This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: yp968 – Homework 5 – Radin – (58415) 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 + 4( x 1) 2 dx . 1. I = 2 sin − 1 parenleftBig x 1 2 parenrightBig + C 2. I = tan − 1 2( x 1) + C 3. I = 1 2 sin − 1 2( x 1) + C 4. I = sin − 1 2( x 1) + C 5. I = 1 2 tan − 1 2( x 1) + C correct 6. I = 2 tan − 1 parenleftBig x 1 2 parenrightBig + C Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the substitution u = 2( x 1) is suggested. For then du = 2 dx , in which case I = 1 2 integraldisplay 1 1 + u 2 du = 1 2 tan − 1 u + C , with C an arbitrary constant. Consequently, I = 1 2 tan − 1 2( x 1) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 2 8 4 + x 2 dx . 1. I = 3 2 π 2. I = 9 8 π 3. I = π correct 4. I = 11 8 π 5. I = 5 4 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 = 4 integraldisplay 1 1 1 + u 2 du . Consequently. I = bracketleftBig 4 tan − 1 u bracketrightBig 1 = π . keywords: 003 10.0 points Determine the integral I = integraldisplay 1 4 √ 4 x 2 dx . 1. I = 4 3 2. I = 1 yp968 – Homework 5 – Radin – (58415) 2 3. I = 2 3 4. I = 2 3 π correct 5. I = π 6. I = 4 3 π 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 = 8 integraldisplay 1 / 2 1 2 √ 1 u 2 du = 4 integraldisplay 1 / 2 1 √ 1 u 2 du . Consequently, I = bracketleftBig 4 sin − 1 u bracketrightBig 1 / 2 = 2 3 π . keywords: 004 10.0 points Determine the integral I = integraldisplay (1 x 2 ) − 1 / 2 4 + arcsin x dx . 1. I = 1 2 ln  4 + arcsin x  + C 2. I = 1 2 (4 + arcsin x ) 2 + C 3. I = 1 2 ln  4 + arcsin x  + C 4. I = (4 + arcsin x ) 2 + C 5. I = ln  4 + arcsin x  + C 6. I = ln  4 + arcsin x  + C correct Explanation: Set u = 4 + arcsin x . Then du = 1 √ 1 x 2 dx = (1 x 2 ) − 1 / 2 dx, so I = integraldisplay 1 u du = ln  u  + C with C an arbitrary constant. Consequently, I = ln  4 + arcsin x  + C . keywords: 005 10.0 points Determine the integral I = integraldisplay π/ 2 6 cos θ 1 + sin 2 θ dθ ....
View
Full
Document
This note was uploaded on 04/29/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.
 Spring '08
 RAdin

Click to edit the document details