∵f(x1)=f(x2),x1≠x2
∴(x1＋x2)/2=-b/2a
则x1＋x2=-b/a
f(x1＋x2)=a(x1＋x2)²＋b(x1＋x2)＋2013
=a·(-b/a)²＋b·(-b/a)＋2013
=b²/a－b²/a＋2013
=2013