admin百科知識 2022-02-13 5:29:38 設函數f(x)在[0,1]上具有連續導數,且f(0)+f(1)=0,証明:|∫ f(x)dx|≤1÷2×∫ |f’ (x) |dx設函數f(x)在[0,1]上具有連續導數,且f(0) f(1)=0,証明:|∫ f(x)dx|≤1÷2×∫ |f’ (x) |dx積分都是上限爲1,下限爲0網友廻答：匿名網友先用分部積分得到∫ f(x)dx = -∫ (x-1/2)f'(x)dx然後|∫ (x-1/2)f'(x)dx| db標簽 生活常識_百科知識_各類知識大全»設函數f(x)在[0,1]上具有連續導數,且f(0)+f(1)=0,証明:|∫ f(x)dx|≤1÷2×∫ |f’ (x) |dx
admin![設函數f(x)在[0,1]上具有連續導數,且f(0)+f(1)=0,証明:|∫ f(x)dx|≤1÷2×∫ |f’ (x) |dx,第1張](data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw== "設函數f(x)在[0,1]上具有連續導數,且f(0)+f(1)=0,証明:|∫ f(x)dx|≤1÷2×∫ |f’ (x) |dx,第1張")
![設函數f(x)在[0,1]上具有連續導數,且f(0)+f(1)=0,証明:|∫ f(x)dx|≤1÷2×∫ |f’ (x) |dx,第2張](/statics/default/images/niming.jpg "設函數f(x)在[0,1]上具有連續導數,且f(0)+f(1)=0,証明:|∫ f(x)dx|≤1÷2×∫ |f’ (x) |dx,第2張")
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