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Unformatted text preview: Version 098 – FINAL – cheng – (58520) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the volume of the paraboloid gener ated by rotating the graph of y = 6 √ x about the xaxis between x = 0 and x = 3. 1. V = 165 π cu.units 2. V = 162 π cu.units correct 3. V = 161 π cu.units 4. V = 164 π cu.units 5. V = 163 π cu.units Explanation: The volume, V , of the solid of revolution generated by rotating the graph of y = f ( x ) about the xaxis between x = a and x = b is given by V = π integraldisplay b a f ( x ) 2 dx. When f ( x ) = 6 √ x and a = 0 , b = 3, there fore, V = π integraldisplay 3 36 x dx = π 2 bracketleftBig 36 x 2 bracketrightBig 3 . Consequently, V = 162 π cu.units . keywords: volume, integral, solid of revolu tion 002 10.0 points Find the bounded area enclosed by the graphs of f and g when f ( x ) = x 2 − x, g ( x ) = 2 x . 1. Area = 11 2 sq. units 2. Area = 7 2 sq. units 3. Area = 9 2 sq. units correct 4. Area = 13 2 sq. units 5. Area = 15 2 sq. units Explanation: The graph of f is a parabola opening up wards, while the graph of g is a straight line of slope 2. Both graphs pass through the origin. Thus the required area is the shaded region in the figure below graph of g graph of f 3 To express the area as a definite integral we need to find where the graphs intersect, i.e . when x 2 − x = 2 x . As the solutions of this equation are x = 0 and x = 3, the area of the shaded region is thus given by Area = integraldisplay 3 ( g ( x ) − f ( x )) dx = integraldisplay 3 (3 x − x 2 ) dx = bracketleftBig 3 2 x 2 − 1 3 x 3 bracketrightBig 3 . Consequently, Area = 9 2 sq.units . Version 098 – FINAL – cheng – (58520) 2 003 10.0 points Evaluate the definite integral I = integraldisplay π/ 3 sec x tan x 5 − sec x dx . 1. I = − ln 7 6 2. I = − ln 4 3 3. I = ln 4 3 correct 4. I = ln 2 5. I = ln 7 6 6. I = − ln 2 Explanation: Since d dx (sec x ) = sec x tan x , use of the substitution u = 5 − sec x is suggested. For then du = − sec x tan xdx , while x = 0 = ⇒ u = 4 , x = π 3 = ⇒ u = 3 . Thus I = − integraldisplay 3 4 1 u du = integraldisplay 4 3 1 u du . Consequently, I = bracketleftBig ln u bracketrightBig 4 3 = ln 4 3 . 004 10.0 points To determine the integral I = integraldisplay 1 ( x 2 + 4) 3 / 2 dx , which one of the following substitutions should be used? 1. x = 2 tan u correct 2. x = 2 sec u 3. x = √ 2 tan u 4. x = √ 2 sec u 5. x = 2 sin u 6. x = √ 2 sin u Explanation: Since sec 2 θ = 1 + tan 2 θ , the substition x = 2 tan u is suggested. For then x 2 + 4 = 4(1 + tan 2 u ) = 4 sec 2 u while dx = 2 sec 2 u du ....
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This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas.
 Fall '08
 RAdin
 Calculus

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