Homework Help is Here – Start Your Trial Now!
Math
Calculus
EXAMPLE 5 Find the area of the region bounded by the curves y = sin(x), y = cos(x), x = 0, and x = 7/2. SOLUTION The points of intersection occur when sin(x) = cos(x), that is, when x = (since 0sxs n/2). The region is sketched in the figure. Observe that cos(x) > sin(x) when , but sin(x) 2 cos(x) when Therefore the required area is 피2 ! Icos(x) – sin(x)| dx = A1 + A2 A = 7/4 /2 | (cos(x) – sin(x)) dx + 7/4 (sin(x) – cos(x)) dx 7/4 7/2 + 피/4 -0 - 1 + -0 - 1 + = 2/2 - 2. In this particular example we could have saved some work by noticing that the region is symmetric about x = n/4 and so /4 A = 2A, = 2 (cos(x) – sin(x)) dx. Jo

EXAMPLE 5 Find the area of the region bounded by the curves y = sin(x), y = cos(x), x = 0, and x = 7/2. SOLUTION The points of intersection occur when sin(x) = cos(x), that is, when x = (since 0sxs n/2). The region is sketched in the figure. Observe that cos(x) > sin(x) when , but sin(x) 2 cos(x) when Therefore the required area is 피2 ! Icos(x) – sin(x)| dx = A1 + A2 A = 7/4 /2 | (cos(x) – sin(x)) dx + 7/4 (sin(x) – cos(x)) dx 7/4 7/2 + 피/4 -0 - 1 + -0 - 1 + = 2/2 - 2. In this particular example we could have saved some work by noticing that the region is symmetric about x = n/4 and so /4 A = 2A, = 2 (cos(x) – sin(x)) dx. Jo

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 67E
See similar textbooks
icon
Related questions
Question
EXAMPLE 5 Find the area of the region bounded by the curves y = sin(x), y = cos(x), x = 0, and x = 7/2. SOLUTION The points of intersection occur when sin(x) = cos(x), that is, when x = (since 0sxs n/2). The region is sketched in the figure. Observe that cos(x) > sin(x) when , but sin(x) 2 cos(x) when Therefore the required area is 피2 ! Icos(x) – sin(x)| dx = A1 + A2 A = 7/4 /2 | (cos(x) – sin(x)) dx + 7/4 (sin(x) – cos(x)) dx 7/4 7/2 + 피/4 -0 - 1 + -0 - 1 + = 2/2 - 2. In this particular example we could have saved some work by noticing that the region is symmetric about x = 7/4 and so /4 A = 2A, = 2 (cos(x) – sin(x)) dx. Jo
Transcribed Image Text:EXAMPLE 5 Find the area of the region bounded by the curves y = sin(x), y = cos(x), x = 0, and x = 7/2. SOLUTION The points of intersection occur when sin(x) = cos(x), that is, when x = (since 0sxs n/2). The region is sketched in the figure. Observe that cos(x) > sin(x) when , but sin(x) 2 cos(x) when Therefore the required area is 피2 ! Icos(x) – sin(x)| dx = A1 + A2 A = 7/4 /2 | (cos(x) – sin(x)) dx + 7/4 (sin(x) – cos(x)) dx 7/4 7/2 + 피/4 -0 - 1 + -0 - 1 + = 2/2 - 2. In this particular example we could have saved some work by noticing that the region is symmetric about x = 7/4 and so /4 A = 2A, = 2 (cos(x) – sin(x)) dx. Jo
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
SEE MORE TEXTBOOKS