# ISM_T11_C05_D - 344 Chapter 5 Integration CHAPTER 5...

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344 Chapter 5 Integration CHAPTER 5 PRACTICE EXERCISES 1. (a) Each time subinterval is of length t 0.4 sec. The distance traveled over each subinterval, using the ? œ midpoint rule, is h v v t, where v is the velocity at the left endpoint and v the velocity at ?? œ± " # ab ii 1 i i 1 the right endpoint of the subinterval. We then add h to the height attained so far at the left endpoint v to ? i arrive at the height associated with velocity v at the right endpoint. Using this methodology we build i1 the following table based on the figure in the text: t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90 76 65 h (ft) 0 2 9 25 56 114 188 257 320 378 432 481 525 564 592 620.2 t (sec) 6.4 6.8 7.2 7.6 8.0 v (fps) 50 37 25 12 0 h (ft) 643.2 660.6 672 679.4 681.8 NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. Remember that some shifting of the graph occurs in the printing process. The total height attained is about 680 ft. (b) The graph is based on the table in part (a). 2. (a) Each time subinterval is of length t 1 sec. The distance traveled over each subinterval, using the ? œ midpoint rule, is s v v t, where v is the velocity at the left, and v the velocity at the " # 1 i i 1 right, endpoint of the subinterval. We then add s to the distance attained so far at the left endpoint v ? i to arrive at the distance associated with velocity v at the right endpoint. Using this methodology we build the table given below based on the figure in the text, obtaining approximately 26 m for the total distance traveled: t (sec) 0 1 2 3 4 5 6 7 8 9 10 v (m/sec) 0 0.5 1.2 2 3.4 4.5 4.8 4.5 3.5 2 0 s (m) 0 0.25 1.1 2.7 5.4 9.35 14 18.65 22.65 25.4 26.4 (b) The graph shows the distance traveled by the moving body as a function of time for 0 t 10. ŸŸ 3. (a) a ( 2) (b) (b 3a ) b 3 a 25 3( 2) 31 !! ! ! ! 10 10 10 10 10 k1 oo o o o a 44 4 k œ œ ²œ ² ² œ ² œ ²²œ """ # kk k k k (c) (a b 1) a b 2 25 (1)(10) 13 ! ! 10 10 10 10 o o k k ± ² œ±² " œ ² ± ²œ

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Chapter 5 Practice Exercises 345 (d) b b (10) 25 0 !! ! ˆ‰ 10 10 10 k1 oo o 55 5 ## # ±œ ± œ ± œ kk 4. (a) 3a 3 a 3(0) 0 (b) (a b ) a b 0 7 7 ! ! ! 20 20 20 20 20 o o o k k k k œ œ œ ² œ ² œ²œ (c) b (20) (7) 8 ! 20 20 20 o "" " # ± œ ± œ 2b 77 7 22 k k (d) a 2 a 2 0 2(20) 40 ! ab 20 20 20 o ± œ± œ ± 5. Let u 2x 1 du 2 dx du dx; x 1 u 1, x 5 u 9 œ±Ê œ Ê œ œÊœ œÊœ " # (2x 1) dx u du u 3 1 2 '' 11 59 ± œ œ œ±œ ±"Î# ±"Î# "Î# " # * " ± 6. Let u x 1 du 2x dx du x dx; x 1 u 0, x 3 u 8 œ±Ê œ Ê œ # " # x x 1 dx u du u (16 0) 6 10 38 ± # "Î\$ %Î\$ "Î\$ " # ) ! œ œ ± œ 33 88 7. Let u 2 du dx; x u , x 0 u 0 œÊ œ œ ±Êœ ± œÊœ x 2 1 1 # cos dx (cos u)(2 du) [2 sin u] 2 sin 0 2 sin 2(0 ( 1)) 2 Î 00 2 ˆ x # # ! ±Î # œœ œ ± ± œ ± ± œ 1 1 8. Let u sin x du cos x dx; x 0 u 0, x u 1 œ œ Ê œ œ Ê œ 1 # (sin x)(cos x) dx u du 21 1 Î œ ’“ u 2 # " ! " # 9. (a) f(x) dx 3 f(x) dx (12) 4 (b) f(x) dx f(x) dx f(x) dx 6 4 2 ' ' ' 2 5 5 2 œ œ œ œ ± (c) g(x) dx g(x) dx 2 (d) ( g(x)) dx g(x) dx (2) 2 ' ' 52 2 2 25 5 5 ± 1 1 (e) dx f(x) dx g(x) dx (6) (2) ' 2 5 Š‹ f(x) g(x) 5 5 5 5 8 ² " " œ ² 10. (a) g(x) dx 7 g(x) dx (7) 1 (b) g(x) dx g(x) dx g(x) dx 1 2 1 ' ' ' 1 0 0 2 2 1 œ œ œ œ ± œ±œ± (c) f(x) dx f(x) dx (d) 2 f(x) dx 2 f(x) dx 2 ( ) 2 ' ' 20 0 0 02 2 2 œ œ œ 1 ÈÈ È È (e) [g(x) 3 f(x)] dx g(x) dx 3 f(x) dx 1 3 ' 0 2 ± œ ± 1 11. x 4x 3 0 (x 3)(x 1) 0 x 3
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ISM_T11_C05_D - 344 Chapter 5 Integration CHAPTER 5...

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