homework 3 solutions - Mcbiles, Emily Homework 3 Due: Sep...

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Mcbiles, Emily – Homework 3 – Due: Sep 19 2006, 3:00 am – Inst: David Fonken 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points The graph o± f is shown in the fgure 2 4 6 8 2 4 - 2 I± the ±unction g is defned by g ( x ) = Z x 2 f ( t ) dt, ±or what value o± x does g ( x ) have a maxi- mum? 1. x = - 8 2. not enough in±ormation given 3. x = 3 . 5 4. x = - 2 5. x = 6 correct Explanation: By the Fundamental theorem o± calculus, i± g ( x ) = Z x 2 f ( t ) dt, then g 0 ( x ) = f ( x ). Thus the critical points g occur at the zeros o± f , i.e. , at the x - intercepts o± the graph o± f . To determine which o± these gives a local maximum o± g we use the sign chart g 0 + - 2 6 8 ±or g 0 . This shows that max g ( x ) at x = 6 , since the sign o± g 0 changes ±rom positive to negative at x = 6. keywords: FTC, integral, sign chart, maxi- mum 002 (part 1 o± 1) 10 points Find g 0 ( x ) when g ( x ) = Z x π (6 + cos t ) dt . 1. g 0 ( x ) = 6 x + sin x 2. g 0 ( x ) = 6 - sin x 3. g 0 ( x ) = 6 + cos x correct 4. g 0 ( x ) = - sin x 5. g 0 ( x ) = 6 x - cos x Explanation: By the Fundamental theorem o± Calculus, g ( x ) = Z x a f ( t ) dt , then g 0 ( x ) = d dx Z x a g ( t ) dt = f ( x ) . In the given example, there±ore, g 0 ( x ) = 6 + cos x . keywords:
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2 003 (part 1 of 1) 10 points If F ( x ) = Z x 0 2 e 12 sin 2 θ dθ , ±nd the value of F 0 ( π/ 4). 1. F 0 ( π/ 4) = 2 e 6 correct 2. F 0 ( π/ 4) = 6 e 12 3. F 0 ( π/ 4) = 6 e 6 4. F 0 ( π/ 4) = 6 e 2 5. F 0 ( π/ 4) = 2 e 12 Explanation: By the Fundamental theorem of calculus, F 0 ( x ) = 2 e 12 sin 2 x . At x = π/ 4, therefore, F 0 ( π/ 4) = 2 e 6 since sin( π 4 ) = 1 2 . keywords: integral, FTC 004 (part 1 of 1) 10 points Evaluate the de±nite integral I = Z π/ 2 0 (3 cos x + sin x ) dx . 1. I = 0 2. I = 2 3. I = 3 4. I = 4 correct 5. I = 1 Explanation: By the Fundamental Theorem of Calculus, I = h F ( x ) i π/ 2 0 = F ( π 2 ) - F (0) for any anti-derivative F of f ( x ) = 3 cos x + sin x . Taking F ( x ) = 3 sin x - cos x and using the fact that cos 0 = sin π 2 = 1 , sin 0 = cos π 2 = 0 , we thus see that I = 4 . keywords: integral, FTC, trig function 005 (part 1 of 1) 10 points Evaluate the de±nite integral I = Z 1 0 (6 + 9 x - 7 x 2 ) dx . Correct answer: 8 . 16667 . Explanation: By linearity, Z 1 0 (6 + 9 x - 7 x 2 ) dx = 6 Z 1 0 dx + 9 Z 1 0 x dx - 7 Z 1 0 x 2 dx . But
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This note was uploaded on 03/19/2008 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

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homework 3 solutions - Mcbiles, Emily Homework 3 Due: Sep...

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