Notes: `y = x^3, y = x` Sketch the region enclosed by the given curves and find its area.

Friday, 6 October 2017

$\displaystyle{y}={{x}}^{{3}},{y}={x}$ Sketch the region enclosed by the given curves and find its area.

You need to determine first the points of intersection between curves $\displaystyle{y}={{x}}^{{3}}$ and $\displaystyle{y}={x}$ , by solving the equation, such that:

$\displaystyle{{x}}^{{3}}={x}\Rightarrow{{x}}^{{3}}-{x}={0}$

Factoring out x yields:

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

Hence, the endpoints of integral are x = -1, x = 0, x...

You need to determine first the points of intersection between curves $\displaystyle{y}={{x}}^{{3}}$ and $\displaystyle{y}={x}$ , by solving the equation, such that:

$\displaystyle{{x}}^{{3}}={x}\Rightarrow{{x}}^{{3}}-{x}={0}$

Factoring out x yields:

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

Hence, the endpoints of integral are x = -1, x = 0, x = 1.

You need to decide what curve is greater than the other on the interval [-1,1]. You need to notice that $\displaystyle{{x}}^{{3}}>{x}$ on the interval [-1,0], and $\displaystyle{{x}}^{{3}}<{x}$ on [0,1] hence, you may evaluate the area of the region enclosed by the given curves, such that:

$\displaystyle{\int_{{a}}^{{b}}}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.},$ where f(x) > g(x) for $\displaystyle{x}\in{\left[{a},{b}\right]}$

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

$\displaystyle{\int_{{-{1}}}^{{1}}}{\left|{{x}}^{{3}}-{x}\right|}{\left.{d}{x}\right.}={\int_{{-{1}}}^{{0}}}{{x}}^{{3}}{\left.{d}{x}\right.}-{\int_{{-{1}}}^{{0}}}{x}{\left.{d}{x}\right.}+{\int_{{0}}^{{1}}}{x}{\left.{d}{x}\right.}-{\int_{{0}}^{{1}}}{{x}}^{{3}}{\left.{d}{x}\right.}$

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

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

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

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

Hence, evaluating the area of the region enclosed by the given curves, yields $\displaystyle{\int_{{-{1}}}^{{1}}}{\left|{{x}}^{{3}}-{x}\right|}{\left.{d}{x}\right.}=\frac{{1}}{{2}}.$