Exam1Review_Answers - Miller, Kierste Review 1 Due: May 6...

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Miller, Kierste – Review 1 – Due: May 6 2007, 6:00 pm – Inst: Gary Berg 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points IF the graph oF f passes through the point (1 , - 4) and the slope oF the tangent line at ( x, f ( x )) is 8 x - 9, fnd the value oF f (2). 1. f (2) = 2 2. f (2) = 1 3. f (2) = - 2 4. f (2) = - 1 correct 5. f (2) = 0 Explanation: By defnition f 0 ( x ) = 8 x - 9 . Thus f ( x ) = 4 x 2 - 9 x + C with C an arbitrary constant. But iF the graph oF f passes through (1 , - 4), then f (1) = 4 - 9 + C = - 4 . Consequently, f ( x ) = 4 x 2 - 9 x + 1 . At x = 2, thereFore, f (2) = - 1 . keywords: antiderivative, tangent line, Func- tion value 002 (part 1 oF 1) 10 points ±ind the value oF the indefnite integral I = Z 1 + x 1 - x 2 dx . 1. I = sin - 1 x - p 1 - x 2 + C correct 2. I = tan - 1 x + p 1 - x 2 + C 3. I = 1 2 sin - 1 x + p 1 - x 2 + C 4. I = 1 2 tan - 1 x + 1 2 p 1 - x 2 + C 5. I = 1 2 tan - 1 x - p 1 - x 2 + C 6. I = sin - 1 x - 1 2 p 1 - x 2 + C Explanation: We deal with the two integrals I 1 = Z 1 1 - x 2 dx, I 2 = x 1 - x 2 dx separately. Now d dx sin - 1 x = 1 1 - x 2 , so we see that I 1 = sin - 1 x + C . On the the other hand, to evaluate I 2 set u = 1 - x 2 . Then du = - 2 x dx , and so I 2 = - 1 2 Z u - 1 / 2 du = - u + C . Consequently, I = sin - 1 x - p 1 - x 2 + C with C an arbitrary constant. keywords: 003 (part 1 oF 1) 10 points
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Miller, Kierste – Review 1 – Due: May 6 2007, 6:00 pm – Inst: Gary Berg 2 Evaluate the defnite integral I = Z ln 3 0 e x 4 + e x dx . 1. I = 7 - 5 2. I = 7 - 4 3. I = 1 2 7 - 5 · 4. I = 1 2 7 - 4 · 5. I = 2( 7 - 5) correct 6. I = 2( 7 - 4) Explanation: Set u = 4 + e x . Then du = e x dx, while x = 0 = u = 5 , x = ln 3 = u = 7 . In this case I = Z 7 5 1 u du = 2 h u i 7 5 . Consequently, I = 2( 7 - 5) . keywords: 004 (part 1 oF 1) 10 points Evaluate the integral I = Z π/ 3 0 4 cos θ - 3 cos 2 θ · dθ . 1. I = - 1 3 correct 2. I = - 7 3 3. I = - 5 3 4. I = 1 3 5. I = 5 3 6. I = 7 3 Explanation: Since 1 cos 2 θ = sec 2 θ , d tan θ = sec 2 θ , we see that I = Z π/ 3 0 4 cos θ - 3 sec 2 θ · = h 4 sin θ - 3 tan θ i π/ 3 0 = 2 3 - 3 3 . Consequently, I = - 1 3 . keywords: integral, trig Function 005 (part 1 oF 1) 10 points ±ind the value oF the integral I = Z 5 0 f f 4 x - x 2 f f dx . 1. I = 13 correct 2. I = 38 3 3. I = 79 6 4. I = 77 6 5. I = 25 2 Explanation: The graph oF f ( x ) = 4 x - x 2 is a parabola
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Miller, Kierste – Review 1 – Due: May 6 2007, 6:00 pm – Inst: Gary Berg 3 4 5 (not drawn to scale) which lies above the x - axis on [0 ,
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This note was uploaded on 05/08/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.

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Exam1Review_Answers - Miller, Kierste Review 1 Due: May 6...

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