# HW4 - momin (rrm497) Homework 04 cheng (58520) This...

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momin (rrm497) – Homework 04 – cheng – (58520) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind 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 = 7 sq.units 2. Area = 5 sq.units 3. Area = 4 sq.units 4. Area = 8 sq.units correct 5. Area = 6 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 fgure below. graph oF f In terms oF defnite integrals, thereFore, the required area is given by i 2 0 (2 x x 2 ) dx i 4 2 (2 x x 2 ) dx . Now i 2 0 (2 x x 2 ) dx = b x 2 1 3 x 3 B 2 0 = 4 3 , while i 4 2 (2 x x 2 ) dx = b x 2 1 3 x 3 B 4 2 = 20 3 . Consequently, Area = 8 sq.units . keywords: integral, graph, area 002 10.0 points ±ind the area enclosed by the graphs oF f ( x ) = cos x , g ( x ) = sin x on [0 , π ]. 1. area = 2 2 correct 2. area = 4( 2 + 1) 3. area = 4 2 4. area = 2 + 1 5. area = 2( 2 + 1) 6. area = 2 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 = i b a | f ( x ) g ( x ) | dx , which For the given Functions is the integral A = i π 0 | cos x sin x | dx . But, as the graphs

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momin (rrm497) – Homework 04 – cheng – (58520) 2 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 b 0 , on [0 , π/ 4], 0 , on [ π/ 4 , π ]. Thus A = i π/ 4 0 { cos θ sin θ } i π π/ 4 { cos θ sin θ } = A 1 A 2 . But by the Fundamental Theorem of Calcu- lus, A 1 = B sin θ + cos θ ± π/ 4 0 = 2 1 , while A 2 = B sin θ + cos θ ± π π/ 4 = (1 + 2) . Consequently, area = A 1 A 2 = 2 2 . 003 10.0 points Compute the area between the graphs of f ( x ) = 8 sin2 x and g ( x ) = 6 sin x + 2 sin2 x on [0 , π/ 2]. 1. Area = 5 2 sq.units 2. Area = 13 4 sq.units 3. Area = 11 4 sq.units 4. Area = 9 4 sq.units 5. Area = 3 sq.units correct Explanation: The required area is given by I = i π/ 2 0 | f ( x ) g ( x ) | dx. Now the graphs of f, g intersect when 8 sin2 x = 6 sin x + 2 sin 2 x , i.e. , when 6(sin 2 x sin x ) = 6 sin x (2 cos x 1) = 0 , since sin 2 x = 2 sin x cos x . Thus the points of intersection are x = 0 , π/ 3.
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## This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.

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HW4 - momin (rrm497) Homework 04 cheng (58520) This...

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