# Sketch the region enclosed by the given curves. Then find the are of the region: 1. y=sin x, y=x, x=PI/2, x=pi

*print*Print*list*Cite

You need to evaluate the functions `y = sin x` and `y = x` at `x = pi/2` and `x = pi` , such that:

`x = pi/2 => y = sin (pi/2) = 1`

`x= pi/2 => y = pi/2 ~~ 1.57`

`x = pi => y = sin pi = 0`

`x = pi => y = pi = 3.14`

Notice that the graph of the function `y = x` is located above the graph of the function `y = sin x` , over the interval `[pi/2, pi]` , hence, evaluating the graph of the region between `y = sin x, y = x` and `x = pi/2, x = pi` , yields:

`A = int_(pi/2)^pi (x - sin x) dx`

Using the property of linearity of integral yields:

`A = int_(pi/2)^pi x dx - int_(pi/2)^pi sin x dx`

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

Using the fundamental theorem of calculus yields:

`A = (pi^2/2 - pi^2/8) + (cos pi - cos (pi/2))`

Since `cos pi = -` 1 and `cos(pi/2) = 0` yields:

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

**Hence, evaluating the area of the region enclosed by the given curves, using the process of integration, yields `A = (3pi^2)/8 - 1.` **