Question

a integral da raiz cubica de 6-2x dx

Answer 1

$\displaystyle\int{\,^{3}\!\!\!\sqrt{6-2x}\,dx}\\ \\ \\ =-\dfrac{1}{2}\cdot (-2)\int{\,^{3}\!\!\!\sqrt{6-2x}\,dx}\\ \\ \\ =-\dfrac{1}{2}\cdot \int{(-2)\,^{3}\!\!\!\sqrt{6-2x}\,dx}\\ \\ \\ =-\dfrac{1}{2}\cdot \int{\,^{3}\!\!\!\sqrt{6-2x}\cdot (-2)\,dx}~~~~~~\mathbf{(i)}$


Substituição:

$6-2x=u~\Rightarrow~-2x\,dx=du$


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

$=-\dfrac{1}{2}\cdot \displaystyle\int{\,^{3}\!\!\!\sqrt{u}\,du}\\ \\ \\ =-\dfrac{1}{2}\cdot \int{u^{1/3}\,du}~~~~~~\mathbf{(ii)}$


Regra da integral da potência:

$\displaystyle\int{u^{n}\,du}=\dfrac{u^{n+1}}{n+1}\,,~~n\neq -1.$


Aplicando a regra acima em $\mathbf{(ii)},$ a integral fica

$=-\dfrac{1}{2}\cdot \displaystyle\int{u^{1/3}\,du}\\ \\ \\ =-\dfrac{1}{2}\cdot \dfrac{u^{(1/3)+1}}{\frac{1}{3}+1}+C\\ \\ \\ =-\dfrac{1}{2}\cdot \dfrac{u^{4/3}}{\frac{4}{3}}+C\\ \\ \\ =-\dfrac{1}{2}\cdot \dfrac{3}{4}\,u^{4/3}+C\\ \\ \\ =-\dfrac{3}{8}\,u^{4/3}+C\\ \\ \\ =-\dfrac{3}{8}\,(6-2x)^{4/3}+C\\ \\ \\ \\ \Rightarrow~\boxed{\begin{array}{c} \displaystyle\int{\,^{3}\!\!\!\sqrt{6-2x}\,dx}=-\dfrac{3}{8}\,(6-2x)^{4/3}+C \end{array}}$