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# solution10_pdf - So(dds785 HW09 radin(54915 This print-out...

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So (dds785) – HW09 – radin – (54915) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. Working through lots oF problems From sec- tion 8.5 is a good way to prepare For Exam 2 (10/26) and the fnal exam. 001 10.0 points Determine the indefnite integral I = i 1 x 2 2 x 24 dx 1. I = sin - 1 p x 5 1 P + C 2. I = sin - 1 p x 1 5 P + C 3. I = ln v v v x + 1 + r x 2 2 x 24 v v v + C 4. I = ln v v v x 1 + r x 2 + 2 x 24 v v v + C 5. I = ln v v v x + 1 + r x 2 + 2 x 24 v v v + C 6. I = ln v v v x 1 + r x 2 2 x 24 v v v + C correct Explanation: By completing the square we see that x 2 2 x 24 = ( x 2 2 x + 1 ) 25 = ( x 1) 2 25 . This suggests the substitution x 1 = 5 sec θ , For then dx = 5 sec θ tan θ dθ , while ( x 1) 2 25 = 25tan 2 θ . In this case I = i 5 sec θ tan θ 5 tan θ = i sec θ dθ = ln | sec θ + tan θ | + C . Now x 1 = 5 sec θ = tan θ = r ( x 1) 2 25 5 , so I = ln v v v x 1 + r ( x 1) 2 25 5 v v v + C . Consequently I = ln v v v x 1 + r x 2 2 x 24 v v v + C . 002 10.0 points Evaluate the defnite integral I = i 1 - 1 e 5 arctan y 1 + y 2 dy . 1. I = 1 6 e 5 π/ 4 + 1 6 e - 5 π/ 4 2. I = 1 5 e 5 π/ 4 + 1 5 e - 5 π/ 4 3. I = 1 6 e 5 π/ 4 + 1 6 e - 5 π/ 4 4. I = 1 6 e 5 π/ 4 1 6 e - 5 π/ 4 5. I = 1 5 e 5 π/ 4 1 5 e - 5 π/ 4 6. I = 1 5 e 5 π/ 4 1 5 e - 5 π/ 4 correct Explanation: Set u = arctan y . Then du = 1 1 + y 2 dy ,

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So (dds785) – HW09 – radin – (54915) 2 in which case I = i π/ 4 - π/ 4 e 5 u du = b e 5 u 5 B π/ 4 - π/ 4 . Consequently, I = 1 5 ( e 5 π/ 4 e - 5 π/ 4 ) . 003 10.0 points Evaluate the defnite integral I = i 2 1 x 2 + 5 x + 1 dx . 1. I = 1 2 6 ln 3 2 2. I = 1 2 + 5 ln3 3. I = 1 2 + 6 ln 3 2 correct 4. I = 1 2 6 ln2 5. I = 1 2 + 5 ln2 6. I = 1 2 5 ln3 Explanation: AFter division x 2 + 5 x + 1 = ( x 2 1) + 6 x + 1 = x 2 1 x + 1 + 6 x + 1 = x 1 + 6 x + 1 . In this case I = i 2 1 p x 1 + 6 x + 1 P dx = b 1 2 x 2 x + 6 ln | x + 1 | B 2 1 = p 0 p 1 2 PP + 6 p ln 3 ln 2 P . Consequently, I = 1 2 + 6 ln 3 2 . 004 10.0 points Evaluate the integral I = i π/ 4 0 sec 2 x (3 sin x ) dx . 1. I = 4 + 1 2 2. I = 2 + 1 2 3. I = 2 2 4. I = 4 2 correct 5. I = 4 + 2 6. I = 2 1 2 Explanation: Since sec 2 x { 3 sin x } = 3 sec 2 x sec x p sin x cos x P , we see that I = i π/ 4 0 { 3 sec 2 x sec x tan x } dx . But d dx tan x = sec 2 x , while d dx sec x = sec x tan x . Consequently, I = b 3 tan x sec x B π/ 4 0 = 4 2 .
So (dds785) – HW09 – radin – (54915) 3 005 10.0 points Evaluate the defnite integral I = i 1 0 6 (1 + x 2 ) 3 / 2 dx . 1. I = 6 2. I = 3 2 2 3. I = 3 2 4. I = 3 2 correct 5. I = 6 2 6. I = 3 Explanation: Set x = tan u . Then dx = sec 2 u du, (1 + x 2 ) 3 / 2 = sec 3 u , while x = 0 = u = 0 , x = 1 = u = π 4 . Thus I = i π/ 4 0 6 sec 2 u sec 3 u du = 6 i π/ 4 0 cos u du = 6 b sin u B π/ 4 0 . Consequently, I = 3 2 . 006 10.0 points Determine the integral I = i sin - 1 x x dx .

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solution10_pdf - So(dds785 HW09 radin(54915 This print-out...

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