Question

integral raiz x.^4raiz3x^3dx

Answer 1

Calcular a integral indefinida:

     $\displaystyle\int\sqrt{x^4}\cdot \sqrt{3x^3}\,dx\\\\\\ =\int\sqrt{(x^2)^2}\cdot \sqrt{3x^3}\,dx\\\\\\ =\int|x^2|\cdot \sqrt{3x^3}\,dx$


Como $x^2\ge 0,$ qualquer que seja $x$ real, podemos dispensar o módulo, e a integral fica

     $=\displaystyle\int x^2\sqrt{3x^3}\,dx\\\\\\ =\int \frac{1}{9}\cdot 9x^2\sqrt{3x^3}\,dx\\\\\\ =\frac{1}{9}\int \sqrt{3x^3}\cdot 9x^2\,dx\\\\\\ =\frac{1}{9}\int (3x^3)^{1/2}\cdot 9x^2\,dx$


Faça a seguinte substituição:

     $\begin{array}{lcl} 3x^3=u&\quad\Rightarrow\quad &3\cdot (3x^{3-1})\,dx=du\\\\ &&9x^2\,dx=du \end{array}$


e a integral fica

     $\displaystyle=\frac{1}{9}\int u^{1/2}\,du\\\\\\ =\frac{1}{9}\cdot \frac{u^{(1/2)+1}}{\frac{1}{2}+1}+C\\\\\\ =\frac{1}{9}\cdot \frac{u^{3/2}}{\frac{3}{2}}+C\\\\\\ =\frac{1}{9}\cdot \frac{2}{3}\,u^{3/2}+C\\\\\\ =\frac{2}{27}\,u^{3/2}+C$

     $=\dfrac{2}{27}\,(3x^3)^{3/2}+C\quad\longleftarrow\quad\textsf{resposta.}$


Bons estudos! :-)