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Unformatted text preview: Cheung, Anthony Exam 2 Due: Nov 7 2006, 11:00 pm Inst: David Benzvi 1 This printout should have 16 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 If f is the function whose graph on [0 , 10] is given by 2 4 6 8 2 4 6 8 use the Trapezoidal Rule with n = 5 to esti mate the definite integral I = Z 8 3 f ( x ) dx. 1. I 23 2. I 47 2 3. I 22 4. I 24 correct 5. I 45 2 Explanation: The Trapezoidal Rule estimates the definite integral I = Z 8 3 f ( x ) dx by I 1 2 h f (3) + 2 { f (4)+ + f (7) } + f (8) i when n = 5. For the given f , therefore, I 1 2 h 7 + 2 { 5 + 5 + 5 + 4 } + 3 i = 24 , reading off the values of f from the graph. keywords: trapezoidal rule, integral, graph 002 (part 1 of 1) 10 points Find Z e 7 x 9 + e 14 x dx. 1. 1 3 arcsin e 7 x + C 2. arcsin e 7 x + C 3. 1 21 arctan 1 3 e 7 x + C correct 4. 1 3 + e 7 x + C 5. None of these. 6. 1 3 arcsec e 7 x + e 7 x + C Explanation: Z e 7 x 9 + e 14 x dx = 1 7 Z 7 e 7 x dx (3) 2 + ( e 7 x ) 2 = 1 7 Z d ( e 7 x ) (3) 2 + ( e 7 x ) 2 = 1 7 1 3 arctan 1 3 e 7 x + C = 1 21 arctan 1 3 e 7 x + C keywords: exponential function, inverse trig function 003 (part 1 of 1) 10 points Cheung, Anthony Exam 2 Due: Nov 7 2006, 11:00 pm Inst: David Benzvi 2 Evaluate the definite integral I = Z 1 x 11 x 2 x 2 dx. 1. I = 7 ln 3 2. I = ln 3 3. I = 7 ln 3 4. I = 7 ln 2 correct 5. I = ln 3 6. I = ln 2 7. I = 7 ln 2 8. I = ln 2 Explanation: After factorization x 2 x 2 = ( x + 1)( x 2) . But then by partial fractions, x 11 x 2 x 2 = 4 x + 1 3 x 2 . Now Z 1 4 x + 1 dx = h 4 ln  ( x + 1)  i 1 = 4 ln 2 , while Z 1 3 x 2 dx = h 3 ln  ( x 2)  i 1 = 3 ln 2 . Consequently, I = 7 ln 2 . keywords: definite integral, rational function, partial fractions, natural log 004 (part 1 of 1) 10 points Evaluate the definite integral I = Z / 4 4 cos x + 6 sin x cos 3 x dx. 1. I = 11 2 2. I = 5 2 3. I = 10 4. I = 7 correct 5. I = 1 Explanation: After division we see that 4 cos x + 6 sin x cos 3 x = 4 sec 2 x + 6 tan x sec 2 x = (4 + 6 tan x ) sec 2 x. Thus I = Z / 4 (4 + 6 tan x ) sec 2 xdx. Let u = tan x ; then du = sec 2 xdx, while x = 0 = u = 0 , x = 4 = u = 1 . In this case I = Z 1 (4 + 6 u ) du = 4 u + 3 u 2 / 1 ....
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 Spring '08
 RAdin
 Calculus

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