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Unformatted text preview: badruddin (ssb776) HW05 Radin (57410) 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 For which one of the following shaded re gions is its area represented by the integral integraldisplay 4 braceleftbigg 1 4 x + ( x + 1) bracerightbigg dx ? 1. 2 4 6 2 2 2 4 6 correct 2. 2 4 6 2 2 2 4 6 3. 2 4 6 2 2 4 6 4. 2 4 6 2 2 4 6 2 5. 2 4 6 2 2 4 6 2 Explanation: If f ( x ) g ( x ) for all x in an interval [ a, b ], then the area between the graphs of f and g is given by Area = integraldisplay b a braceleftBig f ( x ) g ( x ) bracerightBig dx . When f ( x ) = 1 4 x, g ( x ) = ( x + 1) , therefore, the value of integraldisplay 4 braceleftBig 1 4 x + ( x + 1) bracerightBig dx is the area of the shaded region 2 4 6 2 2 2 4 6 . 002 10.0 points badruddin (ssb776) HW05 Radin (57410) 2 Find the area between the graph of f and the xaxis on the interval [0 , 6] when f ( x ) = 3 x x 2 . 1. Area = 30 sq.units 2. Area = 27 sq.units correct 3. Area = 29 sq.units 4. Area = 31 sq.units 5. Area = 28 sq.units Explanation: The graph of f is a parabola opening down wards and crossing the xaxis at x = 0 and x = 3. Thus the required area is similar to the shaded region in the figure below. graph of f In terms of definite integrals, therefore, the required area is given by integraldisplay 3 (3 x x 2 ) dx integraldisplay 6 3 (3 x x 2 ) dx . Now integraldisplay 3 (3 x x 2 ) dx = bracketleftBig 3 2 x 2 1 3 x 3 bracketrightBig 3 = 9 2 , while integraldisplay 6 3 (3 x x 2 ) dx = bracketleftBig 3 2 x 2 1 3 x 3 bracketrightBig 6 3 = 45 2 . Consequently, Area = 27 sq.units . keywords: integral, graph, area 003 10.0 points Find the area enclosed by the graphs of f ( x ) = 1 2 cos x , g ( x ) = 1 2 sin x on [0 , ]. 1. area = 2 correct 2. area = 4 2 3. area = 2 2 4. area = 2( 2 + 1) 5. area = 2 + 1 6. area = 4( 2 + 1) Explanation: The area between the graphs of y = f ( x ) and y = g ( x ) on the interval [ a, b ] is expressed by the integral A = integraldisplay b a  f ( x ) g ( x )  dx , which for the given functions is the integral A = 1 2 integraldisplay  cos x sin x  dx . But, as the graphs y / 2 cos : sin : badruddin (ssb776) HW05 Radin (57410) 3 of y = cos x and y = sin x on [0 , ] show, cos sin braceleftBigg , on [0 , / 4], , on [ / 4 , ]. Thus A = 1 2 integraldisplay / 4 { cos sin } d 1 2 integraldisplay / 4 { cos sin } d = A 1 A 2 ....
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 Fall '09
 RAdin
 Calculus

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