HW05-solutions

# HW05-solutions - kim(tk5895 – HW05 – Henry –(54974 1...

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Unformatted text preview: kim (tk5895) – HW05 – Henry – (54974) 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 ) = 2 x − x 2 and the x-axis. 1. Area = 4 3 sq.units correct 2. Area = 5 3 sq.units 3. Area = 2 sq.units 4. Area = 2 3 sq.units 5. Area = 1 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 2 x y As a definite integral, therefore, the required area is given by integraldisplay 2 (2 x − x 2 ) dx = bracketleftBig x 2 − 1 3 x 3 bracketrightBig 2 . Consequently, Area = 4 3 sq.units . keywords: AreaBetween, AreaBetweenExam, 002 10.0 points Find the area enclosed by the graphs of f ( x ) = 2 sin x , g ( x ) = 2 cos x on [0 , π ]. 1. area = √ 2 2. area = 2 √ 2 3. area = 4 √ 2 correct 4. area = 2( √ 2 + 1) 5. area = 4( √ 2 + 1) 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 = 2 integraldisplay π | sin x − cos x | dx . But, as the graphs y θ π/ 2 π cos θ : sin θ : kim (tk5895) – HW05 – Henry – (54974) 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 = 2 integraldisplay π/ 4 { cos θ − sin θ } dθ − 2 integraldisplay π 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 π/ 4 = − 2(1 + √ 2) . Consequently, area = A 1 − A 2 = 4 √ 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 ) = 5 x 2 − 11 x − 12 , the x-axis, and the line x = 4. 1. A = 313 6 sq.units 2. A = 152 3 sq.units 3. A = 158 3 sq.units 4. A = 307 6 sq.units 5. A = 155 3 sq.units correct Explanation: The graph of f is a parabola opening up- wards and having y-intercept at y = − 12. In addition, since f ( x ) = (5 x + 4)( x − 3) , the graph has x-intercepts at x = − 4 5 and at x = 3. 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 3 f ( x ) dx + integraldisplay 4 3 f ( x ) dx. But integraldisplay 3 f ( x ) dx = integraldisplay 3 5 x 2 − 11 x − 12 dx = bracketleftBig 5 3 x 3 − 11 2 x 2 − 12 x bracketrightBig 3 = − 81 2 , while integraldisplay 4 3 f ( x ) dx = integraldisplay 4 3 5 x 2 − 11 x − 12 dx = bracketleftBig 5 3 x 3 − 11 2 x 2 − 12 x bracketrightBig 4 3 = 67 6 ....
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HW05-solutions - kim(tk5895 – HW05 – Henry –(54974 1...

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