Final Ex 408L

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 14

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online