# `y = sin(x), y = (2x)/pi, x=>0` Sketch the region enclosed by the given curves. Decide whether to integrate with respect to `x` or `y`. Draw a typical approximating rectangle and label its...

`y = sin(x), y = (2x)/pi, x=>0` Sketch the region enclosed by the given curves. Decide whether to integrate with respect to `x` or `y`. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

marizi | Certified Educator

Here is the sketch of the two given functions.

The `y = sin(x) ` is plotted with a red color while `y = (2x)/pi` is plotted with a blue color.

As shown the graph, the two graphs intersect at the following points (approximately):

---> (-1.57 -1)

---> (0,0)

---> (1.57, 1).

The x-values from the intersection points will be used as the limits of integration or boundary values of x for each bounded region.

Using integration with respect to x, we follow the formula for the "Area between Two Curves" as:

A = `int_a^bf(x)-g(x)dx`

such that `f(x)gt=g(x)` for interval [a,b]

Or A =`int_a^b[y_(above)-y_(below)]dx`

Please see the attached file: "graph" to view how a typical approximating rectangle (sky blue in color) is used when using integration with respect to x. In the attached file, the width =dx and the height =f(x) such that

f(x)=` y_(above) - y_(below).`

To find the area of a bounded region, we will solve each bounded region with two separate integral then find the sum for the total bounded area/region.

In the first bounded region (yellow in shade), we have the `y_(above)= (2x)/pi`

and `y_(below) = sin(x)` with limits of integration from x =-1.57 to x=0.

`A_1=`  `int_-1.57^0[(2x)/pi -sin(x)]dx`

=`[cos(x)+x^2/pi]|_(-1.57)^0`

= `[cos(0)+0^2/pi]-[cos(-1.57)+(-1.57)^2/pi]`

= [1+0] - [ 0.00079633+0.7846020385]

= 1 - 0.7853983685

`~~` 0.2146

In the second bounded region (pink in shade), we have the `y_(above)=sin(x)`

and `y_(below) = (2x)/pi` with limits of integration from x =0 to x=1.57.

`A_2`  =`int_0^1.57[sin(x)-(2x)/pi]dx`

=`[-cos(x) - x^2/pi]|_0^(1.57)`

=`[-cos(1.57)-(1.57)^2/pi]-[-cos(0)-0^2/pi]`

=  [ -0.00079633-0.7846020385] -[-1-0]

= - 0.7853983685+1

`~~` ` ` 0.2146

Notice that they are symmetrical about the origin.

We can multiply `A_1` by 2 since the two bounded area is the same.

Total Area= ``

= 0.2146 +0.2146

=0.4292

In the first bounded region (yellow in shade), we have the y_(above)= (2x)/pi

and y_(below) = sin(x) with limits of integration from x =-1.57 to x=1.

In the first bounded region (yellow in shade), we have the y_(above)= (2x)/pi

and y_(below) = sin(x) with limits of integration from x =-1.57 to x=1.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)