③已知f(x)≤g(x)由比较定理 ∫f(x)ts(x)k即jg(x)-f(x)t≥0 而f(x)d=「g(x)→」Ig(x)-f(x)=0 由①得 f(x)≡g(x)③ 已知 f (x)  g(x) 由比较定理    b a b a f (x)dx g(x)dx  −  b a 即 [g(x) f (x)]dx 0   = b a b a 而 f (x)dx g(x)dx   − = b a [g(x) f (x)]dx 0 由①得 f (x)  g(x)