HW4 Solution - Kim, Jin – Homework 4 – Due: Sep 25...

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Unformatted text preview: Kim, Jin – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Diane Radin 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 ‰ ( x + 1)- 1 2 x ¾ dx ? 1. 2 4 6- 2 2- 2- 4- 6 2. 2 4 6- 2 2 4 6 3. 2 4 6- 2 2- 2- 4- 6 4. 2 4 6- 2 2 4 6- 2 correct 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 = Z b a n f ( x )- g ( x ) o dx. When f ( x ) = x + 1 , g ( x ) = 1 2 x, therefore, the value of Z 4 n ( x + 1)- 1 2 x o dx is the area of the shaded region 2 4 6- 2 2 4 6- 2 . keywords: integral, region, area Kim, Jin – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Diane Radin 2 002 (part 1 of 1) 10 points Find the area of the region enclosed by the graphs of f ( x ) = 9- x 2 , g ( x ) = x + 1 , on the interval [0 , 1]. 1. Area = 41 6 sq. units 2. Area = 20 3 sq. units 3. Area = 7 sq. units 4. Area = 43 6 sq. units correct 5. Area = 13 2 sq. units Explanation: The graph of f is a parabola opening down- wards and crossing the x-axis at x = ± 3, while the graph of g is a straight line with slope 1 and y-intercept at y = 1. Now on [0 , 1] we see that f ( x ) = 9- x 2 ≥ x + 1 = g ( x ) , so the area between the graphs of f and g on [0 , 1] is given by Area = Z 1 n f ( x )- g ( x ) o dx = Z 1 n 9- x 2- x- 1 o dx = • 9 x- 1 3 x 3- 1 2 x 2- 1 x ‚ 1 . Consequently, Area = 43 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 = 31 sq.units 2. area = 63 2 sq.units 3. area = 32 sq.units correct 4. area = 65 2 sq.units 5. area = 33 sq.units 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 ( g ( x )- f ( x )) dx = Z 4 (12 x- 3 x 2 ) dx. Kim, Jin – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Diane Radin 3 Now Z 4 (12 x- 3 x 2 ) dx = h 6 x 2- x 3 i 4 = 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....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.

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HW4 Solution - Kim, Jin – Homework 4 – Due: Sep 25...

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