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Unformatted text preview: nanni (arn437) – Assignment 2 – guntel – (54940) 1 This printout should have 7 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Let g ( x ) be the solution of g ′ ( x ) = 3 sin( x )+2 and g (0) = 0. Find g ( π ). 1. 2 π + 6 correct 2. 2 π 6 3. 2 π + 6 4. π + 6 5. π + 3 Explanation: Since an antiderivative of sin( x ) is cos( x ) and an antiderivative of 2 is 2 x , we see that an antiderivative of g ′ ( x ) is 3 cos( x ) + 2 x . Thus g ( x ) must have the form g ( x ) = 3 cos( x ) + 2 x + C where C is a constant. Since g (0) = 0, we have 0 = 3 + C so that C = 3. Thus g ( x ) = 3 cos( x ) + 2 x + 3 . Evaluating at π gives g ( π ) = 3 + 2 π + 3. Consequently, g ( π ) = 2 π + 6 . keywords: antiderivative, trig function 002 10.0 points Consider the following functions: ( A ) F 1 ( x ) = sin 2 x 2 , ( B ) F 2 ( x ) = cos 2 x 4 , ( C ) F 3 ( x ) = cos 2 x 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. F 3 only 2. F 1 only 3. F 2 only 4. F 1 and F 3 only 5. F 2 and F 3 only 6. none of them 7. all of them correct 8. F 1 and...
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 Fall '08
 Hodges
 Trigonometry, F3

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