问答题 设a＜b，证明：[∫ a b f(x)g(x)dx] 2 ≤∫ a b f 2 (x)dx∫ a b g 2 (x)dx．
【正确答案】正确答案：构造辅助函数 F(t)=[∫ a t f(x)g(x)dx] 2 —∫ a t f 2 (x)dx∫ a t g 2 (x)dx， 则F(a)=0，且 F’(t)=2∫ a f t f(x)g(x)dx.f(t)g(t)一f 2 (t)∫ a t g 2 (x)dx—g 2 (t)∫ a t f 2 (x)dx =∫ a t [2f(x)g(x)f(t)g(t)一f 2 (t)g 2 (x)一g 2 (t)f 2 (x)]dx =一∫ a t [f(t)g(x)一g(t)f(x)] 2 dx≤0， 所以F(b)≤0，即[∫ a b f(x)g(x)dx] 2 —∫ a b f 2 (x)dx∫ a b g 2 (x)dx≤0，即 [∫ a b f(x)g(x)dx] 2 ≤∫ a b f 2 (x)dx∫ a b g 2 (x)dx．
【答案解析】