设f(x)为连续函数,且满足∫(上x^3-1,下0)f(t)dt=x,则f(7)=
人气：439 ℃ 时间：2019-10-23 02:05:21
解答
g(x^3-1)-g(0)=x
y=x^3-1,得x=（y+1）^(1/3)
所以,g（y）=（y+1）^(1/3)+g(0)
f(y)=g'(y)=1/3（y+1）^(-2/3)
7带入.f(7)=1/12
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