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Unformatted text preview: wwm364 – Homework 5 – Cepparo – (58400) 1 This printout should have 18 questions. Multiplechoice 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 + 16( x 5) 2 dx . 1. I = tan − 1 4( x 5) + C 2. I = 1 4 sin − 1 4( x 5) + C 3. I = 1 4 tan − 1 4( x 5) + C correct 4. I = 4 tan − 1 parenleftBig x 5 4 parenrightBig + C 5. I = 4 sin − 1 parenleftBig x 5 4 parenrightBig + C 6. I = sin − 1 4( x 5) + C Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the substitution u = 4( x 5) is suggested. For then du = 4 dx , in which case I = 1 4 integraldisplay 1 1 + u 2 du = 1 4 tan − 1 u + C , with C an arbitrary constant. Consequently, I = 1 4 tan − 1 4( x 5) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 3 8 9 + x 2 dx . 1. I = 25 24 π 2. I = 19 24 π 3. I = 2 3 π correct 4. I = 7 6 π 5. I = 11 12 π Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the substitution x = 3 u is suggested. For then dx = 3 du , while x = 0 = ⇒ u = 0 , x = 3 = ⇒ u = 1 . Thus I = 8 3 integraldisplay 1 1 1 + u 2 du . Consequently. I = bracketleftBig 8 3 tan − 1 u bracketrightBig 1 = 2 3 π . keywords: 003 10.0 points Determine the integral I = integraldisplay 3 2 7 √ 9 x 2 dx . 1. I = 7 3 π wwm364 – Homework 5 – Cepparo – (58400) 2 2. I = 7 4 3. I = 7 3 4. I = 7 6 π correct 5. I = 7 4 π 6. I = 7 6 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 = 3 u . Then dx = 3 du while x = 0 = ⇒ u = 0 and x = 3 2 = ⇒ u = 1 2 . In this case I = 21 integraldisplay 1 / 2 1 3 √ 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 1 + 3 arcsin x dx . 1. I = 1 6 (1 + 3 arcsin x ) 2 + C 2. I = 1 6 ln  1 + 3 arcsin x  + C 3. I = 1 6 ln  1 + 3 arcsin x  + C 4. I = 1 3 (1 + 3 arcsin x ) 2 + C 5. I = 1 3 ln  1 + 3 arcsin x  + C correct 6. I = 1 3 ln  1 + 3 arcsin x  + C Explanation: Set u = 1 + 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  1 + 3 arcsin x  + C . keywords: 005 10.0 points Determine the integral I = integraldisplay π/ 2 7 cos θ 1 + sin 2 θ dθ ....
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This note was uploaded on 03/24/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.
 Spring '08
 RAdin

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