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Homework 7 Solutions

# Homework 7 Solutions - Mcbiles Emily Homework 7 Due 3:00 am...

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Mcbiles, Emily – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Fonken 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the definite integral I = Z 3 0 5 x 2 + 1 dx . 1. I = 5 ln(2 - 3 ) 2. I = 5(2 + 3 ) 3. I = 3(2 - 3 ) 4. I = 3 ln(2 + 3 ) 5. I = 5(2 - 3 ) 6. I = 5 ln(2 + 3 ) correct Explanation: Set x = tan u , then dx = sec 2 u du , x 2 + 1 = sec 2 u , while x = 0 = u = 0 , x = 3 = u = π 3 . In this case I = Z 3 0 5 sec 2 u sec u du = Z 3 0 5 sec u du . Now Z sec u du = ln | sec u + tan u | + C 1 , Thus I = 5 h ln | sec u + tan u | i π/ 3 0 . Consequently, I = 5 ln(2 + 3 ) . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 1 0 t 2 - t 4 dt . 1. I = 1 2 2. I = 1 8 π correct 3. I = 1 4 4. I = 1 8 5. I = 1 2 π 6. I = 1 4 π Explanation: Set t 2 = 2 sin u . Then 2 t dt = 2 cos u du , in which case t dt = 1 2 dt . On the other hand, t = 0 = u = 0 , t = 1 = u = π 4 .

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Mcbiles, Emily – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Fonken 2 Thus I = 1 2 Z π/ 4 0 cos u 2 cos u du = 1 2 h u i π/ 4 0 . Consequently, I = 1 8 π . keywords: definite integral, substitution, in- verse sin integral, 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 4 2 6 x 2 - 6 x + 10 dx . 1. I = 7 2 π 2. I = 6 π 3. I = 3 π correct 4. I = 5 4 π 5. I = 5 2 π Explanation: By completing the square we see that x 2 - 6 x + 10 = ( x 2 - 6 x + 9) + 10 - 9 = ( x - 3) 2 + 1 . Thus I = Z 4 2 6 ( x - 3) 2 + 1 . Since d dx tan - 1 x = 1 1 + x 2 , this suggests the substitution x - 3 = u . For then dx = du , while x = 2 = u = - 1 , x = 4 = u = 1 . In this case, I = 6 Z 1 - 1 1 1 + u 2 du = 6 h tan - 1 u i 1 - 1 = 6 £ tan - 1 1 - tan - 1 ( - 1) / . But tan - 1 1 = π 4 while tan - 1 ( - 1) = - π 4 . Consequently, I = 3 π . keywords: trigonometric substitutions, com- plete square 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 / 2 0 6 sin - 1 x + 3 1 - x 2 dx . 1. I = 2 π + 3 2 + 3 · 2. I = 2 π - 3 2 + 3 · 3. I = π + 3 2 - 3 · 4. I = 2 π + 3 2 - 3 · 5. I = π - 3 2 - 3 · correct Explanation: Let x = sin θ ; then dx = cos θ dθ and 1 - x 2 = 1 - sin 2 θ = cos 2 θ , while x = 0 = θ = 0 , x = 1 2 = θ = π 6 .
Mcbiles, Emily – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Fonken 3 In this case I = Z π/ 6 0 6 θ + 3 cos θ cos θ dθ = Z π/ 6 0 (6 θ cos θ + 3) = 6 Z π/ 6 0 θ cos θ dθ + 1 2 π . To evaluate the last integral we now use integration by parts: Z π/ 6 0 θ cos θ dθ = h θ sin θ i π/ 6 0 - Z π/ 6 0 sin θ dθ = h θ sin θ + cos θ i π/ 6 0 = π 12 + 3 2 - 1 . Consequently I = π + 3 3 - 2 · + 1 2 π = π - 3 2 - 3 · . keywords: substitution, inverse trig function, integration by parts 005 (part 1 of 1) 10 points Find the unique function y satisfying the equations dy dx = 3 ( x - 4)(8 - x ) , y (5) = 0 .

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Homework 7 Solutions - Mcbiles Emily Homework 7 Due 3:00 am...

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