Calculus with Analytic Geometry by edwards & Penney soln ch6

Calculus with Analytic Geometry

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Unformatted text preview: Section 6.1 C06S01.001: With a = 0 and b = 1, lim n →∞ n X i =1 2 x ? i ∆ x = Z 1 2 x dx = · x 2 ¸ 1 = 1. C06S01.002: With a = 1 and b = 2, lim n →∞ n X i =1 ∆ x ( x ? i ) 2 = Z 2 1 1 x 2 dx = · − 1 x ¸ 2 1 = 1 2 . C06S01.003: With a = 0 and b = 1, lim n →∞ n X i =1 (sin πx ? i ) ∆ x = Z 1 sin πx dx = · − 1 π cos πx ¸ 1 = 2 π . C06S01.004: With a = − 1 and 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 ¸ 3 − 1 = 24. C06S01.005: With a = 0 and b = 4, lim n →∞ n X i =1 x ? i q ( x ? i ) 2 + 9 ∆ x = Z 4 x p x 2 + 9 dx = · 1 3 ( x 2 + 9) 3 / 2 ¸ 4 = 125 3 − 9 = 98 3 . C06S01.006: The limit is Z 4 2 x 2 dx = · 1 3 x 3 ¸ 4 2 = 64 3 − 8 3 = 56 3 ≈ 18 . 666666666667. C06S01.007: The limit is Z 3 − 1 (2 x − 1) dx = · x 2 − x ¸ 3 − 1 = 6 − 2 = 4. C06S01.008: The limit is Z 4 √ 2 x + 1 dx = · 1 3 (2 x + 1) 3 / 2 ¸ 4 = 9 − 1 3 = 26 3 . C06S01.009: The limit is Z − 3 x √ x 2 + 16 dx = · p x 2 + 16 ¸ − 3 = 4 − 5 = − 1. C06S01.010: The limit is Z √ π x cos x 2 dx = · 1 2 sin x 2 ¸ √ π = 0 − 0 = 0. C06S01.011: With a = 1 and b = 4, lim n →∞ n X i =1 2 πx ? i f ( x ? i ) ∆ x = Z 4 1 2 πxf ( x ) dx. Compare this with Eq. (2) in Section 6.3. C06S01.012: With a = − 1 and b = 1, lim n →∞ n X i =1 [ f ( x ? i )] 2 ∆ x = Z 1 − 1 [ f ( x )] 2 dx . C06S01.013: With a = 0 and b = 10, lim n →∞ n X i =1 q 1 + [ f ( x ? i )] 2 ∆ x = Z 10 q 1 + [ f ( x )] 2 dx . C06S01.014: With a = − 2 and b = 3, we have lim n →∞ n X i =1 2 πm i q 1 + [ f ( m i )] 2 ∆ x = Z 3 − 2 2 πx q 1 + [ f ( x )] 2 dx. 1 C06S01.015: M = Z 100 1 5 x dx = · 1 10 x 2 ¸ 100 = 1000 − 0 = 1000 (grams). C06S01.016: M = Z 25 (60 − 2 x ) dx = · 60 x − x 2 ¸ 25 = 875 − 0 = 875 (grams). C06S01.017: M = Z 10 x (10 − x ) dx = · 5 x 2 − 1 3 x 3 ¸ 10 = 500 3 − 0 = 500 3 (grams). C06S01.018: M = Z 10 10 sin πx 10 dx = · − 100 π cos πx 10 ¸ 10 = 100 π − − 100 π ¶ = 200 π (grams). C06S01.019: The net distance is Z 10 ( − 32) dt = · − 32 t ¸ 10 = − 320 and the total distance is 320. C06S01.020: The net distance is Z 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 5 t 5 5, this is the total distance as well. C06S01.021: The net distance is Z 10 (4 t − 25) dt = · 2 t 2 − 25 t ¸ 10 = 200 − 250 = − 50 . Because v ( t ) = 4 t − 25 5 0 for 0 5 t 5 6 . 25 and v ( t ) = 0 for 6 . 25 5 t 5 10, the total distance is − Z 6 . 25 v ( t ) dt + Z 10 6 . 25 v ( t ) dt = 625 8 + 225 8 = 425 4 = 106 . 25 . C06S01.022: Because v ( t ) = 0 for 0 5 t 5 5, the net and total distance are both equal to Z 5 | 2 t − 5 | dt = Z 2 . 5 (5 − 2 t ) dt + Z 5 2 . 5 (2 t − 5) dt = 25 4 + 25 4 = 25 2 = 12 . 5 ....
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Calculus with Analytic Geometry by edwards & Penney...

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