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Unformatted text preview: Version 010 Review 1 Odell (58340) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the area of the region enclosed by the graphs of f ( x ) = 3 + x x 2 , g ( x ) = x + 1 , on the interval [0 , 1]. 1. Area = 2 sq. units 2. Area = 7 3 sq. units 3. Area = 8 3 sq. units 4. Area = 5 3 sq. units correct 5. Area = 4 3 sq. units Explanation: The graph of f is a parabola opening down wards, and the graph of g is a straight line. On the other hand, f (0) = 3 1 = g (0) , while f (1) = 3 2 = g (1) . Thus f ( x ) g ( x ) on [0 , 1], so the area be tween the graphs of f and g on [0 , 1] is given by integraldisplay 1 ( f ( x ) g ( x )) dx = integraldisplay 1 (2 x 2 ) dx . Consequently, the region has Area = bracketleftBig 2 x 1 3 x 3 bracketrightBig 1 = 5 3 sq. units . keywords: integral, area 002 10.0 points Determine the definite integral I to which the Riemann sum n summationdisplay k =1 parenleftbigg 6 + k n parenrightbigg 2 parenleftbigg 4 n parenrightbigg converges as n . 1. I = integraldisplay 2 (6 + x 2 ) dx 2. I = integraldisplay 4(6 + x ) 2 dx 3. I = integraldisplay 1 4(6 + x 2 ) dx 4. I = integraldisplay 1 4(6 + x ) 2 dx correct 5. I = integraldisplay 2 (6 + x ) 2 dx 6. I = integraldisplay (6 + x 2 ) dx Explanation: By definition, integraldisplay b a f ( x ) dx = lim n n summationdisplay k = 1 f ( x k ) x where x k is a sample point in the subinterval [ x k 1 , x k ] of [ a, b ]. For the given sum n summationdisplay k =1 parenleftbigg 6 + k n parenrightbigg 2 parenleftbigg 4 n parenrightbigg , therefore, x k = k n , x = 1 n , f ( x ) = 4(6 + x ) 2 , [ a, b ] = [0 , 1] . Consequently, I = integraldisplay 1 4(6 + x ) 2 dx . Version 010 Review 1 Odell (58340) 2 003 10.0 points Determine the integral I = integraldisplay 1 x ( x + 5) 4 dx. 1. I = 1 5( x + 5) 5 + C 2. I = 2 3( x + 5) 3 + C correct 3. I = 1 5( x + 5) 5 + C 4. I = 2 5( x 5) 5 + C 5. I = 2 3( x + 5) 3 + C 6. I = 1 3( x 5) 3 + C Explanation: Set u = x + 5. Then du = 1 2 x dx, so I = 2 integraldisplay 1 u 4 du = 2 3 1 u 3 + C . Consequently, I = 2 3( x + 5) 3 + C with C an arbitrary constant. 004 10.0 points The base of a solid S is the bounded region enclosed by the graphs of y = x 2 , y = 3 . Determine the volume of S if crosssections perpendicular to the yaxis are equilateral tri angles. 1. volume = 9 4 2 cu. units 2. volume = 9 4 cu. units 3. volume = 9 2 cu. units 4. volume = 9 2 3 cu. units correct 5. volume = 9 2 2 cu. units 6. volume = 9 4 3 cu. units Explanation: A view of the solid together with a cross section is shown in the figure x y z Now the length of the base of an equilateral triangular crosssection is specified by its po sition on the parabola, so its value is given s = 2 x = 2 y ....
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This note was uploaded on 04/15/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Cepparo

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