zzas +234
Sveiki,
Integralas: [tex]\int_{x=2}^{3}\int_{y=0}^{x-2}\frac{1}{(x-y)(x+y)^{2}}dydx[/tex]
Įsivedu keitinį: [tex]u=x+y; v=x-y \Rightarrow y=\frac{u-v}{2}; x=\frac{u+v}{2}[/tex]
Ieškau ribų: [tex]2\leq x\leq 3; 0\leq y\leq x-2 \Rightarrow 2\leq x+y\leq 4; 2\leq x-y\leq 3 \Rightarrow 2\leq u\leq 4; 2\leq v\leq 3[/tex]
Jakobianas: [tex]J=1/2[/tex]
Taigi, gaunu integralą: [tex]\int_{2}^{3}\int_{2}^{4}\frac{1}{2vu^{2}}dudv=\frac{1}{8}(ln3-ln2)[/tex]
Atsakymas, deja, neteisingas. Kur problema?