Encontre o valor da área A demarcada, entre as curvas.

Anexos:

Respostas

respondido por: solkarped

✅ Após resolver os cálculos, concluímos que a área da figura de cor azul entre as referidas curvas é:

$\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\tt A = 8\,u.\,a.\:\:\:}}\end{gathered}$}$

Sejam as funções:

$\Large\begin{cases}\tt f(x) = -x^{2} + 4x\\\tt g(x) = x^{2}\end{cases}$

Analisando o gráfico fornecido percebemos que a área da figura azul corresponde à soma de duas integrais definidas, que são:

$\Large\displaystyle\text{$\begin{gathered}\tt 1^{\underline{o}}\end{gathered}$}$ $\Large\displaystyle\text{$\begin{gathered}\tt A = \int_{a}^{b}g(x)\,dx + \int_{b}^{c}f(x)\,dx\end{gathered}$}$

Para calcular esta área devemos calcular os limites de integração. Para isso encontrar os limites de integração do primeiro intervalo:

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

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

$\Large\displaystyle\text{$\begin{gathered}\tt -x^{2} - x^{2} + 4x = 0\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt -2x^{2} + 4x = 0\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt -x^{2} + 2x = 0\end{gathered}$}$

Onde as raízes são:

$\Large\displaystyle\text{$\begin{gathered}\tt x' = 0\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt x'' - 2 = 0 \Longrightarrow x'' = 2\end{gathered}$}$

Portanto, o primeiro intervalo é:

$\Large\displaystyle\text{$\begin{gathered}\tt I_{1} = \left[x', \,x''\right] = \left[0,\,2\right]\end{gathered}$}$

Agora devemos encontrar os limites de integração do segundo intervalo:

$\Large\displaystyle\text{$\begin{gathered}\tt -x^{2} + 4x = 0\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt x''' = 0\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt x'''' - 4 = 0\Longrightarrow x''''= 4\end{gathered}$}$

Como os intervalos são complementares, então o segundo intervalo é:

$\Large\displaystyle\text{$\begin{gathered}\tt I_{2} = \left[x'',\,x''''\right] = \left[2, 4\right]\end{gathered}$}$

Substituindo os valores na primeira equação temos:

$\Large\displaystyle\text{$\begin{gathered}\tt A = \int_{0}^{2}x^{2}\,dx + \int_{2}^{4}(-x^{2} + 4x)\,dx\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \bigg(\frac{x^{2 + 1}}{2 + 1}\bigg)\bigg|_{0}^{2} + \bigg(-\frac{x^{2 + 1}}{2 + 1} + \frac{4x^{1 + 1}}{1 + 1}\bigg)\bigg|_{2}^{4}\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \bigg(\frac{x^{3}}{3}\bigg)\bigg|_{0}^{2} + \bigg(-\frac{x^{3}}{3} + \frac{4x^{2}}{2}\bigg)\bigg|_{2}^{4}\end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\tt = \left[\frac{2^{3}}{3} - \frac{0^{3}}{3}\right] + \left[\bigg(\frac{-(4^{3})}{3} + \frac{4\cdot4^{2}}{2}\bigg) - \bigg(\frac{-(2^{3})}{3} + \frac{4\cdot2^{2}}{2}\bigg)\right]\end{gathered}$}$

$\Large\displaystyle\text{$\begin{gathered}\tt = \frac{8}{3} + \left[\bigg(\frac{-64}{3} + \frac{64}{2}\bigg) - \bigg(\frac{-8}{3} + \frac{16}{2}\bigg)\right]\end{gathered}$}$