Question

Calcule a área da região limitada superiormente pela função g(x)=8√x, x≥0
, e inferiormente pela função f(x) = x2.

a) 56/3
b) 75/3
c) 36/3
d) 45/3
e) 64/3

Answer 1

✅ Após resolver os cálculos, concluímos que a área entre as curvas das respectivas funções é:

 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S = \int\limits_{0}^{4}\,\left[8\sqrt{x} - x^{2}\right]\,dx = \frac{64}{3}\,u.\,a.\:\:\:}}\end{gathered} }$  

Portanto, a opção correta é:

                    $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf Alternativa\:E\:\:\:}}\end{gathered} }$

Sejam as funções:

                      $\Large\begin{cases}\tt g(x) = 8\sqrt{x},\:\:\:x\geq0\\ \tt f(x) = x^{2}\end{cases}$

Para resolver esta questão devemos:

• Obter o intervalo de integração. Para isso, devemos encontrar os pontos de interseção das duas curvas, ou seja:

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

                                $\Large\displaystyle\begin{gathered}\tt 8\sqrt{x} = x^{2}\end{gathered}$

                           $\Large\displaystyle\begin{gathered}\tt (8\sqrt{x})^{2} = (x^{2})^{2}\end{gathered}$

                     $\Large\displaystyle\begin{gathered}\tt 8^{2}\cdot(\sqrt{x})^{2} = x^{4}\end{gathered}$

                              $\Large\displaystyle\begin{gathered} \tt 64\cdot x = x^{4}\end{gathered}$

                       $\Large\displaystyle\begin{gathered} \tt x^4 - 64x = 0\end{gathered}$

         Portanto, as interseções são:

                $\Large\displaystyle\begin{gathered}\tt I' = (0,0)\:\:\:e\:\:\:I'' = (4, 16)\end{gathered}$

         Portanto o intervalo de integração é:

                               $\Large\displaystyle\begin{gathered}\tt I = \left[0, 4\right]\end{gathered}$

• Calcular a área limitada pelas funções. Para isso, devemos utilizar a seguinte fórmula:

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

         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:

             $\Large\displaystyle\begin{gathered}\tt S = \int\limits_{0}^{4}\,\left[8\sqrt{x} - x^{2}\right]\,dx\end{gathered}$

                  $\Large\displaystyle\text{ \begin{gathered}\tt = \int\limits_{0}^{4}\,\left[8x^{\frac{1}{2}} - x^2\right]\,dx\end{gathered} }$

                  $\Large\displaystyle\text{ \begin{gathered}\tt = \left[8\cdot\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} -\frac{x^{2 + 1}}{2 + 1} + c \right]\Bigg|_{0}^{4}\end{gathered} }$

                  $\Large\displaystyle\text{ \begin{gathered}\tt = \left[8\cdot\frac{x^{\frac{3}{2}}}{\frac{3}{2}}- \frac{x^{3}}{3} + c\right]\Bigg|_{0}^{4}\end{gathered} }$

                  $\Large\displaystyle\text{ \begin{gathered}\tt = \left[\frac{16x^{\frac{3}{2}}}{3} -\frac{x^{3}}{3} + c\right]\Bigg|_{0}^{4}\end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}\tt = \left[\frac{16\cdot4^{\frac{3}{2}}}{3} - \frac{4^{3}}{3} + c\right] - \left[\frac{16\cdot0^{\frac{3}{2}}}{3} - \frac{0^{3}}{3} + c\right]\end{gathered} }$

                   $\Large\displaystyle\begin{gathered}\tt = \left[\frac{128}{3} - \frac{64}{3} + c\right] - \left[0 - 0 + c\right]\end{gathered}$

                   $\Large\displaystyle\begin{gathered}\tt = \frac{128}{3} - \frac{64}{3} + c - c\end{gathered}$

                   $\Large\displaystyle\begin{gathered}\tt = \frac{128 - 64}{3}\end{gathered}$

                   $\Large\displaystyle\begin{gathered}\tt = \frac{64}{3}\end{gathered}$

✅ Portanto, a área é:

               $\Large\displaystyle\begin{gathered}\tt S = \frac{64}{3}\,u.\,a.\end{gathered}$                  

           

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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