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Unformatted text preview: Version 078 L EXAM 1 kalahurka (55230) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A function h has graph 2 4 2 2 4 on ( 3 , 4). If f ( x ) = integraldisplay x 2 h ( t ) dt, ( x  2) , which of the following is the graph of f on ( 3 , 4)? 1. 2 4 2 2 2 2. 2 4 2 2 2 3. 2 4 2 2 2 correct 4. 2 4 2 2 2 5. 2 4 2 2 2 6. 2 4 2 2 2 Explanation: Since f ( x ) is defined only for x  2, there will be no graph of f on the interval ( 3 , 2). Version 078 L EXAM 1 kalahurka (55230) 2 This already eliminates two of the possible graphs. On the other hand, f ( 2) = integraldisplay 2 2 h ( x ) dx = 0 , eliminating two more graphs. Finally, by the Fundamental Theorem of Calculus, f ( x ) = h ( x ) on ( 2 , 4), so f ( x ) will be increasing on any interval on which h > 0, and decreasing on any interval on which h < 0. To determine which of the remaining two possible graphs is the graph of f we thus have to look at the sign of h on an interval and the slope of a possible graph of f on that interval. Consequently, the only possible graph of f is 2 4 2 2 2 002 10.0 points Determine g ( x ) when g ( x ) = integraldisplay 3 x 2 t 2 sec t dt . 1. g ( x ) = 4 x sec x tan x 2. g ( x ) = 2 x 2 tan x 3. g ( x ) = 2 x 2 tan x 4. g ( x ) = 2 x 2 sec x 5. g ( x ) = 4 x sec 2 x 6. g ( x ) = 4 x sec 2 x 7. g ( x ) = 2 x 2 sec x correct 8. g ( x ) = 4 x sec x tan x Explanation: By Properties of integrals and the Funda mental Theorem of Calculus, d dx parenleftBig integraldisplay a x f ( t ) dt parenrightBig = d dx parenleftBig integraldisplay x a f ( t ) dt parenrightBig = f ( x ) . When g ( x ) = integraldisplay 3 x f ( t ) dt , f ( t ) = 2 t 2 sec t , therefore, g ( x ) = 2 x 2 sec x . 003 10.0 points Evaluate the integral I = integraldisplay 4 3 8 x 2 6 x + 10 dx . 1. I = 4 2. I = 3. I = 2 correct 4. I = 4 5. I = 2 6. I = 1 Explanation: By completing the square we see that x 2 6 x + 10 = ( x 2 6 x + 9) + 10 9 = ( x 3) 2 + 1 . In this case I = integraldisplay 4 3 8 ( x 3) 2 + 1 . Version 078 L EXAM 1 kalahurka (55230) 3 Now set x 3 = u . For then dx = du , while x = 3 = u = 0 , x = 4 = u = 1 , and so I = 8 integraldisplay 1 1 1 + u 2 du = 8 bracketleftBig tan 1 u bracketrightBig 1 ....
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This note was uploaded on 03/26/2012 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas at Austin.
 Spring '10
 ZHENG
 Differential Equations, Equations

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