HW05-solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 12

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online