Question

Bună! Am de calculat o integrală. Mă ajutați, vă rog?

Answer 1

Răspuns:

substitutie

x²=y    2xdx=dy=> xdx=$\frac{dy}{2}$

schimbi limitele fe integrare

x=0    y=0

x=1   1²=1

$I=\frac{1}{2} \int\limits^1_0 {\frac{y}{y+20} } \, dy$

Adui si scazi 2020 la numarator

$I=\frac{1}{2}\int\limits^1_0 }\frac{y+2020-2020}{y+2020} \, dx } =$

$\frac{1}{2}*\int\limits^1_0{\frac{y+2020}{y+2020} } \, dy -\frac{1}{2} *\int\limits^0_b {\frac{1}{y+2020} } \, dy$=

$\frac{1}{2}*\int\limits^1\, dy-\frac{1}{2} ln(y+2020)|o/1=$

$\frac{1}{2}-\frac{1}{2} ln(1+2020)+\frac{1}{2} ln(0+2020)$

$\frac{1}{2} +\frac{1}{2}ln \frac{2020}{2021}$

Explicație pas cu pas: