`y=cos(x) , y=1-cos(x) ,0<=x<=pi`

Refer the attached image, y=cos(x) is plotted in red color and y=1-cos(x) is plotted in blue color.

From the graph,

cos(x) is above (1-cos(x)) from 0 to pi/3 and

(1-cos(x)) is above cos(x) from pi/3 to pi.

Area of the region enclosed by the given curves...

## See

This Answer NowStart your **subscription** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

`y=cos(x) , y=1-cos(x) ,0<=x<=pi`

Refer the attached image, y=cos(x) is plotted in red color and y=1-cos(x) is plotted in blue color.

From the graph,

cos(x) is above (1-cos(x)) from 0 to pi/3 and

(1-cos(x)) is above cos(x) from pi/3 to pi.

Area of the region enclosed by the given curves A=`int_0^(pi/3)(cos(x)-(1-cos(x)))dx+int_(pi/3)^pi((1-cos(x))-cos(x))dx`

`A=int_0^(pi/3)(2cos(x)-1)dx+int_(pi/3)^pi(1-2cos(x))dx`

`A=[2sin(x)-x]_0^(pi/3)+[x-2sin(x)]_(pi/3)^pi`

`A=((2sin(pi/3)-pi/3)-(2sin(0)-0))+(pi-2sin(pi)-(pi/3-2sin(pi/3))`

`A=(2*sqrt(3)/2-pi/3)+pi-pi/3+2*sqrt(3)/2`

`A=sqrt(3)-pi/3+pi-pi/3+sqrt(3)`

`A=pi/3+2sqrt(3)~~4.511`