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) =[*]∫-x20f(u)du=[*]∫0-x2f(u)du, [*] 则n-2=2,n=4,且 [*]f’(0)=1,于是f’(0)=-4.

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