Question

Resolva à integral tripla abaixo

Answer 1

• O resultado dessa integral tripla é:

       $\large\displaystyle\text{ \begin{gathered}\int _0^2 \int^{4-2x}_0\int^{2-x-\frac{y}{2}}_0 dzdydx = \frac{8}{3}\end{gathered} }$

Desejamos resolver a seguinte integral tripla:

      $\large\displaystyle\text{ \begin{gathered}\int _0^2 \int^{4-2x}_0\int^{2-x-\frac{y}{2}}_0 dzdydx \end{gathered} }$

Para resolver integrais duplas // triplas, temos um teorema muito bom, chamado Teorema de Funibi. Com ele, podemos calcular aquela integral tripla sem ligar para a ordem. Logo:

          $\large\displaystyle\text{ \begin{gathered}\int _0^2 \int^{4-2x}_0\int^{2-x-\frac{y}{2}}_0 dzdydx \end{gathered} }$

       $\large\displaystyle\text{ \begin{gathered}\int _0^2 \left\{\int^{4-2x}_0\left[\int^{2-x-\frac{y}{2}}_0 dz\right]dy\right\}dx \end{gathered} }$

      $\large\displaystyle\text{ \begin{gathered}\int _0^2 \left\{\int^{4-2x}_0 \left[ \left.z\right|_0^{2-x-\frac{y}{2}}\right]dy\right\}dx \end{gathered} }$

     $\large\displaystyle\text{ \begin{gathered}\int _0^2 \left\{\int^{4-2x}_0 \left[2-x-\frac{y}{2}\right]dy\right\}dx \end{gathered} }$

   $\large\displaystyle\text{ \begin{gathered}\int _0^2 \left\{\left. \left[2y-xy-\frac{y^2}{4}\right] \right|_0^{4-2x}\right\}dx \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered}\int _0^2 \left\{ \left[2(4-2x)-x(4-2x)-\frac{(4-2x)^2}{4}\right] \right\} dx \end{gathered} }$

 $\large\displaystyle\begin{gathered}\int _0^2 \left(-8x+8+2x^2-\frac{(-2x+4)^2}{4} \right) dx \end{gathered}$

• Para finalizar, só resolver essa simples integral.

     $\large\displaystyle\begin{gathered}\int _0^2 \left(-8x+8+2x^2-\frac{(-2x+4)^2}{4} \right) dx \end{gathered}$

$\large\displaystyle\begin{gathered}\int _0^2 -8x dx +\int _0^28dx+\int _0^22x^2dx-\int_0^2\frac{(-2x+4)^2}{4} dx \end{gathered}$

   $\large\displaystyle\begin{gathered}\left.\frac{-8x^2}{2}\right|_0^2+\left. 8x\right|_0^2+\left. \frac{2x^3}{3}\right|_0^2-\frac{1}{4} \int_0^2 (-2x+4)^2 dx \end{gathered}$

         $\large\displaystyle\begin{gathered}-\frac{32}{2}+16+ \frac{16}{3}-\frac{1}{24} \cdot \left. \frac{u^3}{6} \right|_0^4 \end{gathered}$

           $\large\displaystyle\text{ \begin{gathered}-\frac{32}{2}+16+ \frac{16}{3}-\frac{1}{24} \cdot \frac{32}{3} \\ \vdots \\ \therefore \boxed{\boxed{\green{ \frac{8}{3} }}}\end{gathered} }$  

Opa! Só para fechar com chave de ouro. Irei resolver aquela integral $\large\displaystyle\begin{gathered}\int_0^2 (-2x+4)^2dx \end{gathered}$ novamente. Pois ficou muito ruim de entender anteriormente.

Para começar, apliquei o método da substituição simples. Chamando o -2x+4 de u e -2dx de du. Logo:

           $\large\displaystyle\begin{gathered}\int_0^2 (-2x+4)^2dx \Rightarrow -\frac{1}{2}\int _4^0 u^2du\end{gathered}$

Agora irei retirar esse sinal negativo do 1/2. Mudando então a ordem do intervalo.

            $\large\displaystyle\begin{gathered}\int_0^2 (-2x+4)^2dx \Rightarrow \frac{1}{2}\int _0^4 u^2du\end{gathered}$

Por fim, apliquei a propriedade de integração do monômio. Ficando:

             $\large\displaystyle\begin{gathered}\int_0^2 (-2x+4)^2dx \Rightarrow \frac{1}{2}\cdot \left.\frac{u^3}{3}\right|_0^4\end{gathered}$

               $\large\displaystyle\begin{gathered}\int_0^2 (-2x+4)^2dx \Rightarrow \frac{4^3}{6}\end{gathered}$

               $\large\displaystyle\text{ \begin{gathered}\boxed{\int_0^2 (-2x+4)^2dx \Rightarrow \frac{32}{3}}\end{gathered} }$

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