Răspuns:
substitutie
x²=y 2xdx=dy=> xdx=[tex]\frac{dy}{2}[/tex]
schimbi limitele fe integrare
x=0 y=0
x=1 1²=1
[tex]I=\frac{1}{2} \int\limits^1_0 {\frac{y}{y+20} } \, dy[/tex]
Adui si scazi 2020 la numarator
[tex]I=\frac{1}{2}\int\limits^1_0 }\frac{y+2020-2020}{y+2020} \, dx } =[/tex]
[tex]\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[/tex]=
[tex]\frac{1}{2}*\int\limits^1\, dy-\frac{1}{2} ln(y+2020)|o/1=[/tex]
[tex]\frac{1}{2}-\frac{1}{2} ln(1+2020)+\frac{1}{2} ln(0+2020)[/tex]
[tex]\frac{1}{2} +\frac{1}{2}ln \frac{2020}{2021}[/tex]
Explicație pas cu pas:
