Let $$[x]$$ denote the greatest integer $$\leq x$$. Consider the function $$f(x) = \max\{x^2, 1 + [x]\}$$. Then the value of the integral $$\int_0^2 f(x) dx$$ is:

To evaluate the integral of the function $$f(x) = \max\{x^2, 1 + [x]\}$$ from $$0$$ to $$2$$, we must break the interval $$[0, 2]$$ into subintervals based on the behavior of the greatest integer function,$$[x]$$, and the intersection of the two functions inside the maximum.

Step 1: Analyze the function on subintervals

The value of $$[x]$$ changes at integer values. We analyze $$1 + [x]$$ and compare it to $$x^2$$ over the interval$$[0, 2]$$.

For $$0 \leq x < 1$$:

$$[x] = 0 \implies 1 + [x] = 1$$

We need to find $$\max\{x^2, 1\}$$.

Since $$x^2 < 1$$ for all $$x$$ in this interval, the maximum is $$1$$.

Therefore, $$f(x) = 1$$.

For $$1 \leq x < 2$$:

$$[x] = 1 \implies 1 + [x] = 2$$

We need to find $$\max\{x^2, 2\}$$.

We must determine where $$x^2 = 2$$. Taking the square root gives $$x = \sqrt{2}$$, which falls inside our interval.

Subinterval $$1 \leq x < \sqrt{2}$$: Here, $$x^2 < 2$$, so the maximum is $$2$$. Thus, $$f(x) = 2$$.

Subinterval $$\sqrt{2} \leq x < 2$$: Here, $$x^2 \geq 2$$, so the maximum is $$x^2$$. Thus, $$f(x) = x^2$$.

Step 2: Set up the piecewise integrals

Now we can express the original integral as the sum of three separate definite integrals based on the subintervals we found:

$$\int_0^2 f(x) dx = \int_0^1 1 \, dx + \int_1^{\sqrt{2}} 2 \, dx + \int_{\sqrt{2}}^2 x^2 \, dx$$

Step 3: Evaluate each integral

Calculate the definite integral for each part separately:

1. First part:

$$\int_0^1 1 \, dx = [x]_0^1 = 1 - 0 = 1$$

2. Second part:

$$\int_1^{\sqrt{2}} 2 \, dx = [2x]_1^{\sqrt{2}} = 2\sqrt{2} - 2(1) = 2\sqrt{2} - 2$$

3. Third part:

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

Step 4: Sum the results

Add the evaluated parts together to find the total area:

$$\text{Total Area} = 1 + (2\sqrt{2} - 2) + \left(\frac{8}{3} - \frac{2\sqrt{2}}{3}\right)$$

$$\text{Total Area} = \frac{5 + 4\sqrt{2}}{3}$$

Final Answer

The value of the integral is: $$\frac{5 + 4\sqrt{2}}{3}$$