This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Kim, Jin – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Diane Radin 1 This printout should have 22 questions. Multiplechoice 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 xaxis at x = ± 3, while the graph of g is a straight line with slope 1 and yintercept 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 xaxis at x = 0 and x = 4, while the graph of g is a parabola opening downwards and crossing the xaxis 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....
View
Full
Document
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.
 Spring '09
 Turner

Click to edit the document details