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Unformatted text preview: Version 071 – Homework 3 – Gompf – (58370) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The graph of f is shown in the figure 2 4 6 8 2 4 6 If F is an antiderivative of f and integraldisplay 8 3 f ( x ) dx = 23 , find the value of F (8) F (2). 1. F (8) F (2) = 32 2. F (8) F (2) = 29 3. F (8) F (2) = 28 correct 4. F (8) F (2) = 30 5. F (8) F (2) = 31 Explanation: We already know that the area under the graph on the interval 3 ≤ x ≤ 8 is equal to 23, alternatively, by the Fundamental Theorem of Calculus we can say that F (8) F (3) = 23 . On the other hand, integraldisplay 8 2 f ( x ) dx = integraldisplay 3 2 f ( x ) dx + integraldisplay 8 3 f ( x ) dx. Thus we need to find integraldisplay 3 2 f ( x ) dx = F (3) F (2) . Now integraldisplay 3 2 f ( x ) dx = integraldisplay 3 2 2 x dx = bracketleftBig x 2 bracketrightBig 3 2 = 5 . Consequently, F (8) F (2) = 23 + 5 = 28 . keywords: velocity, distance, graph analysis, fundamental theorem 002 10.0 points Calculate the indefinite integral I = integraldisplay (5 √ x )(4 + √ x ) dx . 1. I = 5 x 2 3 x √ x 1 2 x 2 + C 2. I = 5 x √ x 1 2 x 2 + C 3. I = 20 x + 2 3 x √ x + 1 2 x 2 + C 4. I = 5 x + 2 3 x √ x + 1 2 x 2 + C 5. I = 20 x √ x 1 2 x 2 + C 6. I = 20 x + 2 3 x √ x 1 2 x 2 + C correct Explanation: After expansion (5 √ x )(4 + √ x ) = 20 + √ x x . Thus I = integraldisplay ( 20 + √ x x ) dx = 20 x + 2 3 x √ x 1 2 x 2 + C . Version 071 – Homework 3 – Gompf – (58370) 2 Consequently, I = 20 x + 2 3 x √ x 1 2 x 2 + C . 003 10.0 points Evaluate the definite integral I = integraldisplay π 3 3 sin 2 x 4 cos 2 x cos x dx . 1. I = 3 2 √ 3 correct 2. I = 6 + 2 √ 3 3. I = 3 + 2 √ 3 4. I = 3 4 √ 2 5. I = 6 2 √ 3 6. I = 3 + 4 √ 2 Explanation: Since sin 2 x = 2 sin x cos x , the integrand can be rewritten as 6 sin x cos x 4 cos 2 x cos x = 2(3 sin x 2 cos x ) . Thus I = 2 integraldisplay π 3 (3 sin x 2 cos x ) dx = 2 bracketleftBig 3 cos x 2 sin x bracketrightBig π 3 = 2 parenleftbigg 3 2 √ 3 parenrightbigg + 6 . Consequently, I = 3 2 √ 3 . 004 10.0 points Evaluate the integral I = integraldisplay 4 d dx (3 + 2 x 2 ) 1 / 2 dx. 1. I = √ 35 √ 3 correct 2. I = √ 35 3. I = √ 3 √ 35 4. I = √ 3 5. I = √ 35 + √ 3 Explanation: As an indefinite integral, integraldisplay d dx (3 + 2 x 2 ) 1 / 2 dx = (3 + 2 x 2 ) 1 / 2 + C where C is an arbitrary constant. Thus integraldisplay 4 d dx (3 + 2 x 2 ) 1 / 2 dx = bracketleftBig (3 + 2 x 2 ) 1 / 2 bracketrightBig 4 . Consequently, I = √ 35 √ 3 . 005 10.0 points Determine the indefinite integral I = integraldisplay 3 cos 2 θ cos 2 θ dθ ....
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This note was uploaded on 12/04/2009 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.
 Spring '08
 RAdin
 Calculus

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