Question

Admitere UPTimisoara 2019 AM 185

Answer 1

Răspuns:

Varianta d. 512.

Explicație pas cu pas:

Fie $I=\int_0^{1024}\dfrac{\ln(2017-x)}{\ln[1505^2-(512-x)^2]}dx$.

Cu substitutia $t=1024-x,\:\:dx=-dt$, limitele se schimba in $1024-1024=0$ si $1024-0=1024$, deci:

$I=\int_0^{1024}\dfrac{\ln(2017-x)}{\ln[1505^2-(512-x)^2]}dx\\\\I=\int_{1024}^0\dfrac{\ln(2017-(1024-t))}{\ln\{1505^2-[512-(1024-t)]^2\}}(-1)dt\\\\I=\int_{0}^{1024}\dfrac{\ln(2017-(1024-t))}{\ln\{1505^2-[512-(1024-t)]^2\}}dt\\\\I=\int_{0}^{1024}\dfrac{\ln(993+t)}{\ln[1505^2-(512-t)^2]}dt\\\\\text{Dar stim din ipoteza ca } I=\int_{0}^{1024}\dfrac{\ln(2017-t)}{\ln[1505^2-(512-t)^2]}dt\\\\\\\text{Asadar } 2I=I+I=\int_{0}^{1024}\dfrac{\ln(2017-t)+\ln(993+t)}{\ln[1505^2-(512-t)^2]}dt$

$2I=\int_{0}^{1024}\dfrac{\ln[(2017-t)(993+t)]}{\ln[(1505-512+t)(1505+512-t)]}dt\\\\2I=\int_{0}^{1024}\dfrac{\ln[(2017-t)(993+t)]}{\ln[(993+t)(2017-t)]}dt=\int_{0}^{1024}dt=1024\\\\I=512$