HW04-solutions - wei (jw35975) HW04 kalahurka (55230) 1...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: wei (jw35975) HW04 kalahurka (55230) 1 This print-out should have 23 questions. Multiple-choice 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 x-axis 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 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 (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 , / 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 / 2 | 2 cos x- 2 sin x | dx . But, as the graphs y / 4 / 2 cos : sin : 1 wei (jw35975) HW04 kalahurka (55230) 2 of y = cos x and y = sin x on [0 , / 2] show, cos - sin braceleftBigg , on [0 , / 4], , on [ / 4 , / 2]. Thus A = 2 integraldisplay / 4 { cos - sin } d- 2 integraldisplay / 2 / 4 { cos - sin } d = A 1- A 2 . But by the Fundamental Theorem of Calcu- lus, A 1 = 2 bracketleftBig sin + cos bracketrightBig / 4 = 2( 2- 1) , while A 2 = 2 bracketleftBig sin + cos bracketrightBig / 2 / 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 ....
View Full Document

Page1 / 13

HW04-solutions - wei (jw35975) HW04 kalahurka (55230) 1...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online