# Homework 7 - ejm753 – Homework 7 – Cepparo –(58400 1 This print-out should have 19 questions Multiple-choice questions may continue on the

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Unformatted text preview: ejm753 – Homework 7 – Cepparo – (58400) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay √ 2 4 x 2 √ 4- x 2 dx . 1. I = 2 parenleftBig π 2- 1 parenrightBig 2. I = 4 parenleftBig 2 π 3- √ 3 2 parenrightBig 3. I = π 2- √ 3 2 4. I = 4 parenleftBig π 2- 1 parenrightBig correct 5. I = 2 parenleftBig π 3- √ 3 2 parenrightBig 6. I = 2 π 3- 1 Explanation: Set x = 2 sin θ . Then dx = 2 cos θ dθ , while x = 0 = ⇒ θ = 0 , x = √ 2 = ⇒ θ = π 4 . In this case I = integraldisplay π/ 4 16 sin 2 θ 2 cos θ 2 cos θ dθ = 16 integraldisplay π/ 4 sin 2 θ dθ = 8 integraldisplay π/ 4 (1- cos 2 θ ) dθ = 8 bracketleftBig θ- 1 2 sin 2 θ bracketrightBig π/ 4 . Consequently, I = 4 parenleftBig π 2- 1 parenrightBig . keywords: 002 10.0 points Determine the integral I = integraldisplay x 2 (9- x 2 ) 3 / 2 dx . 1. I = x √ 9- x 2- sin- 1 parenleftBig x 3 parenrightBig + C correct 2. I = 3 x √ 9- x 2- sin- 1 parenleftBig x 9 parenrightBig + C 3. I = 3 x 2 √ 9- x 2 + sin- 1 parenleftBig x 2 9 parenrightBig + C 4. I = 3 x √ 9- x 2- sin- 1 parenleftBig x 2 9 parenrightBig + C 5. I = x √ 9- x 2 + sin- 1 parenleftBig x 3 parenrightBig + C 6. I = x 2 √ 9- x 2 + sin- 1 parenleftBig x 2 3 parenrightBig + C Explanation: Let x = 3 sin θ . Then dx = 3 cos θ dθ , 9- x 2 = 9 cos 2 θ . In this case, I = integraldisplay 9 · 3 sin 2 θ cos θ 3 3 cos 3 θ dθ = integraldisplay sin 2 θ cos 2 θ dθ = integraldisplay tan 2 θ dθ . Now tan 2 θ = sec 2 θ- 1 , d dθ tan θ = sec 2 θ , ejm753 – Homework 7 – Cepparo – (58400) 2 and so I = integraldisplay (sec 2 θ- 1) dθ = tan θ- θ + C . Consequently, I = x √ 9- x 2- sin- 1 parenleftBig x 3 parenrightBig + C with C ann arbitrary constant. 003 10.0 points Determine the integral I = integraldisplay 2 ( x 2 + 4) 3 2 dx . 1. I = 1 2 √ x 2 + 4 + C 2. I = x 2 √ x 2 + 4 + C correct 3. I = √ x 2 + 4 x + C 4. I = √ x 2 + 4 2 x + C 5. I = x √ x 2 + 4 2 + C 6. I = x √ x 2 + 4 + C Explanation: Set x = 2 tan u. Then dx = 2 sec 2 u du , while ( x 2 + 4) 3 2 = ( 4(tan 2 u + 1) ) 3 2 = 8 sec 3 u . Thus I = integraldisplay 4 8 sec 2 u sec 3 u du = 1 2 integraldisplay cos u du , and so I = 1 2 sin u + C = 1 2 sin parenleftBig tan- 1 x 2 parenrightBig + C . But by Pythagoras u radicalbig x 2 + 4 2 x we see that sin parenleftBig tan- 1 x 2 parenrightBig = x √ x 2 + 4 . Consequently, I = x 2 √ x 2 + 4 + C with C an arbitrary constant. keywords: trig substitution 004 10.0 points Determine the indefinite integral I = integraldisplay 4 + x √ x 2- 1 dx ....
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## This note was uploaded on 04/28/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.

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Homework 7 - ejm753 – Homework 7 – Cepparo –(58400 1 This print-out should have 19 questions Multiple-choice questions may continue on the

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