# CalcAnswers5 - nav277 – Homework 5 – Odell –(58340 1...

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Unformatted text preview: nav277 – Homework 5 – Odell – (58340) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the integral I = integraldisplay 1 1 + 25( x- 1) 2 dx . 1. I = 1 5 tan − 1 5( x- 1) + C correct 2. I = 1 5 sin − 1 5( x- 1) + C 3. I = sin − 1 5( x- 1) + C 4. I = tan − 1 5( x- 1) + C 5. I = 5 sin − 1 parenleftBig x- 1 5 parenrightBig + C 6. I = 5 tan − 1 parenleftBig x- 1 5 parenrightBig + C Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the substitution u = 5( x- 1) is suggested. For then du = 5 dx , in which case I = 1 5 integraldisplay 1 1 + u 2 du = 1 5 tan − 1 u + C , with C an arbitrary constant. Consequently, I = 1 5 tan − 1 5( x- 1) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 2 6 4 + x 2 dx . 1. I = 3 4 π correct 2. I = 5 8 π 3. I = 1 4 π 4. I = 3 8 π 5. I = 1 2 π Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the substitution x = 2 u is suggested. For then dx = 2 du , while x = 0 = ⇒ u = 0 , x = 2 = ⇒ u = 1 . Thus I = 3 integraldisplay 1 1 1 + u 2 du . Consequently. I = bracketleftBig 3 tan − 1 u bracketrightBig 1 = 3 4 π . keywords: 003 10.0 points Determine the integral I = integraldisplay 1 7 √ 4- x 2 dx . 1. I = 7 3 nav277 – Homework 5 – Odell – (58340) 2 2. I = 7 3 π 3. I = 7 6 4. I = 7 4 5. I = 7 4 π 6. I = 7 6 π correct Explanation: Since integraldisplay 1 √ 1- x 2 dx = sin − 1 x + C , we need to reduce I to an integal of this form by changing the x variable. Indeed, set x = 2 u . Then dx = 2 du while x = 0 = ⇒ u = 0 and x = 1 = ⇒ u = 1 2 . In this case I = 14 integraldisplay 1 / 2 1 2 √ 1- u 2 du = 7 integraldisplay 1 / 2 1 √ 1- u 2 du . Consequently, I = bracketleftBig 7 sin − 1 u bracketrightBig 1 / 2 = 7 6 π . keywords: 004 10.0 points Determine the integral I = integraldisplay (1- x 2 ) − 1 / 2 4 + 3 arcsin x dx . 1. I =- 1 3 ln | 4 + 3 arcsin x | + C 2. I = 1 6 (4 + 3 arcsin x ) 2 + C 3. I =- 1 6 ln | 4 + 3 arcsin x | + C 4. I = 1 6 ln | 4 + 3 arcsin x | + C 5. I =- 1 3 (4 + 3 arcsin x ) 2 + C 6. I = 1 3 ln | 4 + 3 arcsin x | + C correct Explanation: Set u = 4 + 3 arcsin x . Then du = 3 √ 1- x 2 dx = 3(1- x 2 ) − 1 / 2 dx, so I = 1 3 integraldisplay 1 u du = 1 3 ln | u | + C with C an arbitrary constant. Consequently, I = 1 3 ln | 4 + 3 arcsin x | + C . keywords: 005 10.0 points Determine the integral I = integraldisplay π/ 2 4 cos θ 1 + sin 2 θ dθ ....
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## This note was uploaded on 04/15/2011 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas.

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CalcAnswers5 - nav277 – Homework 5 – Odell –(58340 1...

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