一道定积分题f设f(x)连续,a不等于0,且S[0到1]f(ax+b)dx=S[b到2]f(x)dx成立求a和b

Question

一道定积分题f
设f(x)连续,a不等于0,且S[0到1]f(ax+b)dx=S[b到2]f(x)dx成立
求a和b

Answer 1

设 u = ax + b,du = adx
由于 x的积分上下限为:0 -> 1
因此 u的积分上下限为：b -> a+b
S[0到1]f(ax+b)dx = S[b到a+b] [f(u)/a] du =
= S[b到a+b] [f(x)/a] dx = S[b到2]f(x)dx
当 a = b = 1时,上式一定成立.