Question

Determine a area da regiao limitada pelas curvas Y=x^2+2 y=4-x^2.

Answer 1

$\mathrm{Sejam\ duas\ curvas}\ f:y=x^2+2\ \text{e}\ g:y=4-x^2.$

$\text{Deseja-se}\ \mathrm{calcular\ a\ \acute{a}rea\ da\ regi\tilde{a}o\ delimitada\ entre}\ f\ \text{e}\ g.$

$\mathrm{O\ 1^{\circ}\ passo\ \acute{e}\ calcular\ as\ abscissas\ nas\ quais\ as\ curvas\ se\ interceptam.}$

$f\cap g\Longrightarrow x^2+2=4-x^2\Longrightarrow x^2=1\ \therefore\ \boxed{x=\pm1}$

$\mathrm{Desse\ modo,\ a\ regi\tilde{a}o\ entre}\ f\ \text{e}\ g\ \mathrm{est\acute{a}\ no\ intervalo}\ -1\leq x\leq 1.$

$\mathrm{Dado\ um\ determinado\ valor\ desse\ intervalo,\ p.\ ex.}\ x=0:$

$f(0)=0^2+2\Longrightarrow f(0)=2$

$g(0)=4-0^2\Longrightarrow g(0)=4$

$\mathrm{Quando}\ -1\leq x\leq 1,\ \mathrm{a\ imagem\ de}\ g\ \mathrm{est\acute{a}\ acima\ da\ imagem\ de}\ f.$

$\mathrm{J\acute{a}\ \acute{e}\ poss\acute{\i}vel\ calcular\ a\ \acute{a}rea\ da\ regi\tilde{a}o.}$

$\boxed{S=\int\limits_{x_1}^{x_2}{[g(x)-f(x)]\mathrm{d}x}}$

$\Longrightarrow S=\int\limits^{1}_{-1}{[4-x^2-(x^2+2)]\mathrm{d}x}\Longrightarrow S=\int\limits_{-1}^1{[2-2x^2]\mathrm{d}x}$

$\Longrightarrow S=2\bigg(\int{\mathrm{d}x}-\int{x^2\mathrm{d}x}\bigg)\bigg|_{-1}^{1}\Longrightarrow S=2\bigg(x-\dfrac{x^3}{3}\bigg)\bigg|^{1}_{-1}$

$\Longrightarrow S=2\bigg[\bigg(1-\dfrac{1}{3}\bigg)-\bigg(-1-\bigg(-\dfrac{1}{3}\bigg)\bigg)\bigg]$

$\Longrightarrow S=2\bigg[\dfrac{2}{3}+\dfrac{2}{3}\bigg]=2\bigg[\dfrac{4}{3}\bigg]\ \therefore\ \boxed{S=\dfrac{8}{3}\ \mathrm{u.a.}}$