solution10_pdf - So (dds785) HW09 radin (54915) 1 This...

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Unformatted text preview: So (dds785) HW09 radin (54915) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Working through lots of problems from sec- tion 8.5 is a good way to prepare for Exam 2 (10/26) and the final exam. 001 10.0 points Determine the indefinite integral I = integraldisplay 1 x 2 2 x 24 dx 1. I = sin- 1 parenleftBig x 5 1 parenrightBig + C 2. I = sin- 1 parenleftBig x 1 5 parenrightBig + C 3. I = ln vextendsingle vextendsingle vextendsingle x + 1 + radicalbig x 2 2 x 24 vextendsingle vextendsingle vextendsingle + C 4. I = ln vextendsingle vextendsingle vextendsingle x 1 + radicalbig x 2 + 2 x 24 vextendsingle vextendsingle vextendsingle + C 5. I = ln vextendsingle vextendsingle vextendsingle x + 1 + radicalbig x 2 + 2 x 24 vextendsingle vextendsingle vextendsingle + C 6. I = ln vextendsingle vextendsingle vextendsingle x 1 + radicalbig x 2 2 x 24 vextendsingle vextendsingle vextendsingle + C correct Explanation: By completing the square we see that x 2 2 x 24 = ( x 2 2 x + 1 ) 25 = ( x 1) 2 25 . This suggests the substitution x 1 = 5 sec , for then dx = 5 sec tan d , while ( x 1) 2 25 = 25tan 2 . In this case I = integraldisplay 5 sec tan 5 tan d = integraldisplay sec d = ln | sec + tan | + C . Now x 1 = 5 sec = tan = radicalbig ( x 1) 2 25 5 , so I = ln vextendsingle vextendsingle vextendsingle x 1 + radicalbig ( x 1) 2 25 5 vextendsingle vextendsingle vextendsingle + C . Consequently I = ln vextendsingle vextendsingle vextendsingle x 1 + radicalbig x 2 2 x 24 vextendsingle vextendsingle vextendsingle + C . 002 10.0 points Evaluate the definite integral I = integraldisplay 1- 1 e 5 arctan y 1 + y 2 dy . 1. I = 1 6 e 5 / 4 + 1 6 e- 5 / 4 2. I = 1 5 e 5 / 4 + 1 5 e- 5 / 4 3. I = 1 6 e 5 / 4 + 1 6 e- 5 / 4 4. I = 1 6 e 5 / 4 1 6 e- 5 / 4 5. I = 1 5 e 5 / 4 1 5 e- 5 / 4 6. I = 1 5 e 5 / 4 1 5 e- 5 / 4 correct Explanation: Set u = arctan y . Then du = 1 1 + y 2 dy , So (dds785) HW09 radin (54915) 2 in which case I = integraldisplay / 4- / 4 e 5 u du = bracketleftBig e 5 u 5 bracketrightBig / 4- / 4 . Consequently, I = 1 5 ( e 5 / 4 e- 5 / 4 ) . 003 10.0 points Evaluate the definite integral I = integraldisplay 2 1 x 2 + 5 x + 1 dx . 1. I = 1 2 6 ln 3 2 2. I = 1 2 + 5 ln3 3. I = 1 2 + 6 ln 3 2 correct 4. I = 1 2 6 ln2 5. I = 1 2 + 5 ln2 6. I = 1 2 5 ln3 Explanation: After division x 2 + 5 x + 1 = ( x 2 1) + 6 x + 1 = x 2 1 x + 1 + 6 x + 1 = x 1 + 6 x + 1 ....
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solution10_pdf - So (dds785) HW09 radin (54915) 1 This...

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