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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 = 0 and b = 1, lim n →∞ n i =1 2 x i x = 1 0 2 x dx = x 2 1 0 = 1. 6.1.2: With a = 1 and b = 2, lim n →∞ n i =1 x ( x i ) 2 = 2 1 1 x 2 dx = 1 x 2 1 = 1 2 . 6.1.3: With a = 0 and b = 1, lim n →∞ n i =1 (sin πx i ) ∆ x = 1 0 sin πx dx = 1 π cos πx 1 0 = 2 π . 6.1.4: With a = 1 and b = 3, lim n →∞ n i =1 3 ( x i ) 2 1 x = 3 1 (3 x 2 1) dx = x 3 x 3 1 = 24. 6.1.5: With a = 0 and b = 4, lim n →∞ n i =1 x i ( x i ) 2 + 9 ∆ x = 4 0 x x 2 + 9 dx = 1 3 ( x 2 + 9) 3 / 2 4 0 = 125 3 9 = 98 3 . 6.1.6: The limit is 4 2 1 x dx = ln x 4 2 = ln 4 ln 2 = 2 ln 2 ln 2 = ln 2. 6.1.7: The limit is 1 0 e x dx = e x 1 0 = e 0 e 1 = 1 1 e . 6.1.8: The limit is 4 0 2 x + 1 dx = 1 3 (2 x + 1) 3 / 2 4 0 = 9 1 3 = 26 3 . 6.1.9: The limit is 6 0 x x 2 + 9 dx = 1 2 ln( x 2 + 9) 6 0 = 1 2 ln 45 1 2 ln 9 = 1 2 ln 5. 6.1.10: The limit is 1 0 2 x exp( x 2 ) dx = exp( x 2 ) 1 0 = e 0 e 1 = 1 1 e . 6.1.11: With a = 1 and b = 4, lim n →∞ n i =1 2 πx i f ( x i ) ∆ x = 4 1 2 πxf ( x ) dx. Compare this with Eq. (2) in Section 6.3. 6.1.12: With a = 1 and b = 1, lim n →∞ n i =1 [ f ( x i )] 2 x = 1 1 [ f ( x ) ] 2 dx . 6.1.13: With a = 0 and b = 10, lim n →∞ n i =1 1 + [ f ( x i ) ] 2 x = 10 0 1 + [ f ( x ) ] 2 dx . 6.1.14: With a = 2 and b = 3, we have lim n →∞ n i =1 2 πm i 1 + [ f ( m i ) ] 2 x = 3 2 2 πx 1 + [ f ( x ) ] 2 dx. 524
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6.1.15: M = 100 0 1 5 x dx = 1 10 x 2 100 0 = 1000 0 = 1000 (grams). 6.1.16: M = 25 0 (60 2 x ) dx = 60 x x 2 25 0 = 875 0 = 875 (grams). 6.1.17: M = 10 0 x (10 x ) dx = 5 x 2 1 3 x 3 10 0 = 500 3 0 = 500 3 (grams). 6.1.18: M = 10 0 10 sin πx 10 dx = 100 π cos πx 10 10 0 = 100 π 100 π = 200 π (grams). 6.1.19: The net distance is 10 0 ( 32) dt = 32 t 10 0 = 320 and the total distance is 320. 6.1.20: The net distance is 5 1 (2 t + 10) dt = t 2 + 10 t 5 1 = 75 11 = 64 and because v ( t ) = 2 t + 10 0 for 1 t 5, this is the total distance as well. 6.1.21: The net distance is 10 0 (4 t 25) dt = 2 t 2 25 t 10 0 = 200 250 = 50 . Because v ( t ) = 4 t 25 0 for 0 t 6 . 25 and v ( t ) 0 for 6 . 25 t 10, the total distance is 6 . 25 0 v ( t ) dt + 10 6 . 25 v ( t ) dt = 625 8 + 225 8 = 425 4 = 106 . 25 . 6.1.22: Because v ( t ) 0 for 0 t 5, the net and total distance are both equal to 5 0 | 2 t 5 | dt = 2 . 5 0 (5 2 t ) dt + 5 2 . 5 (2 t 5) dt = 25 4 + 25 4 = 25 2 = 12 . 5 . 6.1.23: The net distance is 3 2 4 t 3 dt = t 4 3 2 = 81 16 = 65 . Because v ( t ) 0 for 2 t 0, the total distance is 0 2 4 t 3 dt + 3 0 4 t 3 dt = 16 + 81 = 97 . 525
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6.1.24: The net distance is 1 0 . 1 t 1 t 2 dt = 1 2 t 2 + 1 t 1 0 . 1 = 3 2 2001 200 = 1701 200 = 8 . 505 . Because v ( t ) 0 for 0 . 1 t 1, the total distance is 8 . 505. 6.1.25: Because v ( t ) = sin 2 t 0 for 0 t π/ 2, the net distance and the total distance are both equal to π/ 2 0 sin 2 t dt = 1 2 cos 2 t π/ 2 0 = 1 2 1 2 = 1 . 6.1.26: The net distance is π/ 2 0 cos 2 t dt = 1 2 sin 2 t π/ 2 0 = 0 0 = 0 . Because v ( t ) = cos 2 t 0 for π/ 4 t π/ 2, the total distance is π/ 4 0 cos 2 t dt π/ 2 π/ 4 cos 2 t dt = 1 2 + 1 2 = 1 . 6.1.27: The net distance is 1 1 cos πt dt = 1 π sin πt 1 1 = 0 0 = 0 . Because v ( t ) = cos πt 0 for 1 t 0 . 5 and for 0 . 5 t 1, the total distance is 0 . 5 1 cos πt dt + 0 . 5 0 . 5 cos πt dt 1 0 . 5 cos πt dt = 1 π + 2 π + 1 π = 4 π .
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