`y = sin(x), y = x, x = pi/2, x = pi` Sketch the region enclosed by the given curves. Decide whether to integrate with respect to `x` or `y`....

`y=sin(x),y=x, x=pi/2 , x=pi`

Refer the attached image. Graph of y=sin(x) is plotted in red color and y=x is plotted in blue color.

Area of the region enclosed by the given curves A=`int_(pi/2)^pi(x-sin(x))dx`

`A=[x^2/2-(-cos(x))]_(pi/2)^(pi)`

`A=[x^2/2+cos(x)]_(pi/2)^(pi)`

`A=((pi)^2/2+cos(pi))-((pi/2)^2/2+cos(pi/2))`

`A=(pi^2/2-1)-(pi^2/8)`

`A=pi^2/2-pi^2/8-1`

`A=(3pi^2)/8-1~~2.7011`

`y=sin(x),y=x, x=pi/2 , x=pi`

Refer the attached image. Graph of y=sin(x) is plotted in red color and y=x is plotted in blue color.

Area of the region enclosed by the given curves A=`int_(pi/2)^pi(x-sin(x))dx`

`A=[x^2/2-(-cos(x))]_(pi/2)^(pi)`

`A=[x^2/2+cos(x)]_(pi/2)^(pi)`

`A=((pi)^2/2+cos(pi))-((pi/2)^2/2+cos(pi/2))`

`A=(pi^2/2-1)-(pi^2/8)`

`A=pi^2/2-pi^2/8-1`

`A=(3pi^2)/8-1~~2.7011`