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# HW 4 - Portillo Tom Homework 4 Due 3:00 am Inst David...

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Portillo, Tom – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Benzvi 1 This print-out should have 21 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 Determine the indefinite integral I = Z 1 θ 2 2 sin 1 θ · - 6 θ 1. I = - 3 θ 2 + 2 sin 1 θ · + C 2. I = 3 θ 2 - 2 sin 1 θ · + C 3. I = - 3 θ 2 + 2 cos 1 θ · + C 4. I = 3 θ 2 - 2 cos 1 θ · + C 5. I = 3 θ 2 + 2 cos 1 θ · + C correct 6. I = 3 θ 2 + 2 sin 1 θ · + C Explanation: Set u = 1 . Then du = - 1 θ 2 dθ , so I = - Z (2 sin u - 6 u ) du = 2 cos u + 3 u 2 + C with C an arbitrary constant. Consequently, I = 3 θ 2 + 2 cos 1 θ · + C . keywords: integral, trig function, substitution 002 (part 1 of 1) 10 points Evaluate the definite integral I = Z π/ 4 0 { 7 sec 2 x - 6 sec 2 x tan 2 x } dx. 1. I = 16 3 2. I = 6 3. I = 5 correct 4. I = 19 3 5. I = 17 3 Explanation: Since the integrand is of the form sec 2 xf (tan x ) , f ( x ) = 7 - 6 x 2 , the substitution u = tan x is needed. For then du = sec 2 x dx, while x = 0 = u = 0 , x = π 4 = u = 1 . Thus I = Z 1 0 (7 - 6 u 2 ) du = h 7 u - 2 u 3 i 1 0 . Consequently, I = 5 . keywords: indefinite integral, trigonometric substitution 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 12 4 x - 7 x - 3 dx. Correct answer: 1 . 33333 .

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Portillo, Tom – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Benzvi 2 Explanation: Set u = x - 3. Then du = dx, so Z 12 4 x - 7 x - 3 dx = Z 9 1 u - 4 u du = Z 9 1 u 1 / 2 - 4 u - 1 / 2 · du. Consequently, I = Z 9 1 u 1 / 2 - 4 u - 1 / 2 · du = h 2 3 u 3 / 2 - 8 u 1 / 2 i 9 1 = 4 3 . keywords: integral, square root, substitution 004 (part 1 of 1) 10 points Find the area between the graph of f and the x -axis on the interval [0 , 6] when f ( x ) = 3 x - x 2 . 1. Area = 27 sq.units correct 2. Area = 25 sq.units 3. Area = 24 sq.units 4. Area = 28 sq.units 5. Area = 26 sq.units Explanation: The graph of f is a parabola opening down- wards and crossing the x -axis at x = 0 and x = 3. Thus the required area is similar to the shaded region in the figure below. graph of f In terms of definite integrals, therefore, the required area is given by Z 3 0 (3 x - x 2 ) dx - Z 6 3 (3 x - x 2 ) dx. Now Z 3 0 (3 x - x 2 ) dx = h 3 2 x 2 - 1 3 x 3 i 3 0 = 9 2 , while Z 6 3 (3 x - x 2 ) dx = h 3 2 x 2 - 1 3 x 3 i 6 3 = - 45 2 . Consequently, Area = 27 sq.units. keywords: integral, graph, area 005 (part 1 of 1) 10 points Find the area enclosed by the graphs of f ( x ) = 1 2 cos x , g ( x ) = 1 2 sin x on [0 , π ]. 1. area = 2 correct 2. area = 2 2 3. area = 2 + 1 4. area = 4( 2 + 1) 5. area = 4 2 6. area = 2( 2 + 1) Explanation:
Portillo, Tom – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Benzvi 3 The area between the graphs of y = f ( x ) and y = g ( x ) on the interval [ a, b ] is expressed by the integral A = Z b a | f ( x ) - g ( x ) | dx , which for the given functions is the integral A = Z π 0 | 1 2 cos x - 1 2 sin x | dx . But, as the graphs y θ π/ 2 π cos θ : sin θ : of y = cos x and y = sin x on [0 , π ] show, cos θ - sin θ ( 0 , on [0 , π/ 4], 0 , on [ π/ 4 , π ].

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