 dx. 2x 9 x 2 2 dx 2 1 1 x 2 1 2x dx 2 1 x 9 1 1 2 1 1 2 73. u 2 u, dx 1. When x 0 du 2, u 2 1 When x 2 0. 0 1 2 x dx u 1 u du 1 u3 2 u1 2 du 25 u 5 2 23 u 3 0 2 1 2 5 2 3 4 15 2 75. 0 cos 2 x dx 3 1, x 0, u 7 3 2 sin x 2 3 u 1, dx 1. When x 2 0 3 2 3 2 33 4 77. u x du 7, u 8 When x Area 8. 1 3 3 x3 x 0 1 dx 1 8 u u4 1 u du 3 u1 du 37 u 7 3 34 u 4 8 3 1 384 7 12 3 7 3 4 1209 28 79. A 0 2 sin x sin 2x dx 2 cos x 1 cos 2x 2 1 dx 2 7 4 0 2 3 81. Area 2 sec2 x dx 2 2 3 2 2 sec2 x 2 2 tan x 2 2 2 3 2 3 1 4 83. 0 x 2x 3 1 dx 3.333 10 3 85. 3 xx 15 3 dx 28.8 144 5 3 87. 0 5 cos 6 d 7.377 1 1 5 0 0 8 1 1 4 89. 2x 2x 1 2 dx 1 2 dx 1 2 2x 4x2 1 2 2 dx 4x 1 dx C1 1 2x 6 43 x 3 1 . 6 1 3 C1 x 43 x 3 C2 2x2 x 1 6 C1 2x2 They differ by a constant: C2 S ection 4.5 Integration by Substitution 203 91. f x 2 x2 x2 x2 x2 1 is even. 2 93. f x x2 dx 8 3 272 15 2 x5 5 x3 3 2 0 2 x x2 x x2 1 3 is odd. 0 1 dx 2 0 x4 32 5 1 3 dx 2 2 2 2 95. 0 x2 dx 0 x3 3 2 0 2 8 ; the function x2 is an even function. 3 x2 dx 0 2 (a) 2 2 x2 dx x2 dx 0 8 3 8 3 4 4 2 2 (b) 2 0 x2 dx 3x2 dx 2 2 0 x2 dx 2 16 3 8 (c) x2 dx 0 (d) 3 0 x2 dx 4 4 4 97. 4 x3 6x2 2x 3 dx 4 x3 2x dx 4 6x2 3 dx 0 2 0 6x2 3 dx 2 2x3 3x 0 232 99. Answers will vary. See Guidelines for Making a Change of Variables on page 292. 2 101. f x x x2 1 2 is odd. Hence, 2 x x2 1 2 dx 0. 103. dV dt Vt V0 V1 k t t k 1 k 2 1 k 1 C C 2 2 dt k t 1 C 500,000 400,000 200,000 and Solving this system yields k C 300,000. Thus, Vt When t b 200,000 t1 4, V 4 300,000. \$340,000. 105. 1 b (a) (b) (c) a 74.50 a 43.75 sin 262.5 262.5 cos cos cos t 6 t 6 t dt 6 3 0 6 3 12 0 1 b a 74.50t 262.5 262.5 cos t 6 b a 1 74.50t 3 1 74.50t 3 1 74.50t 12 1 223.5 3 1 447 3 1 894 12 102.352 thousand units 223.5 262.5 102.352 thousand units 74.5 thousand units 262.5 262.5 262.5 t 6 204 Chapter 4 1 b (a) (b) a b Integration 1 b a 1 cos 60 t 30 1 60 107. 2 sin 60 t a cos 120 t dt 1 120 1 30 0 1 b sin 120 t a 1 1 60 1 1 240 0 0 1 cos 60 t 30 1 cos 60 t 30 1 sin 120 t 120 1 sin 120 t 120 60 0 1 240 1 30 1 120 1.382 amps 1 30 4 1.273 amps 1 30 240 0 30 2 22 2 1 1 30 1 cos 60 t 30 1 sin 120 t 120 1 30 5 1 30 (c) 0 30 0 0 amps 109. False 2x 111. True 10 10 10 10 1 2 dx 1 2 2x 1 2 2 dx 1 2x 6 1 3 C ax3 10 bx2 cx d dx 10 ax3 cx dx 10 bx2 d dx Even 0 2 0 bx2 d dx Odd 113. True 4 sin x cos x dx 2 sin 2x dx cos 2x C 115. Let u b x fx a h, then du b dx. When x h b a, u h a h. When x b, u b h. Thus, h dx a h f u du a h f x dx. Section 4.6 2 Numerical Integration x2 dx 0 2 1. Exact: Trapezoidal: 0 2 13 x 3 1 0 4 1 0 6 2 0 8 3 2 4 1 2 1 2 2 2.6667 21 2 2 x2 dx x2 dx 0 2 4 3 2 3 2 2 2 2 2 11 4 8 3 2.7500 2.6667 Simpsons: 21 2 2 2 2 3. Exact: 0 2 x3 dx x3 dx 0 2 x4 4 1 0 4 1 0 6 2 4.000 0 Trapezoidal: Simpsons: 0 2 4 1 2 1 2 3 21 3 3 2 4 3 2 3 2 3 2 3 3 17 4 24 6 4.2500 4.0000 x3 dx 21 3 2 3 S ection 4.6 Numerical Integration 205 2 5. Exact: 0 2 x3 dx x3 dx 0 2 14 x 4 1 0 8 1 0 12 2 4.0000 0 Trapezoidal: Simpsons: 0 2 4 1 4 1 4 3 2 3 2 4 2 2 4 3 2 3 3 4 4 3 4 3 21 3 3 2 3 5 4 4 5 4 3 2 3 6 4 2 6 4 3 2 3 7 4 4 7 4 3 8 3 4.0625 8 4.0000 x3 dx 21 9 7. Exact: 4 9 x dx x dx 4 23 x 3 5 2 16 9 2 4 18 2 37 8 16 3 2 38 3 21 4 12.6667 2 47 8 2 26 4 2 57 8 2 31 4 2 67 8 3 Trapezoidal: 12.6640 9 Simpsons: 4 x dx 5 2 24 4 37 8 21 4 47 8 26 4 57 8 31 4 67 8 3 12.6667 2 9. Exact: 1 2 1 x 1 x 1 2 1 2 dx dx 1 x 11 84 11 84 1 2 32 81 4 2 1 1 3 1 54 8 25 1 54 8 25 1 1 2 2 1 6 2 1 9 2 1 9 0.1667 1 32 1 2 Trapezoidal: 1 2 1 74 1 2 1 9 32 121 1 2 0.1676 1 32 1 2 2 Simpsons: 1 1 x 1 2 dx 11 12 4 11 12 4 4 1 74 1 2 1 9 64 81 64 121 0.1667 2 11. Trapezoidal: 0 2 1 1 0 x3 dx x3 dx 1 1 4 1 1 6 21 41 18 18 22 22 21 41 27 8 27 8 3 3 3.283 3.240 Simpsons: Graphing utility: 3.241 1 1 13. 0 x1 x dx 0 1 x1 x dx 1 0 8 1 0 12 1 1 4 1 1 4 1 4 1 4 1 1 2 1 1 2 1 2 1 2 3 1 4 3 1 4 3 4 3 4 Trapezoidal: 0 x1 x dx 2 2 2 0.342 1 Simpsons: 0 x1 x dx 4 2 4 0.372 Graphing utility: 0.393 206 Chapter 4 Integration 2 15. Trapezoidal: 0 cos x2 dx 2 8 0.957 cos 0 2 cos 2 4 2 2 cos 2 2 2 2 cos 2 4 2 2 cos 2 2 Simpsons: 0 cos x2 dx 2 12 0.978 cos 0 4 cos 2 4 2 2 cos 2 2 2 4 cos 2 4 2 2 cos 2 Graphing utility: 0.977 1.1 17. Trapezoidal: 1 1.1 sin x2 dx sin x2 dx 1 1 sin 1 80 1 sin 1 120 2 sin 1.025 2 2 sin 1.05 2 2 2 sin 1.075 2 2 sin 1.1 2 2 0.089 2 Simpsons: 4 sin 1.025 2 sin 1.05 4 sin 1.075 sin 1.1 0.089 Graphing utility: 0.089 4 19. Trapezoidal: 0 4 x tan x dx x tan x dx 0 32 48 0 0 2 4 16 16 tan tan 16 16 2 2 2 2 tan 16 16 2 2 tan 16 16 2 4 3 3 tan 16 16 3 3 tan 16 16 4 4 0.194 0.186 Simpsons: Graphing utility: 0.186 x3 3x2 6x 6 0 20 12 42 3 21. (a) y 23. y = f ( x) fx fx fx fx f 4 x x a b (a) Trapezoidal: Error 12 0.5 since 2. The Trapezoidal Rule overestimates the area if the graph of the integrand is concave up. f x is maximum in 0, 2 when x (b) Simpsons: Error f 4 20 0 180 44 5 0 since x 0. 