w03s1 - Math 116 Exam I Department of Mathematics...

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Unformatted text preview: Math 116 - Exam I Department of Mathematics University of Michigan February 6, 2003 Name: SOLUTIONS Instructor: Section Number: 1. Do not open this exam until you are told to begin. 2. This exam has 10 pages including this cover. There are 10 questions. Note that the problems are not of equal difficulty, and it may be to your advantage to skip over and come back to a problem on which you are stuck. 3. Do not separate the pages of the exam. If any pages do become separated, write your name on them and point them out to your instructor when you turn in the exam. 4. Please read the instructions for each individual exercise carefully. One of the skills being tested on this exam is your ability to interpret questions, so instructors will not answer questions about exam problems during the exam. 5. Show an appropriate amount of work (including appropriate explanation) for each exercise so that the graders can see not only the answer but also how you obtained it. Include units in your answers where appropriate. 6. You may use any calculator except a TI-92 (or other calculator with a full numeric keypad). However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a 3” by 5” notecard. 7. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of the graph, and to write out the entries of the table that you use. 8. Turn off all cell phones and pagers , and remove all headphones. PROBLEM POINTS SCORE 1 12 2 6 3 12 4 6 5 10 6 10 7 10 8 10 9 14 10 10 TOTAL 100 2 1. (3 pts each) Circle true or false. No explanation necessary. (a) If 0 < f ( x ) < g ( x ) for all x , then integraltext 4 1 g ( x ) f ( x ) > 3. True integraldisplay 4 1 g ( x ) f ( x ) dx > integraldisplay 4 1 f ( x ) f ( x ) dx = integraldisplay 4 1 1 dx = x | 4 1 = 4- 1 = 3 . (b) integraltext π π sin 2 ( x 5 ) − cos(42 x ) x 2 + x +1 > 2. False The integral is from π to π , so it’s 0. (c) integraltext ∞ 1 1 x 71 dx converges. True integraldisplay ∞ 1 1 x 71 dx = lim b →∞ integraldisplay b 1 x − 71 dx = lim b →∞- 1 70 x − 70 vextendsingle vextendsingle vextendsingle vextendsingle b 1 = lim b →∞- 1 70 b 70 + 1 70 = 1 70 . (d) If g ( x ) + 2 is an antiderivative of f ( x ), then g ( x ) is also an antiderivative of f ( x )....
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w03s1 - Math 116 Exam I Department of Mathematics...

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