Sol:假设f(x)是在一条长度为x的空街上平均能停的汽车数目,当x>1的时候
f(x) = 1+\frac{\int_0^{x-1}[f(t)+f(x-1-t)]dt}{x-1} = 1 + \frac{2}{x-1}\int_0^{x-1}f(t)dt
不难得知- 0<x<1: f(t)=0
- 1<x<2: f(t)=1
- 2<x<3:
f(t)=1+2\frac{\int_0^{t-1}f(u)du}{(t-1)} = 1+\frac{2(t-2}{t-1}
所以
f(4)=1+\frac{2}{3}[\int_0^{1}f(t)dt + \int_1^{2}f(t)dt + \int_2^{3}f(t)dt] = \frac{11}{3} - \frac{4}{3}ln 2
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