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Unformatted text preview: Cheung, Anthony – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Benzvi 1 This printout should have 17 questions. Multiplechoice 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 6 √ x 2 + 1 dx. 1. I = √ 3(2 √ 3 ) 2. I = 6ln(2 √ 3) 3. I = 6ln(2 + √ 3) correct 4. I = 6(2 √ 3 ) 5. I = √ 3 ln(2 + √ 3) 6. I = 6(2 + √ 3 ) Explanation: Set x = tan u, then dx = sec 2 udu, x 2 + 1 = sec 2 u, while x = 0 = ⇒ u = 0 , x = √ 3 = ⇒ u = π 3 . In this case I = Z √ 3 6sec 2 u sec u du = Z √ 3 6sec udu. Now Z sec udu = ln  sec u + tan u  + C 1 , Thus I = 6 h ln  sec u + tan u  i π/ 3 . Consequently, I = 6ln(2 + √ 3 ) . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 1 2 t √ 2 t 4 dt. 1. I = π 2. I = 1 3. I = 1 2 4. I = 1 4 5. I = 1 4 π correct 6. I = 1 2 π Explanation: Set t 2 = √ 2 sin u . Then 2 tdt = √ 2cos udu, in which case tdt = 1 √ 2 dt. On the other hand, t = 0 = ⇒ u = 0 , t = 1 = ⇒ u = π 4 . Cheung, Anthony – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Benzvi 2 Thus I = 1 √ 2 Z π/ 4 2cos u √ 2cos u du = 1 2 h 2 u i π/ 4 . Consequently, I = 1 4 π . keywords: definite integral, substitution, in verse sin integral, 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 4 2 7 x 2 6 x + 10 dx. 1. I = 4 π 2. I = 7 2 π correct 3. I = 7 π 4. I = 3 π 5. I = 3 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 7 ( 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 = 7 Z 1 1 1 1 + u 2 du = 7 h tan 1 u i 1 1 = 7 £ tan 1 1 tan 1 ( 1) / . But tan 1 1 = π 4 while tan 1 ( 1) = π 4 . Consequently, I = 7 2 π . keywords: trigonometric substitutions, com plete square 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z 1 / 2 µ 2 sin 1 x + 5 √ 1 x 2 ¶ dx. 1. I = 2 π + ‡ 2 + √ 3 · 2. I = 2 π ‡ 2 + √ 3 · 3. I = 2 π + ‡ 2 √ 3 · 4. I = π ‡ 2 √ 3 · correct 5. I = π + ‡ 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 . Cheung, Anthony – Homework 7 – Due: Oct 17 2006, 3:00 am – Inst: David Benzvi 3 In this case I = Z π/ 6 µ 2 θ + 5 cos θ ¶ cos θ dθ = Z π/ 6 (2 θ cos θ + 5) dθ = 2 Z π/ 6 θ cos θ dθ + 5 6 π ....
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This note was uploaded on 01/27/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.
 Spring '08
 RAdin
 Calculus

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