设f(x)≥0,在[a,b]上连续且∫(0,1)f(x)dx=0,试证:在[a,b]上有f(x)=0
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设f(x)≥0,在[a,b]上连续且∫(0,1)f(x)dx=0,试证:在[a,b]上有f(x)=0设f(x)≥0,在[a,b]上连续且∫(0,1)f(x)dx=0,试证:在[a,b]上有f(x)=0
设f(x)≥0,在[a,b]上连续且∫(0,1)f(x)dx=0,试证:在[a,b]上有f(x)=0
设f(x)≥0,在[a,b]上连续且∫(0,1)f(x)dx=0,试证:在[a,b]上有f(x)=0
设f(x)≥0,在[a,b]上连续且∫(0,1)f(x)dx=0,试证:在[a,b]上有f(x)=0
反证法:如果存在一点x0属于[a,b],使得f(x0)>0,那么根据f(x)的连续性,存在一δ,使得任一x属于(x0-δ,x0+δ),都有f(x)>0,又根据f(x)>=0,从而有∫(0,1)f(x)dx>∫(x0-δ,x0+δ)f(x)dx>0,这与∫(0,1)f(x)dx=0矛盾,所以不存在一点x0属于[a,b],使得f(x0)>0,故在[a,b]上f(x)恒为0.
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