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homework 4 solutions - Mcbiles Emily Homework 4 Due 3:00 am...

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Mcbiles, Emily – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Fonken 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 4 sin 1 θ · - 2 θ 1. I = - 1 θ 2 + 4 sin 1 θ · + C 2. I = - 1 θ 2 + 4 cos 1 θ · + C 3. I = 1 θ 2 - 4 cos 1 θ · + C 4. I = 1 θ 2 + 4 sin 1 θ · + C 5. I = 1 θ 2 + 4 cos 1 θ · + C correct 6. I = 1 θ 2 - 4 sin 1 θ · + C Explanation: Set u = 1 . Then du = - 1 θ 2 dθ , so I = - Z (4 sin u - 2 u ) du = 4 cos u + u 2 + C with C an arbitrary constant. Consequently, I = 1 θ 2 + 4 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 { 9 sec 2 x - 6 sec 2 x tan 2 x } dx. 1. I = 8 2. I = 22 3 3. I = 25 3 4. I = 23 3 5. I = 7 correct Explanation: Since the integrand is of the form sec 2 xf (tan x ) , f ( x ) = 9 - 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 (9 - 6 u 2 ) du = h 9 u - 2 u 3 i 1 0 . Consequently, I = 7 . keywords: indefinite integral, trigonometric substitution 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 6 3 x + 1 x - 2 dx. Correct answer: 10 . 6667 .
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Mcbiles, Emily – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Fonken 2 Explanation: Set u = x - 2. Then du = dx, so Z 6 3 x + 1 x - 2 dx = Z 4 1 u + 3 u du = Z 4 1 u 1 / 2 + 3 u - 1 / 2 · du. Consequently, I = Z 4 1 u 1 / 2 + 3 u - 1 / 2 · du = h 2 3 u 3 / 2 + 6 u 1 / 2 i 4 1 = 32 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 = 26 sq.units 2. Area = 23 sq.units 3. Area = 25 sq.units 4. Area = 27 sq.units correct 5. Area = 24 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 ) = cos x , g ( x ) = sin x on [0 , π ]. 1. area = 2 + 1 2. area = 2( 2 + 1) 3. area = 2 4. area = 4( 2 + 1) 5. area = 2 2 correct 6. area = 4 2 Explanation:
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Mcbiles, Emily – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Fonken 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 | cos x - 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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