Matéria: Matemática
Autor: damicaeltb
Perguntado 3 anos atrás

1) Trace a área limitada pelas funções usando o Geogebra dada e calcule esta área: y = x² - 2 e y = 2

Respostas

respondido por: solkarped

✅ Após resolver os cálculos, concluímos que a área entre as referidas funções é:

$\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf S = \int_{-2}^{2}(-x^{2} + 4)\,dx = \frac{32}{3}\,u.\,a.\:\:\:}}\end{gathered}$}$

Sejam as funções polinomiais:

$\Large\begin{cases}\tt y = x^{2} - 2\\ \tt y = 2\end{cases}$

Organizando as funções temos:

$\Large\begin{cases}\tt f(x) = x^{2} - 2\\ \tt g(x) = 2\end{cases}$

Para resolver esta questão, devemos:

Obter o intervalo de integração. Para isso fazemos:

$\Large\displaystyle\text{$\begin{gathered}\tt f(x) = g(x)\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt x^{2} - 2 = 2\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt x^{2} = 2 + 2\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt x^{2} = 4\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt x = \pm\sqrt{4}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt x = \pm2\end{gathered}$}$

Portanto, o intervalo de integração é:

$\Large\displaystyle\text{$\begin{gathered}\tt I = (x', x'') = (-2, 2)\end{gathered}$}$

calcular a área limitada pelas funções.

$\Large\displaystyle\text{$\begin{gathered}\tt S = \int_{a}^{b} \left[g(x) - f(x)\right]\,dx\end{gathered}$}$

Onde:

$\Large\begin{cases} \tt S = \acute{A}rea\:entre\:as\:curvas\\\tt a = Limite\:inferior\:intervalo\\\tt b = Limite\:superior\:intervalo\\\tt g(x) = Func_{\!\!,}\tilde{a}o\:mais\:acima\\\tt f(x) = Func_{\!\!,}\tilde{a}o\:mais\:abaixo\end{cases}$

Então, temos:

$\Large\displaystyle\text{$\begin{gathered}\tt S = \int_{-2}^{2}\left[2 - (x^{2} - 2)\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \int_{-2}^{2}\left[2 - x^{2} + 2\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \int_{-2}^{2}\left[-x^{2} + 4\right]\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \bigg(\frac{-x^{2 + 1}}{2 + 1} + 4x + c\bigg)\bigg|_{-2}^{2}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \bigg(\frac{-x^{3}}{3} + 4x + c\bigg)\bigg|_{-2}^{2}\end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\tt = \bigg(\frac{-(2^{3})}{3} + 4\cdot2 + c\bigg) - \bigg(\frac{-(-2)^{3}}{3} + 4\cdot(-2) + c\bigg)\end{gathered}$}$