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Unformatted text preview: wei (jw35975) – HW04 – kalahurka – (55230) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the area between the graph of f and the xaxis on the interval [0 , 4] when f ( x ) = 2 x x 2 . 1. Area = 6 sq.units 2. Area = 8 sq.units correct 3. Area = 7 sq.units 4. Area = 5 sq.units 5. Area = 4 sq.units Explanation: The graph of f is a parabola opening down wards and crossing the xaxis at x = 0 and x = 2. 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 2 (2 x x 2 ) dx integraldisplay 4 2 (2 x x 2 ) dx . Now integraldisplay 2 (2 x x 2 ) dx = bracketleftBig x 2 1 3 x 3 bracketrightBig 2 = 4 3 , while integraldisplay 4 2 (2 x x 2 ) dx = bracketleftBig x 2 1 3 x 3 bracketrightBig 4 2 = 20 3 . Consequently, Area = 8 sq.units . keywords: integral, graph, area 002 10.0 points Find the area enclosed by the graphs of f ( x ) = 2 cos x , g ( x ) = 2 sin x on [0 , π/ 2]. 1. area = 2( √ 2 + 1) 2. area = √ 2 1 3. area = 2( √ 2 1) 4. area = 4( √ 2 + 1) 5. area = 4( √ 2 1) correct 6. area = √ 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 = integraldisplay π/ 2  2 cos x 2 sin x  dx . But, as the graphs y θ π/ 4 π/ 2 cos θ : sin θ : 1 wei (jw35975) – HW04 – kalahurka – (55230) 2 of y = cos x and y = sin x on [0 , π/ 2] show, cos θ sin θ braceleftBigg ≥ , on [0 , π/ 4], ≤ , on [ π/ 4 , π/ 2]. Thus A = 2 integraldisplay π/ 4 { cos θ sin θ } dθ 2 integraldisplay π/ 2 π/ 4 { cos θ sin θ } dθ = A 1 A 2 . But by the Fundamental Theorem of Calcu lus, A 1 = 2 bracketleftBig sin θ + cos θ bracketrightBig π/ 4 = 2( √ 2 1) , while A 2 = 2 bracketleftBig sin θ + cos θ bracketrightBig π/ 2 π/ 4 = 2(1 √ 2) . Consequently, area = A 1 A 2 = 4( √ 2 1) . 003 10.0 points Find the bounded area enclosed by the graphs of f ( x ) = x 2 x, g ( x ) = 2 x . 1. Area = 9 2 sq. units correct 2. Area = 13 2 sq. units 3. Area = 15 2 sq. units 4. Area = 7 2 sq. units 5. Area = 11 2 sq. units Explanation: The graph of f is a parabola opening up wards, while the graph of g is a straight line of slope 2. Both graphs pass through the origin. Thus the required area is the shaded region in the figure below graph of g graph of f 3 To express the area as a definite integral we need to find where the graphs intersect, i.e ....
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 Spring '10
 ZHENG
 Differential Equations, Equations, Royal Flying Corps squadrons, Wei, Royal Air Force aircraft squadrons

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