方程x2+2Xx+Y=0有实根的条件是: (2X)2-4*1*Y≥0,即X2≥Y;则F{X2≥Y}= J {x=0-→1,y=0→x2}dxdy=[J {x=0→1} (x2-0)dx] /(1*1)=1/3
