f(x)=∫(0,2x)f(t/2)dt+ln2,显然f(0)=ln2两边求导f'(x)=f(2x/2)*(2x)'即f'(x)=2f(x)为什么不用将dt配成d(2/t),原式变成f(x)=2∫(0,2x)f(t/2)d2/t)+ln2两边求导f'(x)=2f(2x/2)*(2x)'即f'(x)=4f(x)

Answer 1

如果要d(x/2)的话,注意积分上下限可能有更变的.
将t变为t/2,d(t/2) = (1/2)dt ==> dt = 2 d(t/2)
当t = 0时,t/2 = 0
当t = 2x时,t/2 = 2x/2 = x
所以∫(0→2x) f(t/2) dt = ∫(0→x) f(t/2) * 2 d(t/2) = 2∫(0→x) f(u) du,假设u = t/2
∴f'(x) = 2f(x) * x' = 2f(x)