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Unformatted text preview: wei (jw35975) HW04 kalahurka (55230) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the area between the graph of f and the xaxis on the interval [0 , 4] when f ( x ) = 2 x x 2 . 1. Area = 6 sq.units 2. Area = 8 sq.units correct 3. Area = 7 sq.units 4. Area = 5 sq.units 5. Area = 4 sq.units Explanation: The graph of f is a parabola opening down wards and crossing the xaxis at x = 0 and x = 2. 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 2 (2 x x 2 ) dx integraldisplay 4 2 (2 x x 2 ) dx . Now integraldisplay 2 (2 x x 2 ) dx = bracketleftBig x 2 1 3 x 3 bracketrightBig 2 = 4 3 , while integraldisplay 4 2 (2 x x 2 ) dx = bracketleftBig x 2 1 3 x 3 bracketrightBig 4 2 = 20 3 . Consequently, Area = 8 sq.units . keywords: integral, graph, area 002 10.0 points Find the area enclosed by the graphs of f ( x ) = 2 cos x , g ( x ) = 2 sin x on [0 , / 2]. 1. area = 2( 2 + 1) 2. area = 2 1 3. area = 2( 2 1) 4. area = 4( 2 + 1) 5. area = 4( 2 1) correct 6. area = 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 = integraldisplay / 2  2 cos x 2 sin x  dx . But, as the graphs y / 4 / 2 cos : sin : 1 wei (jw35975) HW04 kalahurka (55230) 2 of y = cos x and y = sin x on [0 , / 2] show, cos  sin braceleftBigg , on [0 , / 4], , on [ / 4 , / 2]. Thus A = 2 integraldisplay / 4 { cos  sin } d 2 integraldisplay / 2 / 4 { cos  sin } d = A 1 A 2 . But by the Fundamental Theorem of Calcu lus, A 1 = 2 bracketleftBig sin + cos bracketrightBig / 4 = 2( 2 1) , while A 2 = 2 bracketleftBig sin + cos bracketrightBig / 2 / 4 = 2(1 2) . Consequently, area = A 1 A 2 = 4( 2 1) . 003 10.0 points Find the bounded area enclosed by the graphs of f ( x ) = x 2 x, g ( x ) = 2 x . 1. Area = 9 2 sq. units correct 2. Area = 13 2 sq. units 3. Area = 15 2 sq. units 4. Area = 7 2 sq. units 5. Area = 11 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 ....
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 Spring '10
 ZHENG
 Differential Equations, Equations

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