solution_pdf8 - Horstman(mdh995 – HW08 – Radin...

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Unformatted text preview: Horstman (mdh995) – HW08 – Radin – (58505) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the indefinite integral I = integraldisplay 1 √ x 2 − 4 x − 21 dx 1. I = ln vextendsingle vextendsingle vextendsingle x − 2 + radicalbig x 2 − 4 x − 21 vextendsingle vextendsingle vextendsingle + C correct 2. I = sin- 1 parenleftBig x − 2 5 parenrightBig + C 3. I = ln vextendsingle vextendsingle vextendsingle x − 2 + radicalbig x 2 + 4 x − 21 vextendsingle vextendsingle vextendsingle + C 4. I = ln vextendsingle vextendsingle vextendsingle x + 2 + radicalbig x 2 − 4 x − 21 vextendsingle vextendsingle vextendsingle + C 5. I = ln vextendsingle vextendsingle vextendsingle x + 2 + radicalbig x 2 + 4 x − 21 vextendsingle vextendsingle vextendsingle + C 6. I = sin- 1 parenleftBig x − 5 2 parenrightBig + C Explanation: By completing the square we see that x 2 − 4 x − 21 = ( x 2 − 4 x + 4 ) − 25 = ( x − 2) 2 − 25 . This suggests the substitution x − 2 = 5 sec θ , for then dx = 5 sec θ tan θ dθ , while ( x − 2) 2 − 25 = 25tan 2 θ . In this case I = integraldisplay 5 sec θ tan θ 5 tan θ dθ = integraldisplay sec θ dθ = ln | sec θ + tan θ | + C . Now x − 2 = 5 sec θ = ⇒ tan θ = radicalbig ( x − 2) 2 − 25 5 , so I = ln vextendsingle vextendsingle vextendsingle x − 2 + radicalbig ( x − 2) 2 − 25 5 vextendsingle vextendsingle vextendsingle + C . Consequently I = ln vextendsingle vextendsingle vextendsingle x − 2 + radicalbig x 2 − 4 x − 21 vextendsingle vextendsingle vextendsingle + C . 002 10.0 points Evaluate the definite integral I = integraldisplay 2 1 x 2 + 2 x + 1 dx . Correct answer: 1 . 7164. Explanation: After division x 2 + 2 x + 1 = ( x 2 − 1) + 3 x + 1 = x 2 − 1 x + 1 + 3 x + 1 = x − 1 + 3 x + 1 . In this case I = integraldisplay 2 1 parenleftBig x − 1 + 3 x + 1 parenrightBig dx = bracketleftBig 1 2 x 2 − x + 3 ln | x + 1 | bracketrightBig 2 1 = parenleftBig 1 − 1 2 parenrightBig + 3 parenleftBig ln 3 − ln 2 parenrightBig . Consequently, I = 1 2 + 3 ln 3 2 = 1 . 7164 . Horstman (mdh995) – HW08 – Radin – (58505) 2 003 10.0 points Evaluate the integral I = integraldisplay π/ 4 sec 2 x { 3 + sin x } dx . 1. I = 4 + √ 2 2. I = 2 − √ 2 3. I = 2 − 1 2 √ 2 4. I = 4 − 1 2 √ 2 5. I = 2 + √ 2 correct 6. I = 4 + 1 2 √ 2 Explanation: Since sec 2 x { 3 + sin x } = 3 sec 2 x + sec x parenleftBig sin x cos x parenrightBig , we see that I = integraldisplay π/ 4 { 3 sec 2 x + sec tan x } dx . But d dx tan x = sec 2 x , while d dx sec x = sec x tan x ....
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solution_pdf8 - Horstman(mdh995 – HW08 – Radin...

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