A=∫ba|=∫π0|sinx−cosx|dx=∫π/40(cosx−sinx)dx+∫ππ/4(sin=[sinx+cosx]|π/40+[−cos=(√2−1)+(1+ The area of the region is 2√2 units2. If R is the region between the graphs of the functions f(x)=sinx and g(x)=cosx over the interval [π/2,2π], find the area of region R.

## Answers ( )

Answer:A=∫ba|=∫π0|sinx−cosx|dx=∫π/40(cosx−sinx)dx+∫ππ/4(sin=[sinx+cosx]|π/40+[−cos=(√2−1)+(1+ The area of the region is 2√2 units2. If R is the region between the graphs of the functions f(x)=sinx and g(x)=cosx over the interval [π/2,2π], find the area of region R.

Step-by-step explanation:Step-by-step explanation:The area enclosed between y

2

=x and y=x

October 15, 2019

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Hashmat Shaikh

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VIDEO EXPLANATION

ANSWER

Intersection point of two curves are O(0,0) and A(1,1)

∴ Required area =∫

0

4

(y

2

−y

1

)dx

∫

0

1

(

x

−x)dx

=[

3/2

x

3/2

−

2

x

2

]

0

1

=

3

2

[1−0]−

2

1

(1−0)

=

3

2

−

2

1

=

6

1

unit

solution