设连续函数f(x)满足x^2=x-∫(上限2x，下限O)f(x)dx,求f(x). 设f(x)为(-∞，+∞)内的连续函数，且满足f(x)=∫f...
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2x=1-f(2x)·(2x)'
2f(2x)=-2x+1
f(2x)=-x+(1/2)
令2x=t,则x=t/2
f(t)=-(t/2)+(1/2)
所以f(x)=-(x/2)+(1/2)
f(x)=∫(0,2x)f(t/2)dt+1 f(0)=∫(0,0)f(t/2)dt+1=0+1=1 f'(x)=f(2x/2)·(2x)'=2f(x) 即y'=2y→dy/y=2dx 两边积分：lny=2x+C→y=e^(2x+C) y(0)=e^C=1→C=0 ∴f(x)=e^2x
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