Cal Rv 2 - baek (sb32676) Review2 gogolev (57440) This...

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baek (sb32676) – Review2 – gogolev – (57440) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Determine the integral I = i 3 2 0 6 9 - x 2 dx . 1. I = π correct 2. I = 1 3. I = 2 π 4. I = 2 5. I = 3 2 π 6. I = 3 2 Explanation: Since i 1 1 - x 2 dx = sin 1 x + C , we need to reduce I to an integal oF this Form by changing the x variable. Indeed, set x = 3 u . Then dx = 3 du while x = 0 = u = 0 and x = 3 2 = u = 1 2 . In this case I = 18 i 1 / 2 0 1 3 1 - u 2 du = 6 i 1 / 2 0 1 1 - u 2 du . Consequently, I = b 6 sin 1 u B 1 / 2 0 = π . keywords: 002 10.0 points ±ind the value oF the defnite integral I = i π/ 6 0 5 cos x 1 + 4 sin x dx . 1. I = 5 4 ln p 4 P 2. I = 5 2 ln p 3 P 3. I = 5 4 ln p 3 P correct 4. I = 5 2 ln p 4 P 5. I = 5 4 ln p 5 P 6. I = 5 2 ln p 5 P Explanation: Since the integrand has the Form cos x f (4 sin x ) , f ( x ) = 5 1 + x , the substitution u = 4 sin x is suggested. ±or then du = 4 cos x dx, while x = 0 = u = 0 , x = π 6 = u = 2 . In this case I = 5 4 i 2 0 1 1 + u du = 5 4 b ln | 1 + u | B 2 0 .
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baek (sb32676) – Review2 – gogolev – (57440) 2 Consequently, I = 5 4 ln p 3 P . 003 10.0 points Determine f xx + f yx when f ( x, y ) = 2 x 2 + xy 3 - 5 y 2 + 4 . 1. f xx + f yx = 2 xy - 10 + 2 y 2 2. f xx + f yx = 2 + y 3. f xx + f yx = 4 + 3 y 2 correct 4. f xx + f yx = 4 x + y 2 - 10 y 5. f xx + f yx = 4 x + 4 + y 3 Explanation: Using the law for diFerentiation of mono- mials we see that f x = 4 x + y 3 , f xx = 4 , while f y = 3 xy 2 - 10 y, f yx = 3 y 2 . Consequently, f xx + f yx = 4 + 3 y 2 . 004 10.0 points Evaluate the de±nite integral I = i π/ 4 0 3 cos x + 5 sin x cos 3 x dx . 1. I = 1 2 2. I = 11 2 correct 3. I = 17 4 4. I = 8 5. I = 7 4 Explanation: After division we see that 3 cos x + 5 sin x cos 3 x = 3 sec 2 x + 5 tan x sec 2 x = (3 + 5 tan x ) sec 2 x . Thus I = i π/ 4 0 (3 + 5 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx , while x = 0 = u = 0 , x = π 4 = u = 1 . In this case I = i 1 0 (3 + 5 u ) du = b 3 u + 5 2 u 2 B 1 0 . Consequently, I = 11 2 . 005 10.0 points ²ind the volume of the solid under the graph of f ( x, y ) = 1 + 6 x 2 + 4 y and above the rectangle A = ± ( x, y ) : 1 x 2 , 0 y 5 ² . 1. volume = 123 cu.units
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baek (sb32676) – Review2 – gogolev – (57440) 3 2. volume = 125 cu.units correct 3. volume = 130 cu.units 4. volume = 132 cu.units 5. volume = 124 cu.units Explanation: The volume is given by the double integral V = i 5 0 i 2 1 (1 + 6 x 2 + 4 y )
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Cal Rv 2 - baek (sb32676) Review2 gogolev (57440) This...

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