25. f x 2 in 1, 3 . x3 1 and f 1 2. 366. (a) f x is maximum when x Trapezoidal: Error f (b) f 4 4 23 12n2 2 < 0.00001, n2 > 133,333.33, n > 365.15; let n x 24 in 1, 3 x5 1 and when f 4 x is maximum when x 25 180n4 1 24. 26. Simpsons: Error 24 < 0.00001, n4 > 426,666.67, n > 25.56; let n S ection 4.6 Numerical Integration 207 27. f x (a) f x 1 x 1 41 x 32 in 0, 2 . 0 and f 0 1 . 4 130. f x is maximum when x Trapezoidal: Error (b) f 4 81 12n2 4 in 0, 2 < 0.00001, n2 > 16,666.67, n > 129.10; let n x 4 16 1 15 x 72 f x is maximum when x 32 15 180n4 16 0 and f 4 0 15 . 16 12. Simpsons: Error < 0.00001, n4 > 16,666.67, n > 11.36; let n 29. f x (a) f x tan x2 2 sec2 x2 1 4x2 tan x2 in 0, 1 . 1 and f 1 3 f x is maximum when x Trapezoidal: Error (b) f 4 49.5305. 643. 10 12n2 3 49.5305 < 0.00001, n2 > 412,754.17, n > 642.46; let n 32x4 tan x2 1 and f 4 x 4 8 sec2 x2 12x2 36x2 tan2 x2 9184.4734. 48x4 tan3 x2 in 0, 1 f x is maximum when x 1 Simpsons: Error 1 05 9184.4734 < 0.00001, n4 > 5,102,485.22, n > 47.53; let n 180n4 Cx 5 48. 31. Let f x Ax3 Bx2 D. Then f 0 4 x 0. Simpsons: Error ba 0 180n4 Therefore, Simpsons Rule is exact when approximating the integral of a cubic polynomial. 1 Example: 0 x3 dx 1 0 6 4 1 2 3 1 1 4 This is the exact value of the integral. 33. f x n 4 8 10 12 16 20 2 Ln 12.7771 14.0868 14.3569 14.5386 14.7674 14.9056 3x2 on 0, 4 . Mn 15.3965 15.4480 15.4544 15.4578 15.4613 15.4628 Rn 18.4340 16.9152 16.6197 16.4242 16.1816 16.0370 Tn 15.6055 15.5010 15.4883 15.4814 15.4745 15.4713 Sn 15.4845 15.4662 15.4658 15.4657 15.4657 15.4657 208 Chapter 4 Integration 35. f x n 4 8 10 12 16 20 sin x on 0, 4 . Ln 2.8163 3.1809 3.2478 3.2909 3.3431 3.3734 2 Mn 3.5456 3.5053 3.4990 3.4952 3.4910 3.4888 Rn 3.7256 3.6356 3.6115 3.5940 3.5704 3.5552 Tn 3.2709 3.4083 3.4296 3.4425 3.4568 3.4643 Sn 3.3996 3.4541 3.4624 3.4674 3.4730 3.4759 37. A 0 x cos x dx 14 84 0.701 y Simpsons Rule: n 2 x cos x dx 0 0 cos 0 4 28 cos 28 2 14 cos 14 4 3 3 cos 28 28 ... 2 cos 2 1 1 2 4 2 x 5 39. W 0 100x 125 x 3 dx 12 x3 dx 5 0 3 12 400 400 5 12 15 12 125 125 5 12 15 12 3 Simpsons Rule: n 5 100x 125 0 200 3 10 12 0 125 10 12 3 ... 10,233.58 ft lb 12 41. 0 6 1 x2 dx Simpsons Rule, n 6 1 0 2 36 6 4 6.0209 3.1416 2 6.0851 4 6.1968 2 6.3640 4 6.6002 6.9282 1 113.098 36 43. Area 1000 125 2 10 2 125 2 120 2 112 2 90 2 90 2 95 2 88 2 75 2 35 89,250 sq m t 45. 0 sin x dx 2, n 10 2.477. By trial and error, we obtain t
