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HW_5_Answers - Miller Kierste Homework 5 Due 3:00 am Inst...

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Miller, Kierste – Homework 5 – Due: Feb 20 2007, 3:00 am – Inst: Gary Berg 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the most general function f such that f 0 ( x ) = 5 - 4 4 - x 2 . 1. f ( x ) = 5 x - 2 tan - 1 x 2 + C 2. f ( x ) = 5 2 + 2 tan - 1 x 2 + C 3. f ( x ) = 5 2 x - 4 sin - 1 x + C 4. f ( x ) = 5 x - 4 sin - 1 x 2 + C correct 5. f ( x ) = 5 x + 4 sin - 1 x 2 + C 6. f ( x ) = 5 + 2 tan - 1 x + C Explanation: Since d dx sin - 1 x 2 · = 1 4 - x 2 , we see that f ( x ) = 5 x - 4 sin - 1 x 2 + C with C an arbitrary constant. keywords: 002 (part 1 of 1) 10 points Find the value of the integral I = Z 5 2 1 9 + ( x - 2) 2 dx . 1. I = 3 2. I = 3 π 3. I = 1 12 π correct 4. I = 1 6 π 5. I = 1 6 6. I = 1 3 π Explanation: Set 3 tan u = x - 2. Then 9 + ( x - 2) 2 = 9 + (3 tan u ) 2 = 9(1 + tan 2 u ) = 9 sec 2 u , while 3 sec 2 u du = dx . Also x = 2 = u = 0 , and x = 5 = u = π 4 . In this case I = Z π/ 4 0 3 sec 2 u 9 sec 2 u du = 1 3 Z π/ 4 0 du. Consequently, I = 1 3 h u i π/ 4 0 = 1 12 π . keywords: 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 7 0 1 1 - 25 x 2 dx . Correct answer: 0 . 159121 . Explanation:
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Miller, Kierste – Homework 5 – Due: Feb 20 2007, 3:00 am – Inst: Gary Berg 2 Since Z 1 1 - x 2 dx = sin - 1 x + C , a change of variable x is needed to reduce I to this form. Set u = 5 x . Then du = 5 dx , and x = 0 = u = 0 , while x = 1 7 = u = 5 7 . In this case I = 1 5 Z 5 7 0 1 1 - u 2 du = 1 5 sin - 1 u 5 7 0 . Consequently, I = 1 5 arcsin 5 7 = 0 . 159121 . keywords: 004 (part 1 of 1) 10 points Find the value of the definite integral I = Z 1 2 0 2 sin - 1 x 1 - x 2 dx . 1. I = 1 25 π 2 2. I = 1 36 π 2 correct 3. I = 1 16 π 2 4. I = 1 12 π 2 5. I = 1 9 π 2 Explanation: Since Z 1 1 - x 2 dx = sin - 1 x + C , this suggests the substitution u = sin - 1 x , for then du = 1 1 - x 2 dx , while x = 0 = u = 0 , x = 1 2 = u = π 6 . Thus I = 2 Z π 6 0 u du = h u 2 i π 6 0 . Consequently, I = 1 36 π 2 . keywords: 005 (part 1 of 1) 10 points Evaluate the definite integral I = Z π π/ 4 2 sin 2 x 1 + cos 2 2 x dx . Correct answer: - 0 . 785397 . Explanation: Set u = cos 2 x . Then du = - 2 sin 2 x dx , while x = π 4 = u = 0 , x = π = u = 1 .
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