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Unformatted text preview: Portillo, Tom Homework 7 Due: Oct 17 2006, 3:00 am Inst: David Benzvi 1 This printout should have 17 questions. Multiplechoice 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 Evaluate the definite integral I = Z 3 1 x 2 + 1 dx. 1. I = 2 3 2. I = ln(2 + 3) correct 3. I = ln(2 3) 4. I = 2 + 3 5. I = 3(2 3) 6. I = 3ln(2 + 3) Explanation: Set x = tan u, then dx = sec 2 udu, x 2 + 1 = sec 2 u, while x = 0 = u = 0 , x = 3 = u = 3 . In this case I = Z 3 sec 2 u sec u du = Z 3 sec udu. Now Z sec udu = ln  sec u + tan u  + C 1 , Thus I = h ln  sec u + tan u  i / 3 . Consequently, I = ln(2 + 3) . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 1 4 t 2 t 4 dt. 1. I = 1 2. I = 2 3. I = 2 4. I = 1 2 correct 5. I = 1 2 6. I = Explanation: Set t 2 = 2sin u . Then 2 tdt = 2cos udu, in which case tdt = 1 2 dt. On the other hand, t = 0 = u = 0 , t = 1 = u = 4 . Thus I = 1 2 Z / 4 4cos u 2cos u du = 1 2 h 4 u i / 4 . Portillo, Tom Homework 7 Due: Oct 17 2006, 3:00 am Inst: David Benzvi 2 Consequently, I = 1 2 . keywords: definite integral, substitution, in verse sin integral, 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 4 2 5 x 2 6 x + 10 dx. 1. I = 2. I = 5 2 correct 3. I = 3 4. I = 2 5. I = 5 Explanation: By completing the square we see that x 2 6 x + 10 = ( x 2 6 x + 9) + 10 9 = ( x 3) 2 + 1 . Thus I = Z 4 2 5 ( x 3) 2 + 1 . Since d dx tan 1 x = 1 1 + x 2 , this suggests the substitution x 3 = u . For then dx = du , while x = 2 = u = 1 , x = 4 = u = 1 . In this case, I = 5 Z 1 1 1 1 + u 2 du = 5 h tan 1 u i 1 1 = 5 tan 1 1 tan 1 ( 1) / . But tan 1 1 = 4 while tan 1 ( 1) = 4 . Consequently, I = 5 2 . keywords: trigonometric substitutions, com plete square 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 / 2 4 sin 1 x + 3 1 x 2 dx. 1. I = 5 3 + 2 2 3 2. I = 5 3 + 2 2 + 3 3. I = 5 6  2 2 3 correct 4. I = 5 3  2 2 + 3 5. I = 5 6 + 2 2 3 Explanation: Let x = sin ; then dx = cos d and 1 x 2 = 1 sin 2 = cos 2 , while x = 0 = = 0 , x = 1 2 = = 6 . In this case I = Z / 6 4 + 3 cos cos d = Z / 6 (4 cos + 3) d = 4 Z / 6 cos d + 1 2 . Portillo, Tom Homework 7 Due: Oct 17 2006, 3:00 am Inst: David Benzvi 3 To evaluate the last integral we now use integration by parts: Z / 6 cos d = h sin i / 6 Z / 6 sin d = h sin + cos i / 6 = 12 + 3 2 1 ....
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This note was uploaded on 02/18/2010 for the course MATH 408L taught by Professor Benzvi during the Spring '06 term at North Texas.
 Spring '06
 Benzvi

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