Question

$\int\limits^2_0 { x^{2} \sqrt{ x^{3}+1 } } \, dx$
alguem resolve pra min

Answer 1

Calcular a integral definida:

$I=\displaystyle\int\limits_{0}^{2}{x^{2}\sqrt{x^{3}+1}\,dx}\\\\\\ =\dfrac{1}{3}\int\limits_{0}^{2}{3x^{2}\sqrt{x^{3}+1}\,dx}\\\\\\ =\dfrac{1}{3}\int\limits_{0}^{2}{\sqrt{x^{3}+1}\cdot 3x^{2}\,dx}~~~~~~\mathbf{(i)}$


Fazendo a seguinte mudança de variável:

$x^{3}+1=u~~\Rightarrow~~3x^{2}\,dx=du$


Mudando os extremos de integração:

$\text{Quando }x=0~~\Rightarrow~~u=1\\\\ \text{Quando }x=2~~\Rightarrow~~u=9$


Substituindo em $\mathbf{(i)},$ a integral fica

$I=\dfrac{1}{3}\displaystyle\int\limits_{1}^{9}{\sqrt{u}\,du}\\\\\\ =\dfrac{1}{3}\int\limits_{1}^{9}{u^{1/2}\,du}\\\\\\ =\dfrac{1}{3}\cdot \left.\left(\dfrac{u^{1/2+1}}{\frac{1}{2}+1} \right )\right|_{1}^{9}\\\\\\ =\dfrac{1}{3}\cdot \left.\left(\dfrac{u^{3/2}}{\frac{3}{2}} \right )\right|_{1}^{9}\\\\\\ =\dfrac{1}{3}\cdot \left.\left(\dfrac{2}{3}\,u^{3/2} \right )\right|_{1}^{9}\\\\\\ =\dfrac{2}{9}\cdot \left.\left(u^{3/2} \right )\right|_{1}^{9}$

$=\dfrac{2}{9}\cdot \left(9^{3/2}-1^{3/2} \right )\\\\\\ =\dfrac{2}{9}\cdot (27-1)\\\\\\ =\dfrac{2}{9}\cdot 26\\\\\\ =\dfrac{52}{9}$

$\therefore~~\boxed{\begin{array}{c} \displaystyle\int\limits_{0}^{2}{x^{2}\sqrt{x^{3}+1}\,dx}=\dfrac{52}{9} \end{array}}$