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Unformatted text preview: Version 189 FINAL EXAM sachse (56620) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find f (1) when f ( x ) = 3 x and f ( 1) = 5 , f ( 1) = 5 . 1. f (1) = 11 2. f (1) = 13 correct 3. f (1) = 15 4. f (1) = 12 5. f (1) = 14 Explanation: When f ( x ) = 3 x the most general anti derivative of f is f ( x ) = 3 2 x 2 + C where the arbitrary constant C is determined by the condition f ( 1) = 5. For f ( 1) = 5 = f ( x ) = 3 2 x 2 + 7 2 . But then f ( x ) = 1 2 x 3 + 7 2 x + D, the arbitrary constant D being given by f ( 1) = 5 =  4 + D = 5 . Thus, f ( x ) = 1 2 x 3 + 7 2 x + 9 . Consequently, f (1) = 13 . 002 10.0 points Determine the definite integral I to which the Riemann sum n summationdisplay k =1 parenleftbigg 5 + k n parenrightbigg 2 parenleftbigg 4 n parenrightbigg converges as n . 1. I = integraldisplay 2 (5 + x ) 2 dx 2. I = integraldisplay 4(5 + x ) 2 dx 3. I = integraldisplay (5 + x 2 ) dx 4. I = integraldisplay 2 (5 + x 2 ) dx 5. I = integraldisplay 1 4(5 + x 2 ) dx 6. I = integraldisplay 1 4(5 + x ) 2 dx correct Explanation: By definition, integraldisplay b a f ( x ) dx = lim n n summationdisplay k = 1 f ( x k ) x where x k is a sample point in the subinterval [ x k 1 , x k ] of [ a, b ]. For the given sum n summationdisplay k =1 parenleftbigg 5 + k n parenrightbigg 2 parenleftbigg 4 n parenrightbigg , therefore, x k = k n , x = 1 n , f ( x ) = 4(5 + x ) 2 , [ a, b ] = [0 , 1] . Consequently, I = integraldisplay 1 4(5 + x ) 2 dx . Version 189 FINAL EXAM sachse (56620) 2 003 10.0 points Use properties of integrals to determine the value of I = integraldisplay 4 f ( x ) dx when integraldisplay 6 f ( x ) dx = 7 , integraldisplay 6 4 f ( x ) dx = 1 . 1. I = 8 2. I = 8 3. I = 9 4. I = 6 5. I = 6 correct 6. I = 9 Explanation: Since integraldisplay a b f ( x ) dx = integraldisplay b a f ( x ) dx and integraldisplay b a f ( x ) dx + integraldisplay c b f ( x ) dx = integraldisplay c a f ( x ) dx, we see that integraldisplay a b f ( x ) dx = parenleftBig integraldisplay c a f ( x ) dx integraldisplay c b f ( x ) dx parenrightBig . Consequently, I = (7 1) = 6 . 004 10.0 points The graph of f is shown in the figure 2 4 6 8 10 2 4 6 2 If the function g is defined by g ( x ) = integraldisplay x 1 f ( t ) dt, for what value of x does g ( x ) have a maxi mum? 1. x = 6 2. not enough information given 3. x = 1 4. x = 5 correct 5. x = 7 6. x = 2 . 5 Explanation: By the Fundamental theorem of calculus, if g ( x ) = integraldisplay x 1 f ( t ) dt, then g ( x ) = f ( x ). Thus the critical points of g occur at the zeros of f , i.e. , at the x intercepts of the graph of f . To determine which of these gives a local maximum of g we use the sign chart g + 1 5 7 Version 189 FINAL EXAM sachse (56620)...
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This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas at Austin.
 Fall '09
 GOGOLEV
 Calculus

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