Question

calcule a area da regiao limitada pelas funções y=2x e y=-x^2-8x

Answer 1

$\Bmatrix y_a =2x\\y_b = -x^2-8x\end$

encontrando os pontos de insersecção
$y_a = y_b\\\\2x=-x^2-8x\\\\0=-x^2-10x\\\\\boxed{0=x(x+10)}$

x= -10 , x=0 este será o intervalo

verificando qual curva limita a area por cima e qual limita por baixo
calculando ya para x=-5
ya = 2*(-5)  = -10

yb para x =-5
yb =(-5)² -8*(-5) = 65

logo a curva yb limita a area por cima e a curva ya limita por baixo

temos
$Area = \int\limits^0_{-10} {(y_b-y_a)} \, dx \\\\\\Area= \int\limits^0_{-10} {-x^2-8x-2x} \, dx \\\\\\Area= \int\limits^0_{-10} {-x^2-10x} \, dx \\\\\\Area = \left [\frac{-x^3}{3}- 10\frac{x^2}{2}\right]^0_{-10} \\\\\\Area = \right\frac{-x^3}{3}-5x^2 \right]^0_{-10} \\\\\\Area= \left( \frac{-0^3}{3}-5*0^2\right) - \left( \frac{-(-10)^3}{3}-5*(-10)^2\right) = \frac{500}{3}$

Answer 2

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

 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S = \int_{-10}^{0}(-x^{2} - 10x)\,dx = \frac{500}{3}\,u.\,a.\:\:\:}}\end{gathered} }$

Sejam as funções polinomiais:

                          $\Large\begin{cases}\tt y = 2x\\ \tt y = -x^{2} - 8x\end{cases}$

Organizando as funções temos:

                          $\Large\begin{cases}\tt f(x) = 2x\\ \tt g(x) = -x^{2} - 8x\end{cases}$

Para resolver esta questão, devemos:

• Obter o intervalo de integração. Para isso fazemos:

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

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

             $\Large\displaystyle\begin{gathered}\tt -x^{2} - 8x - 2x = 0\end{gathered}$

                      $\Large\displaystyle\begin{gathered}\tt -x^{2} - 10x = 0\end{gathered}$

                  $\Large\displaystyle\begin{gathered}\tt x\cdot(-x - 10) = 0\end{gathered}$

           Então:

           $\Large\displaystyle\begin{gathered}\tt -x' - 10 = 0\Longrightarrow x' = -10\end{gathered}$

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

           Portanto, o intervalo de integração é:

                   $\Large\displaystyle\begin{gathered}\tt I = (x', x'') = (-10, 0)\end{gathered}$

• 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:

              $\Large\displaystyle\begin{gathered}\tt S = \int_{-10}^{0}(-x^{2} - 8x - 2x)\,dx\end{gathered}$

                   $\Large\displaystyle\begin{gathered}\tt = \int_{-10}^{0}(-x^{2} - 10x)\,dx\end{gathered}$

                   $\Large\displaystyle\text{ \begin{gathered}\tt = \bigg(-\frac{x^{2 + 1}}{2 + 1} - \frac{10x^{1 + 1}}{1 + 1} + c\bigg)\bigg|_{-10}^{0}\end{gathered} }$

                   $\Large\displaystyle\text{ \begin{gathered}\tt = \bigg(-\frac{x^{3}}{3} - \frac{10x^{2}}{2} + c\bigg)\bigg|_{-10}^{0}\end{gathered} }$

                   $\large\displaystyle\text{ \begin{gathered}\tt = \bigg(-\frac{0^{3}}{3} - \frac{10\cdot0^{2}}{2} + c\bigg) - \bigg(-\frac{(-10)^{3}}{3} - \frac{10\cdot(-10)^{2}}{2} + c\bigg)\end{gathered} }$

                   $\Large\displaystyle\begin{gathered}\tt = c - \frac{1000}{3} + \frac{1000}{2} - c \end{gathered}$

                   $\Large\displaystyle\begin{gathered}\tt = -\frac{1000}{3} + \frac{1000}{2}\end{gathered}$

                  $\Large\displaystyle\begin{gathered}\tt = \frac{-2000 + 3000}{6}\end{gathered}$

                   $\Large\displaystyle\begin{gathered}\tt = \frac{1000}{6}\end{gathered}$

                   $\Large\displaystyle\begin{gathered}\tt = \frac{500}{3}\end{gathered}$

✅ Portanto, a área procurada é:

        $\Large\displaystyle\begin{gathered}\tt S = \int_{-10}^{0}(-x^{2} - 10x)\,dx = \frac{500}{3}\,u.\,a.\end{gathered}$

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

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