Question

calcule a área limitada entre as funções y=3x^2 e y = 9x

Answer 1

Pontos de intersecção:

$\mathsf{3x^2=9x\div3}\\\mathsf{x^2=3x}\\\mathsf{x^2-3x=0\implies~x(x-3)=0}\\\mathsf{x=0}\\\mathsf{x-3=0\implies~x=3}$

$\mathsf{y=9x}\\\mathsf{y=9.0=0}\\\mathsf{y=9.3=27}$

Os pontos são (0,0) (3,27)

A área pedida é dada por

$\displaystyle\mathsf{A=\int\limits_{0}^{3} </p><p>(9x-3x^2)dx}=\displaystyle\mathsf{\left[\dfrac{9}{2}x^2-x^3\right]_{0}^3}$

Note que o limite inferior é nulo. Portanto é muito mais prático substituir somente o limite superior de integração.

Daí

$\mathsf{A=\dfrac{9}{2}\cdot3^2-3^3}\\\mathsf{A=\dfrac{81}{2}-27}\\\mathsf{A=\dfrac{81-54}{2}}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{A=\dfrac{27}{2}~u\cdot a}}}}}$

$\boxed{\boxed{\textsf{Espero~ter~ajudado}}}$

$\dotfill$

$\displaystyle\mathsf{\ell ife=\int\limits_{birth}^{death}\dfrac{happiness}{time}dtime}$

Answer 2

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

 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S = \int_{0}^{3}(-3x^{2} + 9x)\,dx = \frac{27}{2}\,u.\,a.\:\:\:}}\end{gathered} }$

Sejam as funções polinomiais:

                          $\Large\begin{cases}\tt y = 3x^{2}\\ \tt y = 9x\end{cases}$

Organizando as funções temos:

                          $\Large\begin{cases}\tt f(x) = 3x^{2}\\ \tt g(x) = 9x\end{cases}$

Para resolver esta questão, devemos:

• Obter o intervalo de integração. Para isso fazemos:

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

                    $\Large\displaystyle\begin{gathered}\tt 3x^{2} = 9x\end{gathered}$

        $\Large\displaystyle\begin{gathered}\tt 3x^{2} - 9x = 0\end{gathered}$  

  $\Large\displaystyle\begin{gathered}\tt x\cdot(3x - 9) = 0\end{gathered}$

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

                             e

           $\Large\displaystyle\begin{gathered}\tt 3x'' - 9 = 0\end{gathered}$

                       $\Large\displaystyle\begin{gathered}\tt x'' = 3\end{gathered}$

           Portanto, o intervalo de integração é:

                   $\Large\displaystyle\begin{gathered}\tt I = (x', x'') = (0, 3)\end{gathered}$

• calcular a área limitada pelas funções.

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

         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:

              $\Large\displaystyle\begin{gathered}\tt S = \int_{0}^{3}(9x - 3x^{2})\,dx\end{gathered}$

                  $\Large\displaystyle\begin{gathered}\tt = \int_{0}^{3}(-3x^{2} + 9x)\,dx\end{gathered}$

                   $\Large\displaystyle\text{ \begin{gathered}\tt = \bigg(-\frac{3x^{2 + 1}}{2 + 1} + \frac{9x^{1 + 1}}{1 + 1} + c\bigg)\bigg|_{0}^{3}\end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}\tt = \bigg(-\frac{3x^{3}}{3} + \frac{9x^{2}}{2} + c\bigg)\bigg|_{0}^{3}\end{gathered} }$

                    $\large\displaystyle\text{ \begin{gathered}\tt = \bigg(-\frac{3\cdot3^{3}}{3} + \frac{9\cdot3^{2}}{2} + c\bigg) - \bigg(-\frac{3\cdot0^{3}}{3} + \frac{9\cdot0^{2}}{2} + c\bigg)\end{gathered} }$

                    $\Large\displaystyle\begin{gathered}\tt = \bigg(-\frac{81}{3} + \frac{81}{2} + c\bigg) - \bigg(0 + 0 + c\bigg)\end{gathered}$

                    $\Large\displaystyle\begin{gathered}\tt = -\frac{81}{3} + \frac{81}{2} + c - c\end{gathered}$

                     $\Large\displaystyle\begin{gathered}\tt = -\frac{81}{3} + \frac{81}{2}\end{gathered}$

                      $\Large\displaystyle\begin{gathered}\tt = \frac{-162 + 243}{6}\end{gathered}$

                      $\Large\displaystyle\begin{gathered}\tt = \frac{81}{6}\end{gathered}$

                      $\Large\displaystyle\begin{gathered}\tt = \frac{27}{2}\end{gathered}$

✅ Portanto, a área procurada é:

    $\Large\displaystyle\begin{gathered}\tt S = \int_{0}^{3}(-3x^{2} + 9x)\,dx = \frac{27}{2}\,u.\,a.\end{gathered}$

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

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$