Exam 1_1 - Tamayo, Nubia Exam 1 Due: Feb 20 2007, 11:00 pm...

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Tamayo, Nubia – Exam 1 – Due: Feb 20 2007, 11:00 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 be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points When 2 4 6 8 10 2 4 - 2 - 4 is the graph o± a ±unction f , use rectangles to estimate the defnite integral I = Z 10 0 | f ( x ) | dx by subdividing [0 , 10] into 10 equal subin- tervals and taking right endpoints o± these subintervals. 1. I 19 2. I 20 3. I 23 correct 4. I 21 5. I 22 Explanation: The defnite integral I = Z 10 0 | f ( x ) | dx is the area between the graph o± f and the interval [0 , 10]. The area is estimated using the gray-shaded rectangles in 2 4 6 8 10 2 4 - 2 - 4 Each rectangle has base-length 1; and it’s height can be read o² ±rom the graph. Thus Area = 5 + 4 + 1 + 1 + 1 + 3 + 4 + 2 + 2 . Consequently, I 23 . keywords: defnite integrals, graph, absolute value 002 (part 1 o± 1) 10 points For which integral, I , is the expression 1 15 ˆ r 1 15 + r 2 15 + r 3 15 + . . . + r 15 15 ! a Riemann sum approximation. 1. I = 1 15 Z 15 0 x dx 2. I = Z 1 0 x dx correct 3. I = 1 15 Z 1 0 r x 15 dx 4. I = Z 1 0 r x 15 dx 5. I = 1 15 Z 1 0 x dx
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Tamayo, Nubia – Exam 1 – Due: Feb 20 2007, 11:00 pm – Inst: Gary Berg 2 Explanation: When the interval [ a, b ] is divided into n equals intervals, then n X i = 1 f µ a + i ( b - a ) n b - a n is a Riemann sum approximation for the inte- gral Z b a f ( x ) dx of f over [ a, b ] using right endpoints as sample points. Comparing this with 1 15 ˆ r 1 15 + r 2 15 + r 3 15 + . . . + r 15 15 ! we see that [ a, b ] = [0 , 1], n = 15, and [ a, b ] = [0 , 1] , n = 15 , f ( x ) = x . Consequently, the given expression is a Rie- mann sum for I = Z 1 0 x dx . keywords: integral, Riemann sum, square root 003 (part 1 of 1) 10 points If F ( x ) = Z x 0 4 e 6 sin 2 θ dθ , ±nd the value of F 0 ( π/ 4). 1. F 0 ( π/ 4) = 3 e 4 2. F 0 ( π/ 4) = 3 e 3 3. F 0 ( π/ 4) = 3 e 6 4. F 0 ( π/ 4) = 4 e 6 5. F 0 ( π/ 4) = 4 e 3 correct Explanation: By the Fundamental theorem of calculus, F 0 ( x ) = 4 e 6 sin 2 x . At x = π/ 4, therefore, F 0 ( π/ 4) = 4 e 3 since sin( π 4 ) = 1 2 . keywords: integral, FTC 004 (part 1 of 1) 10 points Evaluate the de±nite integral I = Z π 8 0 (4 cos 4 x - sin 4 x ) dx . 1.
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This note was uploaded on 09/23/2009 for the course M 57420 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.

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Exam 1_1 - Tamayo, Nubia Exam 1 Due: Feb 20 2007, 11:00 pm...

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