Let $$f : [-3, 1] \to R$$ be given as $$f(x) = \begin{cases} \min\{(x+6), x^2\}, & -3 \leq x \leq 0 \\ \max\{\sqrt{x}, x^2\}, & 0 \leq x \leq 1 \end{cases}$$. If the area bounded by $$y = f(x)$$ and $$x$$-axis is $$A$$ sq units, then the value of $$6A$$ is equal to ________.

We need to find the area under $$y = f(x)$$ from $$x = -3$$ to $$x = 1$$, where $$f(x) = \min\{x+6, x^2\}$$ for $$-3 \leq x \leq 0$$ and $$f(x) = \max\{\sqrt{x}, x^2\}$$ for $$0 \leq x \leq 1$$.

For the first piece ($$-3 \leq x \leq 0$$): We find where $$x + 6 = x^2$$, giving $$x^2 - x - 6 = 0$$, so $$(x-3)(x+2) = 0$$, hence $$x = -2$$ (taking the root in $$[-3, 0]$$). For $$x \in [-3, -2]$$: $$x + 6 \leq x^2$$ (check $$x = -3$$: $$3$$ vs $$9$$), so $$f(x) = x + 6$$. For $$x \in [-2, 0]$$: $$x + 6 \geq x^2$$ (check $$x = -1$$: $$5$$ vs $$1$$), so $$f(x) = x^2$$.

For the second piece ($$0 \leq x \leq 1$$): We find where $$\sqrt{x} = x^2$$, giving $$x^{1/2} = x^2$$, so $$x^4 = x$$, hence $$x(x^3 - 1) = 0$$, giving $$x = 0$$ or $$x = 1$$. For $$x \in (0, 1)$$: $$\sqrt{x} > x^2$$ (check $$x = 1/4$$: $$1/2$$ vs $$1/16$$), so $$f(x) = \sqrt{x}$$. At $$x = 0$$ and $$x = 1$$, $$f(x) = 0$$ and $$f(x) = 1$$ respectively.

The area is $$A = \int_{-3}^{-2}(x+6)\,dx + \int_{-2}^{0}x^2\,dx + \int_{0}^{1}\sqrt{x}\,dx$$.

Computing each integral: $$\int_{-3}^{-2}(x+6)\,dx = \left[\frac{x^2}{2} + 6x\right]_{-3}^{-2} = (2 - 12) - (\frac{9}{2} - 18) = -10 - (-\frac{27}{2}) = -10 + \frac{27}{2} = \frac{7}{2}$$.

$$\int_{-2}^{0}x^2\,dx = \left[\frac{x^3}{3}\right]_{-2}^{0} = 0 - (-\frac{8}{3}) = \frac{8}{3}$$.

$$\int_{0}^{1}\sqrt{x}\,dx = \left[\frac{2x^{3/2}}{3}\right]_{0}^{1} = \frac{2}{3}$$.

Therefore $$A = \frac{7}{2} + \frac{8}{3} + \frac{2}{3} = \frac{7}{2} + \frac{10}{3} = \frac{21 + 20}{6} = \frac{41}{6}$$.

So $$6A = 6 \cdot \frac{41}{6} = 41$$.