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# HW 7 - Portillo Tom Homework 7 Due 3:00 am Inst David...

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Portillo, Tom – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Benzvi 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 1 x 2 + 1 dx . 1. I = 2 - 3 2. I = ln(2 + 3 ) correct 3. I = ln(2 - 3 ) 4. I = 2 + 3 5. I = 3(2 - 3 ) 6. I = 3 ln(2 + 3 ) 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 sec 2 u sec u du = Z 3 0 sec u du . Now Z sec u du = ln | sec u + tan u | + C 1 , Thus I = h ln | sec u + tan u | i π/ 3 0 . Consequently, I = ln(2 + 3 ) . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 1 0 4 t 2 - t 4 dt . 1. I = 1 2. I = 2 3. I = 2 π 4. I = 1 2 π correct 5. I = 1 2 6. I = 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 . Thus I = 1 2 Z π/ 4 0 4 cos u 2 cos u du = 1 2 h 4 u i π/ 4 0 .

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Portillo, Tom – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Benzvi 2 Consequently, I = 1 2 π . keywords: definite integral, substitution, in- verse sin integral, 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 4 2 5 x 2 - 6 x + 10 dx . 1. I = π 2. I = 5 2 π correct 3. I = 3 π 4. I = 2 π 5. I = 5 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 5 ( 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 = 5 Z 1 - 1 1 1 + u 2 du = 5 h tan - 1 u i 1 - 1 = 5 £ tan - 1 1 - tan - 1 ( - 1) / . But tan - 1 1 = π 4 while tan - 1 ( - 1) = - π 4 . Consequently, I = 5 2 π . keywords: trigonometric substitutions, com- plete square 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 / 2 0 4 sin - 1 x + 3 1 - x 2 dx . 1. I = 5 3 π + 2 2 - 3 · 2. I = 5 3 π + 2 2 + 3 · 3. I = 5 6 π - 2 2 - 3 · correct 4. I = 5 3 π - 2 2 + 3 · 5. I = 5 6 π + 2 2 - 3 · 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 . In this case I = Z π/ 6 0 4 θ + 3 cos θ cos θ dθ = Z π/ 6 0 (4 θ cos θ + 3) = 4 Z π/ 6 0 θ cos θ dθ + 1 2 π .
Portillo, Tom – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Benzvi 3 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 = 1 3 π + 2 3 - 2 · + 1 2 π = 5 6 π - 2 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 = 2 ( x - 4)(9 - x ) , y (5) = 0 .

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HW 7 - Portillo Tom Homework 7 Due 3:00 am Inst David...

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