57319-0136147054_06

# 57319-0136147054_06 - Section 6.1 n 6.1.1: With a = 0 and b...

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Section 6.1 6.1.1: With a =0and b = 1, lim n →∞ n X i =1 2 x ? i x = Z 1 0 2 xdx = ± x 2 ² 1 0 =1. 6.1.2: With a =1and b = 2, lim n →∞ n X i =1 x ( x ? i ) 2 = Z 2 1 1 x 2 dx = ± 1 x ² 2 1 = 1 2 . 6.1.3: With a b = 1, lim n →∞ n X i =1 (sin πx ? i )∆ x = Z 1 0 sin πx dx = ± 1 π cos ² 1 0 = 2 π . 6.1.4: With a = 1and b = 3, lim n →∞ n X i =1 h 3( x ? i ) 2 1 i x = Z 3 1 (3 x 2 1) dx = ± x 3 x ² 3 1 = 24. 6.1.5: With a b =4, lim n →∞ n X i =1 x ? i q ( x ? i ) 2 +9 ∆ x = Z 4 0 x p x 2 +9 dx = ± 1 3 ( x 2 +9) 3 / 2 ² 4 0 = 125 3 9= 98 3 . 6.1.6: The limit is Z 4 2 1 x dx = ± ln x ² 4 2 =ln4 ln 2 = 2 ln 2 ln 2 = ln 2. 6.1.7: The limit is Z 1 0 e x dx = ± e x ² 1 0 = e 0 e 1 =1 1 e . 6.1.8: The limit is Z 4 0 2 x +1 dx = ± 1 3 (2 x +1) 3 / 2 ² 4 0 =9 1 3 = 26 3 . 6.1.9: The limit is Z 6 0 x x 2 dx = ± 1 2 ln( x 2 ² 6 0 = 1 2 ln 45 1 2 ln 9 = 1 2 ln 5. 6.1.10: The limit is Z 1 0 2 x exp( x 2 ) dx = ± exp( x 2 ) ² 1 0 = e 0 e 1 1 e . 6.1.11: With a b = 4, lim n →∞ n X i =1 2 ? i f ( x ? i x = Z 4 1 2 πxf ( x ) dx. Compare this with Eq. (2) in Section 6.3. 6.1.12: With a = b = 1, lim n →∞ n X i =1 [ f ( x ? i )] 2 x = Z 1 1 [ f ( x 2 dx . 6.1.13: With a b = 10, lim n →∞ n X i =1 q 1+[ f ( x ? i 2 x = Z 10 0 q f ( x 2 dx . 6.1.14: With a = 2and b =3,wehave lim n →∞ n X i =1 2 πm i q f ( m i 2 x = Z 3 2 2 q f ( x 2 dx. 524

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6.1.15: M = Z 100 0 1 5 xdx = ± 1 10 x 2 ² 100 0 = 1000 0 = 1000 (grams). 6.1.16: M = Z 25 0 (60 2 x ) dx = ± 60 x x 2 ² 25 0 = 875 0 = 875 (grams). 6.1.17: M = Z 10 0 x (10 x ) dx = ± 5 x 2 1 3 x 3 ² 10 0 = 500 3 0= 500 3 (grams). 6.1.18: M = Z 10 0 10 sin πx 10 dx = ± 100 π cos 10 ² 10 0 = 100 π ³ 100 π ´ = 200 π (grams). 6.1.19: The net distance is Z 10 0 ( 32) dt = ± 32 t ² 10 0 = 320 and the total distance is 320. 6.1.20: The net distance is Z 5 1 (2 t + 10) dt = ± t 2 +10 t ² 5 1 =75 11 = 64 and because v ( t )=2 t = 0for1 5 t 5 5, this is the total distance as well. 6.1.21: The net distance is Z 10 0 (4 t 25) dt = ± 2 t 2 25 t ² 10 0 = 200 250 = 50 . Because v ( t )=4 t 25 5 0for0 5 t 5 6 . 25 and v ( t ) = 0for6 . 25 5 t 5 10, the total distance is Z 6 . 25 0 v ( t ) dt + Z 10 6 . 25 v ( t ) dt = 625 8 + 225 8 = 425 4 = 106 . 25 . 6.1.22: Because v ( t ) = 5 t 5 5, the net and total distance are both equal to Z 5 0 | 2 t 5 | dt = Z 2 . 5 0 (5 2 t ) dt + Z 5 2 . 5 (2 t 5) dt = 25 4 + 25 4 = 25 2 =12 . 5 . 6.1.23: The net distance is Z 3 2 4 t 3 dt = ± t 4 ² 3 2 =81 16 = 65 . Because v ( t ) 5 0for 2 5 t 5 0, the total distance is Z 0 2 4 t 3 dt + Z 3 0 4 t 3 dt =16+81=97 . 525
6.1.24: The net distance is Z 1 0 . 1 ± t 1 t 2 ² dt = ³ 1 2 t 2 + 1 t ´ 1 0 . 1 = 3 2 2001 200 = 1701 200 = 8 . 505 . Because v ( t ) 5 0for0 . 1 5 t 5 1, the total distance is 8 . 505. 6.1.25: Because v ( t )=s in2 t = 5 t 5 π/ 2, the net distance and the total distance are both equal to Z 2 0 sin 2 tdt = ³ 1 2 cos2 t ´ 2 0 = 1 2 ± 1 2 ² =1 . 6.1.26: The net distance is Z 2 0 = ³ 1 2 sin 2 t ´ 2 0 =0 0=0 . Because v ( t )=cos2 t 5 0for 4 5 t 5 2, the total distance is Z 4 0 Z 2 4 = 1 2 + 1 2 . 6.1.27: The net distance is Z 1 1 cos πt dt = ³ 1 π sin πt ´ 1 1 .

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57319-0136147054_06 - Section 6.1 n 6.1.1: With a = 0 and b...

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