Int cal7 - pokharel(yp624 – HW07 – Radin –(57410 1 This print-out should have 22 questions Multiple-choice questions may continue on the next

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: pokharel (yp624) – HW07 – Radin – (57410) 1 This print-out should have 22 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 + 4( x- 7) 2 dx . 1. I = 1 2 sin − 1 2( x- 7) + C 2. I = 2 sin − 1 parenleftBig x- 7 2 parenrightBig + C 3. I = 2 tan − 1 parenleftBig x- 7 2 parenrightBig + C 4. I = sin − 1 2( x- 7) + C 5. I = 1 2 tan − 1 2( x- 7) + C correct 6. I = tan − 1 2( x- 7) + C Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the substitution u = 2( x- 7) 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- 7) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 3 2 5 √ 9- x 2 dx . 1. I = 5 4 2. I = 5 6 3. I = 5 6 π correct 4. I = 5 4 π 5. I = 5 3 π 6. I = 5 3 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 = 15 integraldisplay 1 / 2 1 3 √ 1- u 2 du = 5 integraldisplay 1 / 2 1 √ 1- u 2 du . Consequently, I = bracketleftBig 5 sin − 1 u bracketrightBig 1 / 2 = 5 6 π . keywords: 003 10.0 points pokharel (yp624) – HW07 – Radin – (57410) 2 Evaluate the definite integral I = integraldisplay π/ 2 4 sin θ 1 + cos 2 θ dθ . 1. I = π correct 2. I = 1 4 π 3. I = 3 4 π 4. I = 1 2 π 5. I = 5 4 π Explanation: Since d dθ cos θ =- sin θ , the substitution u = cos θ is suggested. For then du =- sin θ dθ , while θ = 0 = ⇒ u = 1 , θ = π 2 = ⇒ u = 0 , so that I =- 4 integraldisplay 1 1 1 + u 2 du = 4 integraldisplay 1 1 1 + u 2 du , which can now be integrated using the fact that d du tan − 1 u = 1 1 + u 2 . Consequently, I = 4 bracketleftBig tan − 1 u bracketrightBig 1 = π since tan − 1 1 = π 4 . 004 10.0 points Determine the integral I = integraldisplay 1 2 tan − 1 x 1 + x 2 dx . 1. I = π 8 2. I = π 2 8 3. I = π 2 16 correct 4. I = π 16 5. I = π 4 6. I = π 2 4 Explanation: Set u = tan − 1 x . Then du = 1 1 + x 2 dx , while x = 0 = ⇒ u = 0 , x = 1 = ⇒ u = π 4 . In this case, I = integraldisplay π/ 4 2 u du = bracketleftBig u 2 bracketrightBig π/ 4 . Consequently, I = 1 16 π 2 . keywords: 005 10.0 points Determine the integral I = integraldisplay 4- x √ 1- x 2 dx . pokharel (yp624) – HW07 – Radin – (57410) 3 1. I = 2 tan − 1 x- 1 2 radicalbig 1- x 2 + C 2. I = 4 sin − 1 x + radicalbig 1- x 2 + C correct 3. I = 2 sin − 1 x- radicalbig 1- x 2 + C 4. I = 2 tan − 1 x + radicalbig 1- x 2 + C 5. I = 4 tan − 1 x- radicalbig 1- x 2 + C 6. I = 4 sin − 1 x + 1 2 radicalbig 1- x 2 + C Explanation: We deal with the two integrals I 1 = integraldisplay 4 √ 1- x 2 dx, I 2 = x √ 1- x 2 dx separately....
View Full Document

This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

Page1 / 11

Int cal7 - pokharel(yp624 – HW07 – Radin –(57410 1 This print-out should have 22 questions Multiple-choice questions may continue on the next

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online