Multiple choice, I will rate and like. Thank you. y= sin(x)y = cos(x)TT/2ST/2Two students, S and T, considered this problem:Consider the region trapped between the graphs of y = sin x and y = cos x, on the interval 0 < x < .This region is shaded in the figure included above.To find the area of the shaded region, Student S set up the following integral.S (sin æ – cos æ) dxHowever, Student T says this won't find the correct value.What do you think?Student T is correct. A better integral expression isL² (cos x – sin x) dæStudent T is incorrect.Student S has set up the integral correctly.Student T is correct. A better integral expression isL* (sin x – cos x) dx + f7 (cos x – sin æ) dxStudent T is correct. Student S should include Tr² like this:T7 (sin x – cos T)- cos r)°dx

From the given bounded region: Consider the vertical strip; The area of the vertical strip is;…

sy = sin(x)- y = cos(x) T/2 Two students, S and T, considered this problem: Consider the region trapped between the graphs of y = sin x and y = cos x, on the interval 0 < x < 5. This region is shaded in the figure included above. To find the area of the shaded region, Student S set up the following integral. SF (sin z – cos æ) dæ However, Student T says this won't find the correct value. What do you think? Student T is correct. A better integral expression is SF (cos x – sin æ) dr Student T is incorrect. Student S has set up the integral correctly. Student T is correct. A better integral expression is S* (sin æ – cos æ) dx + ƒ? (cos ¤ – sin æ) dx Student Tis correct. Student S should include Tr² like this: TS3 (sin æ – cos æ)² dæ

sy = sin(x)- y = cos(x) T/2 Two students, S and T, considered this problem: Consider the region trapped between the graphs of y = sin x and y = cos x, on the interval 0 < x < 5. This region is shaded in the figure included above. To find the area of the shaded region, Student S set up the following integral. SF (sin z – cos æ) dæ However, Student T says this won't find the correct value. What do you think? Student T is correct. A better integral expression is SF (cos x – sin æ) dr Student T is incorrect. Student S has set up the integral correctly. Student T is correct. A better integral expression is S* (sin æ – cos æ) dx + ƒ? (cos ¤ – sin æ) dx Student Tis correct. Student S should include Tr² like this: TS3 (sin æ – cos æ)² dæ

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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