Let the area enclosed by the lines $$x + y = 2$$, $$y = 0$$, $$x = 0$$ and the curve $$f(x) = \min\left\{x^2 + \frac{3}{4}, 1 + [x]\right\}$$ where $$[x]$$ denotes the greatest integer $$\leq x$$, be $$A$$. Then the value of $$12A$$ is _____.

To find $$A$$, we integrate $$f(x) = \min \{x^2 + \frac{3}{4}, 1 + [x]\}$$ within the region bounded by $$x+y=2$$ (i.e., $$y = 2-x$$) and the axes.

1. Breakdown of $$f(x)$$ for $$x \in [0, 2]$$

• $$x \in [0, \frac{1}{2}]$$: $$x^2 + \frac{3}{4} \le 1$$, so $$f(x) = x^2 + \frac{3}{4}$$
• $$x \in [\frac{1}{2}, 1]$$: $$x^2 + \frac{3}{4} > 1$$, so $$f(x) = 1$$
• $$x \in [1, 2]$$: Here $$1 + [x] = 2$$. Since the line $$y = 2-x$$ is always $$\le 2$$ in this range, the boundary of the enclosed area is $$y = 2-x$$.

2. Integration for Area $$A$$

The area $$A$$ is the integral of the lower boundary:

$$A = \int_{0}^{1/2} (x^2 + \frac{3}{4}) dx + \int_{1/2}^{1} 1 dx + \int_{1}^{2} (2-x) dx$$

Calculations:

1. $$\int_{0}^{1/2} (x^2 + \frac{3}{4}) dx = [\frac{x^3}{3} + \frac{3x}{4}]_0^{1/2} = \frac{1}{24} + \frac{3}{8} = \frac{10}{24} = \frac{5}{12}$$
2. $$\int_{1/2}^{1} 1 dx = 1 - \frac{1}{2} = \frac{1}{2}$$
3. $$\int_{1}^{2} (2-x) dx = [2x - \frac{x^2}{2}]_1^2 = (4 - 2) - (2 - \frac{1}{2}) = 2 - 1.5 = \frac{1}{2}$$

3. Final Value

$$A = \frac{5}{12} + \frac{1}{2} + \frac{1}{2} = \frac{5}{12} + 1 = \frac{17}{12}$$

Therefore:

$$12A = 12 \times \frac{17}{12} = \mathbf{17}$$