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Unformatted text preview: Munoz (gm7794) HW05 Radin (54915) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the bounded area enclosed by the graph of f ( x ) = 6 x x 2 and the xaxis. 1. Area = 36 sq.units correct 2. Area = 35 sq.units 3. Area = 37 sq.units 4. Area = 34 sq.units 5. Area = 33 sq.units Explanation: The graph of f is a parabola opening down wards and crossing the xaxis at x = 0 and x = 6. Thus the required area is similar to the shaded region in 6 x y As a definite integral, therefore, the required area is given by integraldisplay 6 (6 x x 2 ) dx = bracketleftBig 3 x 2 1 3 x 3 bracketrightBig 6 . Consequently, Area = 36 sq.units . keywords: AreaBetween, AreaBetweenExam, 002 10.0 points Find the area enclosed by the graphs of f ( x ) = 1 2 sin x , g ( x ) = 1 2 cos x on [0 , ]. 1. area = 4 2 2. area = 2( 2 + 1) 3. area = 2 2 4. area = 4( 2 + 1) 5. area = 2 + 1 6. area = 2 correct 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  sin x cos x  dx . But, as the graphs y / 2 cos : sin : Munoz (gm7794) HW05 Radin (54915) 2 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 . But by the Fundamental Theorem of Calcu lus, A 1 = 1 2 bracketleftBig sin + cos bracketrightBig / 4 = 1 2 ( 2 1) , while A 2 = 1 2 bracketleftBig sin + cos bracketrightBig / 4 = 1 2 (1 + 2) . Consequently, area = A 1 A 2 = 2 . 003 10.0 points Find the total area, A , of the bounded re gion in the first and fourth quadrants enclosed by the graphs of f ( x ) = 6 x 2 4 x 2 , the xaxis, and the line x = 2. 1. A = 15 2 sq.units 2. A = 8 sq.units correct 3. A = 9 sq.units 4. A = 7 sq.units 5. A = 17 2 sq.units Explanation: The graph of f is a parabola opening up wards and having yintercept at y = 2. In addition, since f ( x ) = (6 x + 2)( x 1) , the graph has xintercepts at x = 1 3 and at x = 1. Thus the graph of f and the vertical line are given by In terms of definite integrals, therefore, the total area A is given by A = integraldisplay 1 f ( x ) dx + integraldisplay 2 1 f ( x ) dx. But integraldisplay 1 f ( x ) dx = integraldisplay 1 6 x 2 4 x 2 dx = bracketleftBig 2 x 3 2 x 2 2 x bracketrightBig 1 = 2 , while integraldisplay 2 1 f ( x ) dx = integraldisplay 2 1 6 x 2 4 x 2 dx = bracketleftBig 2 x 3 2 x 2 2 x bracketrightBig 2 1 = 6 ....
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 Fall '08
 Cepparo
 Calculus

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