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Unformatted text preview: tgo72 – Homework 7 – Gompf – (58370) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay √ 2 x 2 √ 4 x 2 dx . 1. I = 1 2 parenleftBig π 2 1 parenrightBig 2. I = π 2 √ 3 2 3. I = 2 π 3 √ 3 2 4. I = 2 π 3 1 5. I = π 2 1 correct 6. I = 1 2 parenleftBig π 3 √ 3 2 parenrightBig 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 4 sin 2 θ 2 cos θ 2 cos θ dθ = 4 integraldisplay π/ 4 sin 2 θ dθ = 2 integraldisplay π/ 4 (1 cos2 θ ) dθ = 2 bracketleftBig θ 1 2 sin2 θ bracketrightBig π/ 4 . Consequently, I = π 2 1 . keywords: 002 10.0 points Determine the integral I = integraldisplay x 2 (4 x 2 ) 3 / 2 dx . 1. I = 2 x √ 4 x 2 sin 1 parenleftBig x 4 parenrightBig + C 2. I = 2 x 2 √ 4 x 2 + sin 1 parenleftBig x 2 4 parenrightBig + C 3. I = x √ 4 x 2 sin 1 parenleftBig x 2 parenrightBig + C correct 4. I = x √ 4 x 2 + sin 1 parenleftBig x 2 parenrightBig + C 5. I = 2 x √ 4 x 2 sin 1 parenleftBig x 2 4 parenrightBig + C 6. I = x 2 √ 4 x 2 + sin 1 parenleftBig x 2 2 parenrightBig + C Explanation: Let x = 2 sin θ . Then dx = 2 cos θ dθ , 4 x 2 = 4 cos 2 θ . In this case, I = integraldisplay 4 · 2 sin 2 θ cos θ 2 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 , tgo72 – Homework 7 – Gompf – (58370) 2 and so I = integraldisplay (sec 2 θ 1) dθ = tan θ θ + C . Consequently, I = x √ 4 x 2 sin 1 parenleftBig x 2 parenrightBig + C with C ann arbitrary constant. 003 10.0 points Determine the integral I = integraldisplay 1 ( x 2 + 4) 3 2 dx . 1. I = x √ x 2 + 4 4 + C 2. I = x √ x 2 + 4 + C 3. I = x 4 √ x 2 + 4 + C correct 4. I = 1 4 √ x 2 + 4 + C 5. I = √ x 2 + 4 x + C 6. I = √ x 2 + 4 4 x + 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 2 8 sec 2 u sec 3 u du = 1 4 integraldisplay cos u du , and so I = 1 4 sin u + C = 1 4 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 4 √ x 2 + 4 + C with C an arbitrary constant. keywords: trig substitution 004 10.0 points Determine the indefinite integral I = integraldisplay 2 x √ x 2 1 dx . 1. I = 2 ln vextendsingle vextendsingle vextendsingle x radicalbig x 2 1 vextendsingle vextendsingle vextendsingle radicalbig x 2 1 + C 2. I = 2 ln vextendsingle vextendsingle vextendsingle x + radicalbig x 2 1 vextendsingle vextendsingle vextendsingle + radicalbig x 2 1 + C 3. I = ln vextendsingle vextendsingle vextendsingle x radicalbig x 2 1 vextendsingle vextendsingle...
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 Spring '08
 RAdin
 Calculus, Partial fractions in complex analysis, Partial fractions in integration

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