由拉格朗日中值定理,在区间[a,b],存在点ξ使
y'(ξ) = (y(b)-y(a))/(b-a)
y'(ξ) = 2pξ+q
(y(b)-y(a))/(b-a) = p(a+b) +q
2pξ+q = p(a+b) +q
ξ = (a+b)/2