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设f(x)在[0,1]上连续且单调减,试证对任何a∈(0,1)有 ∫0af(x)dx≥a∫01f(x)dx
设f(x)在[0,1]上连续且单调减,试证对任何a∈(0,1)有
∫0af(x)dx≥a∫01f(x)dx
提问人:网友anonymity
发布时间:2022-01-06
设f(x)在[0,1]上连续且单调减,试证对任何a∈(0,1)有
∫0af(x)dx≥a∫01f(x)dx
设f(x)在[0,1]上连续,且∫01f(x)dx=0,∫01xf(x)dx=1,试证:
1)存在x0∈[0,1],使|f(x0)|>4;
2)存在x1∈[0,1],使|f(x1)|=4.
设f(x)在[a,b]上连续,且f(x)≥0,∫abf(x)dx=1,试证
(∫absinλdx)2+(∫abf(x)cosλdx)2≤1
设f(0)=1,f(2)=3,f'(2)=5,求∫01xf"(2x)dx(设f"(x)连续).
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