Hw03-solutions.pdf

# Hw03-solutions.pdf - young(toy68 Hw03 villafuerte...

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young (toy68) – Hw03 – villafuerte altu – (53615) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find the area between the graph of f and the x -axis on the interval [0 , 4] when f ( x ) = 2 x - x 2 . 1. Area = 10 sq.units 2. Area = 11 sq.units 3. Area = 8 sq.units correct 4. Area = 7 sq.units 5. Area = 9 sq.units Explanation: The graph of f is a parabola opening down- wards and crossing the x -axis 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 0 (2 x - x 2 ) dx - integraldisplay 4 2 (2 x - x 2 ) dx . Now integraldisplay 2 0 (2 x - x 2 ) dx = bracketleftBig x 2 - 1 3 x 3 bracketrightBig 2 0 = 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 10.0points Find the bounded area enclosed by the graphs of f and g when f ( x ) = x 2 - x, g ( x ) = 2 x. Correct answer: 4 . 5. 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 3 (not drawn to scale). To express the area as a definite integral we need to find where the graphs intersect: x 2 - x = 2 x, i.e., x ( x - 3) = 0 .

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young (toy68) – Hw03 – villafuerte altu – (53615) 2 The area is thus given by Area = integraldisplay 3 0 ( g ( x ) - f ( x )) dx = integraldisplay 3 0 (3 x - x 2 ) dx = bracketleftbigg 3 2 x 2 - 1 3 x 3 bracketrightbigg 3 0 . Consequently, Area = 9 2 sq.units . 003 10.0points Find the area enclosed by the graphs of f ( x ) = 2 sin x , g ( x ) = 2 cos x on [0 , π ]. 1. area = 4( 2 + 1) 2. area = 2 + 1 3. area = 2 2 4. area = 2( 2 + 1) 5. area = 2 6. area = 4 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 = 2 integraldisplay π 0 | sin x - cos x | dx . But, as the graphs 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 braceleftBigg 0 , on [0 , π/ 4], 0 , on [ π/ 4 , π ]. Thus A = 2 integraldisplay π/ 4 0 { cos θ - sin θ } - 2 integraldisplay π π/ 4 { cos θ - sin θ } = A 1 - A 2 . But by the Fundamental Theorem of Calcu- lus, A 1 = 2 bracketleftBig sin θ + cos θ bracketrightBig π/ 4 0 = 2( 2 - 1) , while A 2 = 2 bracketleftBig sin θ + cos θ bracketrightBig π π/ 4 = - 2(1 + 2) . Consequently, area = A 1 - A 2 = 4 2 . 004 10.0points Find the bounded area enclosed by the graphs of f ( x ) = x 2 - 2 x, g ( x ) = 2 x . 1. Area = 32 3 sq. units correct 2. Area = 41 3 sq. units 3. Area = 29 3 sq. units
young (toy68) – Hw03 – villafuerte altu – (53615) 3 4. Area = 38 3 sq. units 5. Area = 35 3 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.

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