solution_pdf4 - Horstman (mdh995) HW04 Radin (58505) This...

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Horstman (mdh995) – HW04 – Radin – (58505) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind 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 = 26 sq.units 5. Area = 23 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 fgure below. graph oF f In terms oF defnite integrals, thereFore, the required area is given by i 3 0 (3 x x 2 ) dx i 6 3 (3 x x 2 ) dx . Now i 3 0 (3 x x 2 ) dx = b 3 2 x 2 1 3 x 3 B 3 0 = 9 2 , while i 6 3 (3 x x 2 ) dx = b 3 2 x 2 1 3 x 3 B 6 3 = 45 2 . Consequently, Area = 27 sq.units . keywords: integral, graph, area 002 10.0 points ±ind the area enclosed by the graphs oF f ( x ) = sin x , g ( x ) = cos x on [0 , π ]. 1. area = 2 + 1 2. area = 2 3. area = 4 2 4. area = 2 2 correct 5. area = 2( 2 + 1) 6. area = 4( 2 + 1) Explanation: The area between the graphs oF y = f ( x ) and y = g ( x ) on the interval [ a, b ] is expressed by the integral A = i b a | f ( x ) g ( x ) | dx , which For the given Functions is the integral A = i π 0 | sin x cos x | dx . But, as the graphs
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Horstman (mdh995) – HW04 – Radin – (58505) 2 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 b 0 , on [0 , π/ 4], 0 , on [ π/ 4 , π ]. Thus A = i π/ 4 0 { cos θ sin θ } i π π/ 4 { cos θ sin θ } = A 1 A 2 . But by the Fundamental Theorem of Calcu- lus, A 1 = B sin θ + cos θ ± π/ 4 0 = 2 1 , while A 2 = B sin θ + cos θ ± π π/ 4 = (1 + 2) . Consequently, area = A 1 A 2 = 2 2 . 003 10.0 points Compute the area between the graphs of f ( x ) = 11 sin 2 x and g ( x ) = 7 sin x + 4 sin2 x on [0 , π/ 2]. 1. Area = 7 2 sq.units correct 2. Area = 15 4 sq.units 3. Area = 3 sq.units 4. Area = 11 4 sq.units 5. Area = 13 4 sq.units Explanation: The required area is given by I = i π/ 2 0 | f ( x ) g ( x ) | dx. Now the graphs of f, g intersect when 11 sin 2 x = 7 sin x + 4 sin 2 x , i.e. , when 7(sin 2 x sin x ) = 7 sin x (2 cos x 1) = 0 , since sin 2 x = 2 sin x cos x . Thus the points of intersection are x = 0 , π/ 3.
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solution_pdf4 - Horstman (mdh995) HW04 Radin (58505) This...

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