HW07 Solution - badruddin(ssb776 – HW07 – Radin...

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Unformatted text preview: badruddin (ssb776) – 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 + 16( x- 7) 2 dx . 1. I = 4 sin − 1 parenleftBig x- 7 4 parenrightBig + C 2. I = 1 4 sin − 1 4( x- 7) + C 3. I = tan − 1 4( x- 7) + C 4. I = 4 tan − 1 parenleftBig x- 7 4 parenrightBig + C 5. I = sin − 1 4( x- 7) + C 6. I = 1 4 tan − 1 4( x- 7) + C correct Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the substitution u = 4( x- 7) 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- 7) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 1 4 √ 4- x 2 dx . 1. I = 1 2. I = 4 3 π 3. I = 4 3 4. I = 2 3 5. I = π 6. I = 2 3 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 = 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: 003 10.0 points badruddin (ssb776) – HW07 – Radin – (57410) 2 Evaluate the definite integral I = integraldisplay π/ 2 4 sin θ 1 + cos 2 θ dθ . 1. I = 1 2 π 2. I = 3 4 π 3. I = 5 4 π 4. I = 3 2 π 5. I = π correct 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 8 tan − 1 x 1 + x 2 dx . 1. I = π 2 8 2. I = π 16 3. I = π 2 16 4. I = π 2 4 correct 5. I = π 4 6. I = π 8 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 8 u du = bracketleftBig 4 u 2 bracketrightBig π/ 4 . Consequently, I = 1 4 π 2 . keywords: 005 10.0 points Determine the integral I = integraldisplay 4 + x √ 1- x 2 dx . badruddin (ssb776) – HW07 – Radin – (57410) 3 1. I = 4 sin − 1 x- 1 2 radicalbig 1- x 2 + C 2. I = 2 tan − 1 x + 1 2 radicalbig 1- x 2 + C 3. I = 4 sin − 1 x- radicalbig 1- x 2 + C correct 4. I = 4 tan − 1 x + radicalbig 1- x 2 + C 5. I = 2 tan − 1 x- radicalbig 1- x 2 + C 6. I = 2 sin − 1 x + 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....
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This note was uploaded on 11/13/2011 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas.

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HW07 Solution - badruddin(ssb776 – HW07 – Radin...

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