g(x) = ∫(0→1) ƒ(xt) dt
令u = xt,du = x dt
t = 0,u = 0
t = 1,u = x
g(x) = (1/x)∫(0→x) ƒ(u) du
g'(x) = (1/x) * ƒ(x) - (1/x²)∫(0→x) ƒ(u) du
g'(0) = lim(x→0) ƒ(x)/x - lim(x→0) [∫(0→x) ƒ(u) du]/x²
= A - lim(x→0) ƒ(x)/(2x)
= A - (1/2)A
= A/2
既然g'(0)存在,则g(x) = 0处连续,可导则必定连续.