# HW_1_Answers - Miller Kierste Homework 1 Due 3:00 am Inst...

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Miller, Kierste – Homework 1 – Due: Jan 24 2007, 3:00 am – Inst: Gary Berg 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Determine the value oF f (2) when f 0 ( t ) = 6 t - 5 , f ( - 1) = 6 . Correct answer: 0 . Explanation: The most general anti-derivative oF f 0 has the Form F ( t ) = 3 t 2 - 5 t + C , where C is an arbitrary constant whose value is determined by the condition F ( - 1) = 6. ±or then F ( - 1) = 3 + 5 + C = 6 , i . e ., C = - 2 . Consequently, f ( t ) = 3 t 2 - 5 t - 2 , and so at t = 2 , f (2) = +0 . keywords: antiderivatives, a²ne Function, particular values 002 (part 1 oF 1) 10 points Consider the Following Functions: ( A ) F 1 ( x ) = cos2 x 4 , ( B ) F 2 ( x ) = - cos 2 x 2 , ( C ) F 3 ( x ) = sin 2 x 2 . Which are anti-derivatives oF f ( x ) = sin x cos x ? 1. F 1 and F 3 only 2. F 1 only 3. F 3 only 4. F 2 only 5. none oF them 6. F 1 and F 2 only 7. all oF them 8. F 2 and F 3 only correct Explanation: By trig identities, x = 2cos 2 x - 1 = 1 - 2sin 2 x, while sin2 x = 2sin x cos x. But d dx sin x = cos x, d dx cos x = - sin Consequently, by the Chain Rule, ( A ) Not anti-derivative. ( B ) Anti-derivative. ( C ) Anti-derivative. keywords: antiderivative, trig Function, dou- ble angle Formula, T/±, 003 (part 1 oF 1) 10 points ±ind f ( t ) when f 0 ( t ) = cos 1 3 t - 2 3 t and f ( π 2 ) = 4. 1. f ( t ) = 5sin 1 3 t - 3cos 2 3 t + 3 2. f ( t ) = 3sin 1 3 t + cos 2 3 t + 2

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Miller, Kierste – Homework 1 – Due: Jan 24 2007, 3:00 am – Inst: Gary Berg 2 3. f ( t ) = 3sin 1 3 t + 3cos 2 3 t + 1 correct 4. f ( t ) = 5cos 1 3 t - 3sin 2 3 t + 3 5. f ( t ) = 3cos 1 3 t + sin 2 3 t + 2 6. f ( t 1 3 t + 3sin 2 3 t + 1 Explanation: The function f must have the form f ( t 1 3 t 2 3 t + C where the constant C is determined by the condition f π 2 · = 3sin π 6 π 3 + C = 4 . But by known trig values sin π 6 = cos π 3 = 1 2 , so 3 + C = 4. Consequently, f ( t 1 3 t 2 3 t + 1 . keywords: antiderivatives, trigonometric functions 004 (part 1 of 1) 10 points Find the value of f ( - 1) when f 00 ( t ) = 9 t - 6 and f 0 (1) = 4 , f (1) = 5 . Correct answer: - 9 . Explanation: The most general anti-derivative of f 00 has the form f 0 ( t ) = 9 2 t 2 - 6 t + C where C is an arbitrary constant. But if f 0 (1) = 4, then f 0 (1) = 9 2 - 6 + C = 4 , i . e . C = 11 2 . Thus f 0 ( t ) = 9 2 t 2 - 6 t + 11 2 , from which it follows that f ( t ) = 3 2 t 3 - 3 t 2 + 11 2 t + D , where the constant D is determined by the condition f (1) = 3 2 - 3+ 11 2 + D = 5 , i . e . D = 1 . Consequently, f ( t ) = 3 2 t 3 - 3 t 2 + 11 2 t, and so at t = - 1, f ( - 1) = - 9 . keywords: antiderivative, constant 005 (part 1 of 1) 10 points Find the unique anti-derivative F of f ( x ) = 12 e 4 x - 3 such that F (0) = 0. 1. F ( x ) = 3 e 4 x - 3 - 1 2. F ( x ) = 3 e - 3 ( e 4 x - 1 ) correct 3. F ( x ) = e - 3 ( e 4 x - 1 ) 4. F ( x ) = 3 ( e 4 x - 3 - 1 ) 5. F ( x ) = e 4 x - 3 - 1 Explanation:
Miller, Kierste – Homework 1 – Due: Jan 24 2007, 3:00 am – Inst: Gary Berg 3 Since d dx e αx = α e αx , we see that F ( x ) = 3 e 4 x - 3 + C where the arbitrary constant C is specifed by

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HW_1_Answers - Miller Kierste Homework 1 Due 3:00 am Inst...

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