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Unformatted text preview: Munoz (gm7794) – HW05 – Radin – (54915) 1 This printout should have 21 questions. Multiplechoice 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 xaxis. 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 xaxis 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 (6 x − x 2 ) dx = bracketleftBig 3 x 2 − 1 3 x 3 bracketrightBig 6 . 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  sin x − cos x  dx . But, as the graphs y θ π/ 2 π cos θ : sin θ : Munoz (gm7794) – HW05 – Radin – (54915) 2 of y = cos x and y = sin x on [0 , π ] show, cos θ − sin θ braceleftBigg ≥ , on [0 , π/ 4], ≤ , on [ π/ 4 , π ]. Thus A = 1 2 integraldisplay π/ 4 { cos θ − sin θ } dθ − 1 2 integraldisplay π π/ 4 { cos θ − sin θ } dθ = A 1 − A 2 . But by the Fundamental Theorem of Calcu lus, A 1 = 1 2 bracketleftBig sin θ + cos θ bracketrightBig π/ 4 = 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 xaxis, 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 yintercept at y = − 2. In addition, since f ( x ) = (6 x + 2)( x − 1) , the graph has xintercepts 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 f ( x ) dx + integraldisplay 2 1 f ( x ) dx. But integraldisplay 1 f ( x ) dx = integraldisplay 1 6 x 2 − 4 x − 2 dx = bracketleftBig 2 x 3 − 2 x 2 − 2 x bracketrightBig 1 = − 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 ....
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This note was uploaded on 06/10/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.
 Fall '08
 Cepparo
 Calculus

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