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int cal exam 2

# int cal exam 2 - Version 084 EXAM 2 Radin(57410 This...

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Version 084 – EXAM 2 – Radin – (57410) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x → ∞ (ln x ) 2 2 x + 6 ln x exists, and if it does, find its value. 1. limit = 2 2. none of the other answers 3. limit = 0 correct 4. limit = 8 5. limit = -∞ 6. limit = Explanation: Use of L’Hospital’s Rule is suggested, Set f ( x ) = (ln x ) 2 , g ( x ) = 2 x + 6 ln x . Then f, g have derivatives of all orders and lim x → ∞ f ( x ) = , lim x → ∞ g ( x ) = . Thus L’Hospital’s Rule applies: lim x → ∞ f ( x ) g ( x ) = lim x → ∞ f ( x ) g ( x ) . But f ( x ) = 2 ln x x , g ( x ) = 2 + 6 x , so lim x → ∞ f ( x ) g ( x ) = lim x → ∞ 2 ln x 2 x + 6 . We need to apply L’Hospital once again, for then lim x → ∞ 2 ln x 2 x + 6 = lim x → ∞ 2 x slashBig 2 = 0 . Consequently, the limit exists and lim x → ∞ (ln x ) 2 2 x + 6 ln x = 0 . 002 10.0 points Evaluate the definite integral I = integraldisplay π/ 3 0 sec x tan x 5 + sec x dx . 1. I = - ln 7 6 2. I = ln 6 5 3. I = ln 7 10 4. I = ln 7 6 correct 5. I = - ln 6 5 6. I = - ln 7 10 Explanation: Since d dx (sec x ) = sec x tan x , use of the substitution u = 5 + sec x is suggested. For then du = sec x tan xdx , while x = 0 = u = 6 , x = π 3 = u = 7 . Thus I = integraldisplay 7 6 1 u du = bracketleftBig ln u bracketrightBig 7 6 .

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Version 084 – EXAM 2 – Radin – (57410) 2 Consequently, I = ln 7 6 . 003 10.0 points Evaluate the integral I = integraldisplay 1 0 1 4 - x 2 dx . 1. I = 1 3 2. I = 1 6 3. I = 1 4 π 4. I = 1 4 5. I = 1 6 π correct 6. I = 1 3 π Explanation: Set x = 2 sin u ; then dx = 2 cos u du and 4 - x 2 = 4(1 - sin 2 u ) = 2 cos 2 u , while x = 0 = u = 0 , x = 1 = u = π 6 . In this case I = integraldisplay π/ 6 0 cos u cos u du = integraldisplay π/ 6 0 du . Consequently I = 1 6 π . 004 10.0 points Determine the integral I = integraldisplay 2 x (3 + 2 ln x ) 3 dx. 1. I = - 1 4 (3 + 2 ln x ) 4 + C 2. I = - 1 4 ln x (3 + 2 ln x ) 2 + C 3. I = 1 2 ln x (3 + 2 ln x ) 2 + C 4. I = 1 4 ln x (3 + 2 ln x ) 2 + C 5. I = 1 4 (3 + 2 ln x ) 4 + C correct 6. I = - 1 2 (3 + 2 ln x ) 4 + C 7. I = 1 2 (3 + 2 ln x ) 4 + C 8. I = - 1 2 ln x (3 + 2 ln x ) 2 + C Explanation: Set u = 3 + 2 ln x . Then du = 2 x dx, so I = integraldisplay u 3 du = 1 4 u 4 + C . Consequently, I = 1 4 (3 + 2 ln x ) 4 + C with C an arbitrary constant. 005 10.0 points Find the value of the definite integral I = integraldisplay 1 2 0 8 sin 1 x 1 - x 2 dx .
Version 084 – EXAM 2 – Radin – (57410) 3 1. I = 1 3 π 2 2. I = 4 25 π 2 3. I = 4 9 π 2 4. I = 1 4 π 2 5. I = 1 9 π 2 correct Explanation: Since integraldisplay 1 1 - x 2 dx = sin 1 x + C , this suggests the substitution u = sin 1 x , for then du = 1 1 - x 2 dx , while x = 0 = u = 0 ,

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int cal exam 2 - Version 084 EXAM 2 Radin(57410 This...

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