Let $$f: [0, \infty) \to R$$ and $$F(x) = \int_0^x tf(t) \, dt$$. If $$F(x^2) = x^4 + x^5$$, then $$\sum_{r=1}^{12} f(r^2)$$ is equal to:

We are given $$F(x) = \int_0^x t f(t) \, dt$$ and $$F(x^2) = x^4 + x^5$$.

Differentiating both sides with respect to $$x$$:

$$F'(x^2) \cdot 2x = 4x^3 + 5x^4$$

Since $$F'(u) = u \cdot f(u)$$ (by the Fundamental Theorem of Calculus), substituting $$u = x^2$$:

$$x^2 f(x^2) \cdot 2x = 4x^3 + 5x^4$$

$$2x^3 f(x^2) = 4x^3 + 5x^4$$

Dividing by $$2x^3$$ (for $$x \neq 0$$):

$$f(x^2) = 2 + \frac{5x}{2}$$

Let $$x^2 = r^2$$, so $$x = r$$ (taking positive root):

$$f(r^2) = 2 + \frac{5r}{2}$$

Now we compute:

$$\sum_{r=1}^{12} f(r^2) = \sum_{r=1}^{12} \left(2 + \frac{5r}{2}\right) = 24 + \frac{5}{2} \sum_{r=1}^{12} r = 24 + \frac{5}{2} \cdot \frac{12 \cdot 13}{2} = 24 + \frac{5 \cdot 78}{2} = 24 + 195 = 219$$

The answer is $$\boxed{219}$$.