Question

O valor da integral de (1 a e) e de (1 a 1/2) ∫ ∫ lnx/xy dydx é:

Answer 1

Olá!


$\displaystyle\int_1^e \int_1^{\frac{1}{2}}\dfrac{\ln x}{xy}\;dydx= \int_1^e\left[\dfrac{\ln x}{x}\int_1^{\frac{1}{2}}\dfrac{1}{y}\;dy\right]\;dx = \\ \\ \\ = \int_1^e \dfrac{\ln x}{x}\left[\bigg{(}\ln |y|\bigg{)}\bigg{|}_{y=1}^{y=\frac{1}{2}}\right]\;dx=\int_1^e\dfrac{\ln x}{x}\left(\ln\frac{1}{2}-\ln 1\right)\;dx = \\ \\ \\ = \int_1^e\dfrac{\ln x}{x}\left(\ln \frac{1}{2}-0\right)\;dx = \ln\dfrac{1}{2}\int_1^e\dfrac{\ln x}{x}\;dx$


    Seja   $u=\ln{x}.$   Daí,   

$\displaystyle\dfrac{du}{dx} = \dfrac{1}{x}\Rightarrow du=\dfrac{1}{x}dx\;\; \\ \\ \text{e}\;\; x=1\to u=0,\;x=e\to u=1.\\ \\ \therefore \\ \\ \ln\dfrac{1}{2}\int_1^e\dfrac{\ln x}{x}\;dx=\ln\dfrac{1}{2}\int_1^e \ln x\cdot \dfrac{1}{x}\;dx=-\ln2\int_0^1 u\;du = \\ \\ \\ = -\ln2\left(\dfrac{u^2}{2}\right)\bigg{|}_{0}^1=-\ln2\left(\dfrac{1}{2}-0\right) =-\dfrac{\ln 2}{2}.$



     Portanto, resposta (B).



Bons estudos!

Answer 2

Resposta: -ln(2)/2

Explicação passo-a-passo:

Corrigido pelo ava