{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HWK4 - Valenica Daniel Homework 4 Due 3:00 am Inst Cheng...

This preview shows pages 1–4. Sign up to view the full content.

Valenica, Daniel – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Cheng 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 (6 - x ) - 1 2 x dx ? 1. 2 4 6 - 2 2 4 6 - 2 2. 2 4 6 - 2 2 - 2 - 4 - 6 3. 2 4 6 - 2 2 - 2 - 4 - 6 4. 2 4 6 - 2 2 4 6 - 2 5. 2 4 6 - 2 2 4 6 correct 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 ) = 6 - x, g ( x ) = 1 2 x , therefore, the value of Z 4 0 n (6 - x ) - 1 2 x o dx is the area of the shaded region 2 4 6 - 2 2 4 6 . keywords: integral, region, area

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Valenica, Daniel – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Cheng 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 = 22 3 sq. units 2. Area = 7 sq. units 3. Area = 41 6 sq. units 4. Area = 20 3 sq. units 5. Area = 43 6 sq. units correct 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 0 n f ( x ) - g ( x ) o dx = Z 1 0 n 9 - x 2 - x - 1 o dx = 9 x - 1 3 x 3 - 1 2 x 2 - 1 x 1 0 . 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 - 3 x, g ( x ) = 6 x - 2 x 2 . 1. area = 15 sq.units 2. area = 29 2 sq.units 3. area = 14 sq.units 4. area = 31 2 sq.units 5. area = 27 2 sq.units correct Explanation: The graph of f is a parabola opening up- wards and crossing the x -axis at x = 0 and x = 3, while the graph of g is a parabola opening downwards and crossing the x -axis at x = 0 and x = 3. 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 3 0 ( g ( x ) - f ( x )) dx = Z 3 0 (9 x - 3 x 2 ) dx.
Valenica, Daniel – Homework 4 – Due: Sep 25 2007, 3:00 am – Inst: Cheng 3 Now Z 3 0 (9 x - 3 x 2 ) dx = h 9 2 x 2 - x 3 i 3 0 = 27 2 . Thus Area = 27 2 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 = - 2 , 0 , 1 correct 2. x = 0 , 1 , 2 3. x = - 1 , 0 , 2 4. x = - 2 , - 1 , 0 5. x = - 1 , 0 , 3 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 13

HWK4 - Valenica Daniel Homework 4 Due 3:00 am Inst Cheng...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online