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; ∫dA=∫LdxA=∫Ldx Where; ''L'' is the height and dx is the thickness of the strip. Here the length of the strip L must be equal to L=g(x)-f(x), where g(x)>f(x) or we can say the graph of g(x) must be above the graph of f(x). From the shown region, if g(x) = sin(x) and f(x) = cos(x), then sin(x) > cos(x) when x∈0,π4 and sin(x) < cos(x) when x∈π4,π2, which means we should have to integrating by breaking the integral in two parts as below: A1=∫y2-y1dx=∫0π4sinx-cosxdx And A2=∫y2-y1dx=∫π4π2cosx-sinxdx Such that, A=A1+A2 So, A=A1+A2=∫0π4sinx-cosxdx+∫π4π2cosx-sinxdx Hence, the student T is correct, and better expression for find the area is: A=∫0π4sinx-cosxdx+∫π4π2cosx-sinxdx The correct option is: (c).

Math

Calculus

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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Question

100%

Multiple choice, I will rate and like. Thank you.

y= sin(x)
y = cos(x)
TT/2
ST/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 < .
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) dx
However, Student T says this won't find the correct value.
What do you think?
Student T is correct. A better integral expression is
L² (cos x – sin x) dæ
Student T is incorrect.
Student S has set up the integral correctly.
Student T is correct. A better integral expression is
L* (sin x – cos x) dx + f7 (cos x – sin æ) dx
Student T is correct. Student S should include Tr² like this:
T7 (sin x – cos T)
- cos r)°dx

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