Let $$f$$ be a real valued continuous function on $$[0, 1]$$ and $$f(x) = x + \int_0^1 (x-t)f(t)dt$$. Then which of the following points $$(x, y)$$ lies on the curve $$y = f(x)$$?

We are given $$f(x) = x + \int_0^1 (x-t)f(t)\,dt$$. Expanding the integral gives $$f(x) = x + x\int_0^1 f(t)\,dt - \int_0^1 tf(t)\,dt$$. Let $$A = \int_0^1 f(t)\,dt$$ and $$B = \int_0^1 tf(t)\,dt$$, so $$f(x) = x(1 + A) - B$$.

Since $$f(x) = (1+A)x - B$$, substituting into the definitions of $$A$$ and $$B$$ yields $$A = \int_0^1 [(1+A)t - B]\,dt = (1+A)\frac{1}{2} - B$$ and $$B = \int_0^1 t[(1+A)t - B]\,dt = (1+A)\frac{1}{3} - B\frac{1}{2}$$.

The equation for $$A$$ gives $$A = \frac{1+A}{2} - B \Rightarrow 2A = 1 + A - 2B \Rightarrow A = 1 - 2B$$, and the equation for $$B$$ gives $$B = \frac{1+A}{3} - \frac{B}{2} \Rightarrow \frac{3B}{2} = \frac{1+A}{3} \Rightarrow B = \frac{2(1+A)}{9}$$.

Substituting the expression for $$B$$ into $$A = 1 - 2B$$ gives $$A = 1 - 2 \cdot \frac{2(1+A)}{9} = 1 - \frac{4(1+A)}{9}$$. It follows that $$9A = 9 - 4 - 4A$$, hence $$13A = 5$$ and $$A = \frac{5}{13}$$. Then $$B = \frac{2(1 + 5/13)}{9} = \frac{2 \cdot 18/13}{9} = \frac{36}{117} = \frac{4}{13}$$.

Hence $$f(x) = \left(1 + \frac{5}{13}\right)x - \frac{4}{13} = \frac{18x}{13} - \frac{4}{13} = \frac{18x - 4}{13}$$. Checking the provided points, for $$(2, 4)$$ we have $$f(2) = \frac{36-4}{13} = \frac{32}{13} \neq 4$$; for $$(1, 2)$$, $$f(1) = \frac{18-4}{13} = \frac{14}{13} \neq 2$$; for $$(4, 17)$$, $$f(4) = \frac{72-4}{13} = \frac{68}{13} \neq 17$$; and for $$(6, 8)$$, $$f(6) = \frac{108-4}{13} = \frac{104}{13} = 8$$ ✔.

Therefore, the answer is Option D: $$\textbf{(6, 8)}$$.