Question

Me ajudem com essa integral
$\int\limits^3_0 {(x+2) \sqrt{x+1} } \, dx$

Answer 1

Resolução em anexo. Caso não entenda, só perguntar.

Answer 2

Calcular a integral definida:

     $\displaystyle\int_0^3 (x+2)\sqrt{x+1}\,dx$


Faça uma substituição:

     $\sqrt{x+1}=u\quad\Rightarrow\quad\left\{\! \begin{array}{l} x+1=u^2\quad\Rightarrow \quad x=u^2-1\\\\ dx=2u\,du \end{array} \right.$


Novos limites de integração em  u:

     $\begin{array}{lcl} \textsf{Quando~~}x=0&\quad\Rightarrow\quad&u=\sqrt{0+1}\\\\ &&u=\sqrt{1}\\\\ &&u=1 \end{array}$


     $\begin{array}{lcl} \textsf{Quando~~}x=3&\quad\Rightarrow\quad&u=\sqrt{3+1}\\\\ &&u=\sqrt{4}\\\\ &&u=2 \end{array}$


Substituindo tudo, a integral fica

      $\displaystyle=\int_1^2 \big((u^2-1)+2\big)\cdot u\cdot 2u\,du\\\\\\ =2\int_1^2 (u^2+1)\cdot u^2\,du\\\\\\ =2\int_1^2 (u^2\cdot u^2+1\cdot u^2)\,du\\\\\\ =2\int_1^2 (u^4+u^2)\,du$

     $=2\cdot \left(\dfrac{u^{4+1}}{4+1}+\dfrac{u^{2+1}}{2+1}\right)\!\bigg|_1^2\\\\\\ =2\cdot \left(\dfrac{u^5}{5}+\dfrac{u^3}{3}\right)\!\bigg|_1^2\\\\\\ =2\cdot \left(\dfrac{2^5}{5}+\dfrac{2^3}{3}\right)-2\cdot \left(\dfrac{1^5}{5}+\dfrac{1^3}{3}\right)\\\\\\ =2\cdot \left(\dfrac{32}{5}+\dfrac{8}{3}\right)-2\cdot \left(\dfrac{1}{5}+\dfrac{1}{3}\right)$

     $=2\cdot \left(\dfrac{96}{15}+\dfrac{40}{15}\right)-2\cdot \left(\dfrac{3}{15}+\dfrac{5}{15}\right)\\\\\\ =2\cdot \dfrac{136}{15}-2\cdot \dfrac{8}{15}\\\\\\ =\dfrac{272}{15}-\dfrac{16}{15}$

     $=\dfrac{256}{15}\quad\longleftarrow\quad\textsf{resposta.}$
     

Bons estudos! :-)