设f(x)在（0,+∞）上连续,且∫f(t)dt=x³=1,上限为x²,下限为0,则f(2)=?

Question

设f(x)在（0,+∞）上连续,且∫f(t)dt=x³=1,上限为x²,下限为0,则f(2)=?

Answer 1

设不定积分 ∫f(t)dt 的原函数为F(t),即：F‘(t) = f(t),则有：
[0,x²] ∫f(t)dt = F(x²) - F(0) = x³
两边同时对x 求导有：
F'(x²) * (x²)' = (x³)'
==> f(x²) *2x= 3x²
==> f(x²) = 3x/2
∵ f(x)在（0,+∞）上连续
∴ f(2) =(f(√2)²) = 3√2/2