Question

calculo de integral no limite de menos infinito a dois de dx/ (4-x)²

Answer 1

Resolvendo integral por substituição:

$\displaystyle \int_{-\infty}^{2}\dfrac{dx}{(4-x)^2}\\\\ \displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx\\\\\\ Resolvendo:\\\\ \displaystyle \int_{-\infty}^{2}(\underbrace{4-x})^{-2}.dx\\ \ ~~~~~~~~~~~u\\\\ u=4-x\\du=-1.dx\ \ \Rightarrow\ \ dx=-1.du$


$\displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx=\displaystyle \int_{-\infty}^{2}(u)^{-2}.(-1.du)\\\\\\ \displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx=\displaystyle -1\int_{-\infty}^{2} u^{-2}.du\\\\\\ \displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx= -1 \left \left(\dfrac{u^{-2+1}}{-2+1}\right) \right |_{-\infty}^{2}\\\\\\ \displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx= -1 \left \left(\dfrac{u^{-1}}{-1}\right) \right |_{-\infty}^{2}$


$\displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx= \left \dfrac{1}{u}\right |_{-\infty}^{2}\\\\\\ \displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx= \left \dfrac{1}{4-x}\right |_{-\infty}^{2}\\\\\\ \displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx= \dfrac{1}{4-(2)} -\dfrac{1}{4-(-\infty)}\\\\\\ \boxed{\displaystyle \int_{-\infty}^{2}(4-x)^{-2}.dx= \dfrac{1}{2}}$


Espero ter ajudado.
Bons estudos!