设∫(上限x^2下限0)f(t)dt=x^2(1-x^2),求f(x)
问题描述：
设∫(上限x^2下限0)f(t)dt=x^2(1-x^2),求f(x)
答
let
f(x)dx =dF(x)
∫(0->x^2)f(t)dt=x^2.(1-x^2)
F(x^2)-F(0)=x^2.(1-x^2)
[F(x^2)-F(0)]'=[x^2.(1-x^2)]'
2xf(x^2) = x^2(-2x) + 2x(1-x^2)
= 2x(1-2x^2)
f(x^2)=1-2x^2
f(x) =1-2xF(x^2)-F(0)求导那里怎么求的∫(0->x^2)f(t)dt=x^2.(1-x^2)
∫(0->x^2)dF(t)=x^2.(1-x^2)
[F(t)](0->x^2)=x^2.(1-x^2)
F(x^2)-F(0)=x^2.(1-x^2)