Homework 04 - Su Yung Homework 4 Due 3:00 am Inst Shinko...

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Su, Yung – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Shinko Harper 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points For which one of the following shaded re- gions is its area represented by the integral Z 4 0 ( x + 1) - 1 4 x 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 - 2 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 = Z b a n f ( x ) - g ( x ) o dx . When f ( x ) = - 1 4 x, g ( x ) = - ( x + 1) , therefore, the value of Z 4 0 n ( x + 1) - 1 4 x o dx is the area of the shaded region 2 4 6 - 2 2 - 2 - 4 - 6 . keywords: integral, region, area
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Su, Yung – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Shinko Harper 2 002 (part 1 of 1) 10 points Find the area of the region enclosed by the graphs of f ( x ) = 16 - x 2 , g ( x ) = x + 2 , on the interval [0 , 1]. 1. Area = 38 3 sq. units 2. Area = 79 6 sq. units correct 3. Area = 77 6 sq. units 4. Area = 13 sq. units 5. Area = 25 2 sq. units Explanation: The graph of f is a parabola opening down- wards and crossing the x -axis at x = ± 4, while the graph of g is a straight line with slope 1 and y -intercept at y = 2. Now on [0 , 1] we see that f ( x ) = 16 - x 2 x + 2 = g ( x ) , so the area between the graphs of f and g on [0 , 1] is given by Area = Z 1 0 n f ( x ) - g ( x ) o dx = Z 1 0 n 16 - x 2 - x - 2 o dx = 16 x - 1 3 x 3 - 1 2 x 2 - 2 x 1 0 . Consequently, Area = 79 6 sq. units . keywords: integral, area 003 (part 1 of 1) 10 points Find the area bounded by the graphs of f and g when f ( x ) = x 2 - 4 x, g ( x ) = 8 x - 2 x 2 . 1. area = 63 2 sq.units 2. area = 33 sq.units 3. area = 65 2 sq.units 4. area = 31 sq.units 5. area = 32 sq.units correct Explanation: The graph of f is a parabola opening up- wards and crossing the x -axis at x = 0 and x = 4, while the graph of g is a parabola opening downwards and crossing the x -axis at x = 0 and x = 4. Thus the required area is the shaded region in the figure below graph of g graph of f (graphs not drawn to scale). In terms of definite integrals, therefore, the required area is given by Area = Z 4 0 ( g ( x ) - f ( x )) dx = Z 4 0 (12 x - 3 x 2 ) dx.
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Su, Yung – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Shinko Harper 3 Now Z 4 0 (12 x - 3 x 2 ) dx = h 6 x 2 - x 3 i 4 0 = 32 . Thus Area = 32 sq.units . keywords: definite integral, area between graphs, quadratic functions 004 (part 1 of 3) 10 points The shaded region in is bounded by the graphs of f ( x ) = 1 + x - x 2 - x 3 and g ( x ) = 1 - x. (i) Find the x -coordinates of all the points of intersection of the graphs of f and g . 1. x = 0 , 1 , 2 2. x = - 2 , - 1 , 0 3. x = - 1 , 0 , 3 4. x = - 2 , 0 , 1 correct 5. x = - 1 , 0 , 2 Explanation: The x -coordinates of the points of intersec- tion of the two graphs are the solutions of the equation g ( x ) = f ( x ). Now g ( x ) - f ( x ) = x 3 + x 2 - 2 x = x ( x + 2)( x - 1) . Thus the graphs intersect when x = - 2 , 0 , 1 . 005 (part 2 of 3) 10 points (ii) Set up the definite integrals determining the area of this shaded region.
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