# Sol3 - Massaro Michael – Homework 3 – Due 3:00 am –...

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Unformatted text preview: Massaro, Michael – Homework 3 – Due: Sep 19 2007, 3:00 am – Inst: Shinko Harper 1 This print-out should have 22 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 A function h has graph 3 6- 3- 6 3 6 9- 3- 6- 9 If f ( x ) = Z x- 5 h ( t ) dt, ( x ≥ - 5) , which of the following is the graph of f ? 1. 3 6- 3- 6 3 6 9- 3- 6- 9 2. 3 6- 3- 6 3 6 9- 3- 6- 9 3. 3 6- 3- 6 3 6 9- 3- 6- 9 4. 3 6- 3- 6 3 6 9- 3- 6- 9 correct 5. 3 6- 3- 6 3 6 9- 3- 6- 9 Explanation: When f is defined by f ( x ) = Z x- 5 h ( t ) dt, ( x ≥ - 5) , then, first note that f ( x ) is not defined for x <- 5. This eliminates one of the graphs. On the other hand, by the Fundamental theorem of Calculus, f ( x ) = ‰ h ( x ) x >- 5 , , x =- 5; thus f (- 5) = 0 rules out two more of the five given graphs. But- 5 < x <- 3 = ⇒ h ( x ) > , Massaro, Michael – Homework 3 – Due: Sep 19 2007, 3:00 am – Inst: Shinko Harper 2 so the graph of f must be increasing on (- 5 ,- 3). This eliminates one of the two remaining graphs, leaving only 3 6- 3- 6 3 6 9- 3- 6- 9 which must therefore be the graph of f . keywords: graph, integral, Fundamental The- orem Calculus, FTC 002 (part 1 of 1) 10 points Find the value of F 00 ( π/ 6) when F ( x ) = Z x 4 e- 4 cos 2 θ dθ . 1. F 00 ‡ π 6 · = 2 √ 3 e- 3 2. F 00 ‡ π 6 · =- 8 e- 2 3. F 00 ‡ π 6 · = 8 √ 3 e- 3 correct 4. F 00 ‡ π 6 · = 2 √ 3 e- 2 5. F 00 ‡ π 6 · = 2 e- 2 6. F 00 ‡ π 6 · =- 8 e- 3 Explanation: By the Fundamental theorem of calculus, F ( x ) = 4 e- 4 cos 2 x , so after a second differentiation we see that F 00 ( x ) = (8 e- 4 cos 2 x ) 4 sin x cos x. At x = π/ 6, therefore, F 00 ‡ π 6 · = 8 √ 3 e- 3 . keywords: 003 (part 1 of 1) 10 points Determine g ( x ) when g ( x ) = Z 5 x 3 t 2 tan t dt. 1. g ( x ) = 3 x 2 tan x 2. g ( x ) =- 3 x 2 sec x 3. g ( x ) =- 3 x 2 tan x correct 4. g ( x ) = 6 x sec 2 x 5. g ( x ) = 3 x 2 sec x 6. g ( x ) =- 6 x sec 2 x 7. g ( x ) = 6 x sec x tan x 8. g ( x ) =- 6 x sec x tan x Explanation: By Properties of integrals and the Funda- mental Theorem of Calculus, d dx ‡ Z a x f ( t ) dt · = d dx ‡- Z x a f ( t ) dt · =- f ( x ) . When g ( x ) = Z 5 x f ( t ) dt, f ( t ) = 3 t 2 tan t, therefore, g ( x ) =- 3 x 2 tan x . Massaro, Michael – Homework 3 – Due: Sep 19 2007, 3:00 am – Inst: Shinko Harper 3 keywords: Stewart5e, FTC, derivative form, properties of integrals, 004 (part 1 of 1) 10 points Determine F ( x ) when F ( x ) = Z √ x 5 8 cos t t dt. 1. F ( x ) = 8 cos x √ x 2. F ( x ) =- 8 sin( √ x ) √ x 3. F ( x ) =- 8 sin x x 4. F ( x ) = 4 cos( √ x ) x correct 5. F ( x ) = 4 cos x x 6. F ( x ) =- 4 cos( √ x ) √ x 7. F ( x ) = 4 sin( √ x ) x 8. F ( x ) =- 8 sin 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 ) ....
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## This note was uploaded on 07/09/2008 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas.

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Sol3 - Massaro Michael – Homework 3 – Due 3:00 am –...

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