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Unformatted text preview: momin (rrm497) Review1 cheng (58520) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Continuous functions f, g are known to have the properties integraldisplay 3 1 f ( x ) dx = 12 , integraldisplay 3 1 g ( x ) dx = 8 respectively. Use these to find the value of the definite integral I = integraldisplay 3 1 (4 f ( x ) g ( x )) dx. 1. I = 40 correct 2. I = 41 3. I = 42 4. I = 43 5. I = 39 Explanation: By properties of integrals I = integraldisplay 3 1 (4 f ( x ) g ( x )) dx = 4 integraldisplay 3 1 f ( x ) dx integraldisplay 3 1 g ( x ) dx . Consequently, I = 40. 002 10.0 points The acceleration of a particle moving along the xaxis at time t is given by a ( t ) = 6 t 2. If the velocity is 25 when t = 3 and the po sition is 10 when t = 1, then find the position x ( t ). 1. x ( t ) = 36 t 3 4 t 2 77 t + 55 2. x ( t ) = t 3 t 2 + 4 t + 6 correct 3. x ( t ) = 9 t 2 + 1 4. x ( t ) = 3 t 2 2 t + 4 5. x ( t ) = t 3 t 2 + 9 t 20 Explanation: a ( t ) = v = 6 t 2 v = 6 1 2 t 2 2 t + C v (3) = 25 = 25 = 27 6 + C 4 = C Thus v ( t ) = s = 3 t 2 2 t + 4 s = 3 1 3 t 3 2 1 2 t 2 + 4 t + C 1 s (1) = 10 = 10 = 1 1 + 4 + C 1 6 = C 1 Thus s ( t ) = t 3 t 2 + 4 t + 6. 003 10.0 points Evaluate the definite integral I = integraldisplay 4 ( 4 x 3 x ) dx . 1. I = 16 correct 2. I = 17 3. I = 14 4. I = 13 5. I = 15 Explanation: By the Fundamental Theorem of Calculus, I = bracketleftBig F ( x ) bracketrightBig 4 = F (4) F (0) momin (rrm497) Review1 cheng (58520) 2 for any antiderivative F of f ( x ) = 4 x 3 x . Taking F ( x ) = 2 x 2 2 x 3 / 2 , we thus see that I = 32 16 . Consequently, I = 16 . 004 10.0 points Determine the indefinite integral I = integraldisplay x + 2 x + 1 dx . 1. I = 2 3 ( x + 1) 3 / 2 2( x + 1) 1 / 2 + C 2. I = 1 3 ( x + 1) 3 / 2 + 2( x + 1) 1 / 2 + C 3. I = 2 3 ( x + 1) 3 / 2 ( x + 1) 1 / 2 + C 4. I = 2 3 ( x +1) 3 / 2 +2( x +1) 1 / 2 + C correct 5. I = 1 3 ( x + 1) 3 / 2 + ( x + 1) 1 / 2 + C 6. I = 1 3 ( x + 1) 3 / 2 ( x + 1) 1 / 2 + C Explanation: Set u 2 = x + 1. Then 2 u du = dx , so integraldisplay x + 2 x + 1 dx = 2 integraldisplay ( u 2 + 1) du = 2 3 u 3 + 2 u + C . Thus I = 2 3 ( x + 1) 3 / 2 + 2( x + 1) 1 / 2 + C , with C an arbitrary constant. 005 10.0 points Evaluate the integral I = integraldisplay 2 1 4 x 3 x 2 + 3 x 2 dx....
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This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin
 Calculus

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