Review 1 - sl7433 Review 1 Radin(58415 This print-out...

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sl7433 – Review 1 – Radin – (58415) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When f has graph R 1 R 2 a b c express the value of I = integraldisplay c a braceleftBig 3 f ( x ) + | f ( x ) | bracerightBig dx in terms of the areas A 1 = area( R 1 ) , A 2 = area( R 2 ) of the respective lighter shaded regions R 1 and R 2 . 1. I = - 4 A 1 + 2 A 2 2. I = 4 A 1 3. I = 4 A 1 + 2 A 2 4. I = - 4 A 1 - 2 A 2 5. I = 4 A 1 - 2 A 2 6. I = - 2 A 2 002 10.0 points Determine F ( x ) when F ( x ) = integraldisplay x 2 2 cos t t dt . 1. F ( x ) = cos( x ) x 2. F ( x ) = - 2 sin x x 3. F ( x ) = sin( x ) x 4. F ( x ) = cos x x 5. F ( x ) = - 2 sin( x ) x 6. F ( x ) = - cos( x ) x 7. F ( x ) = 2 cos x x 8. F ( x ) = - 2 sin x x 003 10.0 points A particle moving along a straight line has velocity v ( t ) = 3 sin t - 4 cos t at time t . Find the position, s ( t ), of the particle at time t if initially s (0) = 2. (This is the mathematical model of Simple Harmonic Motion .) 1. s ( t ) = - 2 - 3 sin t + 4 cos t 2. s ( t ) = - 1 + 3 cos t - 4 sin t 3. s ( t ) = 5 - 3 cos t + 4 sin t 4. s ( t ) = 6 - 3 sin t - 4 cos t 5. s ( t ) = 5 - 3 cos t - 4 sin t 6. s ( t ) = 6 + 3 sin t - 4 cos t
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sl7433 – Review 1 – Radin – (58415) 2 004 10.0 points Evaluate the definite integral I = integraldisplay 4 0 ( 5 x - 3 x ) dx .
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