若f(x)连续且满足∫x到0 f(x-t)dt=cos(x^2+1),求f(x)

问题描述:

若f(x)连续且满足∫x到0 f(x-t)dt=cos(x^2+1),求f(x)

设u = x-t
t=0时,u=x
t= x时,u=0
∫x到0 f(x-t)dt
= -∫x到0 f(x-t)d(x-t)
= -∫0到x f(u)du
两边求导数,
-f(x) = cos(x^2+1)' = -sin(x^2+1) * 2x
f(x) = 2x sin(x^2+1)