设f’(x)连续,f(0)=0,f’(0)≠0,F(x)=∫0xtf(t2-x2)dt,且当x→0时,F(x)~xn,求n及f’(0).

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问题 设f’(x)连续,f(0)=0,f’(0)≠0,F(x)=∫0xtf(t2-x2)dt,且当x→0时,F(x)~xn,求n及f’(0).

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答案F(x)=∫0xtf(t2-x2)dt=[*]∫0xf(t2-x2)d(t2-x2) [*] 则n-2=2,n=4,且 [*] 于是f’(0)=-4.

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