设f(x)在区间【0,1】上有连续导数,证明x∈【0,1】,有|f(x)|≤∫(|f(t)|+|f′(t)|)dt
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![设f(x)在区间【0,1】上有连续导数,证明x∈【0,1】,有|f(x)|≤∫(|f(t)|+|f′(t)|)dt](/uploads/image/z/3756258-18-8.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%E3%80%900%2C1%E3%80%91%E4%B8%8A%E6%9C%89%E8%BF%9E%E7%BB%AD%E5%AF%BC%E6%95%B0%2C%E8%AF%81%E6%98%8Ex%E2%88%88%E3%80%900%2C1%E3%80%91%2C%E6%9C%89%7Cf%28x%29%7C%E2%89%A4%E2%88%AB%28%7Cf%28t%29%7C%2B%7Cf%E2%80%B2%28t%29%7C%29dt)
设f(x)在区间【0,1】上有连续导数,证明x∈【0,1】,有|f(x)|≤∫(|f(t)|+|f′(t)|)dt
设f(x)在区间【0,1】上有连续导数,证明x∈【0,1】,有|f(x)|≤∫(|f(t)|+|f′(t)|)dt
设f(x)在区间【0,1】上有连续导数,证明x∈【0,1】,有|f(x)|≤∫(|f(t)|+|f′(t)|)dt
存在连续导数,所以 f' 可积
设 x0,0
设f(x)在区间【0,1】上有连续导数,证明x∈【0,1】,有|f(x)|≤∫(|f(t)|+|f′(t)|)dt
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