来自鲁绍基的问题
设f(x)在[a,b]上连续可微,证明maxa≤x≤b|f(x)|≤1b−a∫ba|f(x)|dx+∫ba|f′(x)|dx.
设f(x)在[a,b]上连续可微,证明maxa≤x≤b|f(x)|≤1b−a∫b
a
|f(x)|dx+∫b
a
|f′(x)|dx.
1回答
2020-03-22 23:09
设f(x)在[a,b]上连续可微,证明maxa≤x≤b|f(x)|≤1b−a∫ba|f(x)|dx+∫ba|f′(x)|dx.
设f(x)在[a,b]上连续可微,证明maxa≤x≤b|f(x)|≤1b−a∫b
a
|f(x)|dx+∫b
a
|f′(x)|dx.
因为f(x)在[a,b]上连续可微,故|f(x)|在[a,b]上连续,从而利用闭区间上连续函数的最值定理可得,存在x0∈[a,b],使得|f(x0)|=maxa≤x≤b|f(x)|.对于任意x∈[a,b],f(x)-f(x0)=∫xx0f′(t)dt,①从而|...