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Al Akhawayn University - MATH - 1301
CHAPTER IntegrationSection 4.1 Section 4.2 Section 4.3 Section 4.4 Section 4.5 Section 4.64Antiderivatives and Indefinite Integration . . . . . . . . . 450 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456 Riemann Sums and Definite Int
Al Akhawayn University - MATH - 1301
Review Exercises for Chapter 4209Review Exercises for Chapter 41.y3.f f2x2x1 dx23 x 312 x 2xCx5.x3 x21dxx1 dx x212 x 21 xC7.4x3 sin x dx2x23 cos xC9. f x fx When x y C y2x, 2x dx 1: 1 2 2 x21, 1 x2 C11.at vt v0a a dt 0 a
Al Akhawayn University - MATH - 1301
Review Exercises for Chapter 4483Review Exercises for Chapter 42.y4.fu du3x 3 dx3x2 dx 3x2 33x133 dx3x23Cf6.x32x2 x21dxx 12 x 22 2xx 1 x2dx C8.5 cos x2 sec2 x dx5 sin x2 tan xC10. f x fx6x1 6x 1 dx 3x 1212. 45 mph C
Al Akhawayn University - MATH - 1301
CHAPTER 5 Logarithmic, Exponential, and Other Transcendental FunctionsSection 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 Section 5.6 Section 5.7 Section 5.8 Section 5.9 The Natural Logarithmic Function: Differentiation . . . . 493 The Natural Lo
Al Akhawayn University - MATH - 1301
548Chapter 5Logarithmic, Exponential, and Other Transcendental FunctionsReview Exercises for Chapter 52. f x ln x 33 2 1 x 1 1 2 3 2 4 5 6 y4. ln x2x=31x1ln x21ln x1Horizontal shift 3 units to the right Vertical asymptote: x 36. 3 ln x2 ln
Al Akhawayn University - MATH - 1301
CHAPTER 5 Logarithmic, Exponential, and Other Transcendental FunctionsSection 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 Section 5.6 Section 5.7 Section 5.8 Section 5.9 The Natural Logarithmic Function: Differentiation . . . . 218 The Natural Lo
Al Akhawayn University - MATH - 1301
272Chapter 5Logarithmic, Exponential, and Other Transcendental Functions85. As k increases, the time required for the object to reach the ground increases. ex 2x87. y ycosh x ex 2 eex89.y cosh ycosh x 1 1 sinh y1xsinh xsinh y y y1 cosh2 y
Al Akhawayn University - MATH - 1301
PARTII CHAPTER 6 Applications of IntegrationSection 6.1 Section 6.2 Section 6.3 Section 6.4 Section 6.5 Section 6.6 Section 6.7 Area of a Region Between Two Curves . . . . . . . . . . 264 Volume: The Disk Method . . . . . . . . . . . . . . . . . 271 Vol
Al Akhawayn University - MATH - 1301
Review Exercises for Chapter 6 26. From Exercise 21: F 64 15 1 22299753.98 lb28. h y3y 5x 2 x 2 4y 5 2 4y 5 y330. h y 4 for x, you obtain y. Ly F 4y 5 y y 5 y y12 2 62.40y 7 16 24Solving y x Ly Fy27 16 7 16 y 16y2124y 12y2 dy y2 dy 213
Al Akhawayn University - MATH - 1301
CHAPTER 5 Logarithmic, Exponential, and Other Transcendental FunctionsSection 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 Section 5.6 Section 5.7 Section 5.8 Section 5.9 The Natural Logarithmic Function: Differentiation . . . . 218 The Natural Lo
Al Akhawayn University - MATH - 1301
PARTI CHAPTER 6 Applications of IntegrationSection 6.1 Section 6.2 Section 6.3 Section 6.4 Section 6.5 Section 6.6 Section 6.7 Area of a Region Between Two Curves . . . . . . . . . . .2Volume: The Disk Method . . . . . . . . . . . . . . . . . . 9 Volum
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40Chapter 6Applications of IntegrationReview Exercises for Chapter 651. A1y1 dx x21 x5 14 513. A11 x2 11dxarctan x11(1, 1)45, ,2 3 44y21 25x(1, 0)(5, 0)21,1 211,1 (1, 0)1 2x1 (1, 0)125. A20x 12 x 2x3 dx 14 x
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108Chapter 7 1, x0Integration Techniques, LHpitals Rule, and Improper Integrals83. For n I1x214dxblim1 2bx20142x dxblim1 1 6 x2 1b 3 01 . 6For n &gt; 1, In0x2n x211n3dxblim2n dv 1 6 x2 xx2n 2 2 x2 1 x2bb n20n n 2n1 2
Al Akhawayn University - MATH - 1301
CHAPTER 7 Integration Techniques, LHpitals Rule, and Improper IntegralsSection 7.1 Section 7.2 Section 7.3 Section 7.4 Section 7.5 Section 7.6 Section 7.7 Section 7.8 Basic Integration Rules . . . . . . . . . . . . . . . . . . . 308 Integration by Parts
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358Chapter 7 1 e 32 f x dx500.4Integration Techniques, LHpitals Rule, and Improper Integrals84. (a) f x90x70218(b) P 72 x &lt; (c) 0.50.2525 0.5 0.2475 0.2525P 70 x 721.0These are the same answers because by symmetry, P 70 x &lt; and P 70 x &lt; P 7
Al Akhawayn University - MATH - 1301
CHAPTER 7 Integration Techniques, LHpitals Rule, and Improper IntegralsSection 7.1 Section 7.2 Section 7.3 Section 7.4 Section 7.5 Section 7.6 Section 7.7 Section 7.8 Basic Integration Rules Integration by Parts . . . . . . . . . . . . . . . . . . . 50.
Al Akhawayn University - MATH - 1301
108Chapter 7 1, x0Integration Techniques, LHpitals Rule, and Improper Integrals83. For n I1x214dxblim1 2bx20142x dxblim1 1 6 x2 1b 3 01 . 6For n &gt; 1, In0x2n x211n3dxblim2n dv 1 6 x2 xx2n 2 2 x2 1 x2bb n20n n 2n1 2
Al Akhawayn University - MATH - 1301
CHAPTER Infinite SeriesSection 8.1 Section 8.2 Section 8.3 Section 8.4 Section 8.5 Section 8.6 Section 8.7 Section 8.8 Section 8.98Sequences . . . . . . . . . . . . . . . . . . . . . 369 Series and Convergence . . . . . . . . . . . . . . 373 The Integr
Al Akhawayn University - MATH - 1301
414 66. a2nChapter 8Infinite Series 68. Answers will vary.10 (odd coefficients are zero) 1 ,g 16270. y60 , v0 3x 3x 3x64, k32 1 16 32 x3 3 64 3 1 2 3 1 16 2 32 x 4 4 64 4 1 2 4232x 2 2 64 2 1 2 32 32n.22x 2 2 64 223x3 3 64 316n 224x 4 4 6
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CHAPTER Infinite SeriesSection 8.1 Section 8.2 Section 8.3 Section 8.4 Section 8.5 Section 8.6 Section 8.7 Section 8.8 Section 8.98Sequences . . . . . . . . . . . . . . . . . . . . . 121 Series and Convergence . . . . . . . . . . . . . . 126 The Integr
Al Akhawayn University - MATH - 1301
R eview Exercises for Chapter 8167Review Exercises for Chapter 81. an 1 n! 3. an 2 : 6, 5, 4.67, . . . n Matches (a) 4 5. an 10 0.3n 1:10, 3, . . .Matches (d) n3 n27. an85n n29. limnn n21011. limn1Converges0 012The sequence seems t
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CHAPTER 9 Conics, Parametric Equations, and Polar CoordinatesSection 9.1 Section 9.2 Section 9.3 Section 9.4 Section 9.5 Section 9.6 Conics and Calculus . . . . . . . . . . . . . . . . . . . . 424 Plane Curves and Parametric Equations . . . . . . . . . .
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R eview Exercises for Chapter 9 5 4 3 246150. a r24, c5, b3, e52. a r22, b1, c 1 3 4 cos 23, e19 25 16 cos 2154. A21 22 232 2 sin22d4231 2 sin2d3.3756. (a) r1ed e cos 0, r c a ea a a1 e.(b) The perihelion distance is a When
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CHAPTER 9 Conics, Parametric Equations, and Polar CoordinatesSection 9.1 Section 9.2 Section 9.3 Section 9.4 Section 9.5 Section 9.6 Conics and Calculus . . . . . . . . . . . . . . . . . . . . 177 Plane Curves and Parametric Equations . . . . . . . . . .
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214Chapter 9Conics, Parametric Equations, and Polar Coordinates ed sin63. r11ed and r2 sin1Points of intersection: ed, 0 , ed, dy r1: dx ed sin ed 1 sin 1 dy dx cos sin 1. At ed, cos sin 1. At ed, ed cos sin ed cos 1 sin 1 , dy dx2sin cos2At ed
Al Akhawayn University - MATH - 1301
CHAPTER 10 Vectors and the Geometry of SpaceSection 10.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . 227Section 10.2 Space Coordinates and Vectors in Space . . . . . . . . . . 232 Section 10.3 The Dot Product of Two Vectors . . . . . .
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256 91. x2Chapter 10 y2 z2 z2 16, 16 16 4Vectors and the Geometry of Space 93. x2 y2 z2 z2 2z 2z 0 0, r 2 0, z 12(a) r 2 (b)2(a) r 2 (b)21 0,2 cos 2 cos2 cos95. x2y24y 4r sin , r 4 sin sin , 4 sin 4 sin csc 0,97. x2y29 r 2 sin2 9 cos2 sin2
Arkansas Little Rock - FINC - 3343
Chapter 01: The Goals and Functions of Financial ManagementChapter 1 The Goals and Functions of Financial ManagementDiscussion Questions1-1. How did the recession of 20072009 compare with other recessions since the Great Depression in terms of length?
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Chapter 02: Review of AccountingChapter 2 Review of AccountingDiscussion Questions2-1. Discuss some financial variables that affect the price-earnings ratio. The price-earnings ratio will be influenced by the earnings and sales growth of the firm, the
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Chapter 03: Financial AnalysisChapter 3 Financial AnalysisDiscussion Questions3-1. If we divide users of ratios into short-term lenders, long-term lenders, and stockholders, in which ratios would each group be most interested, and for what reasons? Sho
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Chapter 04: Financial ForecastingChapter 4 Financial ForecastingDiscussion Questions4-1. What are the basic benefits and purposes of developing pro forma statements and a cash budget? The pro-forma financial statements and cash budget enable the firm t
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Chapter 05: Operating and Financial LeverageChapter 5 Operating and Financial LeverageDiscussion Questions5-1. Discuss the various uses for break-even analysis. Such analysis allows the firm to determine at what level of operations it will break even (
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Chapter 06: Working Capital and the Financing DecisionChapter 6 Working Capital and the Financing DecisionDiscussion Questions6-1. Explain how rapidly expanding sales can drain the cash resources of a firm. Rapidly expanding sales will require a buildu
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Chapter 07: Current Asset ManagementChapter 7 Current Asset ManagementDiscussion Questions7-1. In the management of cash and marketable securities, why should the primary concern be for safety and liquidity rather than maximization of profit? Cash and
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Chapter 08: Sources of Short-Term FinancingChapter 8 Sources of Short-Term FinancingDiscussion Questions8-1. Under what circumstances would it be advisable to borrow money to take a cash discount? It is advisable to borrow in order to take a cash disco
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Chapter 09: Time Value of MoneyChapter 9 Time Value of MoneyDiscussion Questions9-1. How is the future value (Appendix A) related to the present value of a single sum (Appendix B)? The future value represents the expected worth of a single amount, wher
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Chapter 10: Valuation and Rates of ReturnChapter 10 Valuation and Rates of ReturnDiscussion Questions10-1. How is valuation of any financial asset related to future cash flows? The valuation of a financial asset is equal to the present value of future
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Chapter 11: Cost of CapitalChapter 11 Cost of CapitalDiscussion Questions11-1. Why do we use the overall cost of capital for investment decisions even when only one source of capital will be used (e.g., debt)? Though an investment financed by low-cost
Clarion - ACCOUNTING - 101
CHAPTER 1 THE ACCOUNTANT'S ROLE IN THE ORGANIZATION ACCOUNTANT'See the front matter of this Solutions Manual for suggestions regarding your choices of assignment material for each chapter.1-1 Management accounting measures, analyzes and reports financia
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CHAPTER 2 AN INTRODUCTION TO COST TERMS AND PURPOSES 2-1 A cost object is anything for which a separate measurement of costs is desired. Examples include a product, a service, a project, a customer, a brand category, an activity, and a department. 2-2 Dir
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CHAPTER 3 COST-VOLUME-PROFIT ANALYSIS NOTATION USED IN CHAPTER 3 SOLUTIONS SP: VCU: CMU: FC: TOI: Selling price Variable cost per unit Contribution margin per unit Fixed costs Target operating income3-1 Cost-volume-profit (CVP) analysis examines the beha
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CHAPTER 4 JOB COSTING 4-1Cost poola grouping of individual cost items. Cost tracingthe assigning of direct costs to the chosen cost object. Cost allocationthe assigning of indirect costs to the chosen cost object. Cost-allocation basea factor that links
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CHAPTER 5 ACTIVITY-BASED COSTING AND ACTIVITY-BASED MANAGEMENT 5-1 Broad averaging (or peanut-butter costing) describes a costing approach that uses broad averages for assigning (or spreading, as in spreading peanut butter) the cost of resources uniformly
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CHAPTER 6 MASTER BUDGET AND RESPONSIBILITY ACCOUNTING 6-1 a. b. c. d. The budgeting cycle includes the following elements: Planning the performance of the company as a whole as well as planning the performance of its subunits. Management agrees on what is
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CHAPTER 7 FLEXIBLE BUDGETS, DIRECT-COST VARIANCES, AND MANAGEMENT CONTROL 7-1 Management by exception is the practice of concentrating on areas not operating as expected and giving less attention to areas operating as expected. Variance analysis helps man
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CHAPTER 8 FLEXIBLE BUDGETS, OVERHEAD COST VARIANCES, AND MANAGEMENT CONTROL 8-1 Effective planning of variable overhead costs involves: 1. Planning to undertake only those variable overhead activities that add value for customers using the product or serv
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CHAPTER 9 INVENTORY COSTING AND CAPACITY ANALYSIS 9-1 No. Differences in operating income between variable costing and absorption costing are due to accounting for fixed manufacturing costs. Under variable costing only variable manufacturing costs are inc
Clarion - ACCOUNTING - 101
Clarion - ACCOUNTING - 101
CHAPTER 11 DECISION MAKING AND RELEVANT INFORMATION 11-1 1. 2. 3. 4. 5. The five steps in the decision process outlined in Exhibit 11-1 of the text are Identify the problem and uncertainties Obtain information Make predictions about the future Make decisi
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CHAPTER 12 PRICING DECISIONS AND COST MANAGEMENT 12-1 The three major influences on pricing decisions are 1. Customers 2. Competitors 3. Costs 12-2 Not necessarily. For a one-time-only special order, the relevant costs are only those costs that will chang
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CHAPTER 13 STRATEGY, BALANCED SCORECARD, AND STRATEGIC PROFITABILITY ANALYSIS 13-1 Strategy specifies how an organization matches its own capabilities with the opportunities in the marketplace to accomplish its objectives. 13-2 The five key forces to cons
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CHAPTER 14 COST ALLOCATION, CUSTOMER-PROFITABILITY ANALYSIS, AND SALES-VARIANCE ANALYSIS 14-1 Disagree. Cost accounting data plays a key role in many management planning and control decisions. The division president will be able to make better operating a
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CHAPTER 15 ALLOCATION OF SUPPORT-DEPARTMENT COSTS, COMMON COSTS, AND REVENUES 15-1 The single-rate (cost-allocation) method makes no distinction between fixed costs and variable costs in the cost pool. It allocates costs in each cost pool to cost objects
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CHAPTER 16 COST ALLOCATION: JOINT PRODUCTS AND BYPRODUCTS 16-1 Exhibit 16-1 presents many examples of joint products from four different general industries. These include: Industry Separable Products at the Splitoff Point Food Processing: Lamb Lamb cuts,
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CHAPTER 17 PROCESS COSTING 17-1 Industries using process costing in their manufacturing area include chemical processing, oil refining, pharmaceuticals, plastics, brick and tile manufacturing, semiconductor chips, beverages, and breakfast cereals. 17-2 Pr
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CHAPTER 18 SPOILAGE, REWORK, AND SCRAP 18-1 Managers have found that improved quality and intolerance for high spoilage have lowered overall costs and increased sales. 18-2 Spoilageunits of production that do not meet the standards required by customers f
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CHAPTER 19 BALANCED SCORECARD: QUALITY, TIME, AND THE THEORY OF CONSTRAINTS 19-1 Quality costs (including the opportunity cost of lost sales because of poor quality) can be as much as 10% to 20% of sales revenues of many organizations. Quality-improvement
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CHAPTER 20 INVENTORY MANAGEMENT, JUST-IN-TIME, AND SIMPLIFIED COSTING METHODS 20-1 Cost of goods sold (in retail organizations) or direct materials costs (in organizations with a manufacturing function) as a percentage of sales frequently exceeds net inco
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CHAPTER 21 CAPITAL BUDGETING AND COST ANALYSIS 21-1 No. Capital budgeting focuses on an individual investment project throughout its life, recognizing the time value of money. The life of a project is often longer than a year. Accrual accounting focuses o
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CHAPTER 22 MANAGEMENT CONTROL SYSTEMS, TRANSFER PRICING, AND MULTINATIONAL CONSIDERATIONS 22-1 A management control system is a means of gathering and using information to aid and coordinate the planning and control decisions throughout an organization an
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CHAPTER 1 THE ACCOUNTANTS ROLE IN THE ORGANIZATION See the front matter of this Solutions Manual for suggestions regarding your choices of assignment material for each chapter. 1-1 Management accounting measures and reports financial and nonfinancial info