Let f be a twice differentiable non-negative function such that $$(f(x))^{2}=25+\int_{0}^{x}\left((f(t))^{2}+(f'(t))^{2}\right)dt$$. Then the mean of $$f(\log_{e}{(1)}),f(\log_{e}{(2)}),.....,f(\log_{e}{(625)})$$ is equal to:

The function $$f$$ satisfies the equation $$(f(x))^{2} = 25 + \int_{0}^{x}\bigl((f(t))^{2} + (f'(t))^{2}\bigr)\,dt \quad-(1)$$ so we begin by differentiating both sides with respect to $$x$$. Using $$\frac{d}{dx}\bigl((f(x))^{2}\bigr) = 2f(x)\,f'(x)$$ and $$\frac{d}{dx}\Bigl(\int_{0}^{x}g(t)\,dt\Bigr) = g(x),$$ we obtain $$2f(x)\,f'(x) = (f(x))^{2} + (f'(x))^{2} \quad-(2).$$

Rearranging (2) gives $$(f'(x))^{2} - 2f(x)\,f'(x) + (f(x))^{2} = 0,$$ which factors as the perfect square $$(f'(x) - f(x))^{2} = 0$$ and hence $$f'(x) - f(x) = 0\,. $$

Therefore $$\frac{df}{dx} = f(x)$$ and separating variables yields $$\frac{1}{f(x)}\,df = dx$$ so integrating both sides leads to $$\int \frac{1}{f(x)}\,df = \int 1\,dx \;\Longrightarrow\; \ln\bigl|f(x)\bigr| = x + C,$$ whence $$f(x) = Ce^{\,x}$$ for some constant $$C$$.

To determine $$C$$ we substitute $$x=0$$ into (1): $$\bigl(f(0)\bigr)^{2} = 25 + \int_{0}^{0}\bigl((f(t))^{2}+(f'(t))^{2}\bigr)\,dt = 25.$$ Since $$f$$ is non-negative it follows that $$f(0)=5$$, so $$C=5$$ and $$f(x)=5e^{\,x}\,. $$

We now require the mean of $$f(\ln 1),\,f(\ln 2),\,\dots,\,f(\ln 625)\,. $$ Noting that for any positive integer $$k$$ we have $$f(\ln k) = 5e^{\,\ln k} = 5k\,, $$ the sum of these 625 values is $$\sum_{k=1}^{625}5k = 5\sum_{k=1}^{625}k = 5\cdot\frac{625\cdot626}{2}\,. $$

Therefore the mean is $$\frac{1}{625}\times5\times\frac{625\times626}{2} = \frac{5\times626}{2} = 5\times313 =1565\,. $$

Final Answer: 1565. Option X