Review012ndtestans

Review012ndtestans - Cheung Anthony – Review 1 – Due...

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Unformatted text preview: Cheung, Anthony – Review 1 – Due: Nov 3 2006, 6:00 pm – Inst: David Benzvi 1 This print-out should have 17 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 A function h has graph 2- 2 2- 2 on (- 4 , 4). If f is defined on (- 4 , 4) by f ( x ) = 1 ,- 4 < x <- 3 , Z x- 3 h ( t ) dt,- 3 ≤ x < 4 , which of the following is the graph of f ? 1. 2- 2 2- 2 2. 2- 2 2 4- 2 3. 2- 2 2- 2 correct 4. 2- 2 2- 2- 4 5. 2- 2 2- 2 Explanation: Cheung, Anthony – Review 1 – Due: Nov 3 2006, 6:00 pm – Inst: David Benzvi 2 Since f (- 3) = Z- 3- 3 h ( t ) dt = 0 , two of the five graphs can be eliminated im- mediately. On the other hand, by the Fun- damental Theorem of Calculus, f ( x ) = h ( x ) on (- 3 , 4); in particular, the critical points of f occur at the x-intercepts of the graph of h . As these x-intercepts occur at- 1 , , 2, this eliminates a third graph. Thus the remain- ing two possible graphs for f both have the same critical points and to decide which one is the graph of f we can use the first derivative test because the graph of f will have a local maximum at an x-intercept of the graph of h where it changes from positive to negative values, and a local minimum at an x-intercept where h changes from negative to positive val- ues. Consequently, the graph of f must be 2- 2 2- 2 keywords: 002 (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 = 31 2 sq.units 2. area = 29 2 sq.units 3. area = 27 2 sq.units correct 4. area = 15 sq.units 5. area = 14 sq.units 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 ( g ( x )- f ( x )) dx = Z 3 (9 x- 3 x 2 ) dx. Now Z 3 (9 x- 3 x 2 ) dx = h 9 2 x 2- x 3 i 3 = 27 2 . Thus Area = 27 2 sq.units . keywords: definite integral, area between graphs, quadratic functions 003 (part 1 of 1) 10 points Cheung, Anthony – Review 1 – Due: Nov 3 2006, 6:00 pm – Inst: David Benzvi 3 Find the volume, V , of the solid generated by rotating about the x-axis the region en- closed by the graphs of y = sec x, x = 0 , y = 0 , x = π 3 . 1. V = π √ 3 cu. units correct 2. V = π ln µ 1 2 + √ 3 ¶ cu. units 3. V = 8 π 3 cu. units 4. V = π cu. units 5. V = π √ 3 cu. units Explanation: The solid is generated by rotating the re- gion π /3 about the x-axis. The cross-section perpen- dicular to the x-axis has area A ( x ) = πy 2 = π sec 2 x....
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This note was uploaded on 01/27/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.

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Review012ndtestans - Cheung Anthony – Review 1 – Due...

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