先把F(x)拆開,
F(x)=∫[0,x] (x^3-t^3)f '''(t)dt
=x^3 ∫[0,x] f '''(t)dt - ∫[0,x] t^3*f '''(t)dt
對(duì)于積分上限函數(shù),其導(dǎo)數(shù)就等于將其上限代入被積函數(shù)即可,
所以 ∫[0,x] f '''(t)dt的導(dǎo)數(shù)為 f '''(x),∫[0,x] t^3*f '''(t)dt的導(dǎo)數(shù)為x^3*f '''(x),
于是
F'(x)=3x^2 ∫[0,x] f '''(t)dt + x^3*f '''(x) - x^3*f '''(x)
=3x^2 * [f ''(x)-f ''(0)] + x^3*f '''(x) - x^3*f '''(x)