Question

A integral $\displaystyle\int_1^e \int_1^{e^2}\int_1^{e^3}\left(\frac{1}{xyz}\right )dx\,dy\,dz$ vale:

a) 4

b) 2

c) 3

d) 1

e) 6

Answer 1

Para o cálculo dessa integral tripla, usamos o Teorema de Fubini (integrais iteradas):

$\displaystyle\int_1^e \int_1^{e^2}\int_1^{e^3}\frac{1}{xyz}\,dx\,dy\,dz\\\\\\ =\int_1^e \int_1^{e^2}\int_1^{e^3}\frac{1}{yz}\cdot \frac{1}{x}\,dx\,dy\,dz\\\\\\ =\int_1^e \int_1^{e^2}\frac{1}{yz}\cdot (\mathrm{\ell n}\,x)\Big|_1^{e^{3}}\,dy\,dz\\\\\\ =\int_1^e \int_1^{e^2}\frac{1}{yz}\cdot \big[\mathrm{\ell n}(e^3)-\mathrm{\ell n}\,1\big]\,dy\,dz$

$=\displaystyle\int_1^e \int_1^{e^2}\frac{1}{yz}\cdot \big[3\cdot 1-0\big]\,dy\,dz\\\\\\ =\int_1^e \int_1^{e^2}\frac{3}{yz}\,dy\,dz\\\\\\ =\int_1^e \int_1^{e^2}\frac{3}{z}\cdot \frac{1}{y}\,dy\,dz\\\\\\ =\int_1^e \frac{3}{z}\cdot (\mathrm{\ell n\,}y)\Big|_1^{e^2}\,dz$

$=\displaystyle\int_1^e \frac{3}{z}\cdot \big[\mathrm{\ell n}(e^2)-\mathrm{\ell n\,}1\big]\,dz\\\\\\ =\int_1^e \frac{3}{z}\cdot \big[2\,\mathrm{\ell n\,}e-\mathrm{\ell n\,}1\big]\,dz\\\\\\ =\int_1^e \frac{3}{z}\cdot \big[2\cdot 1-0\big]\,dz\\\\\\ =\int_1^e \frac{6}{z}\,dz$

$=\displaystyle 6\,\mathrm{\ell n\,}z\Big|_1^e\\\\ =6\cdot (\mathrm{\ell n\,}e-\mathrm{\ell n\,}1)\\\\ =6\cdot (1-0)\\\\ =6\qquad\longleftarrow\quad\textsf{esta \'e a resposta: alternativa e) 6.}$


Caso tenha problemas para visualizar a resposta, experimente abrir pelo navegador: http://brainly.com.br/tarefa/7892693


Dúvidas? Comente.


Bons estudos! :-)