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Final Exam

# Final Exam - Version 189 FINAL EXAM sachse(56620 This...

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Version 189 – FINAL EXAM – sachse – (56620) 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 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 0 (5 + x ) 2 dx 2. I = integraldisplay 0 4(5 + x ) 2 dx 3. I = integraldisplay 0 (5 + x 2 ) dx 4. I = integraldisplay 2 0 (5 + x 2 ) dx 5. I = integraldisplay 1 0 4(5 + x 2 ) dx 6. I = integraldisplay 1 0 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 , Δ x = 1 n , f ( x ) = 4(5 + x ) 2 , [ a, b ] = [0 , 1] . Consequently, I = integraldisplay 1 0 4(5 + x ) 2 dx .

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Version 189 – FINAL EXAM – sachse – (56620) 2 003 10.0 points Use properties of integrals to determine the value of I = integraldisplay 0 4 f ( x ) dx when integraldisplay 6 0 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 -1 0 1 2 3 4 5 6 7 8 9 10 11 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) 3 for g . This shows that the maximum value of g occurs at x = 5 since the sign of g changes from positive to negative at x = 5. 005 10.0 points Determine if the limit lim x 0 sin 1 3 x 2 x exists, and if it does, find its value.

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Final Exam - Version 189 FINAL EXAM sachse(56620 This...

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