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Unformatted text preview: armington (kma786) – 4.9 – Stepp – (55860) 1 This printout should have 10 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 ) = sin 2 t , f parenleftBig π 2 parenrightBig = 5 . 1. f (0) = 1 2. f (0) = 2 3. f (0) = 0 4. f (0) = 4 correct 5. f (0) = 3 Explanation: Since d dx cos mt = − m sin mt , for all m negationslash = 0, we see that f ( t ) = − 1 2 cos 2 t + C where the arbitrary constant C is determined by the condition f ( π/ 2) = 5. But cos 2 t vextendsingle vextendsingle vextendsingle t = π/ 2 = cos π = − 1 . Thus f parenleftBig π 2 parenrightBig = 1 2 + C = 5 , and so f ( t ) = − 1 2 cos 2 t + 9 2 . Consequently, f (0) = 4 . 002 10.0 points Find the value of f ( − 1) when f ′′ ( t ) = 18 t − 2 and f ′ (1) = 5 , f (1) = 6 . Correct answer: 4. Explanation: The most general antiderivative of f ′′ has the form f ′ ( t ) = 9 t 2 − 2 t + C where C is an arbitrary constant. But if f ′ (1) = 5, then f ′ (1) = 9 − 2 + C = 5 , i . e . C = − 2 . Thus f ′ ( t ) = 9 t 2 − 2 t − 2 , from which it follows that f ( t ) = 3 t 3 − t 2 − 2 t + D , where the constant D is determined by the condition f (1) = 3 − 1 − 2 + D = 6 , i . e . D = 6 . Consequently, f ( t ) = 3 t 3 − t 2 − 2 t + 6 , and so at t = − 1, f ( − 1) = 4 . 003 10.0 points Find the unique antiderivative F of f ( x ) = e 3 x + 3 e 2 x + 2 e − x e 2 x for which F (0) = 0. 1. F ( x ) = e x + 3 x − e − x 2. F ( x ) = 1 3 e 3 x + 3 x − 2 3 e − 3 x − 1 3 armington (kma786) – 4.9 – Stepp – (55860) 2 3. F ( x ) = 1 3 e 3 x − 3 x + e − x − 1 3 4. F ( x ) = e x + 3 x − 2 3 e − 3 x − 1 3 correct 5. F ( x ) = e x − 3 x + 2 3 e − 3 x + 5 3 6. F ( x ) = e x − 3 x + 2 3 e − x + 1 3 Explanation: After division, e 3 x + 3 e 2 x + 2 e − x e 2 x = e x + 3 + 2 e − 3 x . Since d dx e αx = αe αx , it thus follows that F ( x ) = e x + 3 x − 2 3 e − 3 x + C where the constant C is determined by the condition F (0) = 0. For then F (0) = 1 − 2 3 + C = 0 ....
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This note was uploaded on 02/11/2011 for the course M 408S taught by Professor Stepp during the Spring '11 term at University of Texas.
 Spring '11
 STEPP
 Calculus

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