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Hw05 - badruddin(ssb776 HW05 Radin(57410 This print-out...

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badruddin (ssb776) – HW05 – Radin – (57410) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points For which one of the following shaded re- gions is its area represented by the integral integraldisplay 4 0 braceleftbigg 1 4 x + ( x + 1) bracerightbigg dx ? 1. 2 4 6 2 2 2 4 6 correct 2. 2 4 6 2 2 2 4 6 3. 2 4 6 2 2 4 6 4. 2 4 6 2 2 4 6 2 5. 2 4 6 2 2 4 6 2 Explanation: If f ( x ) g ( x ) for all x in an interval [ a, b ], then the area between the graphs of f and g is given by Area = integraldisplay b a braceleftBig f ( x ) g ( x ) bracerightBig dx . When f ( x ) = 1 4 x, g ( x ) = ( x + 1) , therefore, the value of integraldisplay 4 0 braceleftBig 1 4 x + ( x + 1) bracerightBig dx is the area of the shaded region 2 4 6 2 2 2 4 6 . 002 10.0 points
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badruddin (ssb776) – HW05 – Radin – (57410) 2 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 = 30 sq.units 2. Area = 27 sq.units correct 3. Area = 29 sq.units 4. Area = 31 sq.units 5. Area = 28 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 integraldisplay 3 0 (3 x x 2 ) dx integraldisplay 6 3 (3 x x 2 ) dx . Now integraldisplay 3 0 (3 x x 2 ) dx = bracketleftBig 3 2 x 2 1 3 x 3 bracketrightBig 3 0 = 9 2 , while integraldisplay 6 3 (3 x x 2 ) dx = bracketleftBig 3 2 x 2 1 3 x 3 bracketrightBig 6 3 = 45 2 . Consequently, Area = 27 sq.units . keywords: integral, graph, area 003 10.0 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 = 4 2 3. area = 2 2 4. area = 2( 2 + 1) 5. area = 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 = integraldisplay b a | f ( x ) g ( x ) | dx , which for the given functions is the integral A = 1 2 integraldisplay π 0 | cos x sin x | dx . But, as the graphs y θ π/ 2 π cos θ : sin θ :
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badruddin (ssb776) – HW05 – Radin – (57410) 3 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 . 004 10.0 points Find the area of the shaded region in x y bounded by the graphs of f ( x ) = x 3 x 2 5 x, g ( x ) = x .
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