设f‘(x)在[a,b]上连续,且f(a)=0,证明:|∫b a f(x)dx|
问题描述：
设f‘(x)在[a,b]上连续,且f(a)=0,证明:|∫b a f(x)dx|
求详细过程
答
设g(x) = ∫ f(t)dt,则g'(x) = f(x),g"(x) = f'(x).
g(x)在[a,b]二阶连续可导,且g(a) = 0,g'(a) = f(a) = 0.
由带Lagrange余项的Taylor展开,存在c∈(a,b)使
g(b) = g(a)+g'(a)(b-a)+g"(c)(b-a)²/2 = f'(c)(b-a)²/2.
即有| ∫ f(t)dt| = |g(b)| = |f'(c)|·(b-a)²/2 ≤ max{|f'(x)|}·(b-a)²/2.