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Unformatted text preview: Version 093 L EXAM 1 radin (54915) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the value of f (0) when f ( t ) = cos 2 t , f parenleftBig 4 parenrightBig = 2 . 1. f (0) = 1 2. f (0) = 3 2 correct 3. f (0) = 1 2 4. f (0) = 0 5. f (0) = 2 Explanation: Since d dx sin mt = m cos mt , for all m negationslash = 0, we see that f ( t ) = 1 2 sin 2 t + C where the arbitrary constant C is determined by the condition f ( / 4) = 2. But sin2 t vextendsingle vextendsingle vextendsingle t = / 4 = sin 2 = 1 . Thus f parenleftBig 4 parenrightBig = 1 2 + C = 2 , and so f ( t ) = 1 2 sin2 t + 3 2 . Consequently, f (0) = 3 2 . 002 10.0 points If an n thRiemann sum approximation to the definite integral I = integraldisplay b a f ( x ) dx is given by n summationdisplay i =1 f ( x i ) x i = 5 n 2 4 n + 3 n 2 , determine the value of I . 1. I = 1 2. I = 5 correct 3. I = 4 4. I = 4 5. I = 3 Explanation: By definition, integraldisplay b a f ( x ) dx = lim n n summationdisplay i =1 f ( x i ) x i . Thus integraldisplay b a f ( x ) dx = lim n 5 n 2 4 n + 3 n 2 . Consequently, I = 5 . 003 10.0 points Continuous functions f, g are known to have the properties integraldisplay 5 1 f ( x ) dx = 7 , integraldisplay 5 1 g ( x ) dx = 19 respectively. Use these to find the value of the definite integral I = integraldisplay 5 1 (2 f ( x ) 3 g ( x )) dx. Version 093 L EXAM 1 radin (54915) 2 1. I = 44 2. I = 43 correct 3. I = 45 4. I = 46 5. I = 42 Explanation: By properties of integrals I = integraldisplay 5 1 (2 f ( x ) 3 g ( x )) dx = 2 integraldisplay 5 1 f ( x ) dx 3 integraldisplay 5 1 g ( x ) dx . Consequently, I = 43. 004 10.0 points For which integral, I , is the expression 1 25 parenleftBigg radicalbigg 1 25 + radicalbigg 2 25 + radicalbigg 3 25 + . . . + radicalbigg 25 25 parenrightBigg a Riemann sum approximation? 1. I = integraldisplay 1 xdx correct 2. I = 1 25 integraldisplay 25 x dx 3. I = 1 25 integraldisplay 1 xdx 4. I = 1 25 integraldisplay 1 radicalbigg x 25 dx 5. I = integraldisplay 1 radicalbigg x 25 dx Explanation: When the interval [ a, b ] is divided into n equals intervals, then n summationdisplay i = 1 f parenleftbigg a + i ( b a ) n parenrightbigg b a n is a Riemann sum approximation for the inte gral integraldisplay b a f ( x ) dx of f over [ a, b ] using right endpoints as sample points. Comparing this with 1 25 parenleftBigg radicalbigg 1 25 + radicalbigg 2 25 + radicalbigg 3 25 + . . . + radicalbigg 25 25 parenrightBigg , we see that [ a, b ] = [0 , 1], and n = 25, while f ( x ) = x ....
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 Fall '08
 Cepparo
 Calculus

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