Cal 2 Test 1

# Cal 2 Test 1 - Version 078 L EXAM 1 Henry(54974 This...

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Version 078 – L EXAM 1 – Henry – (54974) 1 This print-out should have 20 questions. Multiple-choice 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 ) = 5 cos 2 t , f parenleftBig π 4 parenrightBig = 6 . 1. f (0) = 4 2. f (0) = 3 3. f (0) = 2 4. f (0) = 7 2 correct 5. f (0) = 5 2 Explanation: Since d dx sin mt = m cos mt , for all m negationslash = 0, we see that f ( t ) = 5 2 sin 2 t + C where the arbitrary constant C is determined by the condition f ( π/ 4) = 6. But sin 2 t vextendsingle vextendsingle vextendsingle t = π/ 4 = sin π 2 = 1 . Thus f parenleftBig π 4 parenrightBig = 5 2 + C = 6 , and so f ( t ) = 5 2 sin 2 t + 7 2 . Consequently, f (0) = 7 2 . 002 10.0 points If an n th -Riemann sum approximation to the definite integral I = integraldisplay b a f ( x ) dx is given by n summationdisplay i =1 f ( x i ) Δ x i = 3 n 2 6 n + 1 n 2 , determine the value of I . 1. I = 1 2. I = 6 3. I = 2 4. I = 3 correct 5. I = 3 Explanation: By definition, integraldisplay b a f ( x ) dx = lim n → ∞ n summationdisplay i =1 f ( x i ) Δ x i . Thus integraldisplay b a f ( x ) dx = lim n → ∞ 3 n 2 6 n + 1 n 2 . Consequently, I = 3 . 003 10.0 points Continuous functions f, g are known to have the properties integraldisplay 5 2 f ( x ) dx = 28 , integraldisplay 5 2 g ( x ) dx = 14

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Version 078 – L EXAM 1 – Henry – (54974) 2 respectively. Use these to find the value of the definite integral I = integraldisplay 5 2 (2 f ( x ) g ( x )) dx. 1. I = 42 correct 2. I = 44 3. I = 41 4. I = 45 5. I = 43 Explanation: By properties of integrals I = integraldisplay 5 2 (2 f ( x ) g ( x )) dx = 2 integraldisplay 5 2 f ( x ) dx integraldisplay 5 2 g ( x ) dx . Consequently, I = 42. 004 10.0 points For which integral, I , is the expression 1 30 parenleftBigg radicalbigg 1 30 + radicalbigg 2 30 + radicalbigg 3 30 + . . . + radicalbigg 30 30 parenrightBigg a Riemann sum approximation? 1. I = 1 30 integraldisplay 30 0 x dx 2. I = integraldisplay 1 0 radicalbigg x 30 dx 3. I = integraldisplay 1 0 x dx correct 4. I = 1 30 integraldisplay 1 0 radicalbigg x 30 dx 5. I = 1 30 integraldisplay 1 0 x 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 30 parenleftBigg radicalbigg 1 30 + radicalbigg 2 30 + radicalbigg 3 30 + . . . + radicalbigg 30 30 parenrightBigg , we see that [ a, b ] = [0 , 1], and n = 30, while f ( x ) = x . Consequently, the given expression is a Rie- mann sum for I = integraldisplay 1 0 x dx . 005 10.0 points A function h has graph -4 -3 -2 -1 0 1 2 3 4 2 4 2 2 4
Version 078 – L EXAM 1 – Henry – (54974) 3 on ( 3 , 4). If f ( x ) = integraldisplay x 2 h ( t ) dt, ( x ≥ − 2) , which of the following is the graph of f on ( 3 , 4)? 1. -4 -3 -2 -1 0 1 2 3 4 2 4 2 2 2 2.

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