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HW05-solutions - Munoz(gm7794 HW05 Radin(54915 This...

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Munoz (gm7794) – HW05 – Radin – (54915) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the bounded area enclosed by the graph of f ( x ) = 6 x x 2 and the x -axis. 1. Area = 36 sq.units correct 2. Area = 35 sq.units 3. Area = 37 sq.units 4. Area = 34 sq.units 5. Area = 33 sq.units Explanation: The graph of f is a parabola opening down- wards and crossing the x -axis at x = 0 and x = 6. Thus the required area is similar to the shaded region in 6 x y As a definite integral, therefore, the required area is given by integraldisplay 6 0 (6 x x 2 ) dx = bracketleftBig 3 x 2 1 3 x 3 bracketrightBig 6 0 . Consequently, Area = 36 sq.units . keywords: AreaBetween, AreaBetweenExam, 002 10.0 points Find the area enclosed by the graphs of f ( x ) = 1 2 sin x , g ( x ) = 1 2 cos x on [0 , π ]. 1. area = 4 2 2. area = 2( 2 + 1) 3. area = 2 2 4. area = 4( 2 + 1) 5. area = 2 + 1 6. area = 2 correct 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 = integraldisplay b a | f ( x ) g ( x ) | dx , which for the given functions is the integral A = 1 2 integraldisplay π 0 | sin x cos x | dx . But, as the graphs y θ π/ 2 π cos θ : sin θ :
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Munoz (gm7794) – HW05 – Radin – (54915) 2 of y = cos x and y = sin x on [0 , π ] show, cos θ sin θ braceleftBigg 0 , on [0 , π/ 4], 0 , on [ π/ 4 , π ]. Thus A = 1 2 integraldisplay π/ 4 0 { cos θ sin θ } 1 2 integraldisplay π π/ 4 { cos θ sin θ } = A 1 A 2 . But by the Fundamental Theorem of Calcu- lus, A 1 = 1 2 bracketleftBig sin θ + cos θ bracketrightBig π/ 4 0 = 1 2 ( 2 1) , while A 2 = 1 2 bracketleftBig sin θ + cos θ bracketrightBig π π/ 4 = 1 2 (1 + 2) . Consequently, area = A 1 A 2 = 2 . 003 10.0 points Find the total area, A , of the bounded re- gion in the first and fourth quadrants enclosed by the graphs of f ( x ) = 6 x 2 4 x 2 , the x -axis, and the line x = 2. 1. A = 15 2 sq.units 2. A = 8 sq.units correct 3. A = 9 sq.units 4. A = 7 sq.units 5. A = 17 2 sq.units Explanation: The graph of f is a parabola opening up- wards and having y -intercept at y = 2. In addition, since f ( x ) = (6 x + 2)( x 1) , the graph has x -intercepts at x = 1 3 and at x = 1. Thus the graph of f and the vertical line are given by In terms of definite integrals, therefore, the total area A is given by A = integraldisplay 1 0 f ( x ) dx + integraldisplay 2 1 f ( x ) dx. But integraldisplay 1 0 f ( x ) dx = integraldisplay 1 0 6 x 2 4 x 2 dx = bracketleftBig 2 x 3 2 x 2 2 x bracketrightBig 1 0 = 2 , while integraldisplay 2 1 f ( x ) dx = integraldisplay 2 1 6 x 2 4 x 2 dx = bracketleftBig 2 x 3 2 x 2 2 x bracketrightBig 2 1 = 6 . Consequently, A = 8 sq.units. 004 (part 1 of 3) 10.0 points The shaded region in
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Munoz (gm7794) – HW05 – Radin – (54915) 3 is bounded by the graphs of f ( x ) = 1 + x x 2 x 3 and g ( x ) = 1 x.
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