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Unformatted text preview: Su, Yung – Homework 3 – Due: Sep 19 2007, 3:00 am – Inst: Shinko Harper 1 This printout should have 22 questions. Multiplechoice 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 6 h ( t ) dt, ( x ≥  6) , which of the following is the graph of f ? 1. 3 6 3 6 3 6 9 3 6 9 correct 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 5. 3 6 3 6 3 6 9 3 6 9 Explanation: When f is defined by f ( x ) = Z x 6 h ( t ) dt, ( x ≥  6) , then, first note that f ( x ) is not defined for x < 6. This eliminates one of the graphs. On the other hand, by the Fundamental theorem of Calculus, f ( x ) = ‰ h ( x ) x > 6 , , x = 6; thus f ( 6) = 0 rules out two more of the five given graphs. But 6 < x < 4 = ⇒ h ( x ) > , Su, Yung – Homework 3 – Due: Sep 19 2007, 3:00 am – Inst: Shinko Harper 2 so the graph of f must be increasing on ( 6 , 4). 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 5 e 6 cos 2 θ dθ . 1. F 00 ‡ π 6 · = 3 e 3 2. F 00 ‡ π 6 · = 3 √ 3 e 3 3. F 00 ‡ π 6 · = 15 √ 3 e 9 2 correct 4. F 00 ‡ π 6 · = 15 e 3 5. F 00 ‡ π 6 · = 3 √ 3 e 9 2 6. F 00 ‡ π 6 · = 15 e 9 2 Explanation: By the Fundamental theorem of calculus, F ( x ) = 5 e 6 cos 2 x , so after a second differentiation we see that F 00 ( x ) = (10 e 6 cos 2 x ) 6 sin x cos x. At x = π/ 6, therefore, F 00 ‡ π 6 · = 15 √ 3 e 9 2 . keywords: 003 (part 1 of 1) 10 points Determine g ( x ) when g ( x ) = Z 5 x 4 t 2 tan t dt. 1. g ( x ) = 8 x sec x tan x 2. g ( x ) = 4 x 2 tan x 3. g ( x ) = 8 x sec 2 x 4. g ( x ) = 8 x sec x tan x 5. g ( x ) = 8 x sec 2 x 6. g ( x ) = 4 x 2 sec x 7. g ( x ) = 4 x 2 sec x 8. g ( x ) = 4 x 2 tan x correct 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 ) = 4 t 2 tan t, therefore, g ( x ) = 4 x 2 tan x . Su, Yung – 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 2 8 cos t t dt. 1. F ( x ) = 4 cos( √ x ) √ x 2. F ( x ) = 8 sin x √ x 3. F ( x ) = 8 cos x √ x 4. F ( x ) = 4 cos( √ x ) x correct 5. F ( x ) = 4 sin( √ x ) x 6. F ( x ) = 4 cos x x 7. F ( x ) = 8 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 04/15/2009 for the course M 59685 taught by Professor Harper during the Spring '09 term at University of Texas.
 Spring '09
 Harper
 Calculus

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