Question

Determine o valor da integral definida:
∫ 0 até 1
r³ / √4+r² . dr

Escolha uma:a. 16−75–√16−75
b. 13(4−75–√)13(4−75)
c. 13(16−715−−√)13(16−715)
d. 13(16−75–√)13(16−75)
e. 13(75–√)13(75)

Answer 1

Calcular a integral definida:

     $\displaystyle\int_0^1 \frac{r^3}{\sqrt{4+r^2}}\,dr\\\\\\ =\int_0^1 \frac{r^2}{\sqrt{4+r^2}}\cdot r\,dr$


Poderíamos aplicar uma substituição trigonométrica aqui, mas podemos fazer também uma substituição simples nesse caso:

     $u=\sqrt{4+r^2}\\\\\\ \quad\Longrightarrow\quad \left\{ \begin{array}{lcl} u^2=4+r^2&\quad\Longrightarrow\quad&r^2=u^2-4\\\\ 2u\,du=2r\,dr&\quad\Longrightarrow\quad&u\,du=r\,dr \end{array} \right.$


Novos limites de integração em u:

     $\begin{array}{lcl} \mathsf{Quando~~}r=0&\quad\Longrightarrow\quad&u=\sqrt{4+0^2}\\\\ &&u=\sqrt{4}\\\\ &&u=2 \end{array}$


     $\begin{array}{lcl} \mathsf{Quando~~}r=1&\quad\Longrightarrow\quad&u=\sqrt{4+1^2}\\\\ &&u=\sqrt{5} \end{array}$


Substituindo, a integral fica:

     $\displaystyle=\int_2^{\sqrt{5}} \frac{u^2-4}{\diagup\!\!\!\! u}\cdot \diagup\!\!\!\! u\,du\\\\\\ =\int_2^{\sqrt{5}}(u^2-4)\,du\\\\\\ =\left(\frac{u^{2+1}}{2+1}-4u\right)\bigg|_2^{\sqrt{5}}\\\\\\ =\left(\frac{u^3}{3}-4u\right)\bigg|_2^{\sqrt{5}}$

     $=\bigg(\dfrac{(\sqrt{5})^3}{3}-4\cdot (\sqrt{5})\bigg)-\bigg(\dfrac{2^3}{3}-4\cdot 2\bigg)\\\\\\ =\bigg(\dfrac{(\sqrt{5})^2\cdot \sqrt{5}}{3}-4\sqrt{5}\bigg)-\bigg(\dfrac{8}{3}-8\bigg)\\\\\\ =\bigg(\dfrac{5\sqrt{5}}{3}-\dfrac{12\sqrt{5}}{3}\bigg)-\bigg(\dfrac{8}{3}-\dfrac{24}{3}\bigg)\\\\\\ =\dfrac{-7\sqrt{5}}{3}-\bigg(-\dfrac{16}{3}\bigg)$

     $=\dfrac{1}{3}(16-7\sqrt{5})\quad\longleftarrow\quad\mathsf{esta~\acute{e}~a~resposta.}$


Bons estudos! :-)