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Homework 8

# Homework 8 - Rehman(aar638 HW08 sachse(56620 This print-out...

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Rehman (aar638) – HW08 – sachse – (56620) 1 This print-out should have 20 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 0 1 16 - x 2 dx . 1. I = 1 6 2. I = 1 4 π 3. I = 1 3 π 4. I = 1 4 5. I = 1 6 π correct 6. I = 1 3 Explanation: Set x = 4 sin u ; then dx = 4 cos u du and 16 - x 2 = 16(1 - sin 2 u ) = 8 cos 2 u , while x = 0 = u = 0 , x = 2 = u = π 6 . In this case I = integraldisplay π/ 6 0 cos u cos u du = integraldisplay π/ 6 0 du . Consequently I = 1 6 π . 002 10.0 points Evaluate the integral I = integraldisplay 1 0 x 2 (2 - x 2 ) 3 / 2 dx . 1. I = 2 parenleftBig 3 - π 3 parenrightBig 2. I = 2 parenleftBig 3 + π 3 parenrightBig 3. I = 1 - π 4 correct 4. I = 2 parenleftBig 2 + π 3 parenrightBig 5. I = 2 - π 4 6. I = 1 + π 4 Explanation: Let x = 2 sin θ . Then dx = 2 cos θ dθ , 2 - x 2 = 2 cos 2 θ , while x = 0 = θ = 0 , x = 1 = θ = π 4 . In this case, I = integraldisplay π/ 4 0 2 2 sin 2 θ cos θ 2 2 cos 3 θ = integraldisplay π/ 4 0 sin 2 θ cos 2 θ = integraldisplay π/ 4 0 tan 2 θ dθ . Now tan 2 θ = sec 2 θ - 1 , d tan θ = sec 2 θ , and so I = integraldisplay π/ 4 0 (sec 2 θ - 1) = bracketleftBig tan θ - θ bracketrightBig π/ 4 0 . Consequently, I = 1 - π 4 .

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Rehman (aar638) – HW08 – sachse – (56620) 2 003 10.0 points Evaluate the integral I = integraldisplay 1 0 3 ( x 2 + 3) 3 / 2 dx . 1. I = 3 2. I = 3 2 3. I = 1 4. I = 3 4 5. I = 1 2 correct Explanation: Set x = 3 tan u. Then dx = 3 sec 2 u du , while x = 0 = u = 0 , x = 1 = u = π 6 . On the other hand, ( x 2 + 3) 3 / 2 = ( 3(tan 2 u + 1) ) 3 / 2 = 3 3 sec 3 u . Thus I = 3 integraldisplay π/ 6 0 3 sec 2 u 3 3 sec 3 u du = integraldisplay π/ 6 0 cos u du = bracketleftBig sin u bracketrightBig π/ 6 0 . Consequently I = 1 2 . keywords: 004 10.0 points Evaluate the definite integral I = integraldisplay 2 2 4 x 2 x 2 - 1 dx . 1. I = 4( 3 - 2 ) 2. I = 2( 3 + 2 ) 3. I = 3 + 2 4. I = 3 - 2 5. I = 2( 3 - 2 ) correct 6. I = 4( 3 + 2 ) Explanation: Set x = sec u . Then dx = sec u tan u du , x 2 - 1 = tan 2 u , while x = 2 = u = π 4 , x = 2 = u = π 3 . In this case, I = 4 integraldisplay π/ 3 π/ 4 sec u tan u sec 2 u tan u du = integraldisplay π/ 3 π/ 4 4 cos u du = 4 bracketleftBig sin u bracketrightBig π/ 3 π/ 4 . Consequently, I = 2( 3 - 2 ) . 005 10.0 points
Rehman (aar638) – HW08 – sachse – (56620) 3 Which one of the following functions is an antiderivative of f when f ( x ) = 1 x 2 - 6 x + 10 ? 1. F ( x ) = tan - 1 ( x - 3) correct 2. F ( x ) = - 10 ( x - 3) ( x 2 - 6 x + 10) 2 3. F ( x ) = sin - 1 ( x - 3) 4. F ( x ) = ln( x 2 - 6 x + 10) 5. F ( x ) = ln vextendsingle vextendsingle vextendsingle vextendsingle x - 10 x + 3 vextendsingle vextendsingle vextendsingle vextendsingle Explanation: The indefinite integral I = integraldisplay f ( x ) dx = integraldisplay 1 x 2 - 6 x + 10 dx consists of all antiderivatives of f . But by completing the square we see that 1 x 2 - 6 x + 10 = 1 ( x 2 - 6 x + 9) + 1 = 1 ( x - 3) 2 + 1 . Thus I = integraldisplay 1 ( x - 3) 2 + 1 dx = tan - 1 ( x - 3) + C .

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