Notes: `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...

Friday, 4 September 2015

$\displaystyle{y}={\sin{{\left({x}\right)}}},{y}=\frac{{{2}{x}}}{\pi},{x}\Rightarrow{0}$ Sketch the region enclosed by the given curves. Decide whether to integrate with respect to $\displaystyle{x}$ or $\displaystyle{y}$ . Draw a...

Here is the sketch of the two given functions.

The $\displaystyle{y}={\sin{{\left({x}\right)}}}$ is plotted with a red color while $\displaystyle{y}=\frac{{{2}{x}}}{\pi}$ 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 = $\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$

such that $\displaystyle{f{{\left({x}\right)}}}\geq{g{{\left({x}\right)}}}$ for interval [a,b]

Or A = $\displaystyle{\int_{{a}}^{{b}}}{\left[{y}_{{{a}{b}{o}{v}{e}}}-{y}_{{{b}{e}{l}{o}{w}}}\right]}{\left.{d}{x}\right.}$

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)= $\displaystyle{y}_{{{a}{b}{o}{v}{e}}}-{y}_{{{b}{e}{l}{o}{w}}}.$

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 $\displaystyle{y}_{{{a}{b}{o}{v}{e}}}=\frac{{{2}{x}}}{\pi}$

and $\displaystyle{y}_{{{b}{e}{l}{o}{w}}}={\sin{{\left({x}\right)}}}$ with limits of integration from x =-1.57 to x=0.

$\displaystyle{A}_{{1}}=$ $\displaystyle{\int_{{-{{1.57}}}}^{{0}}}{\left[\frac{{{2}{x}}}{\pi}-{\sin{{\left({x}\right)}}}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left[{\cos{{\left({x}\right)}}}+\frac{{{x}}^{{2}}}{\pi}\right]}{{\mid}_{{-{1.57}}}^{{0}}}$

= $\displaystyle{\left[{\cos{{\left({0}\right)}}}+\frac{{{0}}^{{2}}}{\pi}\right]}-{\left[{\cos{{\left(-{1.57}\right)}}}+\frac{{{\left(-{1.57}\right)}}^{{2}}}{\pi}\right]}$

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

= 1 - 0.7853983685

$\displaystyle\approx$ 0.2146

In the second bounded region (pink in shade), we have the $\displaystyle{y}_{{{a}{b}{o}{v}{e}}}={\sin{{\left({x}\right)}}}$

and $\displaystyle{y}_{{{b}{e}{l}{o}{w}}}=\frac{{{2}{x}}}{\pi}$ with limits of integration from x =0 to x=1.57.

$\displaystyle{A}_{{2}}$ = $\displaystyle{\int_{{0}}^{{1.57}}}{\left[{\sin{{\left({x}\right)}}}-\frac{{{2}{x}}}{\pi}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left[-{\cos{{\left({x}\right)}}}-\frac{{{x}}^{{2}}}{\pi}\right]}{{\mid}_{{0}}^{{{1.57}}}}$

= $\displaystyle{\left[-{\cos{{\left({1.57}\right)}}}-\frac{{{\left({1.57}\right)}}^{{2}}}{\pi}\right]}-{\left[-{\cos{{\left({0}\right)}}}-\frac{{{0}}^{{2}}}{\pi}\right]}$