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exam1_solnfall06

# exam1_solnfall06 - Math 116 First Exam Name Instructor 1 Do...

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Math 116 — First Exam October 11, 2006 Name: Instructor: Date: 1. Do not open this exam until you are told to do so. 2. This exam has 10 pages including this cover. There are 9 problems. 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 this exam. If they do become separated, write your name on every page and point this out to your instructor when you hand in the exam. 4. Please read the instructions for each individual problem carefully. One of the skills being tested on this exam is your ability to interpret mathematical 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 problem so that graders can see not only your answer but how you obtained it. Include units in your answer where that is appropriate. 6. You may use any calculator except a TI-92 (or other calculator with a full keyboard). 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 00 × 5 00 note card. 7. If you use graphs or tables to find an answer, be sure 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 12 3 12 4 8 5 8 6 10 7 16 8 12 9 10 Total 100

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Math 116 / Exam 1 (October 11, 2006) page 2 g ( x ) 1 2 3 4 1 2 1. [12 points] For all of parts (a)–(d), let f ( x ) = 2 x - 4 and let g ( x ) be given in the graph to the right. (a) [3 points of 12] Find R 5 1 g 0 ( x ) dx . Solution : By the Fundamental Theorem of Calculus, Z 5 0 g 0 ( x ) dx = g (5) - g (1) = 1 - (2) = - 1 . Alternately, note that g 0 ( x ) = 3 for 0 < x < 1, g 0 ( x ) = - 1 2 for 1 < x < 3, and g 0 ( x ) = 0 for x > 3. Thus R 5 1 g 0 ( x ) dx = R 3 1 - 1 2 dx = - 1 . (b) [3 points of 12] Find R 5 0 g ( x ) dx . Solution : This integral is just the area between the graph of g ( x ) and the x -axis between x = 0 and x = 5, counting area below the axis as negative. This is Z 5 0 g ( x ) dx = - (area between 0 and 2/3) + (area between 0 and 1) + (area between 1 and 3)+ (area between 3 and 5) = - 1 2 ( 1 3 )(1) + 1 2 ( 2 3 )(2) + (3) + (2) = 5 1 2 . (c) [3 points of 12] Find R 4 . 5 2 g ( f ( x )) dx . Solution : Using substitution with w = f ( x ) = 2 x - 4, we have 1 2 dw = dx , so that R 4 . 5 2 g ( f ( x )) dx = 1 2 R w (4 . 5) w (2) g ( w ) dw = 1 2 R 5 0 g ( w ) dw. But the calculation above gives R 5 0 g ( w ) dw = 5 1 2 = 11 2 , so R 4 . 5 2 g ( f ( x )) dx = 1 2 · 11 2 = 11 4 = 2 . 75. (d) [3 points of 12] Find R 5 0 f ( x ) · g 0 ( x ) dx . Solution : Using integration by parts with u = 2 x - 4 and v 0 = g 0 , we have u 0 = 2 and v = g , so that Z 5 0 f ( x ) · g 0 ( x ) dx = (2 x - 4) · g ( x ) fl fl fl fl 5 0 - 2 Z 5 0 g ( x ) dx = (6)(1) - ( - 4)( - 1) - 2( 11 2 ) = 2 - 11 = - 9 . Alternate solution: note that g 0 ( x ) = 3 for 0 < x < 1, g 0 ( x ) = - 1 2 for 1 < x < 3, and g 0 ( x ) = 0 for x > 3. Thus Z 5 0 f ( x ) · g 0 ( x ) dx = Z 1 0 3(2 x - 4) dx + Z 3 1 - 1 2 (2 x - 4) dx = 3( x 2 - 4 x ) fl fl 1 0 + ( - x 2 2 + 2 x ) fl fl 3 1 = - 9 + ( 3 2 - 3 2 ) = - 9 .
Math 116 / Exam 1 (October 11, 2006) page 3 2. [12 points] While working on their team homework, Alex and Chris find that they have evaluated the same integral—but that they each used a different method, and got different answers! Alex found

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