int cal 5 - pokharel (yp624) HW05 Radin (57410) 1 This...

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Unformatted text preview: pokharel (yp624) HW05 Radin (57410) 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 For which one of the following shaded re- gions is its area represented by the integral integraldisplay 4 braceleftbigg ( x + 1) 1 4 x bracerightbigg dx ? 1. 2 4 6 2 2 2 4 6 2. 2 4 6 2 2 4 6 2 3. 2 4 6 2 2 2 4 6 correct 4. 2 4 6 2 2 4 6 2 5. 2 4 6 2 2 4 6 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 ( x + 1) 1 4 x bracerightBig dx is the area of the shaded region 2 4 6 2 2 2 4 6 . 002 10.0 points pokharel (yp624) HW05 Radin (57410) 2 Find the area between the graph of f and the x-axis on the interval [0 , 6] when f ( x ) = 3 x x 2 . 1. Area = 24 sq.units 2. Area = 27 sq.units correct 3. Area = 28 sq.units 4. Area = 25 sq.units 5. Area = 26 sq.units Explanation: The graph of f is a parabola opening down- wards and crossing the x-axis 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 ) = 2 sin x , g ( x ) = 2 cos x on [0 , ]. 1. area = 2 2 2. area = 2 + 1 3. area = 2( 2 + 1) 4. area = 4 2 correct 5. area = 2 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 = 2 integraldisplay | sin x cos x | dx . But, as the graphs y / 2 cos : sin : pokharel (yp624) 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 = 2 integraldisplay / 4 { cos sin } d 2 integraldisplay / 4 { cos sin } d = A 1 A 2 ....
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This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

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int cal 5 - pokharel (yp624) HW05 Radin (57410) 1 This...

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