Notes: `y = |x|, y = x^2 - 2` Sketch the region enclosed by the given curves and find its area.

Wednesday, 3 February 2016

$\displaystyle{y}={\left|{x}\right|},{y}={{x}}^{{2}}-{2}$ Sketch the region enclosed by the given curves and find its area.

You need to find the intersection points beteen the curves, hence, since |x| = y, you need to discuss two cases, y = -x and y = x.

For y = -x, you need to find the intersection between curves by solving the equation, such that:

$\displaystyle-{x}={{x}}^{{2}}-{2}\Rightarrow{{x}}^{{2}}+{x}-{2}={0}$

$\displaystyle{x}_{{{1},{2}}}=\frac{{-{1}\pm\sqrt{{{1}+{8}}}}}{{2}}\Rightarrow{x}_{{{1},{2}}}=\frac{{-{1}\pm{3}}}{{2}}$

$\displaystyle{x}_{{1}}={1};{x}_{{2}}=-{2}$

Hence,...

You need to find the intersection points beteen the curves, hence, since |x| = y, you need to discuss two cases, y = -x and y = x.

For y = -x, you need to find the intersection between curves by solving the equation, such that:

$\displaystyle-{x}={{x}}^{{2}}-{2}\Rightarrow{{x}}^{{2}}+{x}-{2}={0}$

$\displaystyle{x}_{{{1},{2}}}=\frac{{-{1}\pm\sqrt{{{1}+{8}}}}}{{2}}\Rightarrow{x}_{{{1},{2}}}=\frac{{-{1}\pm{3}}}{{2}}$

$\displaystyle{x}_{{1}}={1};{x}_{{2}}=-{2}$

Hence, you need to check what curve is greater on interval [-2,1] and you may notice that $\displaystyle-{x}>{{x}}^{{2}}-{2}$ on [-2,0] and $\displaystyle-{x}<{{x}}^{{2}}-{2}$ on [0,1], hence, you may evaluate the area. such that:

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

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

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

$\displaystyle{\int_{{-{2}}}^{{1}}}{\left|-{x}-{{x}}^{{2}}+{2}\right|}{\left.{d}{x}\right.}={2}-\frac{{8}}{{3}}+{4}-{2}+\frac{{1}}{{3}}+\frac{{1}}{{2}}$

$\displaystyle{\int_{{-{2}}}^{{1}}}{\left|-{x}-{{x}}^{{2}}+{2}\right|}{\left.{d}{x}\right.}={4}-\frac{{7}}{{3}}+\frac{{1}}{{2}}$

$\displaystyle{\int_{{-{2}}}^{{1}}}{\left|-{x}-{{x}}^{{2}}+{2}\right|}{\left.{d}{x}\right.}=\frac{{{24}-{14}+{3}}}{{6}}$

$\displaystyle{\int_{{-{2}}}^{{1}}}{\left|-{x}-{{x}}^{{2}}+{2}\right|}{\left.{d}{x}\right.}=\frac{{13}}{{6}}$

Hence, evaluating the area enclosed by the curves $\displaystyle{y}=-{x}$ and $\displaystyle{y}={{x}}^{{2}}-{2}$ yields $\displaystyle{\int_{{-{2}}}^{{1}}}{\left|-{x}-{{x}}^{{2}}+{2}\right|}{\left.{d}{x}\right.}=\frac{{13}}{{6}}.$