Final Ex 408L - Garza Ernesto – Final 1 – Due 11:00 pm...

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Unformatted text preview: Garza, Ernesto – Final 1 – Due: May 11 2007, 11:00 pm – Inst: JEGilbert 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points When f has graph R 1 R 2 a b c express the sum I = Z c a n 3 f ( x )- 2 | f ( x ) | o dx in terms of the areas A 1 = area( R 1 ) , A 2 = area( R 2 ) of the respective lighter shaded regions R 1 and R 2 . 1. I =- 5 A 2 2. I = A 1 + 5 A 2 3. I = A 1 4. I =- A 1- 5 A 2 5. I = A 1- 5 A 2 correct 6. I =- A 1 + 5 A 2 Explanation: As the graph shows, f takes positive values on [ a, b ] and negative values on [ b, c ], so Z b a f ( x ) dx = A 1 , Z c b f ( x ) dx =- A 2 . while Z c a | f ( x ) | dx = Z b a f ( x ) dx- Z c b f ( x ) dx, and Z c a f ( x ) dx = Z b a f ( x ) dx + Z c b f ( x ) dx, Thus I = 3( A 1- A 2 )- 2( A 1 + A 2 ) . Consequently, I = A 1- 5 A 2 . keywords: definite integral, properties inte- grals, area 002 (part 1 of 1) 10 points If an n th-Riemann sum approximation to the definite integral I = Z b a f ( x ) dx is given by n X i = 1 f ( x * i ) Δ x i = 3 n 2- 5 n + 6 n 2 , determine the value of I . 1. I = 4 2. I =- 2 3. I = 3 correct 4. I = 6 5. I =- 5 Explanation: Garza, Ernesto – Final 1 – Due: May 11 2007, 11:00 pm – Inst: JEGilbert 2 By definition, Z b a f ( x ) dx = lim n → ∞ f ( x * i ) Δ x i . Thus Z b a f ( x ) dx = lim n → ∞ 3 n 2- 5 n + 6 n 2 . Consequently, I = 3 . keywords: Riemann sum, definition definite integral, limit, conceptual 003 (part 1 of 1) 10 points Determine F ( x ) when F ( x ) = Z √ x 5 6 sin t t dt. 1. F ( x ) =- 6 cos x x 2. F ( x ) = 3 cos( √ x ) x 3. F ( x ) = 6 sin x √ x 4. F ( x ) =- 6 cos( √ x ) √ x 5. F ( x ) = 3 sin x x 6. F ( x ) =- 3 sin( √ x ) √ x 7. F ( x ) = 3 sin( √ x ) x correct 8. F ( x ) =- 6 cos x √ x Explanation: By the Fundamental Theorem of Calculus and the Chain Rule, d dx ‡ Z g ( x ) a f ( t ) dt · = f ( g ( x )) g ( x ) . When F ( x ) = Z √ x 5 6 sin t t dt, therefore, F ( x ) = 6 sin( √ x ) √ x ‡ d dx √ x · . Consequently, F ( x ) = 3 sin( √ x ) x , since d dx √ x = 1 2 √ x . keywords: Stewart5e, FTC, Chain Rule 004 (part 1 of 1) 10 points A car heads north from Austin on IH 35. Its velocity t hours after leaving Austin is given in miles per hour by v ( t ) = 12 + 32 3 t- t 2 . Determine how many hours will elapse before the car is next in Austin. 1. 18 hours later correct 2. 19 hours later 3. 20 hours later 4. 17 hours later 5. car never returns to Austin Explanation: Garza, Ernesto – Final 1 – Due: May 11 2007, 11:00 pm – Inst: JEGilbert 3 Since the car leaves Austin at time t = 0, the position of the car t hours later is the anti-derivative s ( t ) = Z (12 + 32 3 t- t 2 ) dt, s (0) = 0 of v ( t ). But Z (12 + 32 3 t- t 2 ) dt = 12 t + 16 3 t 2- 1 3 t 3 = t ( t + 2) ‡ 6- 1 3 t · + C....
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Final Ex 408L - Garza Ernesto – Final 1 – Due 11:00 pm...

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