Notes: `y = sqrt(x - 1), x - y = 1` Sketch the region enclosed by the given curves and find its area.

Friday, 15 April 2016

$\displaystyle{y}=\sqrt{{{x}-{1}}},{x}-{y}={1}$ Sketch the region enclosed by the given curves and find its area.

First, you need to find the point of intersection between the curves $\displaystyle{y}=\sqrt{{{x}-{1}}}$ and $\displaystyle{y}={x}-{1}$ , by solving the equation:

$\displaystyle\sqrt{{{x}-{1}}}={x}-{1}\Rightarrow{x}-{1}={{\left({x}-{1}\right)}}^{{2}}\Rightarrow{{\left({x}-{1}\right)}}^{{2}}-{\left({x}-{1}\right)}={0}$

Factoring out (x-1) yields:

$\displaystyle{\left({x}-{1}\right)}{\left({x}-{1}-{1}\right)}={0}\Rightarrow{\left({x}-{1}\right)}{\left({x}-{2}\right)}={0}\Rightarrow{x}-{1}={0}{\quad\text{or}\quad}{x}-{2}={0}$

Hence, x = 1 and x = 2 and these values...

First, you need to find the point of intersection between the curves $\displaystyle{y}=\sqrt{{{x}-{1}}}$ and $\displaystyle{y}={x}-{1}$ , by solving the equation:

$\displaystyle\sqrt{{{x}-{1}}}={x}-{1}\Rightarrow{x}-{1}={{\left({x}-{1}\right)}}^{{2}}\Rightarrow{{\left({x}-{1}\right)}}^{{2}}-{\left({x}-{1}\right)}={0}$

Factoring out (x-1) yields:

$\displaystyle{\left({x}-{1}\right)}{\left({x}-{1}-{1}\right)}={0}\Rightarrow{\left({x}-{1}\right)}{\left({x}-{2}\right)}={0}\Rightarrow{x}-{1}={0}{\quad\text{or}\quad}{x}-{2}={0}$

Hence, x = 1 and x = 2 and these values are the endpoints of the definite integral you need to evaluate to find the area enclosed by the given curves.

You must check what curve is greater than the other on interval [1,2] and you may notice that $\displaystyle{y}={x}-{1}$ is greater that $\displaystyle{y}=\sqrt{{{x}-{1}}}$ on interval [1,2].

$\displaystyle{x}-{1}>\sqrt{{{x}-{1}}}$

You may evaluate the area such that:

$\displaystyle{\int_{{1}}^{{2}}}{\left|{\left({\left({x}-{1}\right)}-\sqrt{{{x}-{1}}}\right)}\right|}{\left.{d}{x}\right.}={\int_{{1}}^{{2}}}{x}{\left.{d}{x}\right.}-{\int_{{1}}^{{2}}}{\left.{d}{x}\right.}-{\int_{{1}}^{{2}}}\sqrt{{{x}-{1}}}{\left.{d}{x}\right.}$

$\displaystyle{\int_{{1}}^{{2}}}{\left|{\left({\left({x}-{1}\right)}-\sqrt{{{x}-{1}}}\right)}\right|}{\left.{d}{x}\right.}={{\left|\frac{{{{x}}^{{2}}}}{{2}}\right|}_{{1}}^{{2}}}-{x}{{\left|_{{1}}^{{2}}-{\left(\frac{{2}}{{3}}\right)}{{\left({x}-{1}\right)}}^{{\frac{{3}}{{2}}}}\right|}_{{1}}^{{2}}}{\mid}$

$\displaystyle{\int_{{1}}^{{2}}}{\left|{\left({\left({x}-{1}\right)}-\sqrt{{{x}-{1}}}\right)}\right|}{\left.{d}{x}\right.}={\left|\frac{{4}}{{2}}-\frac{{1}}{{2}}-{2}+{1}-{\left(\frac{{2}}{{3}}\right)}+{0}\right|}$

$\displaystyle{\int_{{1}}^{{2}}}{\left|{\left({\left({x}-{1}\right)}-\sqrt{{{x}-{1}}}\right)}\right|}{\left.{d}{x}\right.}={\left|\frac{{3}}{{2}}-{1}-\frac{{2}}{{3}}\right|}$

$\displaystyle{\int_{{1}}^{{2}}}{\left|{\left({\left({x}-{1}\right)}-\sqrt{{{x}-{1}}}\right)}\right|}{\left.{d}{x}\right.}={\left|\frac{{{9}-{6}-{4}}}{{6}}\right|}$

$\displaystyle{\int_{{1}}^{{2}}}{\left|{\left({\left({x}-{1}\right)}-\sqrt{{{x}-{1}}}\right)}\right|}{\left.{d}{x}\right.}={\left|-\frac{{1}}{{6}}\right|}=\frac{{1}}{{6}}$

Hence, evaluating the area enclosed by the curves yields $\displaystyle{\int_{{1}}^{{2}}}{\left|{\left({\left({x}-{1}\right)}-\sqrt{{{x}-{1}}}\right)}\right|}{\left.{d}{x}\right.}={\left|-\frac{{1}}{{6}}\right|}=\frac{{1}}{{6}}.$