CalcReview1 - Version 010 Review 1 Odell (58340) 1 This...

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Unformatted text preview: Version 010 Review 1 Odell (58340) 1 This print-out should have 19 questions. Multiple-choice 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 cross-sections perpendicular to the y-axis 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 cross-section 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.

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CalcReview1 - Version 010 Review 1 Odell (58340) 1 This...